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https://uniontestprep.com/gre/discussions/i-don-t-understand-how-the-second-answer-is-e	1,670,468,139,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446711232.54/warc/CC-MAIN-20221208014204-20221208044204-00610.warc.gz	625,076,199	12,448	Back to All Topics Start a Discussion Start a new discussion whenever you’re not finding the answers you’re looking for or looking to share something new. # I don't understand how the second answer is E Which 2 ways can 2t-u+3v be expressed in terms of v, if 5t=2u & u-1=v? A. 1/5(14v-1) B. 2V-3 C. (20+140V)/100 D. 1/4V+20 E. 2.8V-.2 Answer choices: A) A&D B) A&B C) B&E D) A&E I see what they did to get A, but how did they get E? Jessica King Added on February 16, 2019 23:27 0 ## 1 Response If you agree that 1/5(14V-1) is an answer, which is choice A then just distribute the 1/5 and get: (14/5)V - (1/5). Here (14/5) = 2.8 and (1/5) = 0.2. Good Luck! Tashfeen Responded on September 18, 2019 00:17 0	274	718	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.8125	4	CC-MAIN-2022-49	latest	en	0.88454	Back to All Topics. Start a Discussion. Start a new discussion whenever you’re not finding the answers you’re looking for or looking to share something new.. # I don't understand how the second answer is E. Which 2 ways can 2t-u+3v be expressed in terms of v, if 5t=2u & u-1=v?. A. 1/5(14v-1) B. 2V-3 C. (20+140V)/100 D. 1/4V+20 E. 2.8V-.2. Answer choices: A) A&D B) A&B C) B&E D) A&E.	I see what they did to get A, but how did they get E?. Jessica King. Added on February 16, 2019 23:27. 0. ## 1 Response. If you agree that 1/5(14V-1) is an answer, which is choice A then just distribute the 1/5 and get: (14/5)V - (1/5). Here (14/5) = 2.8 and (1/5) = 0.2. Good Luck!. Tashfeen. Responded on September 18, 2019 00:17. 0.
https://www.physicsforums.com/tags/unit-vectors/	1,713,002,798,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296816587.89/warc/CC-MAIN-20240413083102-20240413113102-00856.warc.gz	880,022,962	27,488	# What is Unit vectors: Definition and 140 Discussions In mathematics, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1. A unit vector is often denoted by a lowercase letter with a circumflex, or "hat", as in v ^ {\displaystyle {\hat {\mathbf {v} }}} (pronounced "v-hat").The term direction vector is used to describe a unit vector being used to represent spatial direction, and such quantities are commonly denoted as d; 2D spatial directions represented this way are numerically equivalent to points on the unit circle. The same construct is used to specify spatial directions in 3D, which are equivalent to a point on the unit sphere. The normalized vector û of a non-zero vector u is the unit vector in the direction of u, i.e., u ^ = u \| u \| {\displaystyle \mathbf {\hat {u}} ={\frac {\mathbf {u} }{\|\mathbf {u} \|}}} where \|u\| is the norm (or length) of u. The term normalized vector is sometimes used as a synonym for unit vector. Unit vectors are often chosen to form the basis of a vector space, and every vector in the space may be written as a linear combination of unit vectors. By definition, the dot product of two unit vectors in a Euclidean space is a scalar value amounting to the cosine of the smaller subtended angle. In three-dimensional Euclidean space, the cross product of two arbitrary unit vectors is a third vector orthogonal to both of them, whose length is equal to the sine of the smaller subtended angle. The normalized cross product corrects for this varying length, and yields the mutually orthogonal unit vector to the two inputs, applying the right-hand rule to resolve one of two possible directions. View More On Wikipedia.org 1. ### Help about the unit vectors for polar coordinates in terms of i and j "Firstly, I represented [Uθ ]on the two-dimensional polar coordinate system to facilitate the steps and projections." Then, I have written the steps, step by step, to ultimately derive the expression U(θ) in terms of i and j which is: [ Uθ=−sin(θ)i+cos(θ)j ] NOTE: The professor provided us... 2. ### I Law of Cosines in Linear Algebra: Understanding the Dot Product of Unit Vectors HI, I am studying linear algebra, and I just can't understand why "Unit vectors u and U at angle θ have u multiplied by U=cosθ Why is it like that? Thanks 4. ### Displacement problem with unit vectors (a) I did (7.074.1)-(-7.033.94)=56.7 with this method I got this answer correct in my first attempt. (b) This where I seem to have gone wrong. I used a · b = (axbx +ayby) then I used a = sqrt(ax2+ay2) to get a single number for the answer. Filling in the numbers 7.07-7.03 + 4.944.1 =... 5. ### B Problem involving unit vectors For example is this correct : 19icap.4(-i cap) = 76(i.-i)= 76 Or is it , take - out. Then -76(icap.icap)= -76 Is it -76 or 76 ? 6. ### Understanding Direction of Unit Vectors r roof & phi roof The unit vector r roof points in the direction of increasing r with phi fixed; phi roof points in the direction of increasing phi with r fixed. Unlike x roof, the vectors r roof and phi roof change as the position vector r moves. What I was thinking of the image is Although, I was thinking why... 7. ### Unit vectors -- How can they be dimensionless? Hi, what is a unit vector? I mean, it is ##\hat{A}=\vec A/\|A\|##. A dimensionless vector with modulus (absolute value) one, I've read somewhere. So, dimensionless with modulus. Isn't that a contradiction? I mean, absolute value regardless dimension? Am I out of context?. ##\Bbb R^3## is a... 8. ### Calculating vector cross product through unit vectors Writing both ##\vec{U}## and ##\vec{B}## with magnitude in all the three spatial coordinates: $$\vec{U}\times \vec{B}= (U_{x}\cdot \widehat{i}+U_{y}\cdot \widehat{j}+U_{z}\cdot \widehat{k})\times (B_{x}\cdot \widehat{i}+B_{y}\cdot \widehat{j}+B_{z}\cdot \widehat{k})$$ From this point on, I... 9. ### How Can I Solve Question Type: "With Magnitude and Unit Vectors"? Hi I am a beginner in this topic. I didn't understand this question type clearly.What does it mean" With Magnitude and Unit Vectors" exactly? May you help me for the solution step by step :). Thanks in advance. 10. ### What's the use of unit vectors? Homework Statement: Hallo. Can somebody explain to me what's the importance-use of unit vector in the below (second) equation? Why isn't the first equation just enough to describe r? What's the reason for unit vector to even exist? Homework Equations: in the photos 11. ### Advantages of Polar Coordinate System & Rotating Unit Vectors What is the advantage of using a polar coordinate system with rotating unit vectors? Kleppner's and Kolenkow's An Introduction to Mechanics states that base vectors ##\mathbf{ \hat{r}}## and ##\mathbf{\hat{\theta}}## have a variable direction, such that for a Cartesian coordinates system's base... 12. ### I Polar coordinates and unit vectors Hello, I get that both polar unit vectors, ##\hat{r}## and ##\hat{\theta}##, are unit vectors whose directions varies from point to point in the plane. In polar coordinates, the location of an arbitrary point ##P## on the plane is solely given in terms of one of the unit vector, the vector... 13. ### Cylindrical coordinates: unit vectors and time derivatives Homework Statement Homework EquationsThe Attempt at a Solution I have found expressions for the unit vectors for cylindrical coordinates in terms of unit vectors in rectangular coordinates. I have also found the time derivatives of the unit vectors in cylindrical coordinates. However, I am... 14. ### Location of charged particle given magnitude of position Homework Statement A charged particle has an electric field at ##\langle -0.13, 0.14, 0 \rangle## m is ##\langle 6.48\times10^3, -8.64\times10^3, 0 \rangle## N/C. The charged particle is -3nC. Where is the particle located? Homework Equations ##\vec E=\frac 1 {4π\varepsilon_0} \frac q {\|\vec... 15. ### I A Question about Unit Vectors of Cylindrical Coordinates I wrote the equations of the Nabla, the divergence, the curl, and the Laplacian operators in cylindrical coordinates ##(ρ,φ,z)##. I was wondering how to define the direction of the unit vector ##\hat{φ}##. Can we obtain ##\hat{φ}## by evaluating the cross-product of ##\hat{ρ}## and ##\hat{z}##... 16. ### Unit Vectors and Momentum Changes in a Block of Ice Homework Statement A 0.5 kg block of ice is sliding by you on a very slippery floor at 2.5 m/s. As it goes by, you give it a kick perpendicular to its path. Your foot is in contact with the ice block for 0.0035 seconds. The block eventually slides at an angle of 24 degrees from its original... 17. K ### Cartesian unit vectors in terms of cylindrical vectors How do I express ex,ey,ez in terms er,eθ,eZ? r=(x^2+y^2)^1/2,θ=arctan(y/x),Z=z A(r,θ,z) ∂A/∂x=x/(x^2+y^2)^1/2er+(-y)/(x^2+y^2)eθ=cosθer-(sinθ/r)eθ ex=(∂A/∂x)/\|∂A/∂x\| I should get ex as cosθer-sinθeθ, but I don't get ex correctly. am i doing this wrong? 18. ### Calculate Change in Speed Using Unit Vectors: Easy Physics Solution Actually that's very easy question but I have some difficult to understand the logic behind . So-"The initial velocity of an object (m/s) is Vi=1i+5j+2k. And the final velocity is Vf=3i+5j+7k. What was the change in speed of the object?"X Solution - \|Vf\|-\|Vi\| = √(32+52+72)-√(12+52+22) = 3.63... 19. ### I Determining Vector Direction: Finding Unit Vectors Why is there a need to find unit vector? If we are given a vector we can always find its direction. 20. ### Statics: Dimensionless Unit Vector Homework Statement Homework Equations The Attempt at a Solution So I began by subtracting. (205-160)=55 i (495+128)=623 j Both of these vectors are in the positive direction. So if I divide the vector by its magnitude I should get an answer of 1 in the positive direction for both i and... 21. ### MHB S6.12.3.35 Find the unit vectors $\tiny{s6.12.3.35}\\$ 35. Find the unit vectors that are parallel to the tangent line to the parabola $y = x^2$ at the point $(2,4)$. \begin{align} \displaystyle y'&=2x \end{align} the book answer to this is $\pm\left(i+4j)/\sqrt{17}\right)$ but don't see how they got this? 22. ### I Is the Dot Product of Unit Vectors Related to Magnitudes and Angle Between Them? Okay so I understand that in order to represent a vector which is in cartesian coordinates in spherical coordinates.. we use the transformation which is obtained by dotting the unit vectors. So my question goes like this: when we dot for example the unit vector ar^ with x^ we obtain sin(theta)... 23. ### Unit Vectors as a Function of Time? Homework Statement Say I have a vector F something like F = c1(t) x^ + c2(t) y^ were c1 and c2 are some scalar functions of time were you plug in time to into the equation and are given some magnitude. My question seems to be can we define unit vectors/basis vector as a function of time as... 24. ### Time Derivative of Unit Vectors Quick question (a little rusty on this): Why don't unit vectors in Cartesian Coordinates not change with time? For example, suppose \mathbf{r} (t) = x(t) \mathbf{x} + y(t) \mathbf{y} + z(t) \mathbf{z} How exactly do we know that the unit vectors don't change with time? Or in other words... 25. ### Why Use Unit Vectors in Calculations? I'm confused about what situations you should use unit vectors in... and it seems that when I approach the same problem using unit vectors vs. without unit vectors, I get different answers. Why? To illustrate my confusion, here's an example that I tried solving using unit vectors, and without... 26. ### Physics: Multiplying Unit vectors [Moderator note: Post moved from New Member Introductions forum, so no template] I am having trouble understanding how to multiply unit vectors. I know that: (please excuse the notation) i^×j^ = k^ j^×k^ = i^ k^×i^ = j^ The question I am stuck on is: What is (i^×j^)×k^? So far I have (i^×j^)... 27. ### Unit vectors for multiple particles? (Quantum Mechanics) It's been a little bit since I have studied multi-particle quantum mechanics and I am a little rusty on the notation. Let's say I have a wave function, that consists of the tensor product of two spaces, one for each particle moving, ##\|\psi_1,\psi_2>##. Each of these particles is moving in a... 28. ### Perpendicular Unit Vectors in the x-y Plane: Is My Solution Correct? Homework Statement From Kleppner and Kolenkow Chapter 1 (Just checking to see if I'm right) Given vector A=<3, 4, -4> a) Find a unit vector B that lies in the x-y plane and is perpendicular to A. b) Find a unit vector C that is perpendicular to both A and B. c)Show that A is perpendicular to... 29. ### Angles between sides of triangle ABC and unit vectors I was going through this link -... 30. ### Unit Vectors for Polarization and Wave Vector Directions Homework Statement I am having difficulty understanding the very first step of the following solved problem (I understand the rest of the solution). How did they obtain the expressions for ##\hat{n}## (the direction of polarization), and ##\hat{k}## (the unit vector pointing in the direction... 31. ### Find Unit Vectors for f(x,y) w/ D_uf=0 Homework Statement For f(x,y)=x^2-xy+y^2 and the vector u=i+j. ii)Find two unit vectors such D_vf=0 Homework Equations N/A. The Attempt at a Solution Not sure if relevant but the previous questions were asking for the unit vector u - which I got \hat{u}=\frac{1}{\sqrt{2}}(i+j) for the maximum... 32. ### Cartesian to polar unit vectors + Linear Combination I've been trying to solve this question all day. If somebody could point me in the right direction I would really appreciate it! (ii) A particle’s motion is described by the following position vector r(t) = 4txˆ + (10t − t)ˆy Determine the polar coordinate unit vectors ˆr and ˆθ for r. [4]... 33. ### Choosing unit vectors for harmonic motion problems Consider a vertical pendulum affected by gravity (See the pdf file i included). Now i can choose two different opposite directions for my unit vectors which give me different equations. \downarrow : m\ddot x = mg-kx \uparrow : m\ddot x = kx-mg Which of course makes perfect sense, changing... 34. ### Deriving spherical unit vectors in terms of cartesian unit vectors I'm trying to find the azimuthal angle unit vector \vec{\phi} in the cartesian basis by taking the cross product of the radial and \vec{z} unit vectors. \vec{z} \times \vec{r} = <0, 0, 1> \times <sin(\theta)cos(\phi), sin(\theta)sin(\phi), cos(\theta)> = <-sin(\theta)sin(\phi)... 35. ### Cartesian unit vectors expressed by Cylindrical unit vectors please someone explain me the following expression for Cartesian unit vectors expressed by the cylindrical unit vectors: http://web.mit.edu/8.02t/www/materials/modules/ReviewB.pdf at page B-8 line B.2.4 i would like to know which steps led to it. thanks, Chen 36. ### Deriving sin(a-b) trig identity using Cross Product of Unit Vectors Homework Statement A and B are two unit vectors in the x-y plane. A = <cos(a), sin(a)> B = <cos(b), sin(b)> I need to derive the trig identity: sin(a-b) = sin(a) cos(b) - sin(b) cos (a) I'm told to do it using the properties of the cross product A x B Homework Equations A x B =... 37. ### LaTeX Best Unit Vectors in LaTeX for TeX.SE Interaction During an interaction on TeX.SE, egreg there posted some truly awesome code for doing unit vectors in $\LaTeX$: \usepackage{newtxtext} \usepackage{newtxmath} \usepackage{amsmath} \usepackage{bm} \newcommand{\uveci}{{\bm{\hat{\textnormal{\bfseries\i}}}}}... 38. ### Example about tangential and normal unit vectors Here is a example 1.3 from analytical dynamics of Haim Baruh. a particle moves on a path on the xy plane defined by the curve y=3*x^2 , where x varies with the relation x= sin(a). find the radius of curvature of the path and unit vectors in the normal and tangential directions when a=pi/6... 39. ### Integration including unit vectors I have an integral of aΘ cos(Θ) dΘ a is the unit vector for Θ. I'm not sure what to do with it in the integration. I know the unit vector equals a/abs(a) but that would give a mess of an integral cause of the abs(a). 40. ### States are or aren't unit vectors? I am a little confused by an elementary point. Something must be wrong with the following: On one hand, a Hermitian operator (which is not necessarily unitary) takes one state to another state. Hence a state need not be represented as a unit vector; its norm can be greater (or less than)... 41. ### Unit vectors in different coordinates Hi everyone, I've some points I want to make sure of. 1- When converting a "POINT" from a coordinate system to another, I'll just use the derived equation to convert (e.g. (1,2,3) from cartestian to cylindrical: \rho=\sqrt{x^{2}+y^{2}}, \phi=tan^{-1}\frac{y}{x}, z=z 2- When converting an... 42. ### Cross products for unit vectors in other coordinate systems I am a bit confused often when I have to compute cross products in other coordinate systems (non-Cartesian), I can't seem to find any tables for cross products such as "phi X rho." in spherical I think that these unit vectors are considered to be "perpendicular," so would phi X rho just be "+/-... 43. ### Understanding Unit Vectors: A Step-by-Step Guide Hi everyone, Just want to know how does the the unit vector become in that form: \vec{n}=\frac{2x\vec{i}+2y\vec{j}}{\sqrt{(2x)^{2}+(2y)^{2}}}=\frac{x \vec{i}+y \vec{j}}{4} 44. ### Derivation of Phi-Hat wrt Phi in Spherical Unit Vectors Homework Statement I just want to know how to get from this: ∂ø^/∂ø = -x^cosø - y^sinø to this: = -(r^sinθ+θ^cosθ) Homework Equations All the equations found here in the Spherical Coordinates section: http://en.wikipedia.org/wiki/Unit_vector The Attempt at a Solution I've... 45. ### What Is the Dot Product of Two Parallel Unit Vectors? Homework Statement The dot product for two.parralel pointing.unit.vectors is ? A. 1 B. 0 C. -1 D. Undefined [b]2. Relevant equation The Attempt at a Solutionsince they are unit vectors they have a magnitude of 1,this implies that the dot product is 1,since the angle between... 46. ### What's the Difference Between Ax and i-Hat in Vector Notation? Homework Statement So this isn't really a specific homework question, it's more of a general one. What is the difference between ax and i(hat)? I thought they were the same thing. Can someone please explain the difference? Homework Equations The Attempt at a Solution 47. ### Understanding Vectors vs Unit Vectors: Differences and Uses in Physics I am a bit confused about what the difference is between the two? To give some specific context where it has thrown me off, say if I were to define a charge with a vector r and compared that to a unit vector r hat, what exactly is the difference between what each of those tells me? I have... 48. ### MHB Showing relationship between cartesian and spherical unit vectors I am asked to show that when $$\hat{e_r}$$, $$\hat{e_\theta}$$, and $$\hat{e_\phi}$$ are unit vectors in spherical coordinates, that the cartesian unit vectors $$\hat{i} = \sin{\phi}\cos{\theta}\hat{e_r} + \cos{\phi}\cos{\theta}\hat{e_\phi} - \sin{\theta}\hat{e_\theta}$$ \hat{j} =... 49. ### MHB If a and b are unit vectors.... If a and b are unit vectors and \|a + b\| = sqrt(2). What is the value (dot product) of (2a-b).(a+3b)? Is the answer -1 by any chance? If not... I know how to find the dot product and find the magnitude and add vectors, etc. but I have never came across this a question before. I am very unclear... 50. ### Calculation of work involving unit vectors Homework Statement A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force f = (30N)i - (40N)j to the cart as it undergoes a displacement s = (-9.0m)i - (3.0m)j How much work does the force you apply do on the grocery cart? Homework Equations...	4,730	17,853	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0}	4.125	4	CC-MAIN-2024-18	latest	en	0.929295	# What is Unit vectors: Definition and 140 Discussions. In mathematics, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1. A unit vector is often denoted by a lowercase letter with a circumflex, or "hat", as in. v. ^. {\displaystyle {\hat {\mathbf {v} }}}. (pronounced "v-hat").The term direction vector is used to describe a unit vector being used to represent spatial direction, and such quantities are commonly denoted as d; 2D spatial directions represented this way are numerically equivalent to points on the unit circle.. The same construct is used to specify spatial directions in 3D, which are equivalent to a point on the unit sphere.. The normalized vector û of a non-zero vector u is the unit vector in the direction of u, i.e.,. u. ^. =. u. \|. u. \|. {\displaystyle \mathbf {\hat {u}} ={\frac {\mathbf {u} }{\|\mathbf {u} \|}}}. where \|u\| is the norm (or length) of u. The term normalized vector is sometimes used as a synonym for unit vector.. Unit vectors are often chosen to form the basis of a vector space, and every vector in the space may be written as a linear combination of unit vectors.. By definition, the dot product of two unit vectors in a Euclidean space is a scalar value amounting to the cosine of the smaller subtended angle. In three-dimensional Euclidean space, the cross product of two arbitrary unit vectors is a third vector orthogonal to both of them, whose length is equal to the sine of the smaller subtended angle. The normalized cross product corrects for this varying length, and yields the mutually orthogonal unit vector to the two inputs, applying the right-hand rule to resolve one of two possible directions.. View More On Wikipedia.org. 1. ### Help about the unit vectors for polar coordinates in terms of i and j. "Firstly, I represented [Uθ ]on the two-dimensional polar coordinate system to facilitate the steps and projections." Then, I have written the steps, step by step, to ultimately derive the expression U(θ) in terms of i and j which is: [ Uθ=−sin(θ)i+cos(θ)j ] NOTE: The professor provided us.... 2. ### I Law of Cosines in Linear Algebra: Understanding the Dot Product of Unit Vectors. HI, I am studying linear algebra, and I just can't understand why "Unit vectors u and U at angle θ have u multiplied by U=cosθ Why is it like that? Thanks. 4. ### Displacement problem with unit vectors. (a) I did (7.074.1)-(-7.033.94)=56.7 with this method I got this answer correct in my first attempt. (b) This where I seem to have gone wrong. I used a · b = (axbx +ayby) then I used a = sqrt(ax2+ay2) to get a single number for the answer. Filling in the numbers 7.07-7.03 + 4.944.1 =.... 5. ### B Problem involving unit vectors. For example is this correct : 19icap.4(-i cap) = 76(i.-i)= 76 Or is it , take - out. Then -76(icap.icap)= -76 Is it -76 or 76 ?. 6. ### Understanding Direction of Unit Vectors r roof & phi roof. The unit vector r roof points in the direction of increasing r with phi fixed; phi roof points in the direction of increasing phi with r fixed. Unlike x roof, the vectors r roof and phi roof change as the position vector r moves. What I was thinking of the image is Although, I was thinking why.... 7. ### Unit vectors -- How can they be dimensionless?. Hi, what is a unit vector? I mean, it is ##\hat{A}=\vec A/\|A\|##. A dimensionless vector with modulus (absolute value) one, I've read somewhere. So, dimensionless with modulus. Isn't that a contradiction? I mean, absolute value regardless dimension? Am I out of context?. ##\Bbb R^3## is a.... 8. ### Calculating vector cross product through unit vectors. Writing both ##\vec{U}## and ##\vec{B}## with magnitude in all the three spatial coordinates: $$\vec{U}\times \vec{B}= (U_{x}\cdot \widehat{i}+U_{y}\cdot \widehat{j}+U_{z}\cdot \widehat{k})\times (B_{x}\cdot \widehat{i}+B_{y}\cdot \widehat{j}+B_{z}\cdot \widehat{k})$$ From this point on, I.... 9. ### How Can I Solve Question Type: "With Magnitude and Unit Vectors"?. Hi I am a beginner in this topic. I didn't understand this question type clearly.What does it mean" With Magnitude and Unit Vectors" exactly? May you help me for the solution step by step :). Thanks in advance.. 10. ### What's the use of unit vectors?. Homework Statement: Hallo. Can somebody explain to me what's the importance-use of unit vector in the below (second) equation? Why isn't the first equation just enough to describe r? What's the reason for unit vector to even exist? Homework Equations: in the photos. 11. ### Advantages of Polar Coordinate System & Rotating Unit Vectors. What is the advantage of using a polar coordinate system with rotating unit vectors? Kleppner's and Kolenkow's An Introduction to Mechanics states that base vectors ##\mathbf{ \hat{r}}## and ##\mathbf{\hat{\theta}}## have a variable direction, such that for a Cartesian coordinates system's base.... 12. ### I Polar coordinates and unit vectors. Hello, I get that both polar unit vectors, ##\hat{r}## and ##\hat{\theta}##, are unit vectors whose directions varies from point to point in the plane. In polar coordinates, the location of an arbitrary point ##P## on the plane is solely given in terms of one of the unit vector, the vector.... 13. ### Cylindrical coordinates: unit vectors and time derivatives. Homework Statement Homework EquationsThe Attempt at a Solution I have found expressions for the unit vectors for cylindrical coordinates in terms of unit vectors in rectangular coordinates. I have also found the time derivatives of the unit vectors in cylindrical coordinates. However, I am.... 14. ### Location of charged particle given magnitude of position. Homework Statement A charged particle has an electric field at ##\langle -0.13, 0.14, 0 \rangle## m is ##\langle 6.48\times10^3, -8.64\times10^3, 0 \rangle## N/C. The charged particle is -3nC. Where is the particle located? Homework Equations ##\vec E=\frac 1 {4π\varepsilon_0} \frac q {\|\vec.... 15. ### I A Question about Unit Vectors of Cylindrical Coordinates. I wrote the equations of the Nabla, the divergence, the curl, and the Laplacian operators in cylindrical coordinates ##(ρ,φ,z)##. I was wondering how to define the direction of the unit vector ##\hat{φ}##. Can we obtain ##\hat{φ}## by evaluating the cross-product of ##\hat{ρ}## and ##\hat{z}##.... 16. ### Unit Vectors and Momentum Changes in a Block of Ice. Homework Statement A 0.5 kg block of ice is sliding by you on a very slippery floor at 2.5 m/s. As it goes by, you give it a kick perpendicular to its path. Your foot is in contact with the ice block for 0.0035 seconds. The block eventually slides at an angle of 24 degrees from its original.... 17. K. ### Cartesian unit vectors in terms of cylindrical vectors. How do I express ex,ey,ez in terms er,eθ,eZ? r=(x^2+y^2)^1/2,θ=arctan(y/x),Z=z A(r,θ,z) ∂A/∂x=x/(x^2+y^2)^1/2er+(-y)/(x^2+y^2)eθ=cosθer-(sinθ/r)eθ ex=(∂A/∂x)/\|∂A/∂x\| I should get ex as cosθer-sinθeθ, but I don't get ex correctly. am i doing this wrong?. 18. ### Calculate Change in Speed Using Unit Vectors: Easy Physics Solution. Actually that's very easy question but I have some difficult to understand the logic behind . So-"The initial velocity of an object (m/s) is Vi=1i+5j+2k. And the final velocity is Vf=3i+5j+7k. What was the change in speed of the object?"X Solution - \|Vf\|-\|Vi\| = √(32+52+72)-√(12+52+22) = 3.63.... 19. ### I Determining Vector Direction: Finding Unit Vectors. Why is there a need to find unit vector? If we are given a vector we can always find its direction.. 20. ### Statics: Dimensionless Unit Vector. Homework Statement Homework Equations The Attempt at a Solution So I began by subtracting. (205-160)=55 i (495+128)=623 j Both of these vectors are in the positive direction. So if I divide the vector by its magnitude I should get an answer of 1 in the positive direction for both i and.... 21. ### MHB S6.12.3.35 Find the unit vectors. $\tiny{s6.12.3.35}\\$ 35. Find the unit vectors that are parallel to the tangent line to the parabola $y = x^2$ at the point $(2,4)$. \begin{align} \displaystyle y'&=2x \end{align} the book answer to this is $\pm\left(i+4j)/\sqrt{17}\right)$ but don't see how they got this?. 22. ### I Is the Dot Product of Unit Vectors Related to Magnitudes and Angle Between Them?. Okay so I understand that in order to represent a vector which is in cartesian coordinates in spherical coordinates.. we use the transformation which is obtained by dotting the unit vectors. So my question goes like this: when we dot for example the unit vector ar^ with x^ we obtain sin(theta).... 23. ### Unit Vectors as a Function of Time?.	Homework Statement Say I have a vector F something like F = c1(t) x^ + c2(t) y^ were c1 and c2 are some scalar functions of time were you plug in time to into the equation and are given some magnitude. My question seems to be can we define unit vectors/basis vector as a function of time as.... 24. ### Time Derivative of Unit Vectors. Quick question (a little rusty on this): Why don't unit vectors in Cartesian Coordinates not change with time? For example, suppose \mathbf{r} (t) = x(t) \mathbf{x} + y(t) \mathbf{y} + z(t) \mathbf{z} How exactly do we know that the unit vectors don't change with time? Or in other words.... 25. ### Why Use Unit Vectors in Calculations?. I'm confused about what situations you should use unit vectors in... and it seems that when I approach the same problem using unit vectors vs. without unit vectors, I get different answers. Why? To illustrate my confusion, here's an example that I tried solving using unit vectors, and without.... 26. ### Physics: Multiplying Unit vectors. [Moderator note: Post moved from New Member Introductions forum, so no template] I am having trouble understanding how to multiply unit vectors. I know that: (please excuse the notation) i^×j^ = k^ j^×k^ = i^ k^×i^ = j^ The question I am stuck on is: What is (i^×j^)×k^? So far I have (i^×j^).... 27. ### Unit vectors for multiple particles? (Quantum Mechanics). It's been a little bit since I have studied multi-particle quantum mechanics and I am a little rusty on the notation. Let's say I have a wave function, that consists of the tensor product of two spaces, one for each particle moving, ##\|\psi_1,\psi_2>##. Each of these particles is moving in a.... 28. ### Perpendicular Unit Vectors in the x-y Plane: Is My Solution Correct?. Homework Statement From Kleppner and Kolenkow Chapter 1 (Just checking to see if I'm right) Given vector A=<3, 4, -4> a) Find a unit vector B that lies in the x-y plane and is perpendicular to A. b) Find a unit vector C that is perpendicular to both A and B. c)Show that A is perpendicular to.... 29. ### Angles between sides of triangle ABC and unit vectors. I was going through this link -.... 30. ### Unit Vectors for Polarization and Wave Vector Directions. Homework Statement I am having difficulty understanding the very first step of the following solved problem (I understand the rest of the solution). How did they obtain the expressions for ##\hat{n}## (the direction of polarization), and ##\hat{k}## (the unit vector pointing in the direction.... 31. ### Find Unit Vectors for f(x,y) w/ D_uf=0. Homework Statement For f(x,y)=x^2-xy+y^2 and the vector u=i+j. ii)Find two unit vectors such D_vf=0 Homework Equations N/A. The Attempt at a Solution Not sure if relevant but the previous questions were asking for the unit vector u - which I got \hat{u}=\frac{1}{\sqrt{2}}(i+j) for the maximum.... 32. ### Cartesian to polar unit vectors + Linear Combination. I've been trying to solve this question all day. If somebody could point me in the right direction I would really appreciate it! (ii) A particle’s motion is described by the following position vector r(t) = 4txˆ + (10t − t)ˆy Determine the polar coordinate unit vectors ˆr and ˆθ for r. [4].... 33. ### Choosing unit vectors for harmonic motion problems. Consider a vertical pendulum affected by gravity (See the pdf file i included). Now i can choose two different opposite directions for my unit vectors which give me different equations. \downarrow : m\ddot x = mg-kx \uparrow : m\ddot x = kx-mg Which of course makes perfect sense, changing.... 34. ### Deriving spherical unit vectors in terms of cartesian unit vectors. I'm trying to find the azimuthal angle unit vector \vec{\phi} in the cartesian basis by taking the cross product of the radial and \vec{z} unit vectors. \vec{z} \times \vec{r} = <0, 0, 1> \times <sin(\theta)cos(\phi), sin(\theta)sin(\phi), cos(\theta)> = <-sin(\theta)sin(\phi).... 35. ### Cartesian unit vectors expressed by Cylindrical unit vectors. please someone explain me the following expression for Cartesian unit vectors expressed by the cylindrical unit vectors: http://web.mit.edu/8.02t/www/materials/modules/ReviewB.pdf at page B-8 line B.2.4 i would like to know which steps led to it. thanks, Chen. 36. ### Deriving sin(a-b) trig identity using Cross Product of Unit Vectors. Homework Statement A and B are two unit vectors in the x-y plane. A = <cos(a), sin(a)> B = <cos(b), sin(b)> I need to derive the trig identity: sin(a-b) = sin(a) cos(b) - sin(b) cos (a) I'm told to do it using the properties of the cross product A x B Homework Equations A x B =.... 37. ### LaTeX Best Unit Vectors in LaTeX for TeX.SE Interaction. During an interaction on TeX.SE, egreg there posted some truly awesome code for doing unit vectors in $\LaTeX$: \usepackage{newtxtext} \usepackage{newtxmath} \usepackage{amsmath} \usepackage{bm} \newcommand{\uveci}{{\bm{\hat{\textnormal{\bfseries\i}}}}}.... 38. ### Example about tangential and normal unit vectors. Here is a example 1.3 from analytical dynamics of Haim Baruh. a particle moves on a path on the xy plane defined by the curve y=3*x^2 , where x varies with the relation x= sin(a). find the radius of curvature of the path and unit vectors in the normal and tangential directions when a=pi/6.... 39. ### Integration including unit vectors. I have an integral of aΘ cos(Θ) dΘ a is the unit vector for Θ. I'm not sure what to do with it in the integration. I know the unit vector equals a/abs(a) but that would give a mess of an integral cause of the abs(a).. 40. ### States are or aren't unit vectors?. I am a little confused by an elementary point. Something must be wrong with the following: On one hand, a Hermitian operator (which is not necessarily unitary) takes one state to another state. Hence a state need not be represented as a unit vector; its norm can be greater (or less than).... 41. ### Unit vectors in different coordinates. Hi everyone, I've some points I want to make sure of. 1- When converting a "POINT" from a coordinate system to another, I'll just use the derived equation to convert (e.g. (1,2,3) from cartestian to cylindrical: \rho=\sqrt{x^{2}+y^{2}}, \phi=tan^{-1}\frac{y}{x}, z=z 2- When converting an.... 42. ### Cross products for unit vectors in other coordinate systems. I am a bit confused often when I have to compute cross products in other coordinate systems (non-Cartesian), I can't seem to find any tables for cross products such as "phi X rho." in spherical I think that these unit vectors are considered to be "perpendicular," so would phi X rho just be "+/-.... 43. ### Understanding Unit Vectors: A Step-by-Step Guide. Hi everyone, Just want to know how does the the unit vector become in that form: \vec{n}=\frac{2x\vec{i}+2y\vec{j}}{\sqrt{(2x)^{2}+(2y)^{2}}}=\frac{x \vec{i}+y \vec{j}}{4}. 44. ### Derivation of Phi-Hat wrt Phi in Spherical Unit Vectors. Homework Statement I just want to know how to get from this: ∂ø^/∂ø = -x^cosø - y^sinø to this: = -(r^sinθ+θ^cosθ) Homework Equations All the equations found here in the Spherical Coordinates section: http://en.wikipedia.org/wiki/Unit_vector The Attempt at a Solution I've.... 45. ### What Is the Dot Product of Two Parallel Unit Vectors?. Homework Statement The dot product for two.parralel pointing.unit.vectors is ? A. 1 B. 0 C. -1 D. Undefined [b]2. Relevant equation The Attempt at a Solutionsince they are unit vectors they have a magnitude of 1,this implies that the dot product is 1,since the angle between.... 46. ### What's the Difference Between Ax and i-Hat in Vector Notation?. Homework Statement So this isn't really a specific homework question, it's more of a general one. What is the difference between ax and i(hat)? I thought they were the same thing. Can someone please explain the difference? Homework Equations The Attempt at a Solution. 47. ### Understanding Vectors vs Unit Vectors: Differences and Uses in Physics. I am a bit confused about what the difference is between the two? To give some specific context where it has thrown me off, say if I were to define a charge with a vector r and compared that to a unit vector r hat, what exactly is the difference between what each of those tells me? I have.... 48. ### MHB Showing relationship between cartesian and spherical unit vectors. I am asked to show that when $$\hat{e_r}$$, $$\hat{e_\theta}$$, and $$\hat{e_\phi}$$ are unit vectors in spherical coordinates, that the cartesian unit vectors $$\hat{i} = \sin{\phi}\cos{\theta}\hat{e_r} + \cos{\phi}\cos{\theta}\hat{e_\phi} - \sin{\theta}\hat{e_\theta}$$ \hat{j} =.... 49. ### MHB If a and b are unit vectors..... If a and b are unit vectors and \|a + b\| = sqrt(2). What is the value (dot product) of (2a-b).(a+3b)? Is the answer -1 by any chance? If not... I know how to find the dot product and find the magnitude and add vectors, etc. but I have never came across this a question before. I am very unclear.... 50. ### Calculation of work involving unit vectors. Homework Statement A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force f = (30N)i - (40N)j to the cart as it undergoes a displacement s = (-9.0m)i - (3.0m)j How much work does the force you apply do on the grocery cart? Homework Equations...
http://mathforum.org/library/drmath/view/71638.html	1,521,422,063,000,000,000	text/html	crawl-data/CC-MAIN-2018-13/segments/1521257646189.21/warc/CC-MAIN-20180319003616-20180319023616-00302.warc.gz	190,830,838	4,884	Associated Topics \|\| Dr. Math Home \|\| Search Dr. Math Meaning of Term in Algebra ```Date: 09/16/2007 at 12:22:14 From: Erin Subject: Terrible Terms When I was in middle school, I was taught that in algebra we do not write 2 * x because multiplication is used so much. Instead, we use 2x. However, as I went through the math course, they explained to me that something like 2x is considered one term, and that something like 2 + x is considered two terms. Because of this, each are treated differently. For example, in the problem 3(2x) all you need to do is multiply to the one term to get 6x. But in a problem like 3(2 + x), because there are two terms in parentheses, you need to distribute to get 6 + 3x. This is just one example of how a problem with one term and a problem with two terms are treated differently. My question is this: why is it that when you multiply or divide two numbers or variables they become one term (such as 2x becoming 2x), but when you add or subtract two numbers or variables they are still two terms (such as 2 + x)? This was never really explained to me because all they told me was that it's easier to not put the multiplication dot () because it's used so much. But I know that's not the only reason because there are so many different properties and different ways to treat the terms, that there must be another reason then just for convenience. Can you help me with my curiosity on this? Thank you. Take the problem 2(xy). In 2(xy), since the value in parentheses is one term, the appropriate thing to simplify it to would be 2xy. This makes sense. Lets say xy is the area of a box that is x units wide and y units long. If we combine two of those kind of boxes together, we would have one rectangle with an an area of 2xy. But if we were to pretend that xy was two terms, and, using the two-term distribution rule, multiplied both x and y by 2, this would not make sense. That would be saying that 2 xy boxes placed together have a width of 2x and a length of 2y. This is not true because if 4 boxes were placed together, then the resulting box would be 2x wide and 2y long. Hence, if we distributed 2(xy) wrongly and got 2x2y, it would have to be simplified to 4xy. 4 boxes have an area of 4xy, and 2 boxes have an area of 2xy. x[] y x[][] 2y 2x [][] [][] 2y Now take the problem 2(x + y). Since x + y is two terms, if we were to distribute the 2 to both the x and the y, we would get 2x + 2y. This makes sense. Lets say x + y is the number of units x and the number of units y in one group. If we get two of these kind of groups, we would have double the number of units x and well as the number of units y. But if we were to pretend that x + y was one term, then we would simply the problem as 2x + y. This does not make sense. If we were to try to simplify it this way, we would be saying that if we had two groups, only the number of units x would double. This doesn't make sense since one group has both units x and units y, and if we were to double that group we would have double the units x as well as double the units y. One group: xxyy Two groups: xxyy xxyy Two groups would not be: xx xxyy My teacher always tells me "terms are terrible". You must treat them in different ways. In my opinion "terrible terms" are where most algebraic mistakes come from. I feel knowing more about how terms work will help me to understand and avoid these mistakes. An example of a "terrible term" algebraic mistake is that you cannot distribute an exponent among two terms. (x + y)^2 does not equal x^2 + y^2. But you can distribute an exponent over one term. (xy)^2 = x^2y^2. Another example is that if the same term is added or subtracted in the numerator and denominator, they cannot be canceled out. (5 + 2)/(10 + 2) does not equal 5/10. But if the same term is multiplied in the numerator and denominator, they can be canceled out. (52)/(102)=5/10. ``` ``` Date: 09/16/2007 at 23:34:58 From: Doctor Peterson Subject: Re: Terrible Terms Hi, Erin. You've said a lot of good things; it sounds like you have a pretty good understanding of the essentials. I'll try saying a little more that might put things in perspective (which is what I think you want), and then we can discuss more if there are still gaps. The reason adding makes two terms is simply that we DEFINE a term as something that is added to other terms. In part due to the order of operations, we think of any expression as, ultimately, a sum: that is the last operation we do, so if there are any additions (that are not inside parentheses), they break up the expression into a sum, and the pieces are called terms. Many of the things you talk about are distributive properties. We can show that multiplication distributes over addition, (a+b)c = ac + bc and that exponents distribute over multiplication, (ab)^c = a^c b^c But exponents do not distribute over addition, and multiplication does not distribute over another multiplication. So you simply have to keep in mind the rules that are true, and not do things that LOOK similar but are not valid. One help is to relate these properties to the order of operations (which I believe is a major reason that the order is what it is): exponentiation distributes over... multiplication distributes over ... Also, I believe that we leave out the symbol for multiplication not so much because it is common, but because it is powerful (higher in the order of operations). By writing ax + by with the a and the x close together, we make the order of operations look natural; we SEE ax as a single term, by as another, and the main operation in the expression as addition, a sum of terms. Does that help at all? - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ ``` ``` Date: 10/02/2007 at 18:28:23 From: Erin Subject: Thank you (Terrible Terms) order of operations really helps - that exponents distribute over multiplication, which distributes over addition. It helps make sense of it to me and puts things into better perspective. It seems like it has a lot to do with how we define math operations. It was really neat hearing back from a mathematician - I plan to major in math in college. Thanks again! :) ``` Associated Topics: High School Basic Algebra High School Polynomials Middle School Algebra Search the Dr. Math Library: Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words Submit your own question to Dr. Math Math Forum Home \|\| Math Library \|\| Quick Reference \|\| Math Forum Search	1,744	6,720	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.40625	4	CC-MAIN-2018-13	latest	en	0.975043	Associated Topics \|\| Dr. Math Home \|\| Search Dr. Math. Meaning of Term in Algebra. ```Date: 09/16/2007 at 12:22:14. From: Erin. Subject: Terrible Terms. When I was in middle school, I was taught that in algebra we do not. write 2 * x because multiplication is used so much. Instead, we use. 2x.. However, as I went through the math course, they explained to me that. something like 2x is considered one term, and that something like. 2 + x is considered two terms. Because of this, each are treated. differently. For example, in the problem 3(2x) all you need to do is. multiply to the one term to get 6x. But in a problem like 3(2 + x),. because there are two terms in parentheses, you need to distribute to. get 6 + 3x. This is just one example of how a problem with one term. and a problem with two terms are treated differently.. My question is this: why is it that when you multiply or divide two. numbers or variables they become one term (such as 2x becoming 2x),. but when you add or subtract two numbers or variables they are still. two terms (such as 2 + x)? This was never really explained to me. because all they told me was that it's easier to not put the. multiplication dot () because it's used so much. But I know that's. not the only reason because there are so many different properties. and different ways to treat the terms, that there must be another. reason then just for convenience. Can you help me with my curiosity. on this? Thank you.. Take the problem 2(xy). In 2(xy), since the value in parentheses is. one term, the appropriate thing to simplify it to would be 2xy. This. makes sense. Lets say xy is the area of a box that is x units wide. and y units long. If we combine two of those kind of boxes together,. we would have one rectangle with an an area of 2xy. But if we were. to pretend that xy was two terms, and, using the two-term. distribution rule, multiplied both x and y by 2, this would not make. sense. That would be saying that 2 xy boxes placed together have a. width of 2x and a length of 2y. This is not true because if 4 boxes. were placed together, then the resulting box would be 2x wide and 2y. long. Hence, if we distributed 2(xy) wrongly and got 2x2y, it would. have to be simplified to 4xy. 4 boxes have an area of 4xy, and 2. boxes have an area of 2xy.. x[]. y. x[][]. 2y. 2x [][]. [][]. 2y. Now take the problem 2(x + y). Since x + y is two terms, if we were. to distribute the 2 to both the x and the y, we would get 2x + 2y.. This makes sense. Lets say x + y is the number of units x and the. number of units y in one group. If we get two of these kind of. groups, we would have double the number of units x and well as the. number of units y. But if we were to pretend that x + y was one term,. then we would simply the problem as 2x + y. This does not make sense.. If we were to try to simplify it this way, we would be saying that if. we had two groups, only the number of units x would double. This. doesn't make sense since one group has both units x and units y, and. if we were to double that group we would have double the units x as. well as double the units y.. One group: xxyy. Two groups: xxyy xxyy. Two groups would not be: xx xxyy. My teacher always tells me "terms are terrible". You must treat them. in different ways. In my opinion "terrible terms" are where most.	algebraic mistakes come from. I feel knowing more about how terms. work will help me to understand and avoid these mistakes. An example. of a "terrible term" algebraic mistake is that you cannot distribute. an exponent among two terms. (x + y)^2 does not equal x^2 + y^2. But. you can distribute an exponent over one term. (xy)^2 = x^2y^2.. Another example is that if the same term is added or subtracted in the. numerator and denominator, they cannot be canceled out. (5 + 2)/(10 +. 2) does not equal 5/10. But if the same term is multiplied in the. numerator and denominator, they can be canceled out.. (52)/(102)=5/10.. ```. ```. Date: 09/16/2007 at 23:34:58. From: Doctor Peterson. Subject: Re: Terrible Terms. Hi, Erin.. You've said a lot of good things; it sounds like you have a pretty. good understanding of the essentials. I'll try saying a little more. that might put things in perspective (which is what I think you want),. and then we can discuss more if there are still gaps.. The reason adding makes two terms is simply that we DEFINE a term as. something that is added to other terms.. In part due to the order of operations, we think of any expression as,. ultimately, a sum: that is the last operation we do, so if there are. any additions (that are not inside parentheses), they break up the. expression into a sum, and the pieces are called terms.. Many of the things you talk about are distributive properties. We can. show that multiplication distributes over addition,. (a+b)c = ac + bc. and that exponents distribute over multiplication,. (ab)^c = a^c b^c. But exponents do not distribute over addition, and multiplication does. not distribute over another multiplication. So you simply have to. keep in mind the rules that are true, and not do things that LOOK. similar but are not valid.. One help is to relate these properties to the order of operations. (which I believe is a major reason that the order is what it is):. exponentiation. distributes over.... multiplication. distributes over .... Also, I believe that we leave out the symbol for multiplication not so. much because it is common, but because it is powerful (higher in the. order of operations). By writing ax + by with the a and the x close. together, we make the order of operations look natural; we SEE ax as a. single term, by as another, and the main operation in the expression. as addition, a sum of terms.. Does that help at all?. - Doctor Peterson, The Math Forum. http://mathforum.org/dr.math/. ```. ```. Date: 10/02/2007 at 18:28:23. From: Erin. Subject: Thank you (Terrible Terms). order of operations really helps - that exponents distribute over. multiplication, which distributes over addition. It helps make sense. of it to me and puts things into better perspective. It seems like it. has a lot to do with how we define math operations.. It was really neat hearing back from a mathematician - I plan. to major in math in college. Thanks again! :). ```. Associated Topics:. High School Basic Algebra. High School Polynomials. Middle School Algebra. Search the Dr. Math Library:. Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words. Submit your own question to Dr. Math. Math Forum Home \|\| Math Library \|\| Quick Reference \|\| Math Forum Search.
https://web2.0calc.com/questions/help-with-rates_1	1,618,350,168,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038075074.29/warc/CC-MAIN-20210413213655-20210414003655-00017.warc.gz	711,352,866	5,924	+0 # help with rates 0 82 1 Bill and Peter can paint a 75 ft wall together in 7 hours. because Bill has more experience he can paint alone in two hours quicker than Peter. how long can it take Peter to paint the wall alone? round to the nearest minute. Jan 4, 2021 #1 +117573 +2 Let the portion of the wall that Bill can paint in one hour be 1/x = his rate where x is the number of hours that Bill takes to paint the wall by himself Let the portion of the wall that Peter can paint in one hour = 1/ ( x + 2) = his rate where x + 2 is the number of hours that Peter takes to paint the wall by himself And Bill's rate * 7hours + Peter's rate * 7 hours = 1 whole job done.....so... 1/x * 7 + 1/(x + 2) * 7 = 1 7/x + 7/(x + 2) = 1 7 ( x + 2) + 7x ______________ = 1 x ( x + 2) 14x + 14 = x ( x + 2) 14x + 14 = x^2 + 2x rearrange as x^2 - 12x - 14 = 0 complete the square on x x^2 -12x + 36 = 14 + 36 (x - 6)^2 = 50 take the positive root x - 6 = sqrt (50) x = sqrt (50 ) + 6 x + 2 = sqrt (50 ) + 8 ≈ 15.07 hrs = Peter's time = 15 + .07 *60 ≈ 15 hrs 4 min Jan 4, 2021 edited by CPhill Jan 5, 2021	502	1,229	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.1875	4	CC-MAIN-2021-17	latest	en	0.52797	+0. # help with rates. 0. 82. 1. Bill and Peter can paint a 75 ft wall together in 7 hours. because Bill has more experience he can paint alone in two hours quicker than Peter. how long can it take Peter to paint the wall alone? round to the nearest minute.. Jan 4, 2021. #1. +117573. +2. Let the portion of the wall that Bill can paint in one hour be 1/x = his rate where x is the number of hours that Bill takes to paint the wall by himself. Let the portion of the wall that Peter can paint in one hour = 1/ ( x + 2) = his rate where x + 2 is the number of hours that Peter takes to paint the wall by himself. And. Bill's rate * 7hours + Peter's rate * 7 hours = 1 whole job done.....so...	1/x * 7 + 1/(x + 2) * 7 = 1. 7/x + 7/(x + 2) = 1. 7 ( x + 2) + 7x. ______________ = 1. x ( x + 2). 14x + 14 = x ( x + 2). 14x + 14 = x^2 + 2x rearrange as. x^2 - 12x - 14 = 0 complete the square on x. x^2 -12x + 36 = 14 + 36. (x - 6)^2 = 50 take the positive root. x - 6 = sqrt (50). x = sqrt (50 ) + 6. x + 2 = sqrt (50 ) + 8 ≈ 15.07 hrs = Peter's time = 15 + .07 *60 ≈ 15 hrs 4 min. Jan 4, 2021. edited by CPhill Jan 5, 2021.
https://stuffsure.com/what-happens-to-the-expected-value-of-m-as-sample-size-increases/	1,723,597,315,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00865.warc.gz	423,439,976	15,008	# The Expected Value of M: What Happens as Sample Size Increases? As sample size increases, the expected value of M (the mean of the distribution of M values) approaches the population mean. This is due to the law of large numbers, which states that as a sample size gets larger, the sample mean gets closer to the population mean. Checkout this video: ## Introduction When we talk about the expected value of M, we are talking about the long-run average value of M. This value can be thought of as the average of all the possible values of M that could be realized. As the sample size increases, the expected value of M will get closer and closer to the population mean. ### What is the Expected Value of M? In probability theory and statistics, the expected value of a random variable is the weighted average of all possible values that this random variable can take. In other words, if you have a six-sided die and you roll it, what is the average of all the numbers that come up? The answer is 3.5. This average value here is known as the expected value. ### What is the Law of Large Numbers? The law of large numbers is a formal statement of the intuitive idea that the average of a sequence of random numbers will tend to be close to the expected value as the number of trials becomes large. More precisely, if X1,…,Xn is a sequence of independent and identically distributed random variables with expected value μ, then the sample mean converges in probability to μ as n approaches infinity: limn→∞P(\|X¯n−μ\|>ε)=0 for all ε>0. ## Sample Size and the Expected Value of M If you’re anything like me, you’ve probably wondered how the expected value of M changes as sample size increases. Does it go up? Does it go down? Does it stay the same? To answer these questions, we’ll need to take a quick look at the definition of expected value. ### How Does Sample Size Affect the Expected Value of M? The Expected Value of M is a statistical tool that measures the center of a data set. It is also known as the mean or average. The Expected Value of M can be affected by two things: the population mean (μ) and the sample size (n). The population mean is the average of all possible values in a population. For example, if you were to take the average height of all American adults, that would be the population mean. The population mean does not change when new data is added. The sample size is the number of values you have in your sample. For example, if you were to take the heights of 100 American adults, that would be your sample size. The sample size can change when new data is added. As the sample size increases, the expected value of M approaches the population mean. This happens because there are more values in the sample and so they better represent the entire population. ### What Happens as Sample Size Increases? As sample size increases, the expected value of M approaches the population mean. This is because, as the number of samples gets larger, the distribution of M becomes more and more Normal. Recall that the Normal distribution is centered at the population mean. Therefore, as the number of samples gets larger, M will tend to be closer and closer to the population mean. ## Conclusion As the sample size increased, the value of M tended to approach 3. This was to be expected, since M is the average of a random sample of numbers drawn from a population with a mean of 3. If the sample size is large enough, the value of M will be very close to 3.	753	3,497	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2024-33	latest	en	0.938374	# The Expected Value of M: What Happens as Sample Size Increases?. As sample size increases, the expected value of M (the mean of the distribution of M values) approaches the population mean. This is due to the law of large numbers, which states that as a sample size gets larger, the sample mean gets closer to the population mean.. Checkout this video:. ## Introduction. When we talk about the expected value of M, we are talking about the long-run average value of M. This value can be thought of as the average of all the possible values of M that could be realized. As the sample size increases, the expected value of M will get closer and closer to the population mean.. ### What is the Expected Value of M?. In probability theory and statistics, the expected value of a random variable is the weighted average of all possible values that this random variable can take. In other words, if you have a six-sided die and you roll it, what is the average of all the numbers that come up? The answer is 3.5. This average value here is known as the expected value.. ### What is the Law of Large Numbers?. The law of large numbers is a formal statement of the intuitive idea that the average of a sequence of random numbers will tend to be close to the expected value as the number of trials becomes large. More precisely, if X1,…,Xn is a sequence of independent and identically distributed random variables with expected value μ, then the sample mean converges in probability to μ as n approaches infinity:. limn→∞P(\|X¯n−μ\|>ε)=0 for all ε>0.. ## Sample Size and the Expected Value of M. If you’re anything like me, you’ve probably wondered how the expected value of M changes as sample size increases. Does it go up? Does it go down? Does it stay the same? To answer these questions, we’ll need to take a quick look at the definition of expected value.. ### How Does Sample Size Affect the Expected Value of M?. The Expected Value of M is a statistical tool that measures the center of a data set.	It is also known as the mean or average. The Expected Value of M can be affected by two things: the population mean (μ) and the sample size (n).. The population mean is the average of all possible values in a population. For example, if you were to take the average height of all American adults, that would be the population mean. The population mean does not change when new data is added.. The sample size is the number of values you have in your sample. For example, if you were to take the heights of 100 American adults, that would be your sample size. The sample size can change when new data is added.. As the sample size increases, the expected value of M approaches the population mean. This happens because there are more values in the sample and so they better represent the entire population.. ### What Happens as Sample Size Increases?. As sample size increases, the expected value of M approaches the population mean. This is because, as the number of samples gets larger, the distribution of M becomes more and more Normal. Recall that the Normal distribution is centered at the population mean. Therefore, as the number of samples gets larger, M will tend to be closer and closer to the population mean.. ## Conclusion. As the sample size increased, the value of M tended to approach 3. This was to be expected, since M is the average of a random sample of numbers drawn from a population with a mean of 3. If the sample size is large enough, the value of M will be very close to 3.
https://aljenan-kw.com/qa/what-is-the-purpose-of-hypothesis.html	1,618,648,748,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038118762.49/warc/CC-MAIN-20210417071833-20210417101833-00291.warc.gz	198,428,307	7,372	# What Is The Purpose Of Hypothesis? ## How do we write a hypothesis? How to Formulate an Effective Research HypothesisState the problem that you are trying to solve. Make sure that the hypothesis clearly defines the topic and the focus of the experiment.Try to write the hypothesis as an if-then statement. Define the variables.. ## What does a hypothesis represent? In science, a hypothesis is an idea or explanation that you then test through study and experimentation. Outside science, a theory or guess can also be called a hypothesis. A hypothesis is something more than a wild guess but less than a well-established theory. … Anyone who uses the word hypothesis is making a guess. ## What are the six steps of hypothesis testing? Step 1: Specify the Null Hypothesis. … Step 2: Specify the Alternative Hypothesis. … Step 3: Set the Significance Level (a) … Step 4: Calculate the Test Statistic and Corresponding P-Value. … Step 5: Drawing a Conclusion. ## Is a hypothesis a prediction? The only interpretation of the term hypothesis needed in science is that of a causal hypothesis, defined as a proposed explanation (and for typically a puzzling observation). A hypothesis is not a prediction. Rather, a prediction is derived from a hypothesis. ## What does P value indicate? In statistics, the p-value is the probability of obtaining results at least as extreme as the observed results of a statistical hypothesis test, assuming that the null hypothesis is correct. … A smaller p-value means that there is stronger evidence in favor of the alternative hypothesis. ## What is the most important step in hypothesis testing? The most important (and often the most challenging) step in hypothesis testing is selecting the test statistic. ## What is the purpose of hypothesis testing? The purpose of hypothesis testing is to determine whether there is enough statistical evidence in favor of a certain belief, or hypothesis, about a parameter. ## What are examples of hypothesis? Examples of Hypothesis:If I replace the battery in my car, then my car will get better gas mileage.If I eat more vegetables, then I will lose weight faster.If I add fertilizer to my garden, then my plants will grow faster.If I brush my teeth every day, then I will not develop cavities.More items… ## What are the two types of hypothesis? A hypothesis is an approximate explanation that relates to the set of facts that can be tested by certain further investigations. There are basically two types, namely, null hypothesis and alternative hypothesis. A research generally starts with a problem. ## What are the 4 steps of hypothesis testing? Step 1: Specify the Null Hypothesis. … Step 2: Specify the Alternative Hypothesis. … Step 3: Set the Significance Level (a) … Step 4: Calculate the Test Statistic and Corresponding P-Value. … Step 5: Drawing a Conclusion. ## What is the purpose of a hypothesis quizlet? What is the purpose of a hypothesis for any study? To provide direction for research by identifying the expected outcome. A hypothesis posed as a declarative statement predicts an expected outcome. ## What is a hypothesis easy definition? A hypothesis is a suggested solution for an unexplained occurrence that does not fit into current accepted scientific theory. The basic idea of a hypothesis is that there is no pre-determined outcome. ## How do you explain hypothesis testing? Key TakeawaysHypothesis testing is used to assess the plausibility of a hypothesis by using sample data.The test provides evidence concerning the plausibility of the hypothesis, given the data.Statistical analysts test a hypothesis by measuring and examining a random sample of the population being analyzed. ## What is Z test used for? A z-test is a statistical test to determine whether two population means are different when the variances are known and the sample size is large. It can be used to test hypotheses in which the z-test follows a normal distribution. A z-statistic, or z-score, is a number representing the result from the z-test.	822	4,055	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5	4	CC-MAIN-2021-17	latest	en	0.927781	# What Is The Purpose Of Hypothesis?. ## How do we write a hypothesis?. How to Formulate an Effective Research HypothesisState the problem that you are trying to solve.. Make sure that the hypothesis clearly defines the topic and the focus of the experiment.Try to write the hypothesis as an if-then statement.. Define the variables... ## What does a hypothesis represent?. In science, a hypothesis is an idea or explanation that you then test through study and experimentation. Outside science, a theory or guess can also be called a hypothesis. A hypothesis is something more than a wild guess but less than a well-established theory. … Anyone who uses the word hypothesis is making a guess.. ## What are the six steps of hypothesis testing?. Step 1: Specify the Null Hypothesis. … Step 2: Specify the Alternative Hypothesis. … Step 3: Set the Significance Level (a) … Step 4: Calculate the Test Statistic and Corresponding P-Value. … Step 5: Drawing a Conclusion.. ## Is a hypothesis a prediction?. The only interpretation of the term hypothesis needed in science is that of a causal hypothesis, defined as a proposed explanation (and for typically a puzzling observation). A hypothesis is not a prediction. Rather, a prediction is derived from a hypothesis.. ## What does P value indicate?. In statistics, the p-value is the probability of obtaining results at least as extreme as the observed results of a statistical hypothesis test, assuming that the null hypothesis is correct. … A smaller p-value means that there is stronger evidence in favor of the alternative hypothesis.. ## What is the most important step in hypothesis testing?. The most important (and often the most challenging) step in hypothesis testing is selecting the test statistic.. ## What is the purpose of hypothesis testing?. The purpose of hypothesis testing is to determine whether there is enough statistical evidence in favor of a certain belief, or hypothesis, about a parameter.	## What are examples of hypothesis?. Examples of Hypothesis:If I replace the battery in my car, then my car will get better gas mileage.If I eat more vegetables, then I will lose weight faster.If I add fertilizer to my garden, then my plants will grow faster.If I brush my teeth every day, then I will not develop cavities.More items…. ## What are the two types of hypothesis?. A hypothesis is an approximate explanation that relates to the set of facts that can be tested by certain further investigations. There are basically two types, namely, null hypothesis and alternative hypothesis. A research generally starts with a problem.. ## What are the 4 steps of hypothesis testing?. Step 1: Specify the Null Hypothesis. … Step 2: Specify the Alternative Hypothesis. … Step 3: Set the Significance Level (a) … Step 4: Calculate the Test Statistic and Corresponding P-Value. … Step 5: Drawing a Conclusion.. ## What is the purpose of a hypothesis quizlet?. What is the purpose of a hypothesis for any study? To provide direction for research by identifying the expected outcome. A hypothesis posed as a declarative statement predicts an expected outcome.. ## What is a hypothesis easy definition?. A hypothesis is a suggested solution for an unexplained occurrence that does not fit into current accepted scientific theory. The basic idea of a hypothesis is that there is no pre-determined outcome.. ## How do you explain hypothesis testing?. Key TakeawaysHypothesis testing is used to assess the plausibility of a hypothesis by using sample data.The test provides evidence concerning the plausibility of the hypothesis, given the data.Statistical analysts test a hypothesis by measuring and examining a random sample of the population being analyzed.. ## What is Z test used for?. A z-test is a statistical test to determine whether two population means are different when the variances are known and the sample size is large. It can be used to test hypotheses in which the z-test follows a normal distribution. A z-statistic, or z-score, is a number representing the result from the z-test.
https://people.richland.edu/james/spring05/m113/tech/tech3.html	1,513,415,480,000,000,000	text/html	crawl-data/CC-MAIN-2017-51/segments/1512948587496.62/warc/CC-MAIN-20171216084601-20171216110601-00525.warc.gz	616,286,535	1,897	# Math 113 - Technology Project 3 Inferential Statistics ## Group Members 1. __________________________ 2. __________________________ 3. __________________________ ## Project Follow the Minitab instructions under the technology exercises link on the website for information about how to do the problems. 1. The First Amendment Center in Nashville, TN, conducted a telephone survey of 1002 adults between May 6 and June 6, 2004, and found that 58% of Americans feel the news media is biased in their reporting. http://www.firstamendmentcenter.org/PDF/SOFA2004results.pdf 1. Simulate 200 samples of size n=1002 and a probability of success of p=0.58. 2. Find the 95% confidence interval for each of the 200 samples found in part a. What percent of them contain the claimed proportion of p=0.58? 3. Find the mean of each of the 200 samples from part a. and create a graphical summary of the means. Compare the results of your 200 samples to the three parts of the sampling distribution on page 340. 2. Demonstrate the Central Limit Theorem. 1. Create a discrete probability distribution with at least 5 unique values for x. 2. Find the mean, variance, and standard deviation for your probability distribution using the formulas from chapter 16. 3. Generate 1000 samples of size 4, 25, and 100 from a population with your distribution. Find the mean of each sample and describe it graphically. Compare the results of your 1000 samples to the three parts of the sampling distribution on page 345. 3. Use Minitab to create graphs demonstrating the graph from a hypothesis test. 1. Duplicate the graph to the right that demonstrates a two tail test with α=0.05. 2. Create a graph demonstrating a right tail test with α=0.05. 3. Create a graph demonstrating a left tail test with α=0.05.	418	1,787	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.796875	4	CC-MAIN-2017-51	latest	en	0.841458	# Math 113 - Technology Project 3. Inferential Statistics. ## Group Members. 1. __________________________. 2. __________________________. 3. __________________________. ## Project. Follow the Minitab instructions under the technology exercises link on the website for information about how to do the problems.. 1. The First Amendment Center in Nashville, TN, conducted a telephone survey of 1002 adults between May 6 and June 6, 2004, and found that 58% of Americans feel the news media is biased in their reporting. http://www.firstamendmentcenter.org/PDF/SOFA2004results.pdf. 1. Simulate 200 samples of size n=1002 and a probability of success of p=0.58.. 2. Find the 95% confidence interval for each of the 200 samples found in part a. What percent of them contain the claimed proportion of p=0.58?. 3. Find the mean of each of the 200 samples from part a.	and create a graphical summary of the means. Compare the results of your 200 samples to the three parts of the sampling distribution on page 340.. 2. Demonstrate the Central Limit Theorem.. 1. Create a discrete probability distribution with at least 5 unique values for x.. 2. Find the mean, variance, and standard deviation for your probability distribution using the formulas from chapter 16.. 3. Generate 1000 samples of size 4, 25, and 100 from a population with your distribution. Find the mean of each sample and describe it graphically. Compare the results of your 1000 samples to the three parts of the sampling distribution on page 345.. 3. Use Minitab to create graphs demonstrating the graph from a hypothesis test.. 1. Duplicate the graph to the right that demonstrates a two tail test with α=0.05.. 2. Create a graph demonstrating a right tail test with α=0.05.. 3. Create a graph demonstrating a left tail test with α=0.05.
https://www.archivemore.com/which-one-is-bigger-cl-or-l/	1,713,995,014,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296819971.86/warc/CC-MAIN-20240424205851-20240424235851-00654.warc.gz	564,972,521	8,879	## Which one is bigger Cl or L? Definition: Centilitre A centilitre (cL or cl) a metric unit of volume that is equal to one hundredth of a litre and is equal to a little more than six tenths (0.6102) of acubic inch, or one third (0.338) of a fluid ounce. “Liter” is American spelling and “Litre” is international spelling. 1 L = 1000 mL. ## Is CL bigger than mL? A centiliter is larger than a milliliter. Simply put, cl is larger than ml. Since a centiliter is 10^1 larger than a milliliter, it means that the conversion factor for cl to ml is 10^1. ## How many moles are in CL? The answer is 35.453. We assume you are converting between grams Cl and mole. You can view more details on each measurement unit: molecular weight of Cl or mol The SI base unit for amount of substance is the mole. 1 grams Cl is equal to 0.028206357713029 mole. 2 mole ## How do you find moles of Cl in AgCl? The mass of silver chloride precipitated is used to calculate: 1. (i) moles of AgCl(s) precipitated. moles = mass ÷ molar mass. 2. (ii) moles of chloride ion, Cl-, present. mole ratio AgCl:Cl- is 1:1. 3. (iii) mass of chloride ion, Cl-, present in the seawater. mass = moles × molar mass. ## How many moles are there in 70.9 g of CL? 39.2 g Cl2(1 mole of Cl270.9 g Cl2)(6.022⋅1023molecules Cl21 mole Cl2)=3.33⋅1023 molecules Cl2 . Hence, the answer is 3.33⋅1023 molecules of Cl2. Note that 6.022⋅1023 is also known as Avogadro’s number, and it can be referred to as the number of molecules in one mole of that substance. ## How many moles of CL are in NaCl? A mole of NaCl contains a mole of sodium ions and a mole of chloride ions. Since a mole of sodium has a mass of 22.990 g and a mole of chlorine has a mass of 35.45 g, the molar mass of NaCl is 58.44 g/mol. ## How many atoms does CL have? Name Chlorine Atomic Number 17 Atomic Mass 35.453 atomic mass units Number of Protons 17 Number of Neutrons 18 12.78g 22.989769 u 24.305 u ## Is sodium a solid? Sodium is a chemical element with symbol Na and atomic number 11. Classified as an alkali metal, Sodium is a solid at room temperature. ## Where do we find sodium? Sodium is the sixth most abundant element on Earth. It is never found in its pure form because it is so reactive. It is only found in compounds such as sodium chloride (NaCL) or table salt. Sodium chloride is found in ocean water (salt water), salt lakes, and underground deposits. ## Is Cl A chlorine? Chlorine (Cl), chemical element, the second lightest member of the halogen elements, or Group 17 (Group VIIa) of the periodic table. Chlorine is a toxic, corrosive, greenish yellow gas that is irritating to the eyes and to the respiratory system. ## How does sodium look like? Sodium is a very soft silvery-white metal. Sodium is the most common alkali metal and the sixth most abundant element on Earth, comprising 2.8 percent of Earth’s crust. ## How do we use sodium in everyday life? Sodium is used as a heat exchanger in some nuclear reactors, and as a reagent in the chemicals industry. But sodium salts have more uses than the metal itself. The most common compound of sodium is sodium chloride (common salt). It is added to food and used to de-ice roads in winter. ## How do you purify sodium? The only purification method is to wait for its natural decay. Sodium hydride and oxide (Na2O, NaH) are crystallising in liquid sodium when cooling, that produces the necessary sodium sur-saturation against hydrogen and oxygen to let the nucleation mechanism happen, and then to allow for crystal growth. ## What makes sodium unique? It’s a soft metal, reactive and with a low melting point, with a relative density of 0,97 at 20ºC (68ºF). From the commercial point of view, sodium is the most important of all the alkaline metals. Sodium reacts quickly with water, and also with snow and ice, to produce sodium hydroxide and hydrogen.	1,035	3,880	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2024-18	latest	en	0.88541	## Which one is bigger Cl or L?. Definition: Centilitre A centilitre (cL or cl) a metric unit of volume that is equal to one hundredth of a litre and is equal to a little more than six tenths (0.6102) of acubic inch, or one third (0.338) of a fluid ounce. “Liter” is American spelling and “Litre” is international spelling. 1 L = 1000 mL.. ## Is CL bigger than mL?. A centiliter is larger than a milliliter. Simply put, cl is larger than ml. Since a centiliter is 10^1 larger than a milliliter, it means that the conversion factor for cl to ml is 10^1.. ## How many moles are in CL?. The answer is 35.453. We assume you are converting between grams Cl and mole. You can view more details on each measurement unit: molecular weight of Cl or mol The SI base unit for amount of substance is the mole. 1 grams Cl is equal to 0.028206357713029 mole.. 2 mole. ## How do you find moles of Cl in AgCl?. The mass of silver chloride precipitated is used to calculate:. 1. (i) moles of AgCl(s) precipitated. moles = mass ÷ molar mass.. 2. (ii) moles of chloride ion, Cl-, present. mole ratio AgCl:Cl- is 1:1.. 3. (iii) mass of chloride ion, Cl-, present in the seawater. mass = moles × molar mass.. ## How many moles are there in 70.9 g of CL?. 39.2 g Cl2(1 mole of Cl270.9 g Cl2)(6.022⋅1023molecules Cl21 mole Cl2)=3.33⋅1023 molecules Cl2 . Hence, the answer is 3.33⋅1023 molecules of Cl2. Note that 6.022⋅1023 is also known as Avogadro’s number, and it can be referred to as the number of molecules in one mole of that substance.. ## How many moles of CL are in NaCl?. A mole of NaCl contains a mole of sodium ions and a mole of chloride ions. Since a mole of sodium has a mass of 22.990 g and a mole of chlorine has a mass of 35.45 g, the molar mass of NaCl is 58.44 g/mol.. ## How many atoms does CL have?. Name Chlorine.	Atomic Number 17. Atomic Mass 35.453 atomic mass units. Number of Protons 17. Number of Neutrons 18. 12.78g. 22.989769 u. 24.305 u. ## Is sodium a solid?. Sodium is a chemical element with symbol Na and atomic number 11. Classified as an alkali metal, Sodium is a solid at room temperature.. ## Where do we find sodium?. Sodium is the sixth most abundant element on Earth. It is never found in its pure form because it is so reactive. It is only found in compounds such as sodium chloride (NaCL) or table salt. Sodium chloride is found in ocean water (salt water), salt lakes, and underground deposits.. ## Is Cl A chlorine?. Chlorine (Cl), chemical element, the second lightest member of the halogen elements, or Group 17 (Group VIIa) of the periodic table. Chlorine is a toxic, corrosive, greenish yellow gas that is irritating to the eyes and to the respiratory system.. ## How does sodium look like?. Sodium is a very soft silvery-white metal. Sodium is the most common alkali metal and the sixth most abundant element on Earth, comprising 2.8 percent of Earth’s crust.. ## How do we use sodium in everyday life?. Sodium is used as a heat exchanger in some nuclear reactors, and as a reagent in the chemicals industry. But sodium salts have more uses than the metal itself. The most common compound of sodium is sodium chloride (common salt). It is added to food and used to de-ice roads in winter.. ## How do you purify sodium?. The only purification method is to wait for its natural decay. Sodium hydride and oxide (Na2O, NaH) are crystallising in liquid sodium when cooling, that produces the necessary sodium sur-saturation against hydrogen and oxygen to let the nucleation mechanism happen, and then to allow for crystal growth.. ## What makes sodium unique?. It’s a soft metal, reactive and with a low melting point, with a relative density of 0,97 at 20ºC (68ºF). From the commercial point of view, sodium is the most important of all the alkaline metals. Sodium reacts quickly with water, and also with snow and ice, to produce sodium hydroxide and hydrogen.
https://convertoctopus.com/49-grams-to-pounds	1,660,272,670,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882571538.36/warc/CC-MAIN-20220812014923-20220812044923-00192.warc.gz	200,447,396	7,469	## Conversion formula The conversion factor from grams to pounds is 0.0022046226218488, which means that 1 gram is equal to 0.0022046226218488 pounds: 1 g = 0.0022046226218488 lb To convert 49 grams into pounds we have to multiply 49 by the conversion factor in order to get the mass amount from grams to pounds. We can also form a simple proportion to calculate the result: 1 g → 0.0022046226218488 lb 49 g → M(lb) Solve the above proportion to obtain the mass M in pounds: M(lb) = 49 g × 0.0022046226218488 lb M(lb) = 0.10802650847059 lb The final result is: 49 g → 0.10802650847059 lb We conclude that 49 grams is equivalent to 0.10802650847059 pounds: 49 grams = 0.10802650847059 pounds ## Alternative conversion We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 9.2569871428571 × 49 grams. Another way is saying that 49 grams is equal to 1 ÷ 9.2569871428571 pounds. ## Approximate result For practical purposes we can round our final result to an approximate numerical value. We can say that forty-nine grams is approximately zero point one zero eight pounds: 49 g ≅ 0.108 lb An alternative is also that one pound is approximately nine point two five seven times forty-nine grams. ## Conversion table ### grams to pounds chart For quick reference purposes, below is the conversion table you can use to convert from grams to pounds grams (g) pounds (lb) 50 grams 0.11 pounds 51 grams 0.112 pounds 52 grams 0.115 pounds 53 grams 0.117 pounds 54 grams 0.119 pounds 55 grams 0.121 pounds 56 grams 0.123 pounds 57 grams 0.126 pounds 58 grams 0.128 pounds 59 grams 0.13 pounds	464	1,658	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.0625	4	CC-MAIN-2022-33	latest	en	0.803458	## Conversion formula. The conversion factor from grams to pounds is 0.0022046226218488, which means that 1 gram is equal to 0.0022046226218488 pounds:. 1 g = 0.0022046226218488 lb. To convert 49 grams into pounds we have to multiply 49 by the conversion factor in order to get the mass amount from grams to pounds. We can also form a simple proportion to calculate the result:. 1 g → 0.0022046226218488 lb. 49 g → M(lb). Solve the above proportion to obtain the mass M in pounds:. M(lb) = 49 g × 0.0022046226218488 lb. M(lb) = 0.10802650847059 lb. The final result is:. 49 g → 0.10802650847059 lb. We conclude that 49 grams is equivalent to 0.10802650847059 pounds:. 49 grams = 0.10802650847059 pounds. ## Alternative conversion. We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 9.2569871428571 × 49 grams.. Another way is saying that 49 grams is equal to 1 ÷ 9.2569871428571 pounds.. ## Approximate result.	For practical purposes we can round our final result to an approximate numerical value. We can say that forty-nine grams is approximately zero point one zero eight pounds:. 49 g ≅ 0.108 lb. An alternative is also that one pound is approximately nine point two five seven times forty-nine grams.. ## Conversion table. ### grams to pounds chart. For quick reference purposes, below is the conversion table you can use to convert from grams to pounds. grams (g) pounds (lb). 50 grams 0.11 pounds. 51 grams 0.112 pounds. 52 grams 0.115 pounds. 53 grams 0.117 pounds. 54 grams 0.119 pounds. 55 grams 0.121 pounds. 56 grams 0.123 pounds. 57 grams 0.126 pounds. 58 grams 0.128 pounds. 59 grams 0.13 pounds.
http://www.algebra.com/algebra/homework/equations/Equations.faq.question.127995.html	1,368,918,464,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368696382920/warc/CC-MAIN-20130516092622-00033-ip-10-60-113-184.ec2.internal.warc.gz	306,048,802	5,049	# SOLUTION: Tell wether the ordered pair is a solution of the linear system. (1,-1); 2x - y = 3 4x + 2y = 2 Any help with this equation would be <b>Greatly</b> appreciated. Algebra -> Algebra -> Equations -> SOLUTION: Tell wether the ordered pair is a solution of the linear system. (1,-1); 2x - y = 3 4x + 2y = 2 Any help with this equation would be <b>Greatly</b> appreciated. Log On Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Algebra: Equations Solvers Lessons Answers archive Quiz In Depth Question 127995: Tell wether the ordered pair is a solution of the linear system. (1,-1); 2x - y = 3 4x + 2y = 2 Any help with this equation would be Greatly appreciated.Answer by JessicaGill(40) (Show Source): You can put this solution on YOUR website!There are two ways you can solve this. By graphing to see if the ordered pair (1, -1) falls on both lines, or by simply substituting the x and y coordinates into both equations and seeing if they prove true. Substitution is shown below To substitute take your first equation and substitute 1 for x and -1 for y would be which simplifies to so the ordered pair is a solution for the first equation. Lets check the second, substituting the values in for the x and y coordinates Simplfy this down to so the ordered pair (1,-1) is a solution for both equations. If you would like me to show you the graphing steps, please let me know. thanks, Ms. G	406	1,545	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4	4	CC-MAIN-2013-20	latest	en	0.883373	# SOLUTION: Tell wether the ordered pair is a solution of the linear system. (1,-1); 2x - y = 3 4x + 2y = 2 Any help with this equation would be <b>Greatly</b> appreciated.. Algebra -> Algebra -> Equations -> SOLUTION: Tell wether the ordered pair is a solution of the linear system. (1,-1); 2x - y = 3 4x + 2y = 2 Any help with this equation would be <b>Greatly</b> appreciated. Log On. Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!. Algebra: Equations Solvers Lessons Answers archive Quiz In Depth. Question 127995: Tell wether the ordered pair is a solution of the linear system. (1,-1); 2x - y = 3 4x + 2y = 2 Any help with this equation would be Greatly appreciated.Answer by JessicaGill(40) (Show Source): You can put this solution on YOUR website!There are two ways you can solve this.	By graphing to see if the ordered pair (1, -1) falls on both lines, or by simply substituting the x and y coordinates into both equations and seeing if they prove true. Substitution is shown below To substitute take your first equation and substitute 1 for x and -1 for y would be which simplifies to so the ordered pair is a solution for the first equation. Lets check the second, substituting the values in for the x and y coordinates Simplfy this down to so the ordered pair (1,-1) is a solution for both equations. If you would like me to show you the graphing steps, please let me know. thanks, Ms. G.
https://answerlic.com/what-is-the-dimension-of-the-eigenspace/	1,686,066,924,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224652959.43/warc/CC-MAIN-20230606150510-20230606180510-00548.warc.gz	117,964,477	20,975	11:15 am Science What Is The Dimension Of The Eigenspace? This is a slightly more complicated question. We know that the eigenvectors of a matrix are the columns of the matrix. So if we want to find out how many of these columns there are, we have to look at the dimension of each column. Now, let’s look at our matrix A: The first column has three elements (1 + i + j), which means that it has three dimensions. The second column has one element (1), so it only has one dimension. The third column has two elements (2 + 3j), so it has two dimensions. In total, our matrix has four dimensions: three for each column and one for the entire row. The eigenspace for a given pair of numbers generally does not have those exact same dimensions as those numbers do individually; it is only guaranteed to be “close” to them in this way. (Visited 3 times, 1 visits today)	204	869	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.609375	4	CC-MAIN-2023-23	latest	en	0.953314	11:15 am Science. What Is The Dimension Of The Eigenspace?. This is a slightly more complicated question. We know that the eigenvectors of a matrix are the columns of the matrix. So if we want to find out how many of these columns there are, we have to look at the dimension of each column.. Now, let’s look at our matrix A:. The first column has three elements (1 + i + j), which means that it has three dimensions.	The second column has one element (1), so it only has one dimension. The third column has two elements (2 + 3j), so it has two dimensions. In total, our matrix has four dimensions: three for each column and one for the entire row.. The eigenspace for a given pair of numbers generally does not have those exact same dimensions as those numbers do individually; it is only guaranteed to be “close” to them in this way.. (Visited 3 times, 1 visits today).
https://metanumbers.com/12063	1,669,541,426,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710218.49/warc/CC-MAIN-20221127073607-20221127103607-00229.warc.gz	438,607,599	7,299	# 12063 (number) 12,063 (twelve thousand sixty-three) is an odd five-digits composite number following 12062 and preceding 12064. In scientific notation, it is written as 1.2063 × 104. The sum of its digits is 12. It has a total of 2 prime factors and 4 positive divisors. There are 8,040 positive integers (up to 12063) that are relatively prime to 12063. ## Basic properties • Is Prime? No • Number parity Odd • Number length 5 • Sum of Digits 12 • Digital Root 3 ## Name Short name 12 thousand 63 twelve thousand sixty-three ## Notation Scientific notation 1.2063 × 104 12.063 × 103 ## Prime Factorization of 12063 Prime Factorization 3 × 4021 Composite number Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 12063 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0 The prime factorization of 12,063 is 3 × 4021. Since it has a total of 2 prime factors, 12,063 is a composite number. ## Divisors of 12063 1, 3, 4021, 12063 4 divisors Even divisors 0 4 2 2 Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 16088 Sum of all the positive divisors of n s(n) 4025 Sum of the proper positive divisors of n A(n) 4022 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 109.832 Returns the nth root of the product of n divisors H(n) 2.99925 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors The number 12,063 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 12,063) is 16,088, the average is 4,022. ## Other Arithmetic Functions (n = 12063) 1 φ(n) n Euler Totient Carmichael Lambda Prime Pi φ(n) 8040 Total number of positive integers not greater than n that are coprime to n λ(n) 4020 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1450 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares There are 8,040 positive integers (less than 12,063) that are coprime with 12,063. And there are approximately 1,450 prime numbers less than or equal to 12,063. ## Divisibility of 12063 m n mod m 2 3 4 5 6 7 8 9 1 0 3 3 3 2 7 3 The number 12,063 is divisible by 3. ## Classification of 12063 • Arithmetic • Semiprime • Deficient • Polite • Square Free ### Other numbers • LucasCarmichael ## Base conversion (12063) Base System Value 2 Binary 10111100011111 3 Ternary 121112210 4 Quaternary 2330133 5 Quinary 341223 6 Senary 131503 8 Octal 27437 10 Decimal 12063 12 Duodecimal 6b93 20 Vigesimal 1a33 36 Base36 9b3 ## Basic calculations (n = 12063) ### Multiplication n×y n×2 24126 36189 48252 60315 ### Division n÷y n÷2 6031.5 4021 3015.75 2412.6 ### Exponentiation ny n2 145515969 1755359134047 21174897234008961 255432785333850096543 ### Nth Root y√n 2√n 109.832 22.9343 10.4801 6.55075 ## 12063 as geometric shapes ### Circle Diameter 24126 75794.1 4.57152e+08 ### Sphere Volume 7.35283e+12 1.82861e+09 75794.1 ### Square Length = n Perimeter 48252 1.45516e+08 17059.7 ### Cube Length = n Surface area 8.73096e+08 1.75536e+12 20893.7 ### Equilateral Triangle Length = n Perimeter 36189 6.30103e+07 10446.9 ### Triangular Pyramid Length = n Surface area 2.52041e+08 2.06871e+11 9849.4 ## Cryptographic Hash Functions md5 887c5a2b0d1bcf5b169ac8d63806bd13 e08e1a4c5e250715736695c3f5f7059dd6ee983d 9aa3cb6ab517a4eac5157d3a2263c1ed4c54b12ab6bfca683cba084ee56b009f ba657ff2d8bbc647d87e8604393d573ecc8564ef4dc9fc2b9c51a6dfcf29b598e5b9445331f64304375a69fa04115fe8557a97bb57f190e12b49f63f08200072 6c0442831911c38b6e2b6ac07836c2fd61725cb7	1,452	4,082	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.8125	4	CC-MAIN-2022-49	latest	en	0.802103	# 12063 (number). 12,063 (twelve thousand sixty-three) is an odd five-digits composite number following 12062 and preceding 12064. In scientific notation, it is written as 1.2063 × 104. The sum of its digits is 12. It has a total of 2 prime factors and 4 positive divisors. There are 8,040 positive integers (up to 12063) that are relatively prime to 12063.. ## Basic properties. • Is Prime? No. • Number parity Odd. • Number length 5. • Sum of Digits 12. • Digital Root 3. ## Name. Short name 12 thousand 63 twelve thousand sixty-three. ## Notation. Scientific notation 1.2063 × 104 12.063 × 103. ## Prime Factorization of 12063. Prime Factorization 3 × 4021. Composite number. Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 12063 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0. The prime factorization of 12,063 is 3 × 4021. Since it has a total of 2 prime factors, 12,063 is a composite number.. ## Divisors of 12063. 1, 3, 4021, 12063. 4 divisors. Even divisors 0 4 2 2. Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 16088 Sum of all the positive divisors of n s(n) 4025 Sum of the proper positive divisors of n A(n) 4022 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 109.832 Returns the nth root of the product of n divisors H(n) 2.99925 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors. The number 12,063 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 12,063) is 16,088, the average is 4,022.. ## Other Arithmetic Functions (n = 12063). 1 φ(n) n. Euler Totient Carmichael Lambda Prime Pi φ(n) 8040 Total number of positive integers not greater than n that are coprime to n λ(n) 4020 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1450 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares. There are 8,040 positive integers (less than 12,063) that are coprime with 12,063. And there are approximately 1,450 prime numbers less than or equal to 12,063.. ## Divisibility of 12063. m n mod m 2 3 4 5 6 7 8 9 1 0 3 3 3 2 7 3. The number 12,063 is divisible by 3.. ## Classification of 12063. • Arithmetic. • Semiprime. • Deficient. • Polite. • Square Free. ### Other numbers. • LucasCarmichael. ## Base conversion (12063).	Base System Value. 2 Binary 10111100011111. 3 Ternary 121112210. 4 Quaternary 2330133. 5 Quinary 341223. 6 Senary 131503. 8 Octal 27437. 10 Decimal 12063. 12 Duodecimal 6b93. 20 Vigesimal 1a33. 36 Base36 9b3. ## Basic calculations (n = 12063). ### Multiplication. n×y. n×2 24126 36189 48252 60315. ### Division. n÷y. n÷2 6031.5 4021 3015.75 2412.6. ### Exponentiation. ny. n2 145515969 1755359134047 21174897234008961 255432785333850096543. ### Nth Root. y√n. 2√n 109.832 22.9343 10.4801 6.55075. ## 12063 as geometric shapes. ### Circle. Diameter 24126 75794.1 4.57152e+08. ### Sphere. Volume 7.35283e+12 1.82861e+09 75794.1. ### Square. Length = n. Perimeter 48252 1.45516e+08 17059.7. ### Cube. Length = n. Surface area 8.73096e+08 1.75536e+12 20893.7. ### Equilateral Triangle. Length = n. Perimeter 36189 6.30103e+07 10446.9. ### Triangular Pyramid. Length = n. Surface area 2.52041e+08 2.06871e+11 9849.4. ## Cryptographic Hash Functions. md5 887c5a2b0d1bcf5b169ac8d63806bd13 e08e1a4c5e250715736695c3f5f7059dd6ee983d 9aa3cb6ab517a4eac5157d3a2263c1ed4c54b12ab6bfca683cba084ee56b009f ba657ff2d8bbc647d87e8604393d573ecc8564ef4dc9fc2b9c51a6dfcf29b598e5b9445331f64304375a69fa04115fe8557a97bb57f190e12b49f63f08200072 6c0442831911c38b6e2b6ac07836c2fd61725cb7.
https://aptitude.gateoverflow.in/2034/cat-2003-question-2-89	1,675,020,167,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764499758.83/warc/CC-MAIN-20230129180008-20230129210008-00013.warc.gz	128,026,163	20,314	220 views If three positive real numbers $x, y, z$ satisfy $y – x = z – y$ and $x y z = 4,$ then what is the minimum possible value of $y?$ 1. $2^{\frac{1}{3}}$ 2. $2^{\frac{2}{3}}$ 3. $2^{\frac{1}{4}}$ 4. $2^{\frac{3}{4}}$ y - x = z - y x + z = 2y$\dots(1)$ xyz =4 $\dots(2)$ We know that, AM >= GM Hence, $\frac{x+y+z}{3}$ >= $(xyz)^{\frac{1}{3}}$ From (1) and (2) $\frac{3y}{3}$ >= $4^{\frac{1}{3}}$ y >= $2^{\frac{2}{3}}$ Hence,Option(B)$2^{\frac{2}{3}}$ is the correct choice. 11.1k points 1 381 views	239	521	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.90625	4	CC-MAIN-2023-06	latest	en	0.4181	220 views. If three positive real numbers $x, y, z$ satisfy $y – x = z – y$ and $x y z = 4,$ then what is the minimum possible value of $y?$. 1. $2^{\frac{1}{3}}$. 2. $2^{\frac{2}{3}}$. 3. $2^{\frac{1}{4}}$. 4. $2^{\frac{3}{4}}$. y - x = z - y. x + z = 2y$\dots(1)$.	xyz =4 $\dots(2)$. We know that, AM >= GM. Hence, $\frac{x+y+z}{3}$ >= $(xyz)^{\frac{1}{3}}$. From (1) and (2). $\frac{3y}{3}$ >= $4^{\frac{1}{3}}$. y >= $2^{\frac{2}{3}}$. Hence,Option(B)$2^{\frac{2}{3}}$ is the correct choice.. 11.1k points. 1. 381 views.
https://meritbatch.com/cbse-sample-papers-for-class-11-chemistry-set-6/	1,680,195,370,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00761.warc.gz	445,109,189	20,806	Students must start practicing the questions from CBSE Sample Papers for Class 11 Chemistry with Solutions Set 6 are designed as per the revised syllabus. ## CBSE Sample Papers for Class 11 Chemistry Set 6 with Solutions Time Allowed: 3 hours Maximum Marks: 70 General Instructions: • All questions are compulsory. • This question paper contains 37 questions. • Questions 1-20 in Section A are objective type-very short answer type questions carrying 1 mark each. • Questions 21 – 27 in Section B are short answer type questions carrying 2 marks each. • Questions 28 – 34 in Section C are long-answer 1 type questions carrying 3 marks each. • Questions 35 – 37 in Section D are long-answer 11 type questions carrying 5 marks each. • There is no overall choice. However, an internal choice has been provided in six questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks. You must attempt only one of the choices in such questions. • Use log tables, if necessary. Use of calculator is not allowed. Section – A Question 1. What is the mass percent of carbon in carbon dioxide? [1] (A) 0.034% (B) 27.27% (C) 3.4% (D) 28.7% OR The empirical formula and molecular mass of a compound are CH2O and 180 g respectively. What will be the molecular formula of the compound? (A) C9H18O9 (B) CH2O (C) C6H12O6 (D) C2H4O2 Option (B) is correct. Explanation: Molecular mass of CO2 = 12 + 2(16) = 12 + 32 = 44 g 44 g of CO2 contains 12 g atoms of carbon. Mass percent of carbon = $$\frac{\text { Mass of carbon in } \mathrm{CO}_2}{\text { Molar mass of } \mathrm{CO}_2}$$ × 100s = $$\frac{12}{44}$$ × 100 = 27.27% OR Option (C) is correct. Explanation: Empirical formula mass (CH2O) = 12 + 2(1) + 16 = 30g Molecular mass = 180g n = $$\frac{\text { Molecular mass }}{\text { Empirical formula mass }}$$ = $$\frac{180}{30}$$ = 6 Molecular formula= n × empirical formula = 6 × CH2O = C6H12O6 Question 2. On the basis of thermochemical equations (i), (ii) and (iii), find out which of the algebraic relationship given in options (A) to (D) is correct. [1] (i) C (graphite) + O2(g) → C02(g); ΔfH = x kj mol-1 (ii) C (graphite) + $$\frac{1}{2}$$O2 (g) → CO(g); ΔrH = y kj mol-1 (iii) CO (g) + $$\frac{1}{2}$$O2 (g) → C02 (g) ;ΔH = z kj mol-1 (A) z = x + y (B) x = y – z (C) x = y + z (D) y = 2z – x Option (C) is correct. Explanation: The algebric relationships of the given reaction is Eq.(i) – Eq.(ii) = Eq.(iii) (i) C(graphite) + O2(g) → CO2(g) ; ΔrH = x kJ mol-1 (ii) C(graphite) + $$\frac{1}{2}$$O2(g) → CO(g) ; ΔrH = V kJ (iii) CO(g) + $$\frac{1}{2}$$O2(g) → CO2(g) ; ΔrH = z kJ mol-1 Hence, x – y = z or x = y + z Question 3. Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of dipoles possess ‘partial charges’. The partial charge is : [1] (A) more than unit electronic charge. (B) equal to unit electronic charge. (C) less than unit electronic charge. (D) double the unit electronic charge. Option (C) is correct. Explanation: Partial charge is the small charge developed by displacement of electrons. Question 4. The first ionisation enthalpies of Na, Mg, Al and Si are in the order : [1] (A) Na < Mg > Al < Si (B) Na > Mg > Al > Si (C) Na < Mg < Al < Si (D) Na > Mg > Al < Si OR Which of the following is the correct order of size of the given species: (A) I > I > I+ (B) I+ > I > I (C) I > I+ > I (D) I > I > I+ Option (A) is correct. Explanation: On moving from left to right in a period first ionisation enthalpy increases normally. But, in Mg, as an electron needs to be removed from fully filled s orbital therefore, first ionisation enthalpy of Mg is more than Al. OR Option (D) is correct. Explanation : Size of cation is smaller while that of anion is bigger than its parent atom. Question 5. What will be correct order of vapour pressure of water, acetone and ether at 30°C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point? [1] (A) Water < ether <acetone (B) Water < acetone < ether (C) Ether < acetone < water (D) Acetone < ether < water Option (B) is correct. Explanation: Greater is the boiling point, less is the vapour pressure. Question 6. Which of the following carbocation is most stable? (B) (CH3)3C+ (C) CH3CH2+CH2 (D) CH3+CHCH2CH3 Option (B) is correct. Explanation: Tertiary carbocation is most stable. The variables Pressure, Volume, Concentration and Temperature may change the State of Equilibrium. The change is governed by the Le-Chatelier’s principle. The decomposition of NH3(g) can be made spontaneous by increasing the temperature and lowering pressure. In the reaction, removal of any product from the reaction mixture makes the reversible reaction irreversible and therefore, reaction proceeds to completion. Answer the questions (7) to (10) given below: Question 7. The equilibrium Solid → Liquid → Gas will shift in forward direction when: (A) temperature is increased (B) temperature is lowered (C) pressure is increased (D) pressure is lowered Option (A) is correct. Explanation: Increase in temperature will favour the forward reaction which is endothermic. Question 8. Change in free energy for the equilibrium, gaseous reaction, PCl5 → PCl3 + Cl2 on addition of an inert gas at constant pressure and at constant volume is respectively: (A) decrease, no change (B) increase, no change (C) no change, no change (D) no change, decrease Option (A) is correct. Explanation: Addition of inert gas at constant volume has no effect on any equilibrium reaction. When inert gas is added at constant pressure in the given reaction, it proceeds in forward direction. DG° remains constant, but DG decreases as the reaction proceeds spontaneously in forward direction to attain equilibrium. Question 9. At 25°C, the equilibrium constant K1 and I(z are for the reactions: Which of the following shows the relation between two equilibrium constants? (A) K2 = K2 (B) K2 = 1/(K) (D) K1 = 1/(K) (D) K1 = 1/K2 Option (C) is correct. Explanation: Question 10. A liquid is in equilibrium with its vapour at its boiling point. On an average, the molecules in the two phases have equal: (A) Intermolecular forces (B) Potential energy (C) Kinetic energy (D) None of these Option (C) is correct. Explanation: At boiling point, liquid and vapour phases both are present. So, the molecules in the two phases have equal kinetic energy. Question 11. The addition of HCl to an alkene proceeds in two steps. The first step is the attack of W ion to >C =C< portion which can be shown as A I Option (A) is correct. Explanation: Since double bond is a source of electrons and the charge flows from source of more electron density, therefore, electrons of the double bond attack the proton. Question 12. How will you convert: ethyne to but-2-yne. [1] Students directly write the products and forget to mention side products and reaction conditions. Make a list of important conversions and learn them. Understand each step and conditions involved during conversion reactions. Question 13. The structure of triphenylmethyl cation is given here. This is very stable and some of its salts can be stored for months. Explain the cause of high stability of this cation. Triphenylmethyl cation is a tertiary carbocation which can show nine possible canonical structures and hence is very stable. The three canonical forms for one benzene ring are shown below. Question 14. Can we separate two liquids A (b.p. 353 K) and B (b.p. 365 K) present in a mixture by simple distillation? [1] No, because in simple distillation, vapours of both the liquids will be formed simultaneously and will condense together in receiver as the difference between the boiling points is very less. They can be separated by fractional distillation. Students sometimes answer in yes or no without giving legit explanation. The alternate method of separation also must be mentioned. Question 15. What do you mean by cracking? [1] The thermal decomposition of higher hydrocarbons into lower hydrocarbons in the presence or absence of a catalyst is called cracking. Question 16. Assign the oxidation number to Cr in K2Cr2O7. Let the oxidation number of Cr be x 2 × (+1) + 2x + 7 × (- 2) = 0 2 + 2x – 14 = 0 2x – 12 = 0 x = $$\frac{12}{2}$$ = +6 Oxidation number of Cr in K2Cr2O7 = +6 Question 17. Assertion (A): Toluene on Friedal-Crafts methylation gives o- and p-xylene. Reason (R): CH3-group bonded to benzene ring increases electron density at o- and p- position. (A) Both A and R are correct and R is the correct explanation of A. (B) Both A and R are correct but R is not the correct explanation of A. (C) Both A and R are not correct. (D) A is not correct but R is correct. Option (A) is correct. Question 18. Assertion (A): Benzene on heating with cone. H2SO4 gives benzene sulphonic acid which when heated with superheated steam under pressure gives benzene. [1] Reason (R): Sulphonation is a reversible process. (A) Both A and R are correct and R is the correct explanation of A. (B) Both A and R are correct but R is not the correct explanation of A. (C) Both A and R are not correct. (D) A is not correct but R is correct. Option (A) is correct. Explanation: Sulphonation of benzene is an electrophilic substitution reaction in which SO3 acts as the electrophile. Question 19. Why there is large number of lines in hydrogen spectrum? [1] Large number of lines are there in hydrogen spec¬trum because different possible transitions can be observed which leads to large number of spectral lines. Question 20. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: 1 Pa = 1 Nm-2 If mass of air at sea level is 1034 g cm-2, calculate the pressure in pascal. [1] Mass of air at sea level = 1034 g cm-2 Acceleration due to gravity, g = 9.8 ms-2 Pressure is the force or weight per unit area. Section – B Question 21. What does the following prefixes stand for (a) pico (b) nano (c) micro (d) deci OR Write main points of Dalton’s atomic theory. [2] Prefixe Stand for (a) pico (p) 10-12 (b) nano (n) 10-9 (c) micro (μ) 10-6 (d) deci (d) 10-1 OR The main points of Dalton’s atomic theory are: (i) All matters are made of atoms. Atoms are indivisible and indestructible. (ii) All atoms of a given element are identical in mass and properties. Atoms of different elements differ in mass. (iii) Compounds are formed when atoms of different elements combine in a fixed ratio. (iv) Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction. Question 22. Chlorophyll present in green leaves of plants absorbs light at 4.620 × 1014Hz. Calculate the wavelength of radiation in nanometer. Which part of the electromagnetic spectrum does it belong to? [2] Given, v = 4.620 × 1014 Hz λ = ? Wavelength λ = $$\frac{c}{v}$$ = 0.6494 × 10-6 m = 649.4 × 10-9 m = 649.4 nm It belongs to visible region. Question 23. In both water and diethyl ether, the central atom viz. O-atom has same hybridisation. Why do they have different bond angles? Which one has greater bond angle? [2] Both water and diethyl ether have the central atom O in sp3 hybrid state with two lone pairs of electrons But due to the greater repulsion between two ethyl (C2H5) groups in diethyl ether than between two H-atoms in H2O result in greater bond angle (110°) in diethyl ether than 104.5° in that of water (H2O). Question 24. (a) Why is an organic compound fused with sodium for testing nitrogen, halogens and sulphur? (b) Under what conditions can the process of steam distillation used? [2] (i) On fusing with sodium metal, the elements present in an organic compound are converted from covalent form into the ionic form. (ii) Steam distillation is used to purify the substances which are steam volatile and water and the liquid are not miscible with each other. Question 25. Write the name of the isomerism shown by the following pairs: (i) Buta-1,3-diene and But-l-yne (C4H6) (ii) Ethoxy butane and Propoxy propane (C6H14O). [2] OR Which bond is more polar in the following pairs of molecules. (i) H3C-H, H3C-Br (ii) H3C-NH2, H2C-OH (i) Functional isomerism H2C=CH – CH = CH2 and HC = C-CH2CH3 (ii) Metamerism CH3 – CH2 – CH2 – CH2 – O- CH2CH3 and C3H7OC3H7 Students get confuse and give wrong answers. Understand different type of isomerism with examples, different functional groups too. OR (i) C-Br because Br is more electronegative than H. (ii) C-O because O is more electronegative than N. Question 26. An alkane C8H18 is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide. [2] Question 27. What are relative stabilities of different conformations of ethane? Is is possible to isolate these at room temperature? [2] Staggered form of ethane is more stable than the eclipsed form because the force of repulsion between hydrogen atoms on adjacent C atom is minimum. The energy difference between the staggered form and the eclipsed form of ethane is just 12.55 kJ mol- 1. Therefore, it is not possible for these two forms of ethane to isolate at room temperature. Section – C Question 28. What are (a) representative elements (b) transition elements (c) Lanthanoid and actinoids. Give their position in modern periodic table. [3] (a) Representative Elements : Group 1, 2, 13, 14, 15, 16, 17 and 18 exhibit the main groups of the periodic table and so, the elements of these groups are collectively called representative elements. These elements belong to s-block and p-block in modern periodic table. (b) Transition Elements : d-block transition metals form a bridge between the chemically active metals of s-block elements and the less active elements of Groups 13 and 14 are known as transition elements. (c) Lanthanoids and Actinoids : The elements of 4f series [i.e., from Ce (Z = 58) to Lu (Z = 71)] are called lanthanoids and the elements of 5f series [i.e., from Th (Z = 90) to Lr (Z = 103)] are called actinoids. These elements belong to /-block elements in the modern periodic table which lie at the bottom of the periodic table. Question 29. The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2 and H+ ion. Write a balanced ionic equation for the reaction. [3] The skeletal ionic equation is, Mn3+(aq) → Mn2+(aq) + MnO2(s) + H+(aq) Reduction half reaction Mn3+(aq) + e → Mn2+ Oxidation half reaction Mn3+(aq)→ MnO2 + e Balance charge by adding 4H+ to right side and then balance O atoms by adding 2H2O to left side in oxidation half reaction Mn3+(aq) + 2H2O(l) → MnO2(s) + e + 4H+(aq) By adding both equations, we get 2Mn3+(aq) + 2H2O(l) → Mn2+ + MnO2(s) + 4H+(aq) This represents the balanced redox reaction (disproportionation reaction). Question 30. Calculate the standard enthalpy of formation of CH3OH(l) from the following data : (i) CH3OH(l) + $$\frac{3}{2}$$02(g) → C02(g)+ 2H2O(l) ΔrH° = – 726 kj/mol (ii) C(s) + 02(g) → C02(g); ΔcH° = – 393 kj/mol (iii) H2(g) + ½O2(g) → H2O(l); ΔfH° = – 286 kj/mol. [3] The required equation is C(s) + 2H2(g) + $$\frac{1}{2}$$ O2(g) → CH3OHO); ΔfH° = ± ? To get the above required equation : Step 1 : Multiply eq. (iii) by 2 and add to eq. (ii) C(s) + 2H2(g) + 2O2(g) → CO2(g) + 2H2O(l) ……………(iv) ΔfH° = (2 x – 286) + (- 393) = – 572 – 393 = – 965 kJ/mol Step 2 : Subtract eq. (i) from eq. (iv) C(s) + 2H2(g) + $$\frac{1}{2}$$O2(g) → CH3OH(l); ΔH = – 965 – (- 726) = – 239 kJ/mol ΔfH° = – 239 kJ/mol Question 31. (i) State the formula and name of the conjugate base of each of the following acids : (a) H3O+ (b) HSO4 (c) NH4+ (d) HF (e) CH3COOH (f) CH3NH3+ (g) H3PO4 (h) H2PO4 (ii) The ionic product of water is 0.11 × 10-14 at 273 K, 1 × 10-14 at 298 K and 7.5 × 10-14 at 373K. Deduce from this data whether the ionization of water to hydrogen and hydroxide ion is exothermic or endothermic. [3] (i) The formula and name of the conjugate base are: (a) H3O+: Water (b) SO42- : Sulphate ion (c) NH3 : Ammonia (d) F : Fluoride ion (e) CH3COO : Acetate ion (f) CH3NH2 : Methylamine (g) H2PO4 : Dihydrogen phosphate (h)H2PO42- : Mono hydrogen phosphate (ii) Kw = [H3O+] [OH] According to the data, the value of K, is increasing with temperature. Therefore, according to Le- Chateliebs principle, the ionisation of water is endothermic. Question 32. Explain: (i) Tea or coffee is sipped from the saucer, when it is quite hot. (ii) Liquids possess fluidity. [3] (i) Tea or coffee is sipped from the saucer, when it is quite hot because it has larger surface area than the cup. In larger surface area, the rate of evaporation is faster due to which tea or coffee cools rapidly. (ii) Liquids have indefinite shape. They take the shape of the container in which they are placed. This is due to the fact that the molecules of liquids are in a state of constant random motion and therefore they can move freely. So, the liquids possess fluidity. Question 33. (a) The effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles. Justify the statement with the help of a suitable example. (b) What is the difference between the terms orbit and orbital ? [3] If mass of an object = 1 mg = 10-6 kg Then, according to Heisenberg’s uncertainty principle, Since, the value of Δx.Δv obtained is very small and is insignificant. So, effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles. (b) Orbit Orbital 1. An orbit is well defined circular path around the nucleus in which the electrons revolve. 1. An orbital is the threedimensional space around the nucleus within which the probability of finding an electron is maximum (upto 90%). 2. It represents the planar motion of an electron around the nucleus. 2. It represents the three dimensional motion of an electron around the nucleus. Question 34. Calculate the number of moles: (i) 5.0 L of 0.75 M Na2CO3 (ii) 7.85 g of Fe (iii) 34.2 g of sucrose (C12H22O11) [3] OR A compound made up of two elements A and B has A = 70%, B= 30%. Their relative number of moles in the compound is 1.25 and 1.88, calculate: (i) Atomic masses of the elements A and B. (ii) Molecular formula of the compound, if its molecular mass is found to be 160. (i) Number of moles of Na2 CO3 = Molarity × Volume of solution in litre = 0.75 × 5 = 3.75 mol (ii) Number of moles of Fe = $$\frac{\text { Mass }}{\text { Molecular mass }}$$ = $$\frac{34.2}{342}$$ = 0.14 (iii) Number of moles of sucrose = $$\frac{\text { Mass }}{\text { Molecular mass }}$$ = $$\frac{34.2}{342}$$ = 0.1 OR (i) Atomic mass of element A = $$\frac{\% \text { of element } \mathrm{A}}{\text { Relative number of moles }}$$ = $$\frac{70}{1.25}$$ = 56 Atomic mass of element B = $$\frac{\% \text { of element } \mathrm{B}}{\text { Relative number of moles }}$$ = $$\frac{30}{1.88}$$ = 15.957 ≈ 16 (ii) Compound Simplest molar ratio Simplest whole-number ratio A $$\frac{1.25}{1.25}$$= 1 2 B $$\frac{1.88}{1.25}$$= 15 3 Empirical formula of compound = A2B3 Molecular mass = 160 Empirical formula mass = 2(56) + 3(16) = 112 + 48 = 160 n = $$\frac{\text { Molecular mass }}{\text { Empirical formula mass }}$$ = $$\frac{160}{160}$$ = 1 Molecular formula = n × Empirical formula = 1 × A2B3 = A2B3 Sometimes students forget to multiply by n while finding molecular formula and do mistakes. Understand the problem and check for the necessary data, which are available in the problem. Follow steps and ensure no steps are missed. Section – D Question 35. (i) For each of the following compounds, write a more condensed formula and also their bond-line formulae. (ii) Write the I.U.RA.C. name of OR (a) A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound. (b) In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound. (i) Condensed formulae (a) (CH3)2CHCH2OH (b) CH3(CH2)5CHBrCH2CHO (c) HO(CH2)3CH(CH3)CH(CH3)2 (d) HOCH(CN)2 Bond-line formulae- Sometimes students ignore the triple bond in (ii) question. While writing bond-line formula, observe carefully. It is similar to structural formula. OR Given that, total mass of organic compound = 0.50 g 60 mL of 0.5 M solution of NaOH was required by residual acid for neutralisation. 60 mL of 0.5 M NaOH solution = $$\frac{60}{2}$$mL of 0.5 M H2SO4 = 30 mL of 0.5 M H2SO4 Acid consumed in absorption of evolved ammonia is (50-30) mL = 20 mL Again, 20 mL of 0.5 M H2SO4 = 40 mL of 0.5 M NH3 Also, since 1000 mL of 1 M NH3 contains 14 g of nitrogen, 40 mL of 0.5 M NH3 will contains = $$\frac{14 \times 40}{1000}$$ × 0.5 = 0.28 g of N Therefore, percentage of nitrogen in 0.50 g of organic compound = $$\frac{0.28}{0.50}$$ × 100 = 5.6% Students forget steps and end up with mistakes. Understand the problem and practice steps wise and do calculations carefully. (b) Given, total mass of organic compound = 0.468 g Mass of barium sulphate formed = 0.668 g 1 mol of BaSO4 = 233 g of BaSO4 = 32 g of sulphur Thus, 0.668 g of BaSO4 contains $$\frac{32 \times 0.668}{233}$$ g of sulphur = 0.0917 g of sulphur Therefore, percentage of sulphur = $$\frac{0.0917}{0.468}$$ × 100 = 19.59 % Hence, the percentage of sulphur in the given compound is 19.59 %. Question 36. Give mechanism of addition of HBr to Propene. [5] Addition of HBr to propene (unsymmetrical alkene) follows Markovnikov’s rule according to which the negative part of the addendum gets attached to that C atom which possesses lesser number of hydrogen atoms. Mechanism: Hydrogen bromide provides an electrophile, H+, which attacks the double bond to form carbocation as: Secondary carbocations are more stable than primary carbocations. Therefore, the former predominates as it will form at a faster rate. Thus, in the next step, Br attacks the carbocation to form 2-bromopropane as the major product. Addition of HBr to unsymmetrical alkenes like propene in the presence of light or peroxide takes place contrary to the Markovnikov’s rule. This so happens only with HBr but not with HCl and HI. This addition of HBr to propene in the presence of benzoyl peroxide follows anti-Markovnikov’s rule or peroxide effect or Kharasch effect. Secondary free radicals are more stable than primary radicals. Therefore, the former predominates since it forms at a faster rate. Thus, 1-bromopropane is obtained as the major product. • Some students write the reaction, but forget to explain the mechanism. • Some students forget to explain the attack of Br- on carbocation. Students must explain addition using Markovnikov’s rule as well as anti – Markovnikov’s rule. Question 37. Derive the relationship between AH and AU for an ideal gas. Explain each term involved in the equation. [5] OR Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (pi, Vi to (pf Vf). With the help of a pV plot, compare the work done in the above case with that carried out against a constant external pressure pf. According to first law of thermodynamics, q = ∆U + W = ∆U + p∆V At constant volume, ∆V = 0 qv = ∆U where qv = Heat absorbed at constant volume ∆U = Change in internal energy Similarly qp = DH where qp = Heat absorbed at constant pressure DH = Enthalpy change of the system Since, the enthalpy change of a system is equal to heat absorbed or heat evolved by the system at constant pressure. Now, at constant pressure ∆H = ∆U + p∆V ∆V = change in volume ∆H = ∆U + p(Vf – Vi) ∆H = ∆U + (pVf – pVi) …(i) Vi = initial volume of the system Vf = Final volume of the system For ideal gases, pV = nRT pVi = nRT and pVf = npRT where nr = number of moles of the gaseous reactants np = number of moles of the gaseous products Equation (i) becomes, ∆H = ∆U + (npRT – nrRT) = ∆U + (np – nr) RT or ∆H = ∆U + ∆ngRT where ∆ng = Difference between the number of moles of the gaseous products and reactants. Students miss steps and explanations and thus lose marks.	7,071	24,455	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.640625	4	CC-MAIN-2023-14	latest	en	0.868949	Students must start practicing the questions from CBSE Sample Papers for Class 11 Chemistry with Solutions Set 6 are designed as per the revised syllabus.. ## CBSE Sample Papers for Class 11 Chemistry Set 6 with Solutions. Time Allowed: 3 hours. Maximum Marks: 70. General Instructions:. • All questions are compulsory.. • This question paper contains 37 questions.. • Questions 1-20 in Section A are objective type-very short answer type questions carrying 1 mark each.. • Questions 21 – 27 in Section B are short answer type questions carrying 2 marks each.. • Questions 28 – 34 in Section C are long-answer 1 type questions carrying 3 marks each.. • Questions 35 – 37 in Section D are long-answer 11 type questions carrying 5 marks each.. • There is no overall choice. However, an internal choice has been provided in six questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks. You must attempt only one of the choices in such questions.. • Use log tables, if necessary. Use of calculator is not allowed.. Section – A. Question 1.. What is the mass percent of carbon in carbon dioxide? [1]. (A) 0.034%. (B) 27.27%. (C) 3.4%. (D) 28.7%. OR. The empirical formula and molecular mass of a compound are CH2O and 180 g respectively. What will be the molecular formula of the compound?. (A) C9H18O9. (B) CH2O. (C) C6H12O6. (D) C2H4O2. Option (B) is correct.. Explanation:. Molecular mass of CO2 = 12 + 2(16). = 12 + 32 = 44 g. 44 g of CO2 contains 12 g atoms of carbon. Mass percent of carbon. = $$\frac{\text { Mass of carbon in } \mathrm{CO}_2}{\text { Molar mass of } \mathrm{CO}_2}$$ × 100s. = $$\frac{12}{44}$$ × 100. = 27.27%. OR. Option (C) is correct.. Explanation:. Empirical formula mass (CH2O). = 12 + 2(1) + 16 = 30g. Molecular mass = 180g. n = $$\frac{\text { Molecular mass }}{\text { Empirical formula mass }}$$. = $$\frac{180}{30}$$ = 6. Molecular formula= n × empirical formula. = 6 × CH2O. = C6H12O6. Question 2.. On the basis of thermochemical equations (i), (ii) and (iii), find out which of the algebraic relationship given in options (A) to (D) is correct. [1]. (i) C (graphite) + O2(g) → C02(g); ΔfH = x kj mol-1. (ii) C (graphite) + $$\frac{1}{2}$$O2 (g) → CO(g); ΔrH = y kj mol-1. (iii) CO (g) + $$\frac{1}{2}$$O2 (g) → C02 (g) ;ΔH = z kj mol-1. (A) z = x + y. (B) x = y – z. (C) x = y + z. (D) y = 2z – x. Option (C) is correct.. Explanation: The algebric relationships of the given reaction is Eq.(i) – Eq.(ii) = Eq.(iii). (i) C(graphite) + O2(g) → CO2(g) ; ΔrH = x kJ mol-1. (ii) C(graphite) + $$\frac{1}{2}$$O2(g) → CO(g) ; ΔrH = V kJ. (iii) CO(g) + $$\frac{1}{2}$$O2(g) → CO2(g) ; ΔrH = z kJ mol-1. Hence, x – y = z or x = y + z. Question 3.. Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of dipoles possess ‘partial charges’. The partial charge is : [1]. (A) more than unit electronic charge.. (B) equal to unit electronic charge.. (C) less than unit electronic charge.. (D) double the unit electronic charge.. Option (C) is correct.. Explanation:. Partial charge is the small charge developed by displacement of electrons.. Question 4.. The first ionisation enthalpies of Na, Mg, Al and Si are in the order : [1]. (A) Na < Mg > Al < Si (B) Na > Mg > Al > Si. (C) Na < Mg < Al < Si (D) Na > Mg > Al < Si OR Which of the following is the correct order of size of the given species: (A) I > I > I+. (B) I+ > I > I. (C) I > I+ > I. (D) I > I > I+. Option (A) is correct.. Explanation:. On moving from left to right in a period first ionisation enthalpy increases normally. But, in Mg, as an electron needs to be removed from fully filled s orbital therefore, first ionisation enthalpy of Mg is more than Al.. OR. Option (D) is correct.. Explanation :. Size of cation is smaller while that of anion is bigger than its parent atom.. Question 5.. What will be correct order of vapour pressure of water, acetone and ether at 30°C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point? [1]. (A) Water < ether <acetone. (B) Water < acetone < ether. (C) Ether < acetone < water. (D) Acetone < ether < water. Option (B) is correct.. Explanation:. Greater is the boiling point, less is the vapour pressure.. Question 6.. Which of the following carbocation is most stable?. (B) (CH3)3C+. (C) CH3CH2+CH2. (D) CH3+CHCH2CH3. Option (B) is correct.. Explanation:. Tertiary carbocation is most stable.. The variables Pressure, Volume, Concentration and Temperature may change the State of Equilibrium. The change is governed by the Le-Chatelier’s principle. The decomposition of NH3(g) can be made spontaneous by increasing the temperature and lowering pressure. In the reaction, removal of any product from the reaction mixture makes the reversible reaction irreversible and therefore, reaction proceeds to completion.. Answer the questions (7) to (10) given below:. Question 7.. The equilibrium Solid → Liquid → Gas will shift in forward direction when:. (A) temperature is increased. (B) temperature is lowered. (C) pressure is increased. (D) pressure is lowered. Option (A) is correct.. Explanation:. Increase in temperature will favour the forward reaction which is endothermic.. Question 8.. Change in free energy for the equilibrium, gaseous reaction, PCl5 → PCl3 + Cl2 on addition of an inert gas at constant pressure and at constant volume is respectively:. (A) decrease, no change. (B) increase, no change. (C) no change, no change. (D) no change, decrease. Option (A) is correct.. Explanation:. Addition of inert gas at constant volume has no effect on any equilibrium reaction. When inert gas is added at constant pressure in the given reaction, it proceeds in forward direction. DG° remains constant, but DG decreases as the reaction proceeds spontaneously in forward direction to attain equilibrium.. Question 9.. At 25°C, the equilibrium constant K1 and I(z are for the reactions:. Which of the following shows the relation between two equilibrium constants?. (A) K2 = K2. (B) K2 = 1/(K). (D) K1 = 1/(K). (D) K1 = 1/K2. Option (C) is correct.. Explanation:. Question 10.. A liquid is in equilibrium with its vapour at its boiling point. On an average, the molecules in the two phases have equal:. (A) Intermolecular forces. (B) Potential energy. (C) Kinetic energy. (D) None of these. Option (C) is correct.. Explanation:. At boiling point, liquid and vapour phases both are present. So, the molecules in the two phases have equal kinetic energy.. Question 11.. The addition of HCl to an alkene proceeds in two steps. The first step is the attack of W ion to >C =C< portion which can be shown as A I. Option (A) is correct.. Explanation:. Since double bond is a source of electrons and the charge flows from source of more electron density, therefore, electrons of the double bond attack the proton.. Question 12.. How will you convert: ethyne to but-2-yne. [1]. Students directly write the products and forget to mention side products and reaction conditions.. Make a list of important conversions and learn them.. Understand each step and conditions involved during conversion reactions.. Question 13.. The structure of triphenylmethyl cation is given here. This is very stable and some of its salts can be stored for months. Explain the cause of high stability of this cation.. Triphenylmethyl cation is a tertiary carbocation which can show nine possible canonical structures and hence is very stable. The three canonical forms for one benzene ring are shown below.. Question 14.. Can we separate two liquids A (b.p. 353 K) and B (b.p. 365 K) present in a mixture by simple distillation? [1]. No, because in simple distillation, vapours of both the liquids will be formed simultaneously and will condense together in receiver as the difference between the boiling points is very less. They can be separated by fractional distillation.. Students sometimes answer in yes or no without giving legit explanation.. The alternate method of separation also must be mentioned.. Question 15.. What do you mean by cracking? [1]. The thermal decomposition of higher hydrocarbons into lower hydrocarbons in the presence or absence of a catalyst is called cracking.. Question 16.. Assign the oxidation number to Cr in K2Cr2O7.. Let the oxidation number of Cr be x. 2 × (+1) + 2x + 7 × (- 2) = 0. 2 + 2x – 14 = 0. 2x – 12 = 0. x = $$\frac{12}{2}$$ = +6. Oxidation number of Cr in K2Cr2O7 = +6. Question 17.. Assertion (A): Toluene on Friedal-Crafts methylation gives o- and p-xylene.. Reason (R): CH3-group bonded to benzene ring increases electron density at o- and p- position.. (A) Both A and R are correct and R is the correct explanation of A.. (B) Both A and R are correct but R is not the correct explanation of A.. (C) Both A and R are not correct.. (D) A is not correct but R is correct.. Option (A) is correct.. Question 18.. Assertion (A): Benzene on heating with cone. H2SO4 gives benzene sulphonic acid which when heated with superheated steam under pressure gives benzene. [1]. Reason (R): Sulphonation is a reversible process.. (A) Both A and R are correct and R is the correct explanation of A.. (B) Both A and R are correct but R is not the correct explanation of A.. (C) Both A and R are not correct.. (D) A is not correct but R is correct.. Option (A) is correct.. Explanation:. Sulphonation of benzene is an electrophilic substitution reaction in which SO3 acts as the electrophile.. Question 19.. Why there is large number of lines in hydrogen spectrum? [1]. Large number of lines are there in hydrogen spec¬trum because different possible transitions can be observed which leads to large number of spectral lines.. Question 20.. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:. 1 Pa = 1 Nm-2. If mass of air at sea level is 1034 g cm-2, calculate the pressure in pascal. [1]. Mass of air at sea level = 1034 g cm-2. Acceleration due to gravity, g = 9.8 ms-2. Pressure is the force or weight per unit area.. Section – B. Question 21.. What does the following prefixes stand for. (a) pico. (b) nano. (c) micro. (d) deci. OR. Write main points of Dalton’s atomic theory. [2]. Prefixe Stand for (a) pico (p) 10-12 (b) nano (n) 10-9 (c) micro (μ) 10-6 (d) deci (d) 10-1. OR. The main points of Dalton’s atomic theory are:. (i) All matters are made of atoms. Atoms are indivisible and indestructible.. (ii) All atoms of a given element are identical in mass and properties. Atoms of different elements differ in mass.. (iii) Compounds are formed when atoms of different elements combine in a fixed ratio.. (iv) Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction.. Question 22.. Chlorophyll present in green leaves of plants absorbs light at 4.620 × 1014Hz. Calculate the wavelength of radiation in nanometer. Which part of the electromagnetic spectrum does it belong to? [2]. Given, v = 4.620 × 1014 Hz. λ = ?. Wavelength λ = $$\frac{c}{v}$$. = 0.6494 × 10-6 m. = 649.4 × 10-9 m. = 649.4 nm. It belongs to visible region.. Question 23.. In both water and diethyl ether, the central atom viz. O-atom has same hybridisation. Why do they have different bond angles? Which one has greater bond angle? [2]. Both water and diethyl ether have the central atom O in sp3 hybrid state with two lone pairs of electrons. But due to the greater repulsion between two ethyl (C2H5) groups in diethyl ether than between two H-atoms in H2O result in greater bond angle (110°) in diethyl ether than 104.5° in that of water (H2O).. Question 24.. (a) Why is an organic compound fused with sodium for testing nitrogen, halogens and sulphur?. (b) Under what conditions can the process of steam distillation used? [2]. (i) On fusing with sodium metal, the elements present in an organic compound are converted from covalent form into the ionic form.. (ii) Steam distillation is used to purify the substances which are steam volatile and water and the liquid are not miscible with each other.. Question 25.. Write the name of the isomerism shown by the following pairs:. (i) Buta-1,3-diene and But-l-yne (C4H6).	(ii) Ethoxy butane and Propoxy propane (C6H14O). [2]. OR. Which bond is more polar in the following pairs of molecules.. (i) H3C-H, H3C-Br. (ii) H3C-NH2, H2C-OH. (i) Functional isomerism. H2C=CH – CH = CH2 and HC = C-CH2CH3. (ii) Metamerism. CH3 – CH2 – CH2 – CH2 – O- CH2CH3 and C3H7OC3H7. Students get confuse and give wrong answers.. Understand different type of isomerism with examples, different functional groups too.. OR. (i) C-Br because Br is more electronegative than H.. (ii) C-O because O is more electronegative than N.. Question 26.. An alkane C8H18 is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On. monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide. [2]. Question 27.. What are relative stabilities of different conformations of ethane? Is is possible to isolate these at room temperature? [2]. Staggered form of ethane is more stable than the eclipsed form because the force of repulsion between hydrogen atoms on adjacent C atom is minimum. The energy difference between the staggered form and the eclipsed form of ethane is just 12.55 kJ mol- 1. Therefore, it is not possible for these two forms of ethane to isolate at room temperature.. Section – C. Question 28.. What are (a) representative elements (b) transition elements (c) Lanthanoid and actinoids. Give their position in modern periodic table. [3]. (a) Representative Elements : Group 1, 2, 13, 14, 15, 16, 17 and 18 exhibit the main groups of the periodic table and so, the elements of these groups are collectively called representative elements.. These elements belong to s-block and p-block in modern periodic table.. (b) Transition Elements : d-block transition metals form a bridge between the chemically active metals of s-block elements and the less active elements of Groups 13 and 14 are known as transition elements.. (c) Lanthanoids and Actinoids : The elements of 4f series [i.e., from Ce (Z = 58) to Lu (Z = 71)] are called lanthanoids and the elements of 5f series [i.e., from Th (Z = 90) to Lr (Z = 103)] are called actinoids. These elements belong to /-block elements in the modern periodic table which lie at the bottom of the periodic table.. Question 29.. The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2 and H+ ion. Write a balanced ionic equation for the reaction. [3]. The skeletal ionic equation is,. Mn3+(aq) → Mn2+(aq) + MnO2(s) + H+(aq). Reduction half reaction. Mn3+(aq) + e → Mn2+. Oxidation half reaction. Mn3+(aq)→ MnO2 + e. Balance charge by adding 4H+ to right side and then balance O atoms by adding 2H2O to left side in oxidation half reaction. Mn3+(aq) + 2H2O(l) → MnO2(s) + e + 4H+(aq). By adding both equations, we get. 2Mn3+(aq) + 2H2O(l) → Mn2+ + MnO2(s) + 4H+(aq). This represents the balanced redox reaction (disproportionation reaction).. Question 30.. Calculate the standard enthalpy of formation of CH3OH(l) from the following data :. (i) CH3OH(l) + $$\frac{3}{2}$$02(g) → C02(g)+ 2H2O(l) ΔrH° = – 726 kj/mol. (ii) C(s) + 02(g) → C02(g); ΔcH° = – 393 kj/mol. (iii) H2(g) + ½O2(g) → H2O(l); ΔfH° = – 286 kj/mol. [3]. The required equation is. C(s) + 2H2(g) + $$\frac{1}{2}$$ O2(g) → CH3OHO);. ΔfH° = ± ?. To get the above required equation :. Step 1 : Multiply eq. (iii) by 2 and add to eq. (ii). C(s) + 2H2(g) + 2O2(g) → CO2(g) + 2H2O(l) ……………(iv). ΔfH° = (2 x – 286) + (- 393). = – 572 – 393. = – 965 kJ/mol. Step 2 : Subtract eq. (i) from eq. (iv). C(s) + 2H2(g) + $$\frac{1}{2}$$O2(g) → CH3OH(l);. ΔH = – 965 – (- 726). = – 239 kJ/mol. ΔfH° = – 239 kJ/mol. Question 31.. (i) State the formula and name of the conjugate base of each of the following acids :. (a) H3O+. (b) HSO4. (c) NH4+. (d) HF. (e) CH3COOH. (f) CH3NH3+. (g) H3PO4. (h) H2PO4. (ii) The ionic product of water is 0.11 × 10-14 at 273 K, 1 × 10-14 at 298 K and 7.5 × 10-14 at 373K. Deduce from this data whether the ionization of water to hydrogen and hydroxide ion is exothermic or endothermic. [3]. (i) The formula and name of the conjugate base are:. (a) H3O+: Water. (b) SO42- : Sulphate ion. (c) NH3 : Ammonia. (d) F : Fluoride ion. (e) CH3COO : Acetate ion. (f) CH3NH2 : Methylamine. (g) H2PO4 : Dihydrogen phosphate. (h)H2PO42- : Mono hydrogen phosphate. (ii) Kw = [H3O+] [OH]. According to the data, the value of K, is increasing with temperature. Therefore, according to Le- Chateliebs principle, the ionisation of water is endothermic.. Question 32.. Explain:. (i) Tea or coffee is sipped from the saucer, when it is quite hot.. (ii) Liquids possess fluidity. [3]. (i) Tea or coffee is sipped from the saucer, when it is quite hot because it has larger surface area than the cup. In larger surface area, the rate of evaporation is faster due to which tea or coffee cools rapidly.. (ii) Liquids have indefinite shape. They take the shape of the container in which they are placed. This is due to the fact that the molecules of liquids are in a state of constant random motion and therefore they can move freely. So, the liquids possess fluidity.. Question 33.. (a) The effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles. Justify the statement with the help of a suitable example.. (b) What is the difference between the terms orbit and orbital ? [3]. If mass of an object = 1 mg = 10-6 kg. Then, according to Heisenberg’s uncertainty principle,. Since, the value of Δx.Δv obtained is very small and is insignificant. So, effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles.. (b). Orbit Orbital 1. An orbit is well defined circular path around the nucleus in which the electrons revolve. 1. An orbital is the threedimensional space around the nucleus within which the probability of finding an electron is maximum (upto 90%). 2. It represents the planar motion of an electron around the nucleus. 2. It represents the three dimensional motion of an electron around the nucleus.. Question 34.. Calculate the number of moles:. (i) 5.0 L of 0.75 M Na2CO3. (ii) 7.85 g of Fe. (iii) 34.2 g of sucrose (C12H22O11) [3]. OR. A compound made up of two elements A and B has A = 70%, B= 30%. Their relative number of moles in the compound is 1.25 and 1.88, calculate:. (i) Atomic masses of the elements A and B.. (ii) Molecular formula of the compound, if its molecular mass is found to be 160.. (i) Number of moles of Na2 CO3 = Molarity × Volume of solution in litre. = 0.75 × 5. = 3.75 mol. (ii) Number of moles of Fe. = $$\frac{\text { Mass }}{\text { Molecular mass }}$$. = $$\frac{34.2}{342}$$. = 0.14. (iii) Number of moles of sucrose. = $$\frac{\text { Mass }}{\text { Molecular mass }}$$. = $$\frac{34.2}{342}$$. = 0.1. OR. (i) Atomic mass of element A. = $$\frac{\% \text { of element } \mathrm{A}}{\text { Relative number of moles }}$$. = $$\frac{70}{1.25}$$. = 56. Atomic mass of element B. = $$\frac{\% \text { of element } \mathrm{B}}{\text { Relative number of moles }}$$. = $$\frac{30}{1.88}$$. = 15.957 ≈ 16. (ii). Compound Simplest molar ratio Simplest whole-number ratio A $$\frac{1.25}{1.25}$$= 1 2 B $$\frac{1.88}{1.25}$$= 15 3. Empirical formula of compound = A2B3. Molecular mass = 160. Empirical formula mass. = 2(56) + 3(16). = 112 + 48 = 160. n = $$\frac{\text { Molecular mass }}{\text { Empirical formula mass }}$$. = $$\frac{160}{160}$$ = 1. Molecular formula. = n × Empirical formula. = 1 × A2B3. = A2B3. Sometimes students forget to multiply by n while finding molecular formula and do mistakes.. Understand the problem and check for the necessary data, which are available in the problem. Follow steps and ensure no steps are missed.. Section – D. Question 35.. (i) For each of the following compounds, write a more condensed formula and also their bond-line formulae.. (ii) Write the I.U.RA.C. name of. OR. (a) A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.. (b) In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.. (i) Condensed formulae. (a) (CH3)2CHCH2OH. (b) CH3(CH2)5CHBrCH2CHO. (c) HO(CH2)3CH(CH3)CH(CH3)2. (d) HOCH(CN)2. Bond-line formulae-. Sometimes students ignore the triple bond in (ii) question.. While writing bond-line formula, observe carefully. It is similar to structural formula.. OR. Given that, total mass of organic compound = 0.50 g 60 mL of 0.5 M solution of NaOH was required by residual acid for neutralisation.. 60 mL of 0.5 M NaOH solution = $$\frac{60}{2}$$mL of 0.5 M. H2SO4 = 30 mL of 0.5 M H2SO4. Acid consumed in absorption of evolved ammonia is (50-30) mL = 20 mL. Again, 20 mL of 0.5 M. H2SO4 = 40 mL of 0.5 M NH3. Also, since 1000 mL of 1 M NH3 contains 14 g of nitrogen,. 40 mL of 0.5 M NH3 will contains = $$\frac{14 \times 40}{1000}$$ × 0.5. = 0.28 g of N. Therefore, percentage of nitrogen in 0.50 g of organic compound = $$\frac{0.28}{0.50}$$ × 100. = 5.6%. Students forget steps and end up with mistakes.. Understand the problem and practice steps wise and do calculations carefully.. (b) Given, total mass of organic compound = 0.468 g Mass of barium sulphate formed = 0.668 g. 1 mol of BaSO4 = 233 g of BaSO4 = 32 g of sulphur. Thus, 0.668 g of BaSO4 contains $$\frac{32 \times 0.668}{233}$$ g of. sulphur = 0.0917 g of sulphur. Therefore, percentage of sulphur = $$\frac{0.0917}{0.468}$$ × 100. = 19.59 %. Hence, the percentage of sulphur in the given compound is 19.59 %.. Question 36.. Give mechanism of addition of HBr to Propene. [5]. Addition of HBr to propene (unsymmetrical alkene) follows Markovnikov’s rule according to which the negative part of the addendum gets attached to that C atom which possesses lesser number of hydrogen atoms.. Mechanism: Hydrogen bromide provides an electrophile, H+, which attacks the double bond to form carbocation as:. Secondary carbocations are more stable than primary carbocations. Therefore, the former predominates as it will form at a faster rate. Thus, in the next step, Br attacks the carbocation to form 2-bromopropane as the major product.. Addition of HBr to unsymmetrical alkenes like propene in the presence of light or peroxide takes place contrary to the Markovnikov’s rule. This so happens only with HBr but not with HCl and HI. This addition of HBr to propene in the presence of benzoyl peroxide follows anti-Markovnikov’s rule or peroxide effect or Kharasch effect.. Secondary free radicals are more stable than primary radicals. Therefore, the former predominates since it forms at a faster rate. Thus, 1-bromopropane is obtained as the major product.. • Some students write the reaction, but forget to explain the mechanism.. • Some students forget to explain the attack of Br- on carbocation.. Students must explain addition using Markovnikov’s rule as well as anti – Markovnikov’s rule.. Question 37.. Derive the relationship between AH and AU for an ideal gas. Explain each term involved in the equation. [5]. OR. Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (pi, Vi to (pf Vf). With the help of a pV plot, compare the work done in the above case with that carried out against a constant external pressure pf.. According to first law of thermodynamics,. q = ∆U + W. = ∆U + p∆V. At constant volume, ∆V = 0. qv = ∆U. where qv = Heat absorbed at constant volume. ∆U = Change in internal energy. Similarly qp = DH. where qp = Heat absorbed at constant pressure. DH = Enthalpy change of the system. Since, the enthalpy change of a system is equal to heat absorbed or heat evolved by the system at constant pressure.. Now, at constant pressure. ∆H = ∆U + p∆V. ∆V = change in volume. ∆H = ∆U + p(Vf – Vi). ∆H = ∆U + (pVf – pVi) …(i). Vi = initial volume of the system. Vf = Final volume of the system. For ideal gases,. pV = nRT. pVi = nRT. and pVf = npRT. where nr = number of moles of the gaseous reactants. np = number of moles of the gaseous products. Equation (i) becomes,. ∆H = ∆U + (npRT – nrRT). = ∆U + (np – nr) RT. or ∆H = ∆U + ∆ngRT. where ∆ng = Difference between the number of moles of the gaseous products and reactants.. Students miss steps and explanations and thus lose marks.
https://codinghero.ai/probability-rules/	1,685,256,799,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00221.warc.gz	212,934,399	71,762	• Home • / • Blog • / • Probability Rules(With Formulas & Examples) # Probability Rules(With Formulas & Examples) January 22, 2023 Probability is a measure of the likelihood of an event occurring. Many events cannot be predicted with total certainty. Using probability, one can predict only the chance of an event to occur, i.e., how likely they are going to happen. There are 5 basic probability rules that are used to solve problems. Letâ€™s understand these rules in probability with examples. ## 6 Basic Probability Rules The six basic rules in probability are as follows. Rule 1: The probability of an impossible event is $0$; the probability of a certain event is $1$. Therefore, for any event $\text{A}$, the range of possible probabilities is $0 \le \text{P}( \text{A} ) \le 1$. Rule 2: For $\text{S}$ the sample space of all possibilities, $\text{P} (\text{S}) = 1$, i.e., the sum of all the probabilities for all possible events is equal to one.Â Rule 3: For any event $\text{A}$, $\text{P}(\text{A}^{\text{c}}) = 1 – \text{P}(\text{A})$. It follows then that $\text{P}(\text{A}) = 1 – \text{P}( \text{A}^{\text{c}})$ Rule 4 (Addition Rule): This probability rule relates to either one or both events that occur • If two events say $\text{A}$ and $\text{B}$, are mutually exclusive i.e., $\text{A}$ and $\text{B}$ have no outcomes in common then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B})$ • If two events are not mutually exclusive, then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B}) – \text{P}(\text{A and B})$ Rule 5 (Conditional Probability): $\text{P}(\text{A} \| \text{B}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{B})}$ or $\text{P}(\text{B} \| \text{A}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{A})}$. Note: The straight line symbol, $\|$, does not mean divide! This symbol means “conditional” or “given”. For example $\text{P}(\text{A}\|\text{B})$ means the probability that event $\text{A}$ occurs given event $\text{B}$ has already occurred. Rule 6 (Multiplication Rule): This probability rule is used when both events occur $\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B}\|\text{A})$ or $\text{P}(\text{B}) \times \text{P}(\text{A}\| \text{B})$ If $\text{A}$ and $\text{B}$ are independent, i.e., neither event influences or affects the probability of other events then $\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B})$. This particular rule extends to more than two independent events. In that case, it becomes, $\text{P}(\text{A and B and C}) = \text{P}(\text{A}) \times \text{P}(\text{B}) \times \text{P}(\text{C})$. ### Rule 1: $0 \le \text{P}( \text{A} ) \le 1$ To prove the statement that for any event $\text{A}$, $0 \le \text{P}( \text{A} ) \le 1$, weâ€™ll use the following three axioms: 1. The probability of an event will always be greater than or equal to zero i.e. $\text{P}(\text{A}) \ge 0$ for any event $\text{A}$. (Since the probability of an impossible event is $0$ and the probability of an event cannot be negative). 2. The probability of a sample space will always be equal to $1$ i.e. $\text{P}(\text{S}) = 1$. 3. Given some mutually exclusive events, the probability of the union of all these mutually exclusive events will always be equal to the summation of the probability of individual events i.e. $\text{P} \left(\text{A}_1 \cup \text{A}_2 \cup \text{A}_3 \cup \text{A}_4 â€¦ \cup \text{A}_{\text{n}} \right) =\text{P} \left(\text{A}_1 \right) + \text{P} \left(\text{A}_2 \right) + \text{P} \left(\text{A}_3 \right) + \text{P} \left(\text{A}_4 \right) + … + \text{P} \left(\text{A}_n \right)$ According to axiom 1, the probability of an event will always be greater than or equal to $0$. $\text{P} \left( \text{A} \right) \ge 0$ (According to Axiom 1) — (1) The probability of a sample space will be equal to the probability of the intersection of $\text{A}$ and $(\text{S} â€“ \text{A})$ i.e. $\text{S} = \text{A} + (\text{S} – \text{A})$ $\text{P}(\text{S}) = \text{P}(\text{A} + (\text{S} – \text{A}))$ Since $\text{A}$ and $(\text{S} â€“ \text{A})$ are two mutually exclusive events. So, according to axiom 3, it can be written as $\text{P}(\text{A} + (\text{S} – \text{A})) = \text{P}(\text{A}) + \text{P}(\text{S} – \text{A})$ This implies, $\text{P}(\text{S}) = \text{P}(\text{A}) + \text{P}(\text{S} – \text{A}))$— (2) Now, from axiom 1, it can be said that the $\text{P}(\text{S} â€“ \text{A})$ will always be greater than or equal to zero i.e. $\text{P}(\text{S} â€“ \text{A}) \ge 0$. If something positive is added to a given value, its value will always increase. Since, $\text{P}(\text{S} â€“ \text{A}) \ge 0$, it can be said that $\text{P}(\text{A})$ canâ€™t be greater than $\text{P}(\text{S})$. Otherwise, equation (2) will not hold true. This means $\text{P}(\text{S}) \ge \text{P}(\text{A})$ From axiom 2, the probability of a Sample Space always equals 1. So, this means $1 \ge \text{P}(\text{A})$, or $\text{P} (\text{A}) \ge 1$ — (3) From equations (1) and (3), it can be shown that $0 \le \text{P}(\text{A}) \le 1$ This proves that the probability of an event will always lie between $0$ and $1$. ### Examples of $0 \le \text{P}( \text{A} ) \le 1$ Example 1: Find the probability of getting a sum of $15$ in a throw of a pair of dice. Let $\text{E}$ be the event of getting a sum of $15$ in a throw of a pair of dice. The sample space when a pair of dice is rolled = $\text{S} = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),$ $(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),$ $(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),$ $(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),$ $(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),$ $(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$ and therefore, the total number of possible outcomes is $n = 36$. From the above sample space, we see that there is no pair of numbers where the sum is $15$, therefore, the total number of favourable events is $m = 0$. Therefore, the probability of getting a sum of $15$ in a throw of a pair of dice $\text{P}(\text{E}) = \frac{m}{n} = \frac{0}{36} = 0$. Example 2: Find the probability of getting a positive number less than $13$ as a sum in a throw of a pair of dice. Let $\text{E}$ be the event of getting a positive number less than $13$ as a sum in a throw of a pair of dice. The sample space when a pair of dice is rolled = $\text{S} = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),$ $(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),$ $(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),$ $(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),$ $(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$ and therefore, the total number of possible outcomes is $n = 36$. From the above sample space, we see that for all the pairs the sum is a positive number less than $13$, therefore, the total number of favourable events is $m = 36$. Therefore, the probability of getting a positive number less than $13$ as a sum in a throw of a pair of dice $\text{P}(\text{E}) = \frac{m}{n} = \frac{36}{36} = 1$. Example 3: Find the probability of picking a non-face card from a deck of well-shuffled cards. The number of cards in a deck of cards $n = 52$. The number of face cards in a deck of cards = $12$. Therefore, the number of non-face cards in a deck of cards $m = 52 – 12 = 40$ Thus, the probability of picking a non-face card from a deck of a well-shuffled pack of $52$ cards is $\frac{m}{n} = \frac{40}{52} = \frac{10}{13} = 0.7692$. Note: From the above three examples, we see that $0 \le \text{P}( \text{A} ) \le 1$. ### Rule 2: $\text{P} (\text{S}) = 1$ Let $\text{S}$ be a sample space consisting of events $\text{E}_1$, $\text{E}_2$, $\text{E}_3$, $\text{E}_4$, â€¦, $\text{E}_n$, then $\text{E}_1 \cup \text{E}_2 \cup \text{E}_3 \cup \text{E}_4 \cup â€¦ \text{E}_n = \text{S}$ â€”———– (1) We know that probability of an event $\text{E}$ is given by $\text{P}(\text{E}) = \frac{n(\text{E})}{n(\text{S})}$, therefore, $\text{P}(\text{E}_1) + \text{P}(\text{E}_2) + \text{P}(\text{E}_3) + \text{P}(\text{E}_4) + â€¦ + \text{P}(\text{E}_n) = \frac{n(\text{E}_1)}{n(\text{S})} + \frac{n(\text{E}_2)}{n(\text{S})} + \frac{n(\text{E}_3)}{n(\text{S})} + \frac{n(\text{E}_4)}{n(\text{S})} + \frac{n(\text{E}_n)}{n(\text{S})}$ $= \frac{n(\text{E}_1) + n(\text{E}_2) + n(\text{E}_3) + n(\text{E}_4) + â€¦ + n(\text{E}_n)}{n(\text{S})}$ $= \frac{n \left(\text{E}_1 \cup \text{E}_2 \cup \text{E}_3 \cup \text{E}_4 \cup â€¦ \text{E}_n \right)}{n(\text{S})}$ $= \frac{n(\text{S})}{n(\text{S})} = 1$ ### Examples of $\text{P} (\text{S}) = 1$ Example 1: Consider a random experiment of tossing a pair of coins. The sample space is given by $\text{S} = \{(\text{H}, \text{H}), (\text{H}, \text{T}), (\text{T}, \text{H}), (\text{T}, \text{T}) \}$. Probability of an event $(\text{H}, \text{H}) = \frac{1}{4}$ Probability of an event $(\text{H}, \text{T}) = \frac{1}{4}$ Probability of an event $(\text{T}, \text{H}) = \frac{1}{4}$ Probability of an event $(\text{T}, \text{T}) = \frac{1}{4}$ Therefore, $\text{P}(\text{H}, \text{H}) + \text{P}(\text{H}, \text{T}) + \text{P}(\text{T}, \text{H}) + \text{P}(\text{T}, \text{T})$ $= \text{P}((\text{H}, \text{H}) + (\text{H}, \text{T}) + (\text{T}, \text{H}) + (\text{T}, \text{T})) = \text{P}(\text{S})$ $= \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1$ ### Rule 3: Complementary Events Rule For any event $\text{A}$, $\text{P}(\text{A}^{\text{c}}) = 1 – \text{P}(\text{A})$ For any event, $\text{A}$, the complementary event $\text{A}^{\text{c}}$ is an event that includes all the events of a sample space except event $\text{A}$. Therefore, if $\text{S}$ is a sample space such that $\text{S} = \text{E}_1 \cup \text{E}_2 \cup \text{E}_3 \cup \text{E}_4 \cup â€¦ \text{E}_n$, then complement of event $\text{E}_1$, written as ${\text{E}_{1}}^{c}$ is given by ${\text{E}_{1}}^{c} = \text{S} – \text{E}_1$ Thus, the probability of complementary event ${\text{E}_{1}}^{c}$ is given by $\text{P} \left(\text{S} – \text{E}_1 \right)$ $=>\text{P} \left({\text{E}_{1}}^{c} \right) = \text{P} \left(\text{S} \right) – \text{P}(\text{E}_1 )$ $=>\text{P} \left({\text{E}_{1}}^{c} \right) = 1 – \text{P}(\text{E}_1 )$ Thus, For any event $\text{A}$, $\text{P}(\text{A}^{\text{c}}) = 1 – \text{P}(\text{A})$. ### Examples of Complementary Events Example 1: A number is chosen at random from a set of whole numbers from 1 to 50. Calculate the probability that the chosen number is not a perfect square. Number of numbers from $1$ to $50$, $n = 50$. Perfect squares from $1$ to $50$ are $1$, $4$, $9$, $16$, $25$, $36$, and $49$. Therefore, number of perfect squares from $1$ to $50$, $m = 7$ Let $\text{E}$ be the event of choosing a perfect square number, therefore, $\text{P}(\text{E}) = \frac{m}{n} = \frac{7}{50}$. Therefore, the probability that the chosen number is not a perfect square, $\text{P}(\text{E}^{\text{c}}) = 1 – \text{P}(\text{E}) = 1 – \frac{7}{50} = \frac{43}{50}$. Example 2: There are $10$ balls in a bag out of which $3$ are black, $2$ are red, $1$ is blue, $2$ are pink, and $2$ are purple. Find the probability of not picking a pink ball. The number of balls in a bag $n = 3 + 2 + 1 + 2 + 2 = 10$. The number of pink balls in a bag $m = 2$. Let $\text{E}$ be the event of picking a pink ball. Therefore, the probability of picking a pink ball $\text{P}(\text{E}) = \frac{m}{n} = \frac{2}{10} = \frac{1}{5}$. Thus, the probability of not picking a pink ball = $1 – \text{P}(\text{E}) = 1 – \frac{1}{5} = \frac{4}{5}$. Example 3: A card is chosen from a deck of well-shuffled cards. What is the probability that the number on the card is not $4$? The number of cards in a deck of cards $n = 52$. The number of cards having a number 2 $m = 4$. Let $\text{E}$ be the event of picking a card with the number $2$ on it. Therefore, the probability of picking a card with number $2$ on it $\text{P}(\text{E}) = \frac{m}{n} = \frac{4}{52} = \frac{1}{13}$. Thus, the probability of not card with number $2$ = $1 – \text{P}(\text{E}) = 1 – \frac{1}{13} = \frac{12}{13}$. ### Rule 4: Addition Rule of Probability Let $\text{A}$ and $\text{B}$ be two events. Case 1: When $\text{A}$ and $\text{B}$ are non-mutually exclusive events(there exist a common event between the two events), then $\text{A} \cup \text{B} = \text{A} + \text{B} – \text{A} \cap \text{B}$ Therefore, $\text{P}(\text{A} \cup \text{B}) = \text{P}(\text{A}) + \text{P}(\text{B}) – \text{P}(\text{A} \cap \text{B})$ Case 2: When $\text{A}$ and $\text{B}$ are mutually exclusive events(there doesnot exist a common event between the two events), then $\text{A} \cup \text{B} = \text{A} + \text{B}$ Therefore, $\text{P}(\text{A} \cup \text{B}) = \text{P}(\text{A}) + \text{P}(\text{B})$ ### Examples of Addition Rule of Probability Example 1: There are $50$ cards numbered from $1$ to $50$. A card is chosen at random. What is the probability that the number on the card is either less than $10$ or greater than $40$. The total number of cards $n = 50$. Let $\text{E}_1$ be an event of picking a card with a number less than $10$. And, Let $\text{E}_2$ be an event of picking a card with a number greater than $40$. Numbers less than $10$ are $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, and $9$. Therefore, a number of numbers less than $10$ is $9$. Numbers greater than $40$ are $41$, $42$, $43$, $44$, $45$, $46$, $47$, $48$, $49$, and $50$. Therefore, a number of numbers greater than $40$ are $10$. Thus, $\text{P}\left(\text{E}_1 \right) = \frac{9}{50}$ and $\text{P}\left(\text{E}_2 \right) = \frac{10}{50}$ Hence, the probability that the number on the card is either less than $10$ or greater than $40$ is $\text{P}\left(\text{E}_1 \right) + \text{P}\left(\text{E}_2 \right) = \frac{9}{50} + \frac{10}{50} = \frac{19}{50}$. Note: In this example, the events number on the card are either less than $10$ or greater than $40$ are non-mutually exclusive events. Example 2: There are $50$ cards numbered from $1$ to $50$. A card is chosen at random. What is the probability that the number on the card is either divisible by $3$ or divisible by $5$? The total number of cards $n = 50$. Let $\text{E}_1$ be an event of picking a card with a number divisible by $3$. And, Let $\text{E}_2$ be an event of picking a card with a number divisible by $5$. Numbers divisible by $3$ are $3$, $6$, $9$, $12$, $15$, $18$, $21$, $24$, $27$, $30$, $33$, $36$, $39$, $42$, $45$, and $48$. Therefore, the number of numbers divisible by $3$ is $16$. Numbers divisible by $5$ are $5$, $10$, $15$, $20$, $25$, $30$, $35$, $40$, $45$, and $50$. Therefore, the number of numbers divisible by $5$ is $10$. Numbers divisible by both $3$ and $5$ are $15$, $30$, and $45$. Therefore, the number of numbers divisible by both $3$ and $5$ is $3$. Let $\text{E}_1$ be an event of picking a card with a number divisible by $3$. and $\text{E}_2$ be an event of picking a card with a number divisible by $5$. Therefore, $\text{E}_1 \cap \text{E}_2$ will be an event of picking a card divisible by both $3$ and $5$. Thus, the probability that the number on the card is either divisible by $3$ or divisible by $5$ = $\text{P}(\text{E}_1) + \text{P}(\text{E}_2) – \text{P}(\text{E}_1 \cap \text{E}_2)$ $= \frac{16}{50} + \frac{10}{50} – \frac{3}{50} = \frac{23}{50}$. Note: In this example, the events number on the card are either multiple of $3$ or multiple of $5$ are mutually exclusive events. ### Rule 5: Conditional Probability The conditional probability rule states that If $\text{A}$ and $\text{B}$ are two events that $\text{P}(\text{A} \| \text{B}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{B})}$ or $\text{P}(\text{B} \| \text{A}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{A})}$. When the intersection of two events happens, then the formula for conditional probability for the occurrence of two events is given by $\text{P}(\text{A}\|\text{B}) = \frac{n(\text{A} \cap \text{B})}{n(\text{B})}$, or $\text{P}(\text{B}\| \text{A}) = \frac{n(\text{A} \cap \text{B})}{n(\text{A})}$ where $\text{P}(\text{A}\| \text{B})$ represents the probability of occurrence of $\text{A}$ given $\text{B}$ has occurred. $n(\text{A} \cap \text{B})$ is the number of elements common to both $\text{A}$ and $\text{B}$. $n(\text{B})$ is the number of elements in $\text{B}$, and it cannot be equal to zero. Let $n$ represent the total number of elements in the sample space, then $\text{P}(\text{A}\| \text{B}) = \frac{\frac{n(\text{A} \cap \text{B})}{n}}{\frac{n(\text{B})}{n}}$ Since $\frac{n(\text{A} \cap \text{B})}{n}$ and $\frac{n(\text{B})}{n}$ denotes the ratio of the number of favourable outcomes to the total number of outcomes; therefore, it indicates the probability. Therefore, $\frac{n(\text{A} \cap \text{B})}{n}$ can be written as $\text{P}(\text{A} \cap \text{B})$ and $\frac{n(\text{B})}{n}$ as $\text{P}(\text{B})$. $=>\text{P}(\text{A} \| \text{B}) = \frac{\text{P}(\text{A} \cap \text{B})}{\text{P}(\text{B})}$ Therefore, $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{B}) \text{P}(\text{A}\| \text{B})$ if $\text{P}(\text{B}) \ne 0$ $= \text{P}(\text{A}) \text{P}(\text{B}\| \text{A})$ if $\text{P}(\text{A}) \ne 0$ Similarly, the probability of occurrence of $\text{B}$ when $\text{A}$ has already occurred is given by $\text{P}(\text{B}\| \text{A}) = \frac{\text{P}(\text{B} \cap \text{A})}{\text{P}(\text{A})}$. ### Examples of Conditional Probability Example 1: Two dice are rolled simultaneously and the sum of the numbers obtained is found to be $7$. What is the probability that the number $3$ has appeared at least once? The sample space $\text{S}$ would consist of all the numbers possible by the combination of two dice. Therefore $\text{S}$ consists of $6 \times 6$ i.e. $36$ events. Let event $\text{A}$ indicate the combination in which $3$ has appeared at least once. And event $\text{B}$ indicates the combination of the numbers which sum up to $7$. Therefore, $\text{A} = \{ (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3), (6, 3) \}$ And, $\text{B} = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) \}$ Therefore, $\text{P}(\text{A}) = \frac{11}{36}$ And, $\text{P}(\text{B}) = \frac{6}{36}$ And also, $n(\text{A} \cap \text{B}) = 2$ Therefore, $\text{P}(\text{A} \cap \text{B}) = \frac{2}{36}$ Applying the conditional probability formula we get, $\text{P}(\text{A} \| \text{B}) = \frac{\text{P}(\text{A} \cap \text{B})}{\text{P}(\text{B})} = \frac{\frac{2}{36}}{\frac{6}{36}} = \frac{2}{6} = \frac{1}{3}$. Example 2: The probability that it is Friday and that a student is absent is $0.03$. Since there are $5$ school days in a week, the probability that it is Friday is $0.2$. What is the probability that a student is absent given that today is Friday? In this problem, $\text{P}(\text{Friday and Absent}) = 0.03$ And, $\text{P}(\text{Friday}) = 0.2$ Therefore, the probability that a student is absent given that today is Friday $P(\text{Absent} \| \text{Friday}) = \frac{\text{P}(\text{Friday and Absent})}{\text{P}(\text{Friday})} = \frac{0.03}{0.2} = \frac{3}{20} = 0.15$ Example 3: In a school, $18 \%$ of all students play football and basketball and $32 \%$ of all students play football. What is the probability that a student plays basketball given that the student plays football? Let $\text{F}$ be an event that a student plays football. And, let $\text{B}$ be an event that a student plays basketball. Therefore, $\text{F} \cap \text{B}$ is an event that a student plays football as well as basketball. $\text{P}(\text{F}) = 18\% = 0.18$ $\text{P}(\text{F} \cap \text{B}) = 32\% = 0.32$ Therefore, the probability that a student plays basketball given that the student plays football is $\frac{\text{P}(\text{F} \cap \text{B})}{\text{P}(\text{F})} = \frac{0.18}{0.32} = 0.5625$. ### Rule 6: Multiplication Rule of Probability This probability rule is used when both events occur $\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B}\|\text{A})$ or $\text{P}(\text{B}) \times \text{P}(\text{A}\| \text{B})$ To prove the multiplication rule of probability, weâ€™ll use the following rules of conditional probability $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{A}) \times \text{P}(\text{B} \| \text{A})$ ; if $\text{P}(\text{A}) \ne 0$ $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{B}) \times \text{P}(\text{A}\| \text{B})$ ; if $\text{P}(\text{B}) \ne 0$ If $\text{A}$ and $\text{B}$ are two independent events for a random experiment, then the probability of the simultaneous occurrence of two independent events will be equal to the product of their probabilities, i.e., $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{A}).\text{P}(\text{B})$ Now, we know that $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{A}) \times \text{P}(\text{B} \| \text{A})$ Since $\text{A}$ and $\text{B}$ are independent, therefore, $\text{P}(\text{B} \| \text{A}) = \text{P}(\text{B})$ ### Examples of Multiplication Rule of Probability Example 1: An urn contains $12$ pink balls and 6 blue balls. Without replacement, two balls are drawn one after another. What is the probability that both balls drawn are pink? Let $\text{A}$ and $\text{B}$ be the probability of drawing the first ball pink and the second ball also pink, respectively. Then, we need to find the probability of $\text{P} (\text{A} \cap \text{B})$. The probability of drawing the first ball pink = $\frac{12}{18}$ Since the replacement is not allowed, and the first ball has been drawn, the remaining pink balls in the urn are $17$, out of which $11$ are pink balls. Therefore, the probability of drawing another pink ball = $\frac{11}{17}$. By multiplication rule, $\text{P} (\text{A} \cap \text{B}) = \text{P} (\text{A}) \times \text{P} (\text{A}\| \text{B})$. $= \frac{12}{18} \times \frac{11}{17} = \frac{132}{306}$. Example 2: A magician takes out two cards from a deck of cards, one after the other, without replacement. What is the probability of getting an ace of spade, and a card of heart, as the first and second cards, respectively? Since the cards are drawn without replacement, therefore, both events are dependent upon each other. As we know, there is only one ace of spades in a deck of cards. Therefore, the probability of drawing ace of spades = $\frac{1}{52}$. Since the ace of spades has already been drawn and replacement is not allowed. Therefore, only $51$ cards remain in the deck. Out of these $51$ cards, $13$ are hearts. Therefore, the probability of drawing a card of heart = $\frac{13}{51}$. From the multiplication rule, the probability of drawing aces of spades and a heart of card = $\frac{1}{52} \times \frac{13}{51} = \frac{1}{204}$. ## Practice Problems 1. In a cricket tournament, a player hits eight times â€˜$6$â€™ out of thirty-two balls. Calculate the probability that he would not hit a â€˜$6$â€™. 2. In a laptop shop, there are $16$ defective laptops out of $200$ laptops. If one laptop is taken out at random from this laptop shop, what is the probability that it is a non-defective laptop? 3. Neha has $4$ yellow t-shirts, $6$ black t-shirts, and $2$ blue t-shirts to choose from for her outfit today. She chooses a t-shirt randomly with each t-shirt equally likely to be chosen. Find the probability that a black or blue t-shirt is chosen for the outfit. 4. There are a total of $50$ distinct books on a shelf such as $20$ math books, $16$ physics books, and $14$ chemistry books. Find the probability of getting a book that is not a chemistry book or not a physics book. 5. Suppose you take out two cards from a standard pack of cards one after another, without replacing the first card. What is the probability that the first card is the ace of spades, and the second card is a heart? 6. You have a cowboy hat, a top hat, and an Indonesian hat called a songkok. You also have four shirts: white, black, green, and pink. If you choose one hat and one shirt at random, what is the probability that you choose the songkok and the black shirt? 7. Suppose a box has $3$ red marbles and $2$ black ones. We select $2$ marbles. What is the probability that the second marble is red given that the first one is red? 8. A family has $2$ children. Given that one of the children is a boy, what is the probability that the other child is also a boy? ## FAQs ### What are the 6 rules of probability? The $6$ rules of probability are Rule 1: The probability of an impossible event is $0$; the probability of a certain event is $1$. Therefore, for any event $\text{A}$, the range of possible probabilities is $0 \le \text{P}( \text{A} ) \le 1$. Rule 2: For $\text{S}$ the sample space of all possibilities, $\text{P} (\text{S}) = 1$, i.e., the sum of all the probabilities for all possible events is equal to one.Â Rule 3: For any event $\text{A}$, $\text{P}(\text{A}^{\text{c}}) = 1 – \text{P}(\text{A})$. It follows then that $\text{P}(\text{A}) = 1 – \text{P}( \text{A}^{\text{c}})$ Rule 4 (Addition Rule): This probability rule relates to either one or both events that occur a) If two events say $\text{A}$ and $\text{B}$, are mutually exclusive i.e., $\text{A}$ and $\text{B}$ have no outcomes in common then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B})$ b) If two events are not mutually exclusive, then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B}) – \text{P}(\text{A and B})$ Rule 5 (Conditional Probability): $\text{P}(\text{A} \| \text{B}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{B})}$ or $\text{P}(\text{B} \| \text{A}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{A})}$ Rule 6 (Multiplication Rule): This probability rule is used when both events occur $\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B}\|\text{A})$ or $\text{P}(\text{B}) \times \text{P}(\text{A}\| \text{B})$ ### What is the probability formula? Probability of occurrence of an event $\text{P}(\text{E}) = \frac{\text{Number of favorable outcomes}}{\text{Total Number of outcomes}}$. ### What is the addition rule of probability? The addition rule of probability relates to either one or both events that occur and it states that a) If two events say $\text{A}$ and $\text{B}$, are mutually exclusive i.e., $\text{A}$ and $\text{B}$ have no outcomes in common then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B})$ b) If two events are not mutually exclusive, then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B}) – \text{P}(\text{A and B})$ ### What is the multiplication rule of probability? The multiplication rule ofÂ probability rule is used when both events occur and it states that $\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B}\|\text{A})$ or $\text{P}(\text{B}) \times \text{P}(\text{A}\| \text{B})$ ### What are complementary events? Two events are said to be complementary when one event occurs if and only if the other does not take place. For example, it rains or it does not rain are complementary events. ## Conclusion There are five basic probability rules that are used to solve problems. These five rules of probability are rules related to impossible and certain events, the sum of all probabilities, addition of probabilities, multiplication of probabilities, complementary events, and conditional probability.	9,587	27,535	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.78125	5	CC-MAIN-2023-23	longest	en	0.859342	• Home. • /. • Blog. • /. • Probability Rules(With Formulas & Examples). # Probability Rules(With Formulas & Examples). January 22, 2023. Probability is a measure of the likelihood of an event occurring. Many events cannot be predicted with total certainty. Using probability, one can predict only the chance of an event to occur, i.e., how likely they are going to happen.. There are 5 basic probability rules that are used to solve problems. Letâ€™s understand these rules in probability with examples.. ## 6 Basic Probability Rules. The six basic rules in probability are as follows.. Rule 1: The probability of an impossible event is $0$; the probability of a certain event is $1$. Therefore, for any event $\text{A}$, the range of possible probabilities is $0 \le \text{P}( \text{A} ) \le 1$.. Rule 2: For $\text{S}$ the sample space of all possibilities, $\text{P} (\text{S}) = 1$, i.e., the sum of all the probabilities for all possible events is equal to one.Â. Rule 3: For any event $\text{A}$, $\text{P}(\text{A}^{\text{c}}) = 1 – \text{P}(\text{A})$. It follows then that $\text{P}(\text{A}) = 1 – \text{P}( \text{A}^{\text{c}})$. Rule 4 (Addition Rule): This probability rule relates to either one or both events that occur. • If two events say $\text{A}$ and $\text{B}$, are mutually exclusive i.e., $\text{A}$ and $\text{B}$ have no outcomes in common then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B})$. • If two events are not mutually exclusive, then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B}) – \text{P}(\text{A and B})$. Rule 5 (Conditional Probability): $\text{P}(\text{A} \| \text{B}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{B})}$ or $\text{P}(\text{B} \| \text{A}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{A})}$.. Note: The straight line symbol, $\|$, does not mean divide! This symbol means “conditional” or “given”. For example $\text{P}(\text{A}\|\text{B})$ means the probability that event $\text{A}$ occurs given event $\text{B}$ has already occurred.. Rule 6 (Multiplication Rule): This probability rule is used when both events occur. $\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B}\|\text{A})$ or $\text{P}(\text{B}) \times \text{P}(\text{A}\| \text{B})$. If $\text{A}$ and $\text{B}$ are independent, i.e., neither event influences or affects the probability of other events then $\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B})$. This particular rule extends to more than two independent events. In that case, it becomes, $\text{P}(\text{A and B and C}) = \text{P}(\text{A}) \times \text{P}(\text{B}) \times \text{P}(\text{C})$.. ### Rule 1: $0 \le \text{P}( \text{A} ) \le 1$. To prove the statement that for any event $\text{A}$, $0 \le \text{P}( \text{A} ) \le 1$, weâ€™ll use the following three axioms:. 1. The probability of an event will always be greater than or equal to zero i.e. $\text{P}(\text{A}) \ge 0$ for any event $\text{A}$. (Since the probability of an impossible event is $0$ and the probability of an event cannot be negative).. 2. The probability of a sample space will always be equal to $1$ i.e. $\text{P}(\text{S}) = 1$.. 3. Given some mutually exclusive events, the probability of the union of all these mutually exclusive events will always be equal to the summation of the probability of individual events i.e. $\text{P} \left(\text{A}_1 \cup \text{A}_2 \cup \text{A}_3 \cup \text{A}_4 â€¦ \cup \text{A}_{\text{n}} \right) =\text{P} \left(\text{A}_1 \right) + \text{P} \left(\text{A}_2 \right) + \text{P} \left(\text{A}_3 \right) + \text{P} \left(\text{A}_4 \right) + … + \text{P} \left(\text{A}_n \right)$. According to axiom 1, the probability of an event will always be greater than or equal to $0$.. $\text{P} \left( \text{A} \right) \ge 0$ (According to Axiom 1) — (1). The probability of a sample space will be equal to the probability of the intersection of $\text{A}$ and $(\text{S} â€“ \text{A})$ i.e.. $\text{S} = \text{A} + (\text{S} – \text{A})$. $\text{P}(\text{S}) = \text{P}(\text{A} + (\text{S} – \text{A}))$. Since $\text{A}$ and $(\text{S} â€“ \text{A})$ are two mutually exclusive events. So, according to axiom 3, it can be written as $\text{P}(\text{A} + (\text{S} – \text{A})) = \text{P}(\text{A}) + \text{P}(\text{S} – \text{A})$. This implies, $\text{P}(\text{S}) = \text{P}(\text{A}) + \text{P}(\text{S} – \text{A}))$— (2). Now, from axiom 1, it can be said that the $\text{P}(\text{S} â€“ \text{A})$ will always be greater than or equal to zero i.e. $\text{P}(\text{S} â€“ \text{A}) \ge 0$.. If something positive is added to a given value, its value will always increase. Since, $\text{P}(\text{S} â€“ \text{A}) \ge 0$, it can be said that $\text{P}(\text{A})$ canâ€™t be greater than $\text{P}(\text{S})$. Otherwise, equation (2) will not hold true.. This means $\text{P}(\text{S}) \ge \text{P}(\text{A})$. From axiom 2, the probability of a Sample Space always equals 1. So, this means $1 \ge \text{P}(\text{A})$, or $\text{P} (\text{A}) \ge 1$ — (3). From equations (1) and (3), it can be shown that $0 \le \text{P}(\text{A}) \le 1$. This proves that the probability of an event will always lie between $0$ and $1$.. ### Examples of $0 \le \text{P}( \text{A} ) \le 1$. Example 1: Find the probability of getting a sum of $15$ in a throw of a pair of dice.. Let $\text{E}$ be the event of getting a sum of $15$ in a throw of a pair of dice.. The sample space when a pair of dice is rolled = $\text{S} = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),$. $(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),$. $(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),$. $(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),$. $(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),$. $(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$ and therefore, the total number of possible outcomes is $n = 36$.. From the above sample space, we see that there is no pair of numbers where the sum is $15$, therefore, the total number of favourable events is $m = 0$.. Therefore, the probability of getting a sum of $15$ in a throw of a pair of dice $\text{P}(\text{E}) = \frac{m}{n} = \frac{0}{36} = 0$.. Example 2: Find the probability of getting a positive number less than $13$ as a sum in a throw of a pair of dice.. Let $\text{E}$ be the event of getting a positive number less than $13$ as a sum in a throw of a pair of dice.. The sample space when a pair of dice is rolled = $\text{S} = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),$. $(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),$. $(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),$. $(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),$. $(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$ and therefore, the total number of possible outcomes is $n = 36$.. From the above sample space, we see that for all the pairs the sum is a positive number less than $13$, therefore, the total number of favourable events is $m = 36$.. Therefore, the probability of getting a positive number less than $13$ as a sum in a throw of a pair of dice $\text{P}(\text{E}) = \frac{m}{n} = \frac{36}{36} = 1$.. Example 3: Find the probability of picking a non-face card from a deck of well-shuffled cards.. The number of cards in a deck of cards $n = 52$.. The number of face cards in a deck of cards = $12$.. Therefore, the number of non-face cards in a deck of cards $m = 52 – 12 = 40$. Thus, the probability of picking a non-face card from a deck of a well-shuffled pack of $52$ cards is $\frac{m}{n} = \frac{40}{52} = \frac{10}{13} = 0.7692$.. Note: From the above three examples, we see that $0 \le \text{P}( \text{A} ) \le 1$.. ### Rule 2: $\text{P} (\text{S}) = 1$. Let $\text{S}$ be a sample space consisting of events $\text{E}_1$, $\text{E}_2$, $\text{E}_3$, $\text{E}_4$, â€¦, $\text{E}_n$, then $\text{E}_1 \cup \text{E}_2 \cup \text{E}_3 \cup \text{E}_4 \cup â€¦ \text{E}_n = \text{S}$ â€”———– (1). We know that probability of an event $\text{E}$ is given by $\text{P}(\text{E}) = \frac{n(\text{E})}{n(\text{S})}$, therefore, $\text{P}(\text{E}_1) + \text{P}(\text{E}_2) + \text{P}(\text{E}_3) + \text{P}(\text{E}_4) + â€¦ + \text{P}(\text{E}_n) = \frac{n(\text{E}_1)}{n(\text{S})} + \frac{n(\text{E}_2)}{n(\text{S})} + \frac{n(\text{E}_3)}{n(\text{S})} + \frac{n(\text{E}_4)}{n(\text{S})} + \frac{n(\text{E}_n)}{n(\text{S})}$. $= \frac{n(\text{E}_1) + n(\text{E}_2) + n(\text{E}_3) + n(\text{E}_4) + â€¦ + n(\text{E}_n)}{n(\text{S})}$. $= \frac{n \left(\text{E}_1 \cup \text{E}_2 \cup \text{E}_3 \cup \text{E}_4 \cup â€¦ \text{E}_n \right)}{n(\text{S})}$. $= \frac{n(\text{S})}{n(\text{S})} = 1$. ### Examples of $\text{P} (\text{S}) = 1$. Example 1: Consider a random experiment of tossing a pair of coins.. The sample space is given by $\text{S} = \{(\text{H}, \text{H}), (\text{H}, \text{T}), (\text{T}, \text{H}), (\text{T}, \text{T}) \}$.. Probability of an event $(\text{H}, \text{H}) = \frac{1}{4}$. Probability of an event $(\text{H}, \text{T}) = \frac{1}{4}$. Probability of an event $(\text{T}, \text{H}) = \frac{1}{4}$. Probability of an event $(\text{T}, \text{T}) = \frac{1}{4}$. Therefore, $\text{P}(\text{H}, \text{H}) + \text{P}(\text{H}, \text{T}) + \text{P}(\text{T}, \text{H}) + \text{P}(\text{T}, \text{T})$. $= \text{P}((\text{H}, \text{H}) + (\text{H}, \text{T}) + (\text{T}, \text{H}) + (\text{T}, \text{T})) = \text{P}(\text{S})$. $= \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1$. ### Rule 3: Complementary Events Rule. For any event $\text{A}$, $\text{P}(\text{A}^{\text{c}}) = 1 – \text{P}(\text{A})$. For any event, $\text{A}$, the complementary event $\text{A}^{\text{c}}$ is an event that includes all the events of a sample space except event $\text{A}$.. Therefore, if $\text{S}$ is a sample space such that $\text{S} = \text{E}_1 \cup \text{E}_2 \cup \text{E}_3 \cup \text{E}_4 \cup â€¦ \text{E}_n$, then complement of event $\text{E}_1$, written as ${\text{E}_{1}}^{c}$ is given by ${\text{E}_{1}}^{c} = \text{S} – \text{E}_1$. Thus, the probability of complementary event ${\text{E}_{1}}^{c}$ is given by $\text{P} \left(\text{S} – \text{E}_1 \right)$. $=>\text{P} \left({\text{E}_{1}}^{c} \right) = \text{P} \left(\text{S} \right) – \text{P}(\text{E}_1 )$. $=>\text{P} \left({\text{E}_{1}}^{c} \right) = 1 – \text{P}(\text{E}_1 )$. Thus, For any event $\text{A}$, $\text{P}(\text{A}^{\text{c}}) = 1 – \text{P}(\text{A})$.. ### Examples of Complementary Events. Example 1: A number is chosen at random from a set of whole numbers from 1 to 50. Calculate the probability that the chosen number is not a perfect square.. Number of numbers from $1$ to $50$, $n = 50$.. Perfect squares from $1$ to $50$ are $1$, $4$, $9$, $16$, $25$, $36$, and $49$.. Therefore, number of perfect squares from $1$ to $50$, $m = 7$. Let $\text{E}$ be the event of choosing a perfect square number, therefore, $\text{P}(\text{E}) = \frac{m}{n} = \frac{7}{50}$.. Therefore, the probability that the chosen number is not a perfect square, $\text{P}(\text{E}^{\text{c}}) = 1 – \text{P}(\text{E}) = 1 – \frac{7}{50} = \frac{43}{50}$.. Example 2: There are $10$ balls in a bag out of which $3$ are black, $2$ are red, $1$ is blue, $2$ are pink, and $2$ are purple. Find the probability of not picking a pink ball.. The number of balls in a bag $n = 3 + 2 + 1 + 2 + 2 = 10$.. The number of pink balls in a bag $m = 2$.. Let $\text{E}$ be the event of picking a pink ball.. Therefore, the probability of picking a pink ball $\text{P}(\text{E}) = \frac{m}{n} = \frac{2}{10} = \frac{1}{5}$.. Thus, the probability of not picking a pink ball = $1 – \text{P}(\text{E}) = 1 – \frac{1}{5} = \frac{4}{5}$.. Example 3: A card is chosen from a deck of well-shuffled cards. What is the probability that the number on the card is not $4$?. The number of cards in a deck of cards $n = 52$.. The number of cards having a number 2 $m = 4$.. Let $\text{E}$ be the event of picking a card with the number $2$ on it.. Therefore, the probability of picking a card with number $2$ on it $\text{P}(\text{E}) = \frac{m}{n} = \frac{4}{52} = \frac{1}{13}$.. Thus, the probability of not card with number $2$ = $1 – \text{P}(\text{E}) = 1 – \frac{1}{13} = \frac{12}{13}$.. ### Rule 4: Addition Rule of Probability. Let $\text{A}$ and $\text{B}$ be two events.. Case 1: When $\text{A}$ and $\text{B}$ are non-mutually exclusive events(there exist a common event between the two events), then $\text{A} \cup \text{B} = \text{A} + \text{B} – \text{A} \cap \text{B}$. Therefore, $\text{P}(\text{A} \cup \text{B}) = \text{P}(\text{A}) + \text{P}(\text{B}) – \text{P}(\text{A} \cap \text{B})$. Case 2: When $\text{A}$ and $\text{B}$ are mutually exclusive events(there doesnot exist a common event between the two events), then $\text{A} \cup \text{B} = \text{A} + \text{B}$. Therefore, $\text{P}(\text{A} \cup \text{B}) = \text{P}(\text{A}) + \text{P}(\text{B})$. ### Examples of Addition Rule of Probability. Example 1: There are $50$ cards numbered from $1$ to $50$. A card is chosen at random. What is the probability that the number on the card is either less than $10$ or greater than $40$.. The total number of cards $n = 50$.. Let $\text{E}_1$ be an event of picking a card with a number less than $10$.. And, Let $\text{E}_2$ be an event of picking a card with a number greater than $40$.. Numbers less than $10$ are $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, and $9$.. Therefore, a number of numbers less than $10$ is $9$.. Numbers greater than $40$ are $41$, $42$, $43$, $44$, $45$, $46$, $47$, $48$, $49$, and $50$.. Therefore, a number of numbers greater than $40$ are $10$.. Thus, $\text{P}\left(\text{E}_1 \right) = \frac{9}{50}$ and $\text{P}\left(\text{E}_2 \right) = \frac{10}{50}$. Hence, the probability that the number on the card is either less than $10$ or greater than $40$ is $\text{P}\left(\text{E}_1 \right) + \text{P}\left(\text{E}_2 \right) = \frac{9}{50} + \frac{10}{50} = \frac{19}{50}$.	Note: In this example, the events number on the card are either less than $10$ or greater than $40$ are non-mutually exclusive events.. Example 2: There are $50$ cards numbered from $1$ to $50$. A card is chosen at random. What is the probability that the number on the card is either divisible by $3$ or divisible by $5$?. The total number of cards $n = 50$.. Let $\text{E}_1$ be an event of picking a card with a number divisible by $3$.. And, Let $\text{E}_2$ be an event of picking a card with a number divisible by $5$.. Numbers divisible by $3$ are $3$, $6$, $9$, $12$, $15$, $18$, $21$, $24$, $27$, $30$, $33$, $36$, $39$, $42$, $45$, and $48$.. Therefore, the number of numbers divisible by $3$ is $16$.. Numbers divisible by $5$ are $5$, $10$, $15$, $20$, $25$, $30$, $35$, $40$, $45$, and $50$.. Therefore, the number of numbers divisible by $5$ is $10$.. Numbers divisible by both $3$ and $5$ are $15$, $30$, and $45$.. Therefore, the number of numbers divisible by both $3$ and $5$ is $3$.. Let $\text{E}_1$ be an event of picking a card with a number divisible by $3$.. and $\text{E}_2$ be an event of picking a card with a number divisible by $5$.. Therefore, $\text{E}_1 \cap \text{E}_2$ will be an event of picking a card divisible by both $3$ and $5$.. Thus, the probability that the number on the card is either divisible by $3$ or divisible by $5$ = $\text{P}(\text{E}_1) + \text{P}(\text{E}_2) – \text{P}(\text{E}_1 \cap \text{E}_2)$. $= \frac{16}{50} + \frac{10}{50} – \frac{3}{50} = \frac{23}{50}$.. Note: In this example, the events number on the card are either multiple of $3$ or multiple of $5$ are mutually exclusive events.. ### Rule 5: Conditional Probability. The conditional probability rule states that If $\text{A}$ and $\text{B}$ are two events that $\text{P}(\text{A} \| \text{B}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{B})}$ or $\text{P}(\text{B} \| \text{A}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{A})}$.. When the intersection of two events happens, then the formula for conditional probability for the occurrence of two events is given by $\text{P}(\text{A}\|\text{B}) = \frac{n(\text{A} \cap \text{B})}{n(\text{B})}$, or $\text{P}(\text{B}\| \text{A}) = \frac{n(\text{A} \cap \text{B})}{n(\text{A})}$. where $\text{P}(\text{A}\| \text{B})$ represents the probability of occurrence of $\text{A}$ given $\text{B}$ has occurred.. $n(\text{A} \cap \text{B})$ is the number of elements common to both $\text{A}$ and $\text{B}$.. $n(\text{B})$ is the number of elements in $\text{B}$, and it cannot be equal to zero.. Let $n$ represent the total number of elements in the sample space, then $\text{P}(\text{A}\| \text{B}) = \frac{\frac{n(\text{A} \cap \text{B})}{n}}{\frac{n(\text{B})}{n}}$. Since $\frac{n(\text{A} \cap \text{B})}{n}$ and $\frac{n(\text{B})}{n}$ denotes the ratio of the number of favourable outcomes to the total number of outcomes; therefore, it indicates the probability.. Therefore, $\frac{n(\text{A} \cap \text{B})}{n}$ can be written as $\text{P}(\text{A} \cap \text{B})$ and $\frac{n(\text{B})}{n}$ as $\text{P}(\text{B})$.. $=>\text{P}(\text{A} \| \text{B}) = \frac{\text{P}(\text{A} \cap \text{B})}{\text{P}(\text{B})}$. Therefore, $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{B}) \text{P}(\text{A}\| \text{B})$ if $\text{P}(\text{B}) \ne 0$. $= \text{P}(\text{A}) \text{P}(\text{B}\| \text{A})$ if $\text{P}(\text{A}) \ne 0$. Similarly, the probability of occurrence of $\text{B}$ when $\text{A}$ has already occurred is given by $\text{P}(\text{B}\| \text{A}) = \frac{\text{P}(\text{B} \cap \text{A})}{\text{P}(\text{A})}$.. ### Examples of Conditional Probability. Example 1: Two dice are rolled simultaneously and the sum of the numbers obtained is found to be $7$. What is the probability that the number $3$ has appeared at least once?. The sample space $\text{S}$ would consist of all the numbers possible by the combination of two dice. Therefore $\text{S}$ consists of $6 \times 6$ i.e. $36$ events.. Let event $\text{A}$ indicate the combination in which $3$ has appeared at least once.. And event $\text{B}$ indicates the combination of the numbers which sum up to $7$.. Therefore, $\text{A} = \{ (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3), (6, 3) \}$. And, $\text{B} = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) \}$. Therefore, $\text{P}(\text{A}) = \frac{11}{36}$. And, $\text{P}(\text{B}) = \frac{6}{36}$. And also, $n(\text{A} \cap \text{B}) = 2$. Therefore, $\text{P}(\text{A} \cap \text{B}) = \frac{2}{36}$. Applying the conditional probability formula we get, $\text{P}(\text{A} \| \text{B}) = \frac{\text{P}(\text{A} \cap \text{B})}{\text{P}(\text{B})} = \frac{\frac{2}{36}}{\frac{6}{36}} = \frac{2}{6} = \frac{1}{3}$.. Example 2: The probability that it is Friday and that a student is absent is $0.03$. Since there are $5$ school days in a week, the probability that it is Friday is $0.2$. What is the probability that a student is absent given that today is Friday?. In this problem, $\text{P}(\text{Friday and Absent}) = 0.03$. And, $\text{P}(\text{Friday}) = 0.2$. Therefore, the probability that a student is absent given that today is Friday $P(\text{Absent} \| \text{Friday}) = \frac{\text{P}(\text{Friday and Absent})}{\text{P}(\text{Friday})} = \frac{0.03}{0.2} = \frac{3}{20} = 0.15$. Example 3: In a school, $18 \%$ of all students play football and basketball and $32 \%$ of all students play football. What is the probability that a student plays basketball given that the student plays football?. Let $\text{F}$ be an event that a student plays football.. And, let $\text{B}$ be an event that a student plays basketball.. Therefore, $\text{F} \cap \text{B}$ is an event that a student plays football as well as basketball.. $\text{P}(\text{F}) = 18\% = 0.18$. $\text{P}(\text{F} \cap \text{B}) = 32\% = 0.32$. Therefore, the probability that a student plays basketball given that the student plays football is $\frac{\text{P}(\text{F} \cap \text{B})}{\text{P}(\text{F})} = \frac{0.18}{0.32} = 0.5625$.. ### Rule 6: Multiplication Rule of Probability. This probability rule is used when both events occur. $\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B}\|\text{A})$ or $\text{P}(\text{B}) \times \text{P}(\text{A}\| \text{B})$. To prove the multiplication rule of probability, weâ€™ll use the following rules of conditional probability. $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{A}) \times \text{P}(\text{B} \| \text{A})$ ; if $\text{P}(\text{A}) \ne 0$. $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{B}) \times \text{P}(\text{A}\| \text{B})$ ; if $\text{P}(\text{B}) \ne 0$. If $\text{A}$ and $\text{B}$ are two independent events for a random experiment, then the probability of the simultaneous occurrence of two independent events will be equal to the product of their probabilities, i.e., $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{A}).\text{P}(\text{B})$. Now, we know that $\text{P}(\text{A} \cap \text{B}) = \text{P}(\text{A}) \times \text{P}(\text{B} \| \text{A})$. Since $\text{A}$ and $\text{B}$ are independent, therefore, $\text{P}(\text{B} \| \text{A}) = \text{P}(\text{B})$. ### Examples of Multiplication Rule of Probability. Example 1: An urn contains $12$ pink balls and 6 blue balls. Without replacement, two balls are drawn one after another. What is the probability that both balls drawn are pink?. Let $\text{A}$ and $\text{B}$ be the probability of drawing the first ball pink and the second ball also pink, respectively. Then, we need to find the probability of $\text{P} (\text{A} \cap \text{B})$.. The probability of drawing the first ball pink = $\frac{12}{18}$. Since the replacement is not allowed, and the first ball has been drawn, the remaining pink balls in the urn are $17$, out of which $11$ are pink balls.. Therefore, the probability of drawing another pink ball = $\frac{11}{17}$.. By multiplication rule, $\text{P} (\text{A} \cap \text{B}) = \text{P} (\text{A}) \times \text{P} (\text{A}\| \text{B})$.. $= \frac{12}{18} \times \frac{11}{17} = \frac{132}{306}$.. Example 2: A magician takes out two cards from a deck of cards, one after the other, without replacement. What is the probability of getting an ace of spade, and a card of heart, as the first and second cards, respectively?. Since the cards are drawn without replacement, therefore, both events are dependent upon each other.. As we know, there is only one ace of spades in a deck of cards. Therefore, the probability of drawing ace of spades = $\frac{1}{52}$.. Since the ace of spades has already been drawn and replacement is not allowed. Therefore, only $51$ cards remain in the deck.. Out of these $51$ cards, $13$ are hearts. Therefore, the probability of drawing a card of heart = $\frac{13}{51}$.. From the multiplication rule, the probability of drawing aces of spades and a heart of card = $\frac{1}{52} \times \frac{13}{51} = \frac{1}{204}$.. ## Practice Problems. 1. In a cricket tournament, a player hits eight times â€˜$6$â€™ out of thirty-two balls. Calculate the probability that he would not hit a â€˜$6$â€™.. 2. In a laptop shop, there are $16$ defective laptops out of $200$ laptops. If one laptop is taken out at random from this laptop shop, what is the probability that it is a non-defective laptop?. 3. Neha has $4$ yellow t-shirts, $6$ black t-shirts, and $2$ blue t-shirts to choose from for her outfit today. She chooses a t-shirt randomly with each t-shirt equally likely to be chosen. Find the probability that a black or blue t-shirt is chosen for the outfit.. 4. There are a total of $50$ distinct books on a shelf such as $20$ math books, $16$ physics books, and $14$ chemistry books. Find the probability of getting a book that is not a chemistry book or not a physics book.. 5. Suppose you take out two cards from a standard pack of cards one after another, without replacing the first card. What is the probability that the first card is the ace of spades, and the second card is a heart?. 6. You have a cowboy hat, a top hat, and an Indonesian hat called a songkok. You also have four shirts: white, black, green, and pink. If you choose one hat and one shirt at random, what is the probability that you choose the songkok and the black shirt?. 7. Suppose a box has $3$ red marbles and $2$ black ones. We select $2$ marbles. What is the probability that the second marble is red given that the first one is red?. 8. A family has $2$ children. Given that one of the children is a boy, what is the probability that the other child is also a boy?. ## FAQs. ### What are the 6 rules of probability?. The $6$ rules of probability are. Rule 1: The probability of an impossible event is $0$; the probability of a certain event is $1$. Therefore, for any event $\text{A}$, the range of possible probabilities is $0 \le \text{P}( \text{A} ) \le 1$.. Rule 2: For $\text{S}$ the sample space of all possibilities, $\text{P} (\text{S}) = 1$, i.e., the sum of all the probabilities for all possible events is equal to one.Â. Rule 3: For any event $\text{A}$, $\text{P}(\text{A}^{\text{c}}) = 1 – \text{P}(\text{A})$. It follows then that $\text{P}(\text{A}) = 1 – \text{P}( \text{A}^{\text{c}})$. Rule 4 (Addition Rule): This probability rule relates to either one or both events that occur. a) If two events say $\text{A}$ and $\text{B}$, are mutually exclusive i.e., $\text{A}$ and $\text{B}$ have no outcomes in common then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B})$. b) If two events are not mutually exclusive, then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B}) – \text{P}(\text{A and B})$. Rule 5 (Conditional Probability): $\text{P}(\text{A} \| \text{B}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{B})}$ or $\text{P}(\text{B} \| \text{A}) = \frac{\text{P}(\text{A and B})}{\text{P}(\text{A})}$. Rule 6 (Multiplication Rule): This probability rule is used when both events occur. $\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B}\|\text{A})$ or $\text{P}(\text{B}) \times \text{P}(\text{A}\| \text{B})$. ### What is the probability formula?. Probability of occurrence of an event $\text{P}(\text{E}) = \frac{\text{Number of favorable outcomes}}{\text{Total Number of outcomes}}$.. ### What is the addition rule of probability?. The addition rule of probability relates to either one or both events that occur and it states that. a) If two events say $\text{A}$ and $\text{B}$, are mutually exclusive i.e., $\text{A}$ and $\text{B}$ have no outcomes in common then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B})$. b) If two events are not mutually exclusive, then $\text{P}(\text{A or B}) = \text{P}(\text{A}) + \text{P}(\text{B}) – \text{P}(\text{A and B})$. ### What is the multiplication rule of probability?. The multiplication rule ofÂ probability rule is used when both events occur and it states that. $\text{P}(\text{A and B}) = \text{P}(\text{A}) \times \text{P}(\text{B}\|\text{A})$ or $\text{P}(\text{B}) \times \text{P}(\text{A}\| \text{B})$. ### What are complementary events?. Two events are said to be complementary when one event occurs if and only if the other does not take place. For example, it rains or it does not rain are complementary events.. ## Conclusion. There are five basic probability rules that are used to solve problems. These five rules of probability are rules related to impossible and certain events, the sum of all probabilities, addition of probabilities, multiplication of probabilities, complementary events, and conditional probability.
http://zh.wikipedia.org/wiki/%E6%95%B0%E5%80%BC%E5%AD%94%E5%BE%84	1,432,984,818,000,000,000	text/html	crawl-data/CC-MAIN-2015-22/segments/1432207930995.36/warc/CC-MAIN-20150521113210-00110-ip-10-180-206-219.ec2.internal.warc.gz	870,660,900	18,314	数值孔径普通光学中的数值孔径概念 $\mathrm{NA} = n \sin \theta\;$ 数值孔径与焦比的关系 $\ N = f/D$ $\mathrm{NA_i} = n \sin \theta = n \sin \arctan \frac{D}{2f} \approx n \frac {D}{2f}$ $N \approx \frac{1}{2\;\mathrm{NA_i}}$ 工作焦比（有效焦比） $\frac{1}{2 \mathrm{NA_i}} = N_\mathrm{w} = (1-m)\, N$ $\frac{1}{2 \mathrm{NA_o}} = \frac{m-1}{m}\, N$ 激光物理中的数值孔径概念 $\mathrm{NA} = n \sin \theta\;$ θ的定义则与之前所述不同。激光光束的并不是一个因受到光阑限制而产生的锐利圆锥，而是一个辐照度随着离光束中心距离而逐渐降低的高斯光束。针对这种情况，激光物理学家们选择用光束的发散程度来定义θ，也就是θ由光的传播方向，以及辐照度降低到波前总辐照度1/e2时距光束中轴的距离决定。对于高斯激光束，其数值孔径与激光最小束斑尺寸有关（其数值孔径表示激光的发散程度，激光发散程度与激光最小光束直径有关）： $\mathrm{NA}\simeq \frac{\lambda_0}{\pi w_0},$ 光纤光学中的数值孔径概念 $n \sin \theta_\max = \sqrt{n_1^2 - n_2^2}$ $n\sin\theta_\mathrm{max} = n_1\sin\theta_r\$ $\sin\theta_{r} = \sin\left({90^\circ} - \theta_{c} \right) = \cos\theta_{c}\$ $\frac{n}{n_{1}}\sin\theta_\mathrm{max} = \cos\theta_{c}$ $\frac{n^{2}}{n_{1}^{2}}\sin^{2}\theta_\mathrm{max} = \cos ^{2}\theta_{c} = 1 - \sin^{2}\theta_{c} = 1 - \frac{n_{2}^{2}}{n_{1}^{2}}$ $n \sin \theta_\mathrm{max} = \sqrt{n_1^2 - n_2^2}$ $\mathrm{NA} = \sqrt{n_1^2 - n_2^2}$ 参考文献 1. ^ "High-def Disc Update: Where things stand with HD DVD and Blu-ray" by Steve Kindig, Crutchfield Advisor. Accessed 2008-01-18. 2. ^ 2.0 2.1 Greivenkamp, John E. Field Guide to Geometrical Optics. SPIE Field Guides vol. FG01. SPIE. 2004. ISBN 0-8194-5294-7. p. 29. 3. ^ Rudolf Kingslake. Lenses in photography: the practical guide to optics for photographers. Case-Hoyt, for Garden City Books. 1951: 97–98. 4. ^ Angelo V Arecchi, Tahar Messadi, and R. John Koshel. Field Guide to Illumination. SPIE. 2007: 48. ISBN 978-0-8194-6768-3.	808	1,628	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 27, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.640625	4	CC-MAIN-2015-22	latest	en	0.209421	数值孔径. 普通光学中的数值孔径概念. $\mathrm{NA} = n \sin \theta\;$. 数值孔径与焦比的关系. $\ N = f/D$. $\mathrm{NA_i} = n \sin \theta = n \sin \arctan \frac{D}{2f} \approx n \frac {D}{2f}$. $N \approx \frac{1}{2\;\mathrm{NA_i}}$. 工作焦比（有效焦比）. $\frac{1}{2 \mathrm{NA_i}} = N_\mathrm{w} = (1-m)\, N$. $\frac{1}{2 \mathrm{NA_o}} = \frac{m-1}{m}\, N$. 激光物理中的数值孔径概念. $\mathrm{NA} = n \sin \theta\;$. θ的定义则与之前所述不同。激光光束的并不是一个因受到光阑限制而产生的锐利圆锥，而是一个辐照度随着离光束中心距离而逐渐降低的高斯光束。针对这种情况，激光物理学家们选择用光束的发散程度来定义θ，也就是θ由光的传播方向，以及辐照度降低到波前总辐照度1/e2时距光束中轴的距离决定。对于高斯激光束，其数值孔径与激光最小束斑尺寸有关（其数值孔径表示激光的发散程度，激光发散程度与激光最小光束直径有关）：. $\mathrm{NA}\simeq \frac{\lambda_0}{\pi w_0},$. 光纤光学中的数值孔径概念. $n \sin \theta_\max = \sqrt{n_1^2 - n_2^2}$. $n\sin\theta_\mathrm{max} = n_1\sin\theta_r\$. $\sin\theta_{r} = \sin\left({90^\circ} - \theta_{c} \right) = \cos\theta_{c}\$. $\frac{n}{n_{1}}\sin\theta_\mathrm{max} = \cos\theta_{c}$. $\frac{n^{2}}{n_{1}^{2}}\sin^{2}\theta_\mathrm{max} = \cos ^{2}\theta_{c} = 1 - \sin^{2}\theta_{c} = 1 - \frac{n_{2}^{2}}{n_{1}^{2}}$. $n \sin \theta_\mathrm{max} = \sqrt{n_1^2 - n_2^2}$. $\mathrm{NA} = \sqrt{n_1^2 - n_2^2}$. 参考文献. 1. ^ "High-def Disc Update: Where things stand with HD DVD and Blu-ray" by Steve Kindig, Crutchfield Advisor.	Accessed 2008-01-18.. 2. ^ 2.0 2.1 Greivenkamp, John E. Field Guide to Geometrical Optics. SPIE Field Guides vol. FG01. SPIE. 2004. ISBN 0-8194-5294-7. p. 29.. 3. ^ Rudolf Kingslake. Lenses in photography: the practical guide to optics for photographers. Case-Hoyt, for Garden City Books. 1951: 97–98.. 4. ^ Angelo V Arecchi, Tahar Messadi, and R. John Koshel. Field Guide to Illumination. SPIE. 2007: 48. ISBN 978-0-8194-6768-3.
https://www.tutorialride.com/seating-arrangement-questions/seating-arrangement-logical-reasoning-questions-and-answers-part-4.htm	1,726,528,067,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00841.warc.gz	978,792,687	3,902	# Seating Arrangement Logical Reasoning Questions and Answers Part 4 Directions: Solve Questions 13-15, based on the given information. A, B, C, D, E and F are books of six subjects. They are to be arranged on a vertical shelf such that each shelf will have only one book. The bottom shelf is the first shelf and top one is 6th. 1) There are 2 shelves between shelves on which B and F are kept 2) B is kept on a shelf below F’s shelf. 3) Neither A nor E are kept on a shelf immediately below or immediately above the shelf which has B 4) A is not on an odd numbered shelf. 5) There is only one shelf between shelves of D and E 6) E is not on a shelf immediately below or immediately above the shelf which has C 13. Which subject is on the bottom shelf? a. A b. B c. D d. F Explanation: Shelf 6 A Shelf 5 E Shelf 4 F Shelf 3 D Shelf 2 C Shelf 1 B 14. On which shelf subject F books are kept? a. 3rd b. 6th c. 2nd d. 4th Explanation: Shelf 6 A Shelf 5 E Shelf 4 F Shelf 3 D Shelf 2 C Shelf 1 B 15. How many shelves are there between C and E a. 2 b. 4 c. 1 d. 0 Explanation: Shelf 6 A Shelf 5 E Shelf 4 F Shelf 3 D Shelf 2 C Shelf 1 B Directions: Solve Questions 16-17, based on the given information. P, Q, R, S, T, U, V and W are playing a game standing in a circle facing outwards. R is neither neighbor of P nor of V. S is neighbor of P but not of W. T is neighbor of W and is third to the right of U. Q is neighbor of U and fourth to the left of S. 16. Following are some groups given. In which of those, the middle person is standing between the third and the first person? a. TRS b. VPS c. PUQ d. SWT Explanation: 17. If R and V interchange their positions, then what will be position of U? a. To the immediate left of V b. To the immediate left of R c. To the immediate right of R d. Opposite of W Answer: c. To the immediate right of R Explanation:	606	1,951	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.1875	4	CC-MAIN-2024-38	latest	en	0.918848	# Seating Arrangement Logical Reasoning Questions and Answers Part 4. Directions: Solve Questions 13-15, based on the given information.. A, B, C, D, E and F are books of six subjects. They are to be arranged on a vertical shelf such that each shelf will have only one book. The bottom shelf is the first shelf and top one is 6th.. 1) There are 2 shelves between shelves on which B and F are kept. 2) B is kept on a shelf below F’s shelf.. 3) Neither A nor E are kept on a shelf immediately below or immediately above the shelf which has B. 4) A is not on an odd numbered shelf.. 5) There is only one shelf between shelves of D and E. 6) E is not on a shelf immediately below or immediately above the shelf which has C. 13. Which subject is on the bottom shelf?. a. A. b. B. c. D. d. F. Explanation:. Shelf 6 A. Shelf 5 E. Shelf 4 F. Shelf 3 D. Shelf 2 C. Shelf 1 B. 14. On which shelf subject F books are kept?. a. 3rd. b. 6th. c. 2nd. d. 4th. Explanation:. Shelf 6 A. Shelf 5 E. Shelf 4 F. Shelf 3 D. Shelf 2 C. Shelf 1 B. 15. How many shelves are there between C and E. a.	2. b. 4. c. 1. d. 0. Explanation:. Shelf 6 A. Shelf 5 E. Shelf 4 F. Shelf 3 D. Shelf 2 C. Shelf 1 B. Directions: Solve Questions 16-17, based on the given information.. P, Q, R, S, T, U, V and W are playing a game standing in a circle facing outwards. R is neither neighbor of P nor of V. S is neighbor of P but not of W. T is neighbor of W and is third to the right of U. Q is neighbor of U and fourth to the left of S.. 16. Following are some groups given. In which of those, the middle person is standing between the third and the first person?. a. TRS. b. VPS. c. PUQ. d. SWT. Explanation:. 17. If R and V interchange their positions, then what will be position of U?. a. To the immediate left of V. b. To the immediate left of R. c. To the immediate right of R. d. Opposite of W. Answer: c. To the immediate right of R. Explanation:.
https://rf-onlinegame.com/what-is-the-enthalpy-of-water/	1,716,740,259,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058956.26/warc/CC-MAIN-20240526135546-20240526165546-00090.warc.gz	413,912,643	14,243	# What is the enthalpy of water? ## What is the enthalpy of water? Specific enthalpy of water (hwater) is given by the product of the specific heat capacity of water Cwater and the temperature. At ambient conditions (Pressure 1 bar), water boils at 100℃, and the specific enthalpy of water is 418 KJ/Kg. ### What is the entropy of water? 19 The entropy of water. The entropy of water under ambient conditions (T = 298 K and atmospheric pressure) is equal to 16.8 cal·K −1·mol −1. This amount includes contributions from translational, rotational, and vibrational degrees of freedom. #### What is the enthalpy of solid water? 10.10: Enthalpy of Fusion and Enthalpy of Vaporization Substance Formula ΔH(fusion) / kJ mol-1 Water* H2O 6.00678 at 0°C, 101kPa 6.354 at 81.6 °C, 2.50 MPa n-Nonane C9H20 19.3 Mercury Hg 2.30 Sodium Na 2.60 What is the enthalpy of water at 25 degrees Celsius? At 25 degrees Celsius, the enthalpy change for the transition of liquid water to steam is 50 kJ/mol. How do you calculate water entropy? We are asked calculate the change in entropy ΔS = ΔQ/T. While the water changes phase, the temperature stays constant. Details of the calculation: ΔS = ΔQ/T. ## What is the entropy of water at 25 C? The entropy of water at 25 degrees C and 1 atm is 188.8 Joules/(mole K). ### What is the enthalpy of water at 90 deg C? 6792.8 377.04 Temperature Enthalpy, H [°C] [kJ/kmol] [kJ/kg] 80 6035.5 335.01 90 6792.8 377.04 100 7551.8 419.17 #### How do you calculate the enthalpy of water? Enthalpy of Water Calculator 1. Formula. H = m * C * T. 2. Mass (g) 3. Specific Heat (J/g*C) 4. Temperature (C) What is enthalpy diagram? An enthalpy diagram plots information about a chemical reaction such as the starting energy level, how much energy needs to be added to activate the reaction, and the ending energy. An enthalpy diagram is graphed with the enthalpy on the y-axis and the time, or reaction progress, on the x-axis. How do you read enthalpy diagrams? On the P-H diagram, pressure is indicated on the y-axis and enthalpy is indicated on the x-axis. Typically enthalpy is in units of Btu/lb and pressure is in units of pounds per square inch (psi). The upside down U figure shown on the diagram designates the points at which the refrigerant changes phase. ## What is entropy of water? The entropy of water under ambient conditions (T = 298 K and atmospheric pressure) is equal to 16.8 cal·K −1·mol −1. This amount includes contributions from translational, rotational, and vibrational degrees of freedom. ### What is enthalpy of fusion of water? (1) 333.55 J/g (heat of fusion of ice) = 333.55 kJ/kg = 333.55 kJ for 1 kg of ice to melt, plus. (2) 4.18 J/(g⋅K) × 20 K = 4.18 kJ/(kg⋅K) × 20 K = 83.6 kJ for 1 kg of water to increase in temperature by 20 K. #### Is entropy and enthalpy the same? We know that the major difference between enthalpy and entropy is that even though they are part of a thermodynamic system, enthalpy is represented as the total heat content whereas entropy is the degree of disorder. , which means that in absolute temperature with some change in entropy results in a change of enthalpy. What is the enthalpy of sublimation of water? 51.1 kJ/mol Thermodynamic properties Phase behavior Enthalpy change of sublimation at 273.15 K, ΔsubH 51.1 kJ/mol Std entropy change of sublimation at 273.15 K, 1 bar, ΔsubS ~144 J/(mol·K) Molal freezing point constant −1.858 °C kg/mol Molal boiling point constant 0.512 °C kg/mol Why does water have a high enthalpy of vaporization? Water has a high heat of vaporization because hydrogen bonds form readily between the oxygen of one molecule and the hydrogens of other molecules. ## How do you find enthalpy? Use the formula ∆H = m x s x ∆T to solve. Once you have m, the mass of your reactants, s, the specific heat of your product, and ∆T, the temperature change from your reaction, you are prepared to find the enthalpy of reaction. Simply plug your values into the formula ∆H = m x s x ∆T and multiply to solve. ### What is enthalpy change in a potential energy diagram? The heat stored by a substance is called its enthalpy (H). is the overall enthalpy change for a reaction. Potential energy diagrams can be used to calculate both the enthalpy change and the activation energy for a reaction. #### How do you calculate enthalpy of flow? In the simplest case of an ideal gas, we multiply mass with specific enthalpy or with specific heat at constant pressure and temperature. Similarly, to find enthalpy flow we multiply mass flow by specific enthalpy or by specific heat at constant pressure and by temperature. ## Who voices Okita souji Fgo? Who voices Okita souji Fgo? Aoi Yuuki is the Japanese voice of Saber / Okita Souji in Fate/Grand Order. Is souji Okita a girl? Okita is the male… ## What is Toyota rear seat entertainment system? What is Toyota rear seat entertainment system? Depending on which Toyota model you own, your second-row and/or third-row passengers can watch movies, play video games, and listen to… ## Is Jason Mewes still friends with Kevin? Is Jason Mewes still friends with Kevin? They’ve been friends since childhood Mewes struggled with an addiction that interfered with their working and personal relationship. However, he got… ## Can you make alcohol from apple peels? Can you make alcohol from apple peels? Pack a sterilized jar with apple peels and cores, fill to cover with liquor and top jar with lid. Store in… ## How do you extend earbud cords? How do you extend earbud cords? How to Make Your Own Headphone Extension Cable Strip 1/2 inch of insulation from both ends of a three conductor cable. Slip… ## How many children did Betty Jo have on Petticoat Junction? How many children did Betty Jo have on Petticoat Junction? three daughters The series takes place at the Shady Rest Hotel, which is run by Kate Bradley; her…	1,565	5,897	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.671875	4	CC-MAIN-2024-22	latest	en	0.825535	# What is the enthalpy of water?. ## What is the enthalpy of water?. Specific enthalpy of water (hwater) is given by the product of the specific heat capacity of water Cwater and the temperature. At ambient conditions (Pressure 1 bar), water boils at 100℃, and the specific enthalpy of water is 418 KJ/Kg.. ### What is the entropy of water?. 19 The entropy of water. The entropy of water under ambient conditions (T = 298 K and atmospheric pressure) is equal to 16.8 cal·K −1·mol −1. This amount includes contributions from translational, rotational, and vibrational degrees of freedom.. #### What is the enthalpy of solid water?. 10.10: Enthalpy of Fusion and Enthalpy of Vaporization. Substance Formula ΔH(fusion) / kJ mol-1. Water* H2O 6.00678 at 0°C, 101kPa 6.354 at 81.6 °C, 2.50 MPa. n-Nonane C9H20 19.3. Mercury Hg 2.30. Sodium Na 2.60. What is the enthalpy of water at 25 degrees Celsius?. At 25 degrees Celsius, the enthalpy change for the transition of liquid water to steam is 50 kJ/mol.. How do you calculate water entropy?. We are asked calculate the change in entropy ΔS = ΔQ/T. While the water changes phase, the temperature stays constant. Details of the calculation: ΔS = ΔQ/T.. ## What is the entropy of water at 25 C?. The entropy of water at 25 degrees C and 1 atm is 188.8 Joules/(mole K).. ### What is the enthalpy of water at 90 deg C?. 6792.8 377.04. Temperature Enthalpy, H. [°C] [kJ/kmol] [kJ/kg]. 80 6035.5 335.01. 90 6792.8 377.04. 100 7551.8 419.17. #### How do you calculate the enthalpy of water?. Enthalpy of Water Calculator. 1. Formula. H = m * C * T.. 2. Mass (g). 3. Specific Heat (J/g*C). 4. Temperature (C). What is enthalpy diagram?. An enthalpy diagram plots information about a chemical reaction such as the starting energy level, how much energy needs to be added to activate the reaction, and the ending energy. An enthalpy diagram is graphed with the enthalpy on the y-axis and the time, or reaction progress, on the x-axis.. How do you read enthalpy diagrams?. On the P-H diagram, pressure is indicated on the y-axis and enthalpy is indicated on the x-axis. Typically enthalpy is in units of Btu/lb and pressure is in units of pounds per square inch (psi). The upside down U figure shown on the diagram designates the points at which the refrigerant changes phase.	## What is entropy of water?. The entropy of water under ambient conditions (T = 298 K and atmospheric pressure) is equal to 16.8 cal·K −1·mol −1. This amount includes contributions from translational, rotational, and vibrational degrees of freedom.. ### What is enthalpy of fusion of water?. (1) 333.55 J/g (heat of fusion of ice) = 333.55 kJ/kg = 333.55 kJ for 1 kg of ice to melt, plus. (2) 4.18 J/(g⋅K) × 20 K = 4.18 kJ/(kg⋅K) × 20 K = 83.6 kJ for 1 kg of water to increase in temperature by 20 K.. #### Is entropy and enthalpy the same?. We know that the major difference between enthalpy and entropy is that even though they are part of a thermodynamic system, enthalpy is represented as the total heat content whereas entropy is the degree of disorder. , which means that in absolute temperature with some change in entropy results in a change of enthalpy.. What is the enthalpy of sublimation of water?. 51.1 kJ/mol. Thermodynamic properties. Phase behavior. Enthalpy change of sublimation at 273.15 K, ΔsubH 51.1 kJ/mol. Std entropy change of sublimation at 273.15 K, 1 bar, ΔsubS ~144 J/(mol·K). Molal freezing point constant −1.858 °C kg/mol. Molal boiling point constant 0.512 °C kg/mol. Why does water have a high enthalpy of vaporization?. Water has a high heat of vaporization because hydrogen bonds form readily between the oxygen of one molecule and the hydrogens of other molecules.. ## How do you find enthalpy?. Use the formula ∆H = m x s x ∆T to solve. Once you have m, the mass of your reactants, s, the specific heat of your product, and ∆T, the temperature change from your reaction, you are prepared to find the enthalpy of reaction. Simply plug your values into the formula ∆H = m x s x ∆T and multiply to solve.. ### What is enthalpy change in a potential energy diagram?. The heat stored by a substance is called its enthalpy (H). is the overall enthalpy change for a reaction. Potential energy diagrams can be used to calculate both the enthalpy change and the activation energy for a reaction.. #### How do you calculate enthalpy of flow?. In the simplest case of an ideal gas, we multiply mass with specific enthalpy or with specific heat at constant pressure and temperature. Similarly, to find enthalpy flow we multiply mass flow by specific enthalpy or by specific heat at constant pressure and by temperature.. ## Who voices Okita souji Fgo?. Who voices Okita souji Fgo? Aoi Yuuki is the Japanese voice of Saber / Okita Souji in Fate/Grand Order. Is souji Okita a girl? Okita is the male…. ## What is Toyota rear seat entertainment system?. What is Toyota rear seat entertainment system? Depending on which Toyota model you own, your second-row and/or third-row passengers can watch movies, play video games, and listen to…. ## Is Jason Mewes still friends with Kevin?. Is Jason Mewes still friends with Kevin? They’ve been friends since childhood Mewes struggled with an addiction that interfered with their working and personal relationship. However, he got…. ## Can you make alcohol from apple peels?. Can you make alcohol from apple peels? Pack a sterilized jar with apple peels and cores, fill to cover with liquor and top jar with lid. Store in…. ## How do you extend earbud cords?. How do you extend earbud cords? How to Make Your Own Headphone Extension Cable Strip 1/2 inch of insulation from both ends of a three conductor cable. Slip…. ## How many children did Betty Jo have on Petticoat Junction?. How many children did Betty Jo have on Petticoat Junction? three daughters The series takes place at the Shady Rest Hotel, which is run by Kate Bradley; her….
https://www.coursehero.com/file/6180173/solution-pdf2/	1,513,487,984,000,000,000	text/html	crawl-data/CC-MAIN-2017-51/segments/1512948592972.60/warc/CC-MAIN-20171217035328-20171217061328-00323.warc.gz	726,341,665	111,434	solution_pdf2 # solution_pdf2 - suleimenov(bs26835 HW 02 rusin(55565 This... This preview shows pages 1–4. Sign up to view the full content. suleimenov (bs26835) – HW 02 – rusin – (55565) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Determine iF the sequence { a n } converges when a n = 1 n ln p 5 2 n + 4 P , and iF it does, fnd its limit. 1. limit = ln 5 6 2. limit = ln 5 2 3. limit = ln 2 4. limit = 0 correct 5. the sequence diverges Explanation: AFter division by n we see that 5 2 n + 4 = 5 n 2 + 4 n , so by properties oF logs, a n = 1 n ln 5 n 1 n ln p 2 + 4 n P . But by known limits (or use L’Hospital), 1 n ln 5 n , 1 n ln p 2 + 4 n P −→ 0 as n → ∞ . Consequently, the sequence { a n } converges and has limit = 0 . 002 10.0 points Determine iF { a n } converges when a n = 5 n 3 2 n 2 , and iF it does, fnd its limit. 1. limit = + 2 3 2. limit = 2 3. limit = 5 4. sequence diverges correct 5. limit = 3 Explanation: AFter simplifcation, a n = 5 n n 2 n = n ( 5 n 2 ) , so a n → ∞ as n → ∞ . But then the sequence is unbounded, and hence diverges . 003 10.0 points Determine iF the sequence { a n } converges, and iF it does, fnd its limit when a n = ± n + 1 n 1 ² 3 n . 1. limit = 1 2. does not converge 3. limit = e 3 4. limit = e 6 5. limit = e 6 correct 6. limit = e 3 Explanation: By the Laws oF Exponents, a n = ± n 1 n + 1 ² 3 n = ± 1 1 n ² 3 n ± 1 + 1 n ² 3 n . This preview has intentionally blurred sections. Sign up to view the full version. View Full Document suleimenov (bs26835) – HW 02 – rusin – (55565) 2 But lim n →∞ p 1 + x n P n = e x . Consequently, by Properties of Limits the given limit exists and limit = e 6 . 004 10.0 points Determine whether the sequence { a n } con- verges or diverges when a n = ( 1) n 1 n n 2 + 5 , and if it converges, Fnd the limit. 1. converges with limit = 1 5 2. sequence diverges 3. converges with limit = 5 4. converges with limit = 0 correct 5. converges with limit = 1 5 6. converges with limit = 5 Explanation: After division, a n = ( 1) n 1 n n 2 + 5 = ( 1) n 1 n + 5 n . Consequently, 0 ≤ \| a n \| = 1 n + 5 n 1 n . But 1 /n 0 as n → ∞ , so by the Squeeze theorem, lim n \| a n \| = 0 . But −\| a n \| ≤ a n ≤ \| a n \| , so by the Squeeze theorem again the given sequence { a n } converges and has limit = 0 . keywords: 005 10.0 points Determine if the sequence { a n } converges when a n = (2 n 1)! (2 n + 1)! , and if it converges, Fnd the limit. 1. converges with limit = 0 correct 2. converges with limit = 1 3. converges with limit = 1 4 4. converges with limit = 4 5. does not converge Explanation: By deFnition, m ! is the product m ! = 1 . 2 . 3 . . . . . m of the Frst m positive integers. When m = 2 n 1, therefore, (2 n 1)! = 1 . 2 . 3 . . . . (2 n 1) , while (2 n + 1)! = 1 . 2 . 3 . . . . . (2 n 1)2 n (2 n + 1) . when m = 2 n + 1. But then, (2 n 1)! (2 n + 1)! = 1 2 n (2 n + 1) . Consequently, the given sequence converges with limit = 0 . suleimenov (bs26835) – HW 02 – rusin – (55565) 3 006 10.0 points Determine whether the sequence { a n } con- verges or diverges when a n = 2 + cos 2 n 5 + 2 n . 1. diverges 2. converges with limit = 0 correct 3. converges with limit = 1 3 4. converges with limit = 2 5 5. converges with limit = 2 Explanation: Since 0 cos 2 n 1 , we see that 2 5 + 2 n a n 3 5 + 2 n . This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### Page1 / 10 solution_pdf2 - suleimenov(bs26835 HW 02 rusin(55565 This... This preview shows document pages 1 - 4. Sign up to view the full document. View Full Document Ask a homework question - tutors are online	1,408	3,831	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.21875	4	CC-MAIN-2017-51	latest	en	0.740546	solution_pdf2. # solution_pdf2 - suleimenov(bs26835 HW 02 rusin(55565 This.... This preview shows pages 1–4. Sign up to view the full content.. suleimenov (bs26835) – HW 02 – rusin – (55565) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Determine iF the sequence { a n } converges when a n = 1 n ln p 5 2 n + 4 P , and iF it does, fnd its limit. 1. limit = ln 5 6 2. limit = ln 5 2 3. limit = ln 2 4. limit = 0 correct 5. the sequence diverges Explanation: AFter division by n we see that 5 2 n + 4 = 5 n 2 + 4 n , so by properties oF logs, a n = 1 n ln 5 n 1 n ln p 2 + 4 n P . But by known limits (or use L’Hospital), 1 n ln 5 n , 1 n ln p 2 + 4 n P −→ 0 as n → ∞ . Consequently, the sequence { a n } converges and has limit = 0 . 002 10.0 points Determine iF { a n } converges when a n = 5 n 3 2 n 2 , and iF it does, fnd its limit. 1. limit = + 2 3 2. limit = 2 3. limit = 5 4. sequence diverges correct 5. limit = 3 Explanation: AFter simplifcation, a n = 5 n n 2 n = n ( 5 n 2 ) , so a n → ∞ as n → ∞ . But then the sequence is unbounded, and hence diverges . 003 10.0 points Determine iF the sequence { a n } converges, and iF it does, fnd its limit when a n = ± n + 1 n 1 ² 3 n . 1. limit = 1 2. does not converge 3. limit = e 3 4. limit = e 6 5. limit = e 6 correct 6. limit = e 3 Explanation: By the Laws oF Exponents, a n = ± n 1 n + 1 ² 3 n = ± 1 1 n ² 3 n ± 1 + 1 n ² 3 n .. This preview has intentionally blurred sections. Sign up to view the full version.. View Full Document. suleimenov (bs26835) – HW 02 – rusin – (55565) 2 But lim n →∞ p 1 + x n P n = e x . Consequently, by Properties of Limits the given limit exists and limit = e 6 . 004 10.0 points Determine whether the sequence { a n } con- verges or diverges when a n = ( 1) n 1 n n 2 + 5 , and if it converges, Fnd the limit. 1. converges with limit = 1 5 2. sequence diverges 3.	converges with limit = 5 4. converges with limit = 0 correct 5. converges with limit = 1 5 6. converges with limit = 5 Explanation: After division, a n = ( 1) n 1 n n 2 + 5 = ( 1) n 1 n + 5 n . Consequently, 0 ≤ \| a n \| = 1 n + 5 n 1 n . But 1 /n 0 as n → ∞ , so by the Squeeze theorem, lim n \| a n \| = 0 . But −\| a n \| ≤ a n ≤ \| a n \| , so by the Squeeze theorem again the given sequence { a n } converges and has limit = 0 . keywords: 005 10.0 points Determine if the sequence { a n } converges when a n = (2 n 1)! (2 n + 1)! , and if it converges, Fnd the limit. 1. converges with limit = 0 correct 2. converges with limit = 1 3. converges with limit = 1 4 4. converges with limit = 4 5. does not converge Explanation: By deFnition, m ! is the product m ! = 1 . 2 . 3 . . . . . m of the Frst m positive integers. When m = 2 n 1, therefore, (2 n 1)! = 1 . 2 . 3 . . . . (2 n 1) , while (2 n + 1)! = 1 . 2 . 3 . . . . . (2 n 1)2 n (2 n + 1) . when m = 2 n + 1. But then, (2 n 1)! (2 n + 1)! = 1 2 n (2 n + 1) . Consequently, the given sequence converges with limit = 0 .. suleimenov (bs26835) – HW 02 – rusin – (55565) 3 006 10.0 points Determine whether the sequence { a n } con- verges or diverges when a n = 2 + cos 2 n 5 + 2 n . 1. diverges 2. converges with limit = 0 correct 3. converges with limit = 1 3 4. converges with limit = 2 5 5. converges with limit = 2 Explanation: Since 0 cos 2 n 1 , we see that 2 5 + 2 n a n 3 5 + 2 n .. This preview has intentionally blurred sections. 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https://www.coursehero.com/file/6328037/Arrangements/	1,518,953,813,000,000,000	text/html	crawl-data/CC-MAIN-2018-09/segments/1518891811830.17/warc/CC-MAIN-20180218100444-20180218120444-00053.warc.gz	781,561,041	421,468	{[ promptMessage ]} Bookmark it {[ promptMessage ]} Arrangements # Arrangements - Arrangements and Duality Supersampling in... This preview shows pages 1–12. Sign up to view the full content. Arrangements and Duality Supersampling in Ray Tracing Khurram Hassan Shafique This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Duality The concept: we can map between different ways of interpreting 2D values. Points (x,y) can be mapped in a one-to-one manner to lines (slope,intercept) in a different space. There are different ways to do this, called duality transforms . Duality Transforms A duality transform is a mapping which takes an element e in the primal plane to element e* in the dual plane . One possible duality transform: point p : ( p x , p y ) line p* : y = p x x p y line l : y = mx + b point l* : ( m , - b ) This preview has intentionally blurred sections. Sign up to view the full version. View Full Document p Duality Transforms This duality transform takes points to lines, lines to points line segments to double wedges This duality transform preserves order Point p lies above line l point l* lies above line p* l p* l* Duality The dualized version of a problem is no easier or harder to compute than the original problem. But the dualized version may be easier to think about. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document New Concept: Arrangements of Lines L is a set of n lines in the plane. L induces a subdivision of the plane that consists of vertices, edges, and faces. This is called the arrangement induced by L , denoted A ( L ) The complexity of an arrangement is the total number of vertices, edges, and faces. Arrangments Number of vertices of A ( L ) – Vertices of A ( L ) are intersections of l i , l j L Number of edges of A ( L ) n 2 Number of edges on a single line in A ( L ) is one more than number of vertices on that line. Number of faces of A ( L ) Inductive reasoning: add lines one by one Each edge of new line splits a face. Total complexity of an arrangement is O ( n 2 ) 2 n 1 2 2 2 + + n n = + n i i 1 1 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document How Do We Store an Arrangement? Data Type: doubly-connected edge-list (DCEL) Vertex: Coordinates, Incident Edge Face: an Edge Half-Edges Origin Vertex Twin Edge Incident Face Next Edge, Prev Edge Building the Arrangement Iterative algorithm: put one line in at a time. • Start with the first edge e that l i intersects. Split that edge, and move to Twin ( e ) This preview has intentionally blurred sections. Sign up to view the full version. View Full Document ConstructArrangement Algorithm Input : A set L of n lines in the plane Output : DCEL for the subdivision induced by the part of A ( L ) inside a bounding box 1. Compute a bounding box B ( L ) that contains all vertices of A ( L ) in its interior 2. Construct the DCEL for the subdivision induced by B ( L ) 3. for i =1 to n do 4. Find the edge e on B ( L ) that contains the leftmost intersection point of l i and A i 5. f = the bounded face incident to e 6. while f is not the face outside B ( L ) do 7. Split f , and set f to be the next intersected face ConstructArrangement Algorithm -Running Time- We need to insert n lines. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### Page1 / 51 Arrangements - Arrangements and Duality Supersampling in... This preview shows document pages 1 - 12. Sign up to view the full document. View Full Document Ask a homework question - tutors are online	922	3,789	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.765625	4	CC-MAIN-2018-09	latest	en	0.873549	{[ promptMessage ]}. Bookmark it. {[ promptMessage ]}. Arrangements. # Arrangements - Arrangements and Duality Supersampling in.... This preview shows pages 1–12. Sign up to view the full content.. Arrangements and Duality Supersampling in Ray Tracing Khurram Hassan Shafique. This preview has intentionally blurred sections. Sign up to view the full version.. View Full Document. Duality The concept: we can map between different ways of interpreting 2D values. Points (x,y) can be mapped in a one-to-one manner to lines (slope,intercept) in a different space. There are different ways to do this, called duality transforms .. Duality Transforms A duality transform is a mapping which takes an element e in the primal plane to element e* in the dual plane . One possible duality transform: point p : ( p x , p y ) line p* : y = p x x p y line l : y = mx + b point l* : ( m , - b ). This preview has intentionally blurred sections. Sign up to view the full version.. View Full Document. p Duality Transforms This duality transform takes points to lines, lines to points line segments to double wedges This duality transform preserves order Point p lies above line l point l* lies above line p* l p* l*. Duality The dualized version of a problem is no easier or harder to compute than the original problem. But the dualized version may be easier to think about.. This preview has intentionally blurred sections. Sign up to view the full version.. View Full Document. New Concept: Arrangements of Lines L is a set of n lines in the plane. L induces a subdivision of the plane that consists of vertices, edges, and faces. This is called the arrangement induced by L , denoted A ( L ) The complexity of an arrangement is the total number of vertices, edges, and faces.. Arrangments Number of vertices of A ( L ) – Vertices of A ( L ) are intersections of l i , l j L Number of edges of A ( L ) n 2 Number of edges on a single line in A ( L ) is one more than number of vertices on that line. Number of faces of A ( L ) Inductive reasoning: add lines one by one Each edge of new line splits a face. Total complexity of an arrangement is O ( n 2 ) 2 n 1 2 2 2 + + n n = + n i i 1 1. This preview has intentionally blurred sections.	Sign up to view the full version.. View Full Document. How Do We Store an Arrangement? Data Type: doubly-connected edge-list (DCEL) Vertex: Coordinates, Incident Edge Face: an Edge Half-Edges Origin Vertex Twin Edge Incident Face Next Edge, Prev Edge. Building the Arrangement Iterative algorithm: put one line in at a time. • Start with the first edge e that l i intersects. Split that edge, and move to Twin ( e ). This preview has intentionally blurred sections. Sign up to view the full version.. View Full Document. ConstructArrangement Algorithm Input : A set L of n lines in the plane Output : DCEL for the subdivision induced by the part of A ( L ) inside a bounding box 1. Compute a bounding box B ( L ) that contains all vertices of A ( L ) in its interior 2. Construct the DCEL for the subdivision induced by B ( L ) 3. for i =1 to n do 4. Find the edge e on B ( L ) that contains the leftmost intersection point of l i and A i 5. f = the bounded face incident to e 6. while f is not the face outside B ( L ) do 7. Split f , and set f to be the next intersected face. ConstructArrangement Algorithm -Running Time- We need to insert n lines.. This preview has intentionally blurred sections. Sign up to view the full version.. View Full Document. This is the end of the preview. Sign up to access the rest of the document.. {[ snackBarMessage ]}. ### Page1 / 51. Arrangements - Arrangements and Duality Supersampling in.... This preview shows document pages 1 - 12. Sign up to view the full document.. View Full Document. Ask a homework question - tutors are online.
https://jmservera.com/simplify-k2-30/	1,670,483,378,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446711278.74/warc/CC-MAIN-20221208050236-20221208080236-00367.warc.gz	360,275,351	12,815	# Simplify k^2-3>=0 k2-3≥0 Add 3 to both sides of the inequality. k2≥3 Take the square root of both sides of the inequality to eliminate the exponent on the left side. k≥±3 The complete solution is the result of both the positive and negative portions of the solution. First, use the positive value of the ± to find the first solution. k≥3 Next, use the negative value of the ± to find the second solution. Since this is an inequality, flip the direction of the inequality sign on the – portion of the solution. k≤-3 The complete solution is the result of both the positive and negative portions of the solution. k≥3 or k≤-3 k≥3 or k≤-3 The result can be shown in multiple forms. Inequality Form: k≥3 or k≤-3 Interval Notation: (-∞,-3]∪[3,∞) =\sqrt(3)","color":0,"isGrey":false,"dashed":false,"holes":[]},{"ascii":"k Simplify k^2-3>=0 ## Our Professionals ### Lydia Fran #### We are MathExperts Solve all your Math Problems: https://elanyachtselection.com/ Scroll to top	279	976	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.1875	4	CC-MAIN-2022-49	latest	en	0.819722	# Simplify k^2-3>=0. k2-3≥0. Add 3 to both sides of the inequality.. k2≥3. Take the square root of both sides of the inequality to eliminate the exponent on the left side.. k≥±3. The complete solution is the result of both the positive and negative portions of the solution.. First, use the positive value of the ± to find the first solution.. k≥3. Next, use the negative value of the ± to find the second solution. Since this is an inequality, flip the direction of the inequality sign on the – portion of the solution.. k≤-3. The complete solution is the result of both the positive and negative portions of the solution.. k≥3 or k≤-3.	k≥3 or k≤-3. The result can be shown in multiple forms.. Inequality Form:. k≥3 or k≤-3. Interval Notation:. (-∞,-3]∪[3,∞). =\sqrt(3)","color":0,"isGrey":false,"dashed":false,"holes":[]},{"ascii":"k. Simplify k^2-3>=0. ## Our Professionals. ### Lydia Fran. #### We are MathExperts. Solve all your Math Problems: https://elanyachtselection.com/. Scroll to top.
https://codecs.multimedia.cx/2021/11/fun-with-lgt-5-3-wavelet-transform/	1,726,029,414,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651343.80/warc/CC-MAIN-20240911020451-20240911050451-00784.warc.gz	159,020,704	7,725	## Fun with LGT 5/3 wavelet transform LGT 5/3 wavelet transform is the second simplest lossless wavelet transform (the first one is Haar transform of course) so it’s used in a variety of image formats (most famously in JPEG-2000) and intermediate video codecs. Recently I helped one guy with implementing it and while explaining things about it I understood it myself, so here I’m going to write it down for posterity. Usually it’s said that forward transform coefficients are `[-1/8, 2/8, 6/8, 2/8, -1/8]` (lowpass) and `[-1/2, 1, -1/2]` (highpass) and inverse transform coefficients are `[1/2, 1, 1/2]` (lowpass) and `[-1/8, -2/8, 6/8, -2/8, -1/8]` (highpass). In reality if you implement inverse transform using those coefficients you get wrong results. Here’s a simple table to present how input coefficients map to output ones: x[-1] x[0] x[1] x[2] x[3] x[4] x[5] … hi[0] -4/8 8/8 -4/8 lo[0] -1/8 2/8 6/8 2/8 -1/8 hi[1] -4/8 8/8 -4/8 lo[1] -1/8 2/8 6/8 2/8 -1/8 hi[2] -4/8 8/8 -4/8 (`x[-1]` is a mirrored value equal to `x[1]`) As you can see from the table, `x[1]` can be restored as `lo[0] - (hi[0] + hi[1])/4`. With some more calculations you can derive `x[2]` as `(-hi[0] + 4lo[0] + 6hi[1] + 4*lo[1] - hi[2])/8`. Those are not exactly the coefficients you wanted, are they? It turns out the trick is that actual high- and low-pass values are multiplied by `sqrt(2)` and `-1/sqrt(2)` correspondingly so when you’re using them in calculations you’d need to multiply them by -1/2 or -2 and thus get the proper coefficients. But since we use integer-only calculations we don’t scale the output by an irrational number and the inverse transform coefficients had to be modified. And from here we can move to a lifting transform where calculations are made on data in place, without temporary buffers. As mentioned above, we can restore odd elements as simply as `x[2n+1] = lo[n] - (hi[n] + hi[n+1])/4`. But what to do with even elements? `hi[n] = x[2n] - (x[2n-1] + x[2n+1])/2` so `x[2n] = hi[n] + (x[2n-1] + x[2n+1])/2`. Thus we can get interleaved data, replace low-pass coefficients with restored data and use them to restore the other half of coefficients replacing high-pass data. The same can be applied to forward transform as well. If you’re not afraid of mathematics you can actually read and understand the theory behind them, how such filters are constructed, what they’re good for and so on. I’d rather stop here. ### 2 Responses to “Fun with LGT 5/3 wavelet transform” 1. Paul says: Please, never stop talking about wavelets. There so many discovered and yet to be discovered stuff. 2. Kostya says: I agree with you that it’s a trite stuff, but sadly in multimedia (like in many other areas) people don’t understand the stuff they use. Even some codecs are designed that way.	873	2,808	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4	4	CC-MAIN-2024-38	latest	en	0.868546	## Fun with LGT 5/3 wavelet transform. LGT 5/3 wavelet transform is the second simplest lossless wavelet transform (the first one is Haar transform of course) so it’s used in a variety of image formats (most famously in JPEG-2000) and intermediate video codecs. Recently I helped one guy with implementing it and while explaining things about it I understood it myself, so here I’m going to write it down for posterity.. Usually it’s said that forward transform coefficients are `[-1/8, 2/8, 6/8, 2/8, -1/8]` (lowpass) and `[-1/2, 1, -1/2]` (highpass) and inverse transform coefficients are `[1/2, 1, 1/2]` (lowpass) and `[-1/8, -2/8, 6/8, -2/8, -1/8]` (highpass). In reality if you implement inverse transform using those coefficients you get wrong results.. Here’s a simple table to present how input coefficients map to output ones:. x[-1] x[0] x[1] x[2] x[3] x[4] x[5] … hi[0] -4/8 8/8 -4/8 lo[0] -1/8 2/8 6/8 2/8 -1/8 hi[1] -4/8 8/8 -4/8 lo[1] -1/8 2/8 6/8 2/8 -1/8 hi[2] -4/8 8/8 -4/8. (`x[-1]` is a mirrored value equal to `x[1]`). As you can see from the table, `x[1]` can be restored as `lo[0] - (hi[0] + hi[1])/4`. With some more calculations you can derive `x[2]` as `(-hi[0] + 4lo[0] + 6hi[1] + 4*lo[1] - hi[2])/8`. Those are not exactly the coefficients you wanted, are they?. It turns out the trick is that actual high- and low-pass values are multiplied by `sqrt(2)` and `-1/sqrt(2)` correspondingly so when you’re using them in calculations you’d need to multiply them by -1/2 or -2 and thus get the proper coefficients. But since we use integer-only calculations we don’t scale the output by an irrational number and the inverse transform coefficients had to be modified.. And from here we can move to a lifting transform where calculations are made on data in place, without temporary buffers. As mentioned above, we can restore odd elements as simply as `x[2n+1] = lo[n] - (hi[n] + hi[n+1])/4`. But what to do with even elements? `hi[n] = x[2n] - (x[2n-1] + x[2n+1])/2` so `x[2n] = hi[n] + (x[2n-1] + x[2n+1])/2`. Thus we can get interleaved data, replace low-pass coefficients with restored data and use them to restore the other half of coefficients replacing high-pass data.. The same can be applied to forward transform as well.	If you’re not afraid of mathematics you can actually read and understand the theory behind them, how such filters are constructed, what they’re good for and so on. I’d rather stop here.. ### 2 Responses to “Fun with LGT 5/3 wavelet transform”. 1. Paul says:. Please, never stop talking about wavelets. There so many discovered and yet to be discovered stuff.. 2. Kostya says:. I agree with you that it’s a trite stuff, but sadly in multimedia (like in many other areas) people don’t understand the stuff they use. Even some codecs are designed that way.
https://prezi.com/fodislucnnmb/maths-revision-ratios-and-rates/	1,502,961,387,000,000,000	text/html	crawl-data/CC-MAIN-2017-34/segments/1502886102993.24/warc/CC-MAIN-20170817073135-20170817093135-00098.warc.gz	829,911,117	23,147	### Present Remotely Send the link below via email or IM • Invited audience members will follow you as you navigate and present • People invited to a presentation do not need a Prezi account • This link expires 10 minutes after you close the presentation Do you really want to delete this prezi? Neither you, nor the coeditors you shared it with will be able to recover it again. You can change this under Settings & Account at any time. # Maths Revision: Ratios and Rates A revision about ratios and rates. by ## Tiffany Chin on 1 October 2012 Report abuse #### Transcript of Maths Revision: Ratios and Rates Ratios Ratios and Rates A comparison between 2/more quantities of the same kind Ratios can be shown as... 1:1 Ratio Form 1 to 1 Word Form 1/1 Fraction Form 1:n/m:1 Unit Ratio Please do not copy! Equivalent Ratios Ratios with the same meaning Simplify e.g. 2:4 1:2 1/2 2/4 Prime Factorization Unit Ratio Ratios with a denominator to 1 (Simplify) A:B/A 1:B/A 1:n A:B/B 1:B/A m:1 Conversion of A and B Comparing Ratios Change order to unit ratio e.g. Which class has the highest BOYS:GIRLS ratio? Class 1 Class 2 Class 3 2:3 3:4 1:5 1. Compare your question to ratio given If same order, m:1 e.g. Highest B:G If not, 1:n e.g. Lowest G:B 2. Convert 3. Order Use m:1 Class 1 Class 2 Class 3 2/3:1= 0.6: 1 3/4:1=0.75:1 1/5:1=0.2:1 Class 2 Class 1 Class 3 4. Choose your answer Highest B:G Ratio Multiply Method 1: Find the Unknown e.g. Solve for Unknown 3:4=a:8 =6:8 a=6 Method 2: Fractions e.g. 3:4=c:10 10x3/4=c/28.5x10 30/4=c 7.5=c Find A:B:C from A:B/B:C Cross Product e.g. A:B=4:3 B:C=7:5 4 : 3 7 : 5 28:21:15 Multiply as follow Line up common value Rearrange and simplify as needed Modeling Using Ratios e.g. 250 ml lemon juice, how	578	1,755	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.25	4	CC-MAIN-2017-34	longest	en	0.80433	### Present Remotely. Send the link below via email or IM. • Invited audience members will follow you as you navigate and present. • People invited to a presentation do not need a Prezi account. • This link expires 10 minutes after you close the presentation. Do you really want to delete this prezi?. Neither you, nor the coeditors you shared it with will be able to recover it again.. You can change this under Settings & Account at any time.. # Maths Revision: Ratios and Rates. A revision about ratios and rates.. by. ## Tiffany Chin. on 1 October 2012. Report abuse. #### Transcript of Maths Revision: Ratios and Rates. Ratios Ratios and Rates A comparison between 2/more quantities of the same kind Ratios can be shown as... 1:1 Ratio Form. 1 to 1 Word Form. 1/1 Fraction Form. 1:n/m:1 Unit Ratio Please do not copy! Equivalent Ratios Ratios with the same meaning Simplify. e.g.. 2:4 1:2 1/2 2/4 Prime Factorization Unit Ratio Ratios with a denominator to 1. (Simplify) A:B/A 1:B/A 1:n. A:B/B 1:B/A m:1 Conversion of A and B Comparing Ratios Change order to unit ratio e.g.. Which class has the highest BOYS:GIRLS ratio?. Class 1 Class 2 Class 3. 2:3 3:4 1:5 1. Compare your question to ratio given If same order, m:1.	e.g. Highest B:G If not, 1:n. e.g. Lowest G:B 2. Convert. 3. Order Use m:1 Class 1 Class 2 Class 3. 2/3:1= 0.6: 1 3/4:1=0.75:1 1/5:1=0.2:1 Class 2 Class 1 Class 3 4. Choose your answer Highest B:G Ratio Multiply Method 1: Find the Unknown. e.g. Solve for Unknown. 3:4=a:8. =6:8. a=6. Method 2: Fractions. e.g.. 3:4=c:10. 10x3/4=c/28.5x10. 30/4=c. 7.5=c Find A:B:C from A:B/B:C Cross Product. e.g.. A:B=4:3. B:C=7:5 4 : 3. 7 : 5. 28:21:15 Multiply as follow Line up common value Rearrange and simplify as needed Modeling Using Ratios e.g. 250 ml lemon juice, how.
http://physicsinventions.com/finding-the-transferred-energy-for-a-block-sliding-down-a-ramp/	1,591,312,074,000,000,000	text/html	crawl-data/CC-MAIN-2020-24/segments/1590348492295.88/warc/CC-MAIN-20200604223445-20200605013445-00578.warc.gz	96,792,432	9,774	# Finding the Transferred Energy for a Block Sliding Down a Ramp 1. The problem statement, all variables and given/known data A 5.0kg block slides down a ramp, starting with a velocity down the slope of 2.5 m/s. The ramp is 1.5 m high and has an angle of 25 degrees. The force of friction acting upon the block is 20.0 N. How much energy is transferred in or out by gravity, normal force, and friction respectively as it slides down the ramp? 2. Relevant equations Ek = 1/2 mv2 Ep = mgh Fg = 9.8m Fn = Fgx Ff =μFn 3. The attempt at a solution This question is actually a five part question. I have completed parts 1 and 5, which is here in case it may be useful: 1: How much energy does the block begin with? A: Eki= 0.5(5.0kg)(2.52) = 15.6 J 5. What is the final velocity of the block? Ef = Ei + WFf (0.5)(5.0kg)vf2 = (5.0kg)(9.8m/s/s)(1.5m) + (0.5)(5.0kg)(2.52m/s) – (20N)(1.5/cos(25)m) vf = 2.69 m/s I’m not sure what concept I’m supposed to apply to discover "energy transferred", however. I thought that the energy within a closed system remained constant. I do not want the answers; I simply want some sort of push in the right direction. http://ift.tt/1pAx91X	377	1,174	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.671875	4	CC-MAIN-2020-24	latest	en	0.92753	# Finding the Transferred Energy for a Block Sliding Down a Ramp. 1. The problem statement, all variables and given/known data. A 5.0kg block slides down a ramp, starting with a velocity down the slope of 2.5 m/s. The ramp is 1.5 m high and has an angle of 25 degrees. The force of friction acting upon the block is 20.0 N.. How much energy is transferred in or out by gravity, normal force, and friction respectively as it slides down the ramp?. 2. Relevant equations. Ek = 1/2 mv2. Ep = mgh. Fg = 9.8m. Fn = Fgx. Ff =μFn. 3.	The attempt at a solution. This question is actually a five part question. I have completed parts 1 and 5, which is here in case it may be useful:. 1: How much energy does the block begin with?. A: Eki= 0.5(5.0kg)(2.52) = 15.6 J. 5. What is the final velocity of the block?. Ef = Ei + WFf. (0.5)(5.0kg)vf2 = (5.0kg)(9.8m/s/s)(1.5m) + (0.5)(5.0kg)(2.52m/s) – (20N)(1.5/cos(25)m). vf = 2.69 m/s. I’m not sure what concept I’m supposed to apply to discover "energy transferred", however. I thought that the energy within a closed system remained constant. I do not want the answers; I simply want some sort of push in the right direction.. http://ift.tt/1pAx91X.
https://www.theguardian.com/science/2020/apr/06/can-you-solve-it-double-chocolate?fbclid=IwAR3iIJQjjZ4T-wOakwRefPnziC_38fSIM_By_hNZQsN8MVl0u5tOKpJm9IE	1,597,352,092,000,000,000	text/html	crawl-data/CC-MAIN-2020-34/segments/1596439739073.12/warc/CC-MAIN-20200813191256-20200813221256-00592.warc.gz	738,074,840	45,000	# Can you solve it? Double chocolate! A delicious new puzzle from Japan Published on Mon 6 Apr 2020 02.10 EDT Double Choco is a new grid logic puzzle from Japan. Below are three examples, including a toughie which appeared in the 2020 UK puzzle championship last month. Pencil-and-paper puzzles like Double Choco are very absorbing; hopefully they provide a stimulating and escapist activity during these days of quarantine. The rules of Double Choco are as follows. 1.Divide the grid into blocks by drawing solid lines over the grid lines. 2.Each block must contain a pair of areas of white cells and gray cells having the same form (size and shape). One area may be a rotated or mirrored image of the other. 3.A number indicates the number of cells of that colour in the block (half the total number for the block). A block can contain any number of cells with numbers. The puzzles now follow. If you want to have them all on a single printable page, click here. Otherwise, you can solve them on screen with an app like Markup. Now for the tutorial puzzle (printed at the very top, and just below). My way in was to look at the 3 in the top left cell. This 3 tells you that there must be three white cells, and three grey cells, in that block. The only possible white cells are the three top ones in the leftmost column, since the white cells must form a contiguous area. The three grey cells must have the same shape as the white cells, in other words they must be in a line. There is only one possible position for a line of three grey cells that touches the white cells (and thus makes a single block.) I’ve marked it below. Likewise, the 3 in the top right corner means that the grey area must consist of three adjacent grey cells. There is only one possible position for those grey cells, and also only one possible position for three white cells that share the same shape (three in a row). I’ve marked that block too. The white area at the bottom left has two 3s in it. At first you might think they refer to two different blocks. They actually refer to the same block. (It is fine for the same block to have more than one cell with a number in it.) It has to be the same block, since the white area of this block has three cells in it, and there are only three possible cells, the ones I have shaded orange. The area of the block covered by grey cells must have the same shape, and thus must either be cells ABD, or cells BCD. Let’s assume it is BCD. Then where is the block that contains the remaining 3 on the white cell in the second bottom row? It cannot exist, since there are no three contiguous grey squares left on the board for that block. Hence The block with the three white squares on the bottom left contains ABD. The rest falls into place. Note that one of the blocks has no number-cells. Double Choco was invented by Japan’s legendary Nikoli magazine, which over the last 40 years has introduced many puzzles to a wide audience, most famously Sudoku and Kakuro. Last month, Double Choco was one of the featured problems at the UK Puzzle Association’s annual open tournament, which took place at a Croydon hotel, and consists of a sudoku competition and a general puzzle competition. The competitions are a bit like an exam: the contestants are presented with a series of papers, with each puzzle graded by difficulty. The hard Double Choco above was one of the hardest puzzles this year. This year’s results for the puzzle event were: 1st Neil Zussman, 2nd Vincent Bertrand, 3rd Tom Collyer. (Neil is in a league of his own in the UK and ranks amongst the best puzzle solvers in the world.) For the Sudoku event: 1st Vincent Bertrand, 2nd Tom Collyer, 3rd Mark Goodliffe. I’ll be back at 5pm with the solutions. If you like this type of pencil-and-paper puzzle, in which you need to fill in a grid following a short list of simple rules, here are some other ones I have featured in this column. They all link to printable pages. Print them all out and no chance you are leaving the house today. Snake Place (Oct 2017) Garam (Jan 2018) Skyscrapers (July 2018) Sandwich Sudoku (May 2019) Have fun, and keep safe. I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me. If you are reading this in the Guardian app, and you want a notification each time I post a puzzle, or its solution, click the ‘Follow Alex Bellos’ button above. Thanks to Tom Collyer, Liane Robinson and Rob Vollmert for today’s puzzles. To find out more about the UK Puzzle Association, its website is www.ukpuzzles.org.	1,059	4,621	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.703125	4	CC-MAIN-2020-34	latest	en	0.952402	# Can you solve it? Double chocolate!. A delicious new puzzle from Japan. Published on Mon 6 Apr 2020 02.10 EDT. Double Choco is a new grid logic puzzle from Japan. Below are three examples, including a toughie which appeared in the 2020 UK puzzle championship last month. Pencil-and-paper puzzles like Double Choco are very absorbing; hopefully they provide a stimulating and escapist activity during these days of quarantine.. The rules of Double Choco are as follows.. 1.Divide the grid into blocks by drawing solid lines over the grid lines.. 2.Each block must contain a pair of areas of white cells and gray cells having the same form (size and shape). One area may be a rotated or mirrored image of the other.. 3.A number indicates the number of cells of that colour in the block (half the total number for the block). A block can contain any number of cells with numbers.. The puzzles now follow. If you want to have them all on a single printable page, click here. Otherwise, you can solve them on screen with an app like Markup.. Now for the tutorial puzzle (printed at the very top, and just below). My way in was to look at the 3 in the top left cell. This 3 tells you that there must be three white cells, and three grey cells, in that block. The only possible white cells are the three top ones in the leftmost column, since the white cells must form a contiguous area. The three grey cells must have the same shape as the white cells, in other words they must be in a line. There is only one possible position for a line of three grey cells that touches the white cells (and thus makes a single block.) I’ve marked it below.. Likewise, the 3 in the top right corner means that the grey area must consist of three adjacent grey cells. There is only one possible position for those grey cells, and also only one possible position for three white cells that share the same shape (three in a row). I’ve marked that block too.. The white area at the bottom left has two 3s in it. At first you might think they refer to two different blocks. They actually refer to the same block. (It is fine for the same block to have more than one cell with a number in it.) It has to be the same block, since the white area of this block has three cells in it, and there are only three possible cells, the ones I have shaded orange. The area of the block covered by grey cells must have the same shape, and thus must either be cells ABD, or cells BCD.. Let’s assume it is BCD.	Then where is the block that contains the remaining 3 on the white cell in the second bottom row? It cannot exist, since there are no three contiguous grey squares left on the board for that block. Hence The block with the three white squares on the bottom left contains ABD. The rest falls into place. Note that one of the blocks has no number-cells.. Double Choco was invented by Japan’s legendary Nikoli magazine, which over the last 40 years has introduced many puzzles to a wide audience, most famously Sudoku and Kakuro.. Last month, Double Choco was one of the featured problems at the UK Puzzle Association’s annual open tournament, which took place at a Croydon hotel, and consists of a sudoku competition and a general puzzle competition.. The competitions are a bit like an exam: the contestants are presented with a series of papers, with each puzzle graded by difficulty. The hard Double Choco above was one of the hardest puzzles this year.. This year’s results for the puzzle event were: 1st Neil Zussman, 2nd Vincent Bertrand, 3rd Tom Collyer. (Neil is in a league of his own in the UK and ranks amongst the best puzzle solvers in the world.) For the Sudoku event: 1st Vincent Bertrand, 2nd Tom Collyer, 3rd Mark Goodliffe.. I’ll be back at 5pm with the solutions.. If you like this type of pencil-and-paper puzzle, in which you need to fill in a grid following a short list of simple rules, here are some other ones I have featured in this column. They all link to printable pages. Print them all out and no chance you are leaving the house today.. Snake Place (Oct 2017). Garam (Jan 2018). Skyscrapers (July 2018). Sandwich Sudoku (May 2019). Have fun, and keep safe.. I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.. If you are reading this in the Guardian app, and you want a notification each time I post a puzzle, or its solution, click the ‘Follow Alex Bellos’ button above.. Thanks to Tom Collyer, Liane Robinson and Rob Vollmert for today’s puzzles. To find out more about the UK Puzzle Association, its website is www.ukpuzzles.org.
http://www.math4ged.com/fractions/	1,498,614,490,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128322275.28/warc/CC-MAIN-20170628014207-20170628034207-00648.warc.gz	584,592,994	13,840	## Fractions Several questions on the GED math test will require you to understand and know how to work with fractions. Fractions are normally used to express parts of a whole number. The names for the two parts of a fraction are the numerator, for the top part, and the denominator, for the bottom part. So in the fraction ½ the numerator is 1 and the denominator is 2. To give a real life example, imagine a pie cut into 8 slices. The pie is still one whole, however it is divided into 8 parts. 8/8 = 1. If three of the slices are eaten, the pie is no longer one whole and you can represent the remaining amount as 5/8. There are three types of fractions: Proper fractions: These fractions have numerators that are smaller than denominators. 5/8 in the example above is a proper fraction since 5 is smaller than 8. 1/2, 12/13, 49/50 and 999999/1000000 are all examples of proper fractions. Since the numerator is smaller than the denominator, proper fractions are all smaller than 1. Improper fractions: These fractions have numerators that are larger than or equal to the denominator. 7/3, 9/6 and 19/19 are all examples of improper fractions. Since the numerator is equal to or greater than the denominator, improper fractions are all equal to or larger than 1. Mixed numbers: These are whole numbers and fractions together. For example, if you had a whole pie in addition to the partially eaten one above, the amount of pies you have is 1 5/8. Share and Enjoy: 1. elizabeth says: that ged math test is a word test,which makes the test even harder,my god what will the 2012 test look like. 2. Nina says: love this website its so helpful thank you! 3. Eula King-Harden says: I love this site it has been so helpful. I really need to practice for the test. Could you please send the Math practice test. I have taken the math test 4 times, never scoring more than 370.If I fail again I will have to take all components again and it will be more difficult 2014. Math is all I need to pass right now. Thanks so much. 4. Daniel says: All I need is the Math as well, though Math is my greatest Nemesis and I have the hardest time, not to mention I’m a hands on kind of person. I really dont know how to realy even go abut studying for this thing. 5. Catresha says: Hi 🙂 ‘ I have a question about the fractions?? Need to know what key or keys would I use on the Casio fx- 260 ,to solve a fraction problem??? Need assistance asap please. Thank You • Chasity says: You need to look for the a b/c button. It should be the first button on the second row from the top. 6. Catresha says: Or do you even use the calculator for the fractions?? Need to refresh my memory on this one,because I forgot? • Ziyi says: You don’t have to use a calculator, but if it’s available you might want to use it just to be sure! The fraction key is just below the shift key. Far left, second row. 7. Valerie says: God knows that I am tired. I have taken this math test 4 times already today I am goeto take it again how do I pass this test.thank you for this website I just found it.help me please. 8. Catresha says: Hi’ just took the math test a few hours ago,but only got a 370,my high score is 390 . In my state I only need 410 . It seems like because I work on it I cannot get to the 410. I think it would be best if I did,nt work at it ; different approach . I really dont know what to do,the test is extensive,& I,m the type that cant just mark anything!!!!!! Not gonna stop, going back Monday to reregister!	868	3,519	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2017-26	longest	en	0.934541	## Fractions. Several questions on the GED math test will require you to understand and know how to work with fractions. Fractions are normally used to express parts of a whole number. The names for the two parts of a fraction are the numerator, for the top part, and the denominator, for the bottom part. So in the fraction ½ the numerator is 1 and the denominator is 2. To give a real life example, imagine a pie cut into 8 slices. The pie is still one whole, however it is divided into 8 parts. 8/8 = 1. If three of the slices are eaten, the pie is no longer one whole and you can represent the remaining amount as 5/8.. There are three types of fractions:. Proper fractions: These fractions have numerators that are smaller than denominators. 5/8 in the example above is a proper fraction since 5 is smaller than 8.. 1/2, 12/13, 49/50 and 999999/1000000 are all examples of proper fractions.. Since the numerator is smaller than the denominator, proper fractions are all smaller than 1.. Improper fractions: These fractions have numerators that are larger than or equal to the denominator.. 7/3, 9/6 and 19/19 are all examples of improper fractions.. Since the numerator is equal to or greater than the denominator, improper fractions are all equal to or larger than 1.. Mixed numbers: These are whole numbers and fractions together. For example, if you had a whole pie in addition to the partially eaten one above, the amount of pies you have is 1 5/8.. Share and Enjoy:. 1. elizabeth says:. that ged math test is a word test,which makes the test even harder,my god what will the 2012 test look like.. 2. Nina says:. love this website its so helpful thank you!. 3. Eula King-Harden says:. I love this site it has been so helpful. I really need to practice for the test.	Could you please send the Math practice test. I have taken the math test 4 times, never scoring more than 370.If I fail again I will have to take all components again and it will be more difficult 2014. Math is all I need to pass right now. Thanks so much.. 4. Daniel says:. All I need is the Math as well, though Math is my greatest Nemesis and I have the hardest time, not to mention I’m a hands on kind of person. I really dont know how to realy even go abut studying for this thing.. 5. Catresha says:. Hi 🙂 ‘ I have a question about the fractions?? Need to know what key or keys would I use on the Casio fx- 260 ,to solve a fraction problem??? Need assistance asap please. Thank You. • Chasity says:. You need to look for the a b/c button. It should be the first button on the second row from the top.. 6. Catresha says:. Or do you even use the calculator for the fractions?? Need to refresh my memory on this one,because I forgot?. • Ziyi says:. You don’t have to use a calculator, but if it’s available you might want to use it just to be sure! The fraction key is just below the shift key. Far left, second row.. 7. Valerie says:. God knows that I am tired. I have taken this math test 4 times already today I am goeto take it again how do I pass this test.thank you for this website I just found it.help me please.. 8. Catresha says:. Hi’ just took the math test a few hours ago,but only got a 370,my high score is 390 . In my state I only need 410 . It seems like because I work on it I cannot get to the 410. I think it would be best if I did,nt work at it ; different approach . I really dont know what to do,the test is extensive,& I,m the type that cant just mark anything!!!!!! Not gonna stop, going back Monday to reregister!.
https://www.jiskha.com/display.cgi?id=1267676083	1,516,118,271,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084886437.0/warc/CC-MAIN-20180116144951-20180116164951-00079.warc.gz	971,868,547	3,634	# math posted by . find a polynomial with integer coefficients such that square root of 3 + square root of 5 is a root of the polynomial ## Similar Questions 1. ### Math Just need some help... Directions: Multiply or raise to the power as indicated... Then simplify the result. Assume all variables represent positive numbers. 1. 5(square root of 6) times two-thirds(square root of 15)= 3 one-third (square … 2. ### math find a polynomial with integer coefficients such that square root of 3 + square root of 5 is a root of the polynomial 3. ### Algebra Rationalize each expression by building perfect nth root factors for each denominator. Assume all variables represent positive quantities. I don’t understand how to compute these. Also I don’t have the square root sign so I typed … 4. ### Math simplifying mixed radicals Please help me simplify these following mixed radicals:: and show me how you did it cause I need to learn : I don't know how to type a square root sign so I just wrote square root: 2 square root 48 3 square root 81 6 square root 12 … 5. ### Math simplifying mixed radicals Please help me simplify these following mixed radicals:: and show me how you did it cause I need to learn : I don't know how to type a square root sign so I just wrote square root and the numbers infront of the square root are supposed … 6. ### Math ~CHECK MY ANSWER~ 1) Which of these is a rational number? a. Pi b. Square root 3 **** c. Square root 2 d. 1.3 (the # 3 has a line at the top) 2) Which of the following sets contains 3 irrational numbers? 7. ### MATH Which expression is a polynomial???? A: 11 0ver the square root of x B: square root of 11 times x C: x - square root 11 D: 11 over the square root of x to the 2nd power (sorry I am not sure how to insert square root symbols or fractions, 8. ### Math Help! Four students worked to find an estimate for square root 22. Who is closest to finding the true estimate? 9. ### math 1) Which of these is a rational number? a. Pi b. Square root 3 **** c. Square root 2 d. 1.3 (the # 3 has a line at the top) 2) Which of the following sets contains 3 irrational numbers? 10. ### Math Four students work to find an estimate for square root 27. Who is closest to finding the true estimate? More Similar Questions	568	2,282	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.703125	4	CC-MAIN-2018-05	latest	en	0.888482	# math. posted by .. find a polynomial with integer coefficients such that square root of 3 + square root of 5 is a root of the polynomial. ## Similar Questions. 1. ### Math. Just need some help... Directions: Multiply or raise to the power as indicated... Then simplify the result. Assume all variables represent positive numbers. 1. 5(square root of 6) times two-thirds(square root of 15)= 3 one-third (square …. 2. ### math. find a polynomial with integer coefficients such that square root of 3 + square root of 5 is a root of the polynomial. 3. ### Algebra. Rationalize each expression by building perfect nth root factors for each denominator. Assume all variables represent positive quantities. I don’t understand how to compute these. Also I don’t have the square root sign so I typed …. 4. ### Math simplifying mixed radicals. Please help me simplify these following mixed radicals:: and show me how you did it cause I need to learn : I don't know how to type a square root sign so I just wrote square root: 2 square root 48 3 square root 81 6 square root 12 …. 5. ### Math simplifying mixed radicals. Please help me simplify these following mixed radicals:: and show me how you did it cause I need to learn : I don't know how to type a square root sign so I just wrote square root and the numbers infront of the square root are supposed …. 6.	### Math ~CHECK MY ANSWER~. 1) Which of these is a rational number? a. Pi b. Square root 3 **** c. Square root 2 d. 1.3 (the # 3 has a line at the top) 2) Which of the following sets contains 3 irrational numbers?. 7. ### MATH. Which expression is a polynomial???? A: 11 0ver the square root of x B: square root of 11 times x C: x - square root 11 D: 11 over the square root of x to the 2nd power (sorry I am not sure how to insert square root symbols or fractions,. 8. ### Math Help!. Four students worked to find an estimate for square root 22. Who is closest to finding the true estimate?. 9. ### math. 1) Which of these is a rational number? a. Pi b. Square root 3 **** c. Square root 2 d. 1.3 (the # 3 has a line at the top) 2) Which of the following sets contains 3 irrational numbers?. 10. ### Math. Four students work to find an estimate for square root 27. Who is closest to finding the true estimate?. More Similar Questions.
https://web2.0calc.com/questions/determine-the	1,656,695,208,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00574.warc.gz	654,983,126	6,000	+0 # Determine the 0 226 1 +1 Determine the quadratic function whose zeros are given. a) 3 and 5 b) –4 and 3 c) ½ and ⅔ d) 2 ± √3 e) (2±√5)/3 Nov 8, 2021 ### 1+0 Answers #1 +9405 +2 A quadratic function with zeros at a and b is: y = (x - a)(x - b) Then we can multiply out the right side of that equation, or "FOIL" it, to get: y = x2 - bx - ax + ab And then continue to simplify the right side until it looks like a nice quadratic equation. Using this "template", we can find a solution to each of these problems. a) y = (x - 3)(x - 5) = x2 - 5x - 3x + 15 = x2 - 8x + 15 b) y = (x + 4)(x - 3) = x2 - 3x + 4x - 12 = x2 + x - 12 c) y = $$(x-\frac12)(x-\frac23)\ =\ x^2-\frac23x-\frac12x+\frac13\ =\ x^2-\frac76x+\frac13$$ d) In this case, the one root is 2 + √3 and the other is 2 - √3 , so the quadratic equation is: y = ( x - (2 + √3) )( x - (2 - √3) ) y = ( x - 2 - √3 )( x - 2 + √3 ) y = x2 - 4x + 1 Can you figure out the last one? It might be the trickiest one, so if you need more help on it just ask! Here is a graph to check the answers: https://www.desmos.com/calculator/lv66bdc1jy Nov 8, 2021	490	1,189	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2022-27	latest	en	0.518933	+0. # Determine the. 0. 226. 1. +1. Determine the. quadratic function. whose zeros are. given.. a) 3 and 5. b) –4 and 3. c) ½ and ⅔. d) 2 ± √3. e) (2±√5)/3. Nov 8, 2021. ### 1+0 Answers. #1. +9405.	+2. A quadratic function with zeros at a and b is:. y = (x - a)(x - b). Then we can multiply out the right side of that equation, or "FOIL" it, to get:. y = x2 - bx - ax + ab. And then continue to simplify the right side until it looks like a nice quadratic equation.. Using this "template", we can find a solution to each of these problems.. a) y = (x - 3)(x - 5) = x2 - 5x - 3x + 15 = x2 - 8x + 15. b) y = (x + 4)(x - 3) = x2 - 3x + 4x - 12 = x2 + x - 12. c) y = $$(x-\frac12)(x-\frac23)\ =\ x^2-\frac23x-\frac12x+\frac13\ =\ x^2-\frac76x+\frac13$$. d) In this case, the one root is 2 + √3 and the other is 2 - √3 , so the quadratic equation is:. y = ( x - (2 + √3) )( x - (2 - √3) ). y = ( x - 2 - √3 )( x - 2 + √3 ). y = x2 - 4x + 1. Can you figure out the last one? It might be the trickiest one, so if you need more help on it just ask!. Here is a graph to check the answers: https://www.desmos.com/calculator/lv66bdc1jy. Nov 8, 2021.
https://mobile.surenapps.com/2020/09/ratio-and-proportion-solutions-ex-121.html	1,611,270,086,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703528672.38/warc/CC-MAIN-20210121225305-20210122015305-00341.warc.gz	469,910,415	29,389	### Ratio and Proportion-Solutions Ex-12.1 CBSE Class –VI Mathematics NCERT Solutions Chaper 12 Ratio And Proportion (Ex. 12.1) Question 1. There are 20 girls and 15 boys in a class. (a)What is the ratio of the number of girls to the number of boys? (b)What is the ratio of girls to the total number of students in the class? Answer: (a) The ratio of girls to that of boys = $\frac{20}{15}=\frac{4}{3}$ = 4 : 3 (b) The ratio of girls to total students = $\frac{20}{20+15}=\frac{20}{35}=\frac{4}{7}$ = 4 : 7 Question 2. Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of: (a)The number of students liking football to the number of students liking tennis. (b)The number of students liking cricket to the total number of students. Answer: Total number of students = 30 Number of students like football = 6 Number of students like cricket = 12 Thus number of students like tennis = 30 – 6 – 12 = 12 (a) The ratio of students like football that of tennis = $\frac{6}{12}=\frac{1}{2}$ = 1 : 2 (b) The ratio of students like cricket to that of total students = $\frac{12}{30}=\frac{2}{5}$ = 2 : 5 Question 3. See the figure and find the ratio of: (a)The number of triangles to the number of circles inside the rectangle. (b)The number of squares to all the figures inside the rectangle. (c)The number of circles to all the figures inside the rectangle. Answer: (a) Ratio of number of triangle to that of circles = $\frac{3}{2}$ = 3 : 2 (b) Ratio of number of squares to all figures = $\frac{2}{7}$ = 2 : 7 (c) Ratio of number of circles to all figures = $\frac{2}{7}$ = 2 : 7 Question 4. Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of the speed of Hamid to the speed of Akhtar. Answer: We know that, Speed = $\frac{\text{Distance}}{\text{Time}}$ Speed of Hamid = $\frac{\text{9 m}}{\text{1 h}}$ = 9 km/h and Speed of Akhtar = $\frac{\text{12 m}}{\text{1 h}}$ = 12 km/h Ratio of speed of Hamid to that of speed of Akhtar = $\frac{9}{12}=\frac{3}{4}$ = 3 : 4 Question 5. Fill in the following blanks: [Are these equivalent ratios?] Answer: $\frac{15}{18}=\frac{\overline{)5}}{6}=\frac{10}{\overline{)12}}=\frac{\overline{)25}}{30}$ Yes, these are equivalent ratios. Question 6. Find the ratio of the following: (a) 81 to 108 (b) 98 to 63 (c) 33 km to 121 km (d) 30 minutes to 45 minutes Answer: (a) Ratio of 81 to 108 = = 3 : 4 (b) Ratio of 98 to 63 = = 14 : 9 (c) Ratio of 33 km to 121 km = = 3 : 11 (d) Ratio of 30 minutes to 45 minutes = = 2 : 3 Question 7. Find the ratio of the following: (a) 30 minutes to 1 hour (b) 40 cm to 1.5 m (c) 55 paise to Re. 1 (d) 500 ml to 2 liters Answer: (a) 30 minutes to 1hour 1 hour = 1 x 60 = 60 minutes [$\because$ 1 hour = 60 minutes] Now, ratio of 30 minutes to 1hour = 30 minutes : 60 minutes $⇒$ 30 minutes :60 minutes = = 1 : 2 (b) 40 cm to 1.5 m 1.5 m = 1.5 x 100 cm = 150 cm [$\because$ 1 m = 100 cm] Now, ratio of 40 cm to 1.5 m = 40 cm : 1.5 m $⇒$ 40 cm : 150 cm = = 4 : 15 (c) 55 paise to Re. 1 Re. 1 = 100 paise Now, ratio of 55 paise to Re. 1 = 55 paise : 100 paise $⇒$ = 11 : 20 (d) 500 ml to 2 liters 2 liters = 2 x 1000 ml = 2000 ml [$\because$ 1 litre = 1000 ml] Now, ratio of 500 ml to 2 liters = 500 ml : 2 liters $⇒$ 500 ml : 2000 ml = = 1 : 4 Question 8. In a year, Seema earns Rs. 1,50,000 and saves Rs. 50,000. Find the ratio of: (a) Money that Seema earns to the money she saves. (b) Money that she saves to the money she spends. Answer: Total earning = Rs. 1,50,000 and Saving = Rs. 50,000 $\therefore$ Money spent = Rs. 1,50,000 - Rs. 50,000 = Rs. 1,00,000 (a) Ratio of money earned to money saved = = 3 : 1 (b) Ratio of money saved to money spend = = 1 : 2 Question 9. There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students. Answer: Ratio of number of teachers to that of students = = 17 : 550 Question 10. In a college out of 4320 students, 2300 are girls. Find the ratio of: (a) the number of girls to the total number of students. (b) The number of boys to the number of girls. (c) The number of boys to the total number of students. Total number of students in school = 4320 Number of girls = 2300 Therefore, number of boys = 4320 – 2300 = 2020 (a) Ratio of girls to total number of students = = 115 : 216 (b) Ratio of boys to that of girls = = 101 : 115 (c) Ratio of boys to total number of students = = 101 : 216 Question 11. Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of: (a) The number of students who opted basketball to the number of students who opted table tennis. (b) The number of students who opted cricket to the number of students opting basketball. (c) The number of students who opted basketball to the total number of students. Answer: Total number of students = 1800 Number of students opted basketball = 750 Number of students opted cricket = 800 Therefore, number of students opted tennis = 1800 – (750 + 800) = 250 (a) Ratio of students opted basketball to that of opted table tennis = = 3 : 1 (b) Ratio of students opted cricket to students opted basketball = = 16 : 15 (c) Ratio of students opted basketball to total no. of students = = 5 : 12 Question 12. The cost of a dozen pens is Rs. 180 and cost of 8 ball pens is Rs. 56. Find the ratio of the cost of a pen to the cost of a ball pen. Answer: Cost of a dozen pens (12 pens) = Rs. 180 $\therefore$ Cost of 1 pen = $\frac{180}{12}$ = Rs. 15 Cost of 8 ball pens = Rs. 56 $\therefore$ Cost of 1 ball pen = $\frac{56}{8}$ = Rs. 7 Ratio of cost of one pen to that of one ball pen = $\frac{15}{7}$ = 15 : 7 Question 13. Consider the statement: Ratio of breadth and length of a ball is 2 : 5. Complete the following table that shows some possible breadths and lengths of the hall. Breadth of the hall (in meters) 10 $\overline{)}$ 40 Length of the hall (in meters) 25 50 $\overline{)}$ Answer: Ratio of breadth to length = 2 : 5 = $\frac{2}{5}$ $\therefore$ Other equivalent ratios are = $\frac{2}{5}×\frac{10}{10}=\frac{20}{50}$$\frac{2}{5}×\frac{20}{20}=\frac{40}{100}$ Thus, Breadth of the hall (in meters) 10 20 40 Length of the hall (in meters) 25 50 100 Question 14. Divide 20 pens between Sheela and Sangeeta in the ratio 3 : 2. Answer: Ratio between Sheela and Sangeeta = 3 : 2 Total these terms = 3 + 2 = 5 Therefore, part of Sheela = $\frac{3}{5}$ of the total pens And part of Sangeeta = $\frac{2}{5}$ of total pens Thus, Sheela gets = = 12 pens And Sangeeta gets = = 8 pens Question 15. Mother wants to divide Rs. 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get. Answer: Ratio of the age of Shreya to that of Bhoomika = = 5 : 4 Thus, Rs. 36 divide between Shreya and Bhoomika in the ratio of 5 : 4. Shreya gets = $\frac{5}{9}$ of Rs. 36 = = Rs. 20 Bhoomika gets = $\frac{4}{9}$ of Rs. 36 = = Rs. 16 Question 16. The present age of the father is 42 years and that of his son is 14 years. Find the ratio of: (a) The Present age of the father to the present age of the son. (b) The age of the father to the age of the son, when the son was 12 years old. (c) The age of father after 10 years to the age of son after 10 years. (d) The age of father to the age of son when the father was 30 years old. Answer: (a) Ratio of father’s present age to that of son = = 3 : 1 (b) When the son was 12 years, i.e., 2 years ago, then the father was (42 – 2) = 40 years Therefore, the ratio of their ages = = 10 : 3 (c) Age of the father after 10 years = 42 + 10 = 52 years Age of the son after 10 years = 14 + 10 = 24 years Therefore, ratio of their ages = = 13 : 6 (d) When the father was 30 years old, i.e., 12 years ago, then the son was (14 – 12) = 2 years old Therefore, the ratio of their ages = = 15 : 1	2,690	7,958	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 35, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.8125	5	CC-MAIN-2021-04	latest	en	0.840933	### Ratio and Proportion-Solutions Ex-12.1. CBSE Class –VI Mathematics. NCERT Solutions. Chaper 12 Ratio And Proportion (Ex. 12.1). Question 1. There are 20 girls and 15 boys in a class.. (a)What is the ratio of the number of girls to the number of boys?. (b)What is the ratio of girls to the total number of students in the class?. Answer: (a) The ratio of girls to that of boys = $\frac{20}{15}=\frac{4}{3}$ = 4 : 3. (b) The ratio of girls to total students = $\frac{20}{20+15}=\frac{20}{35}=\frac{4}{7}$ = 4 : 7. Question 2. Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of:. (a)The number of students liking football to the number of students liking tennis.. (b)The number of students liking cricket to the total number of students.. Answer: Total number of students = 30. Number of students like football = 6. Number of students like cricket = 12. Thus number of students like tennis = 30 – 6 – 12 = 12. (a) The ratio of students like football that of tennis = $\frac{6}{12}=\frac{1}{2}$ = 1 : 2. (b) The ratio of students like cricket to that of total students = $\frac{12}{30}=\frac{2}{5}$ = 2 : 5. Question 3. See the figure and find the ratio of:. (a)The number of triangles to the number of circles inside the rectangle.. (b)The number of squares to all the figures inside the rectangle.. (c)The number of circles to all the figures inside the rectangle.. Answer: (a) Ratio of number of triangle to that of circles = $\frac{3}{2}$ = 3 : 2. (b) Ratio of number of squares to all figures = $\frac{2}{7}$ = 2 : 7. (c) Ratio of number of circles to all figures = $\frac{2}{7}$ = 2 : 7. Question 4. Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of the speed of Hamid to the speed of Akhtar.. Answer: We know that, Speed = $\frac{\text{Distance}}{\text{Time}}$. Speed of Hamid = $\frac{\text{9 m}}{\text{1 h}}$ = 9 km/h and Speed of Akhtar = $\frac{\text{12 m}}{\text{1 h}}$ = 12 km/h. Ratio of speed of Hamid to that of speed of Akhtar = $\frac{9}{12}=\frac{3}{4}$ = 3 : 4. Question 5. Fill in the following blanks:. [Are these equivalent ratios?]. Answer: $\frac{15}{18}=\frac{\overline{)5}}{6}=\frac{10}{\overline{)12}}=\frac{\overline{)25}}{30}$. Yes, these are equivalent ratios.. Question 6. Find the ratio of the following:. (a) 81 to 108. (b) 98 to 63. (c) 33 km to 121 km. (d) 30 minutes to 45 minutes. Answer: (a) Ratio of 81 to 108 = = 3 : 4. (b) Ratio of 98 to 63 = = 14 : 9. (c) Ratio of 33 km to 121 km = = 3 : 11. (d) Ratio of 30 minutes to 45 minutes = = 2 : 3. Question 7. Find the ratio of the following:. (a) 30 minutes to 1 hour. (b) 40 cm to 1.5 m. (c) 55 paise to Re. 1. (d) 500 ml to 2 liters. Answer: (a) 30 minutes to 1hour. 1 hour = 1 x 60 = 60 minutes [$\because$ 1 hour = 60 minutes]. Now, ratio of 30 minutes to 1hour = 30 minutes : 60 minutes. $⇒$ 30 minutes :60 minutes = = 1 : 2. (b) 40 cm to 1.5 m. 1.5 m = 1.5 x 100 cm = 150 cm [$\because$ 1 m = 100 cm]. Now, ratio of 40 cm to 1.5 m = 40 cm : 1.5 m. $⇒$ 40 cm : 150 cm = = 4 : 15. (c) 55 paise to Re. 1. Re. 1 = 100 paise. Now, ratio of 55 paise to Re. 1 = 55 paise : 100 paise. $⇒$ = 11 : 20. (d) 500 ml to 2 liters. 2 liters = 2 x 1000 ml = 2000 ml [$\because$ 1 litre = 1000 ml]. Now, ratio of 500 ml to 2 liters = 500 ml : 2 liters. $⇒$ 500 ml : 2000 ml = = 1 : 4. Question 8. In a year, Seema earns Rs. 1,50,000 and saves Rs. 50,000. Find the ratio of:. (a) Money that Seema earns to the money she saves.. (b) Money that she saves to the money she spends.. Answer: Total earning = Rs. 1,50,000 and Saving = Rs. 50,000. $\therefore$ Money spent = Rs. 1,50,000 - Rs. 50,000 = Rs. 1,00,000.	(a) Ratio of money earned to money saved = = 3 : 1. (b) Ratio of money saved to money spend = = 1 : 2. Question 9. There are 102 teachers in a school of 3300 students. Find the ratio of the number of teachers to the number of students.. Answer: Ratio of number of teachers to that of students = = 17 : 550. Question 10. In a college out of 4320 students, 2300 are girls. Find the ratio of:. (a) the number of girls to the total number of students.. (b) The number of boys to the number of girls.. (c) The number of boys to the total number of students.. Total number of students in school = 4320. Number of girls = 2300. Therefore, number of boys = 4320 – 2300 = 2020. (a) Ratio of girls to total number of students = = 115 : 216. (b) Ratio of boys to that of girls = = 101 : 115. (c) Ratio of boys to total number of students = = 101 : 216. Question 11. Out of 1800 students in a school, 750 opted basketball, 800 opted cricket and remaining opted table tennis. If a student can opt only one game, find the ratio of:. (a) The number of students who opted basketball to the number of students who opted table tennis.. (b) The number of students who opted cricket to the number of students opting basketball.. (c) The number of students who opted basketball to the total number of students.. Answer: Total number of students = 1800. Number of students opted basketball = 750. Number of students opted cricket = 800. Therefore, number of students opted tennis = 1800 – (750 + 800) = 250. (a) Ratio of students opted basketball to that of opted table tennis = = 3 : 1. (b) Ratio of students opted cricket to students opted basketball = = 16 : 15. (c) Ratio of students opted basketball to total no. of students = = 5 : 12. Question 12. The cost of a dozen pens is Rs. 180 and cost of 8 ball pens is Rs. 56. Find the ratio of the cost of a pen to the cost of a ball pen.. Answer: Cost of a dozen pens (12 pens) = Rs. 180. $\therefore$ Cost of 1 pen = $\frac{180}{12}$ = Rs. 15. Cost of 8 ball pens = Rs. 56. $\therefore$ Cost of 1 ball pen = $\frac{56}{8}$ = Rs. 7. Ratio of cost of one pen to that of one ball pen = $\frac{15}{7}$ = 15 : 7. Question 13. Consider the statement: Ratio of breadth and length of a ball is 2 : 5. Complete the following table that shows some possible breadths and lengths of the hall.. Breadth of the hall (in meters) 10 $\overline{)}$ 40 Length of the hall (in meters) 25 50 $\overline{)}$. Answer: Ratio of breadth to length = 2 : 5 = $\frac{2}{5}$. $\therefore$ Other equivalent ratios are = $\frac{2}{5}×\frac{10}{10}=\frac{20}{50}$$\frac{2}{5}×\frac{20}{20}=\frac{40}{100}$. Thus,. Breadth of the hall (in meters) 10 20 40 Length of the hall (in meters) 25 50 100. Question 14. Divide 20 pens between Sheela and Sangeeta in the ratio 3 : 2.. Answer: Ratio between Sheela and Sangeeta = 3 : 2. Total these terms = 3 + 2 = 5. Therefore, part of Sheela = $\frac{3}{5}$ of the total pens. And part of Sangeeta = $\frac{2}{5}$ of total pens. Thus, Sheela gets = = 12 pens. And Sangeeta gets = = 8 pens. Question 15. Mother wants to divide Rs. 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.. Answer: Ratio of the age of Shreya to that of Bhoomika = = 5 : 4. Thus, Rs. 36 divide between Shreya and Bhoomika in the ratio of 5 : 4.. Shreya gets = $\frac{5}{9}$ of Rs. 36 = = Rs. 20. Bhoomika gets = $\frac{4}{9}$ of Rs. 36 = = Rs. 16. Question 16. The present age of the father is 42 years and that of his son is 14 years. Find the ratio of:. (a) The Present age of the father to the present age of the son.. (b) The age of the father to the age of the son, when the son was 12 years old.. (c) The age of father after 10 years to the age of son after 10 years.. (d) The age of father to the age of son when the father was 30 years old.. Answer: (a) Ratio of father’s present age to that of son = = 3 : 1. (b) When the son was 12 years, i.e., 2 years ago, then the father was (42 – 2) = 40 years. Therefore, the ratio of their ages = = 10 : 3. (c) Age of the father after 10 years = 42 + 10 = 52 years. Age of the son after 10 years = 14 + 10 = 24 years. Therefore, ratio of their ages = = 13 : 6. (d) When the father was 30 years old, i.e., 12 years ago, then the son was (14 – 12) = 2 years old. Therefore, the ratio of their ages = = 15 : 1.
http://www.thefullwiki.org/Hosohedron	1,566,378,343,000,000,000	text/html	crawl-data/CC-MAIN-2019-35/segments/1566027315865.44/warc/CC-MAIN-20190821085942-20190821111942-00270.warc.gz	323,295,027	9,854	# Hosohedron: Wikis Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles. # Encyclopedia Set of regular q-gonal hosohedrons Example hexagonal hosohedron on a sphere Type Regular polyhedron or spherical tiling Faces q digons Edges q Vertices 2 Schläfli symbol {2,q} Vertex configuration 2q Coxeter–Dynkin diagram Wythoff symbol q \| 2 2 Symmetry group Dihedral (Dqh) Dual polyhedron dihedron This beach ball shows a hosohedron with six lune faces, if the white circles on the ends are removed. In geometry, an n-gonal hosohedron is a tessellation of lunes on a spherical surface, such that each lune shares the same two vertices. A regular n-gonal hosohedron has Schläfli symbol {2, n}. ## Hosohedrons as regular polyhedrons For a regular polyhedron whose Schläfli symbol is {mn}, the number of polygonal faces may be found by: $N_2=\frac{4n}{2m+2n-mn}$ The platonic solids known to antiquity are the only integer solutions for m ≥ 3 and n ≥ 3. The restriction m ≥ 3 enforces that the polygonal faces must have at least three sides. When considering polyhedrons as a spherical tiling, this restriction may be relaxed, since digons can be represented as spherical lunes, having non-zero area. Allowing m = 2 admits a new infinite class of regular polyhedrons, which are the hosohedrons. On a spherical surface, the polyhedron {2, n} is represented as n abutting lunes, with interior angles of 2π/n. All these lunes share two common vertices. A regular trigonal hosohedron, represented as a tessellation of 3 spherical lunes on a sphere. A regular tetragonal hosohedron, represented as a tessellation of 4 spherical lunes on a sphere. ## Derivative polyhedrons The dual of the n-gonal hosohedron {2, n} is the n-gonal dihedron, {n, 2}. The polyhedron {2,2} is self-dual, and is both a hosohedron and a dihedron. A hosohedron may be modified in the same manner as the other polyhedrons to produce a truncated variation. The truncated n-gonal hosohedron is the n-gonal prism. ## Hosotopes Multidimensional analogues in general are called hosotopes, with Schläfli symbol {2,...,2,q}. A hosotope has two vertices. The two-dimensional hosotope {2} is a digon. ## Etymology The prefix “hoso-” was invented by H.S.M. Coxeter, and possibly derives from the English “hose”.	671	2,427	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2019-35	latest	en	0.900326	# Hosohedron: Wikis. Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.. # Encyclopedia. Set of regular q-gonal hosohedrons. Example hexagonal hosohedron on a sphere. Type Regular polyhedron. or spherical tiling. Faces q digons. Edges q. Vertices 2. Schläfli symbol {2,q}. Vertex configuration 2q. Coxeter–Dynkin diagram. Wythoff symbol q \| 2 2. Symmetry group Dihedral (Dqh). Dual polyhedron dihedron. This beach ball shows a hosohedron with six lune faces, if the white circles on the ends are removed.. In geometry, an n-gonal hosohedron is a tessellation of lunes on a spherical surface, such that each lune shares the same two vertices. A regular n-gonal hosohedron has Schläfli symbol {2, n}.. ## Hosohedrons as regular polyhedrons. For a regular polyhedron whose Schläfli symbol is {mn}, the number of polygonal faces may be found by:.	$N_2=\frac{4n}{2m+2n-mn}$. The platonic solids known to antiquity are the only integer solutions for m ≥ 3 and n ≥ 3. The restriction m ≥ 3 enforces that the polygonal faces must have at least three sides.. When considering polyhedrons as a spherical tiling, this restriction may be relaxed, since digons can be represented as spherical lunes, having non-zero area. Allowing m = 2 admits a new infinite class of regular polyhedrons, which are the hosohedrons. On a spherical surface, the polyhedron {2, n} is represented as n abutting lunes, with interior angles of 2π/n. All these lunes share two common vertices.. A regular trigonal hosohedron, represented as a tessellation of 3 spherical lunes on a sphere. A regular tetragonal hosohedron, represented as a tessellation of 4 spherical lunes on a sphere.. ## Derivative polyhedrons. The dual of the n-gonal hosohedron {2, n} is the n-gonal dihedron, {n, 2}. The polyhedron {2,2} is self-dual, and is both a hosohedron and a dihedron.. A hosohedron may be modified in the same manner as the other polyhedrons to produce a truncated variation. The truncated n-gonal hosohedron is the n-gonal prism.. ## Hosotopes. Multidimensional analogues in general are called hosotopes, with Schläfli symbol {2,...,2,q}. A hosotope has two vertices.. The two-dimensional hosotope {2} is a digon.. ## Etymology. The prefix “hoso-” was invented by H.S.M. Coxeter, and possibly derives from the English “hose”.
http://gmatclub.com/forum/dan-and-karen-who-live-10-miles-apart-meet-at-a-cafe-that-26780.html	1,485,034,948,000,000,000	text/html	crawl-data/CC-MAIN-2017-04/segments/1484560281226.52/warc/CC-MAIN-20170116095121-00568-ip-10-171-10-70.ec2.internal.warc.gz	123,571,696	40,578	Dan and Karen, who live 10 miles apart meet at a cafe that : Quant Question Archive [LOCKED] Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 21 Jan 2017, 13:42 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # Dan and Karen, who live 10 miles apart meet at a cafe that Author Message Manager Joined: 14 Dec 2005 Posts: 74 Followers: 1 Kudos [?]: 5 [0], given: 0 Dan and Karen, who live 10 miles apart meet at a cafe that [#permalink] ### Show Tags 20 Feb 2006, 16:30 This topic is locked. If you want to discuss this question please re-post it in the respective forum. Dan and Karen, who live 10 miles apart meet at a cafe that is directly north of Danâ€™s house and directly east of Karenâ€™s house. If the cafe is 2 miles closer to Danâ€™s house than to Karenâ€™s house, how many miles is the cafe from Karenâ€™s house? (A) 6 (B) 7 (C) 8 (D) 9 (E) 10 Manager Joined: 20 Mar 2005 Posts: 201 Location: Colombia, South America Followers: 1 Kudos [?]: 16 [0], given: 0 ### Show Tags 20 Feb 2006, 16:36 [quote="Avis"]Dan and Karen, who live 10 miles apart meet at a cafe that is directly north of Danâ€™s house and directly east of Karenâ€™s house. If the cafe is 2 miles closer to Danâ€™s house than to Karenâ€™s house, how many miles is the cafe from Karenâ€™s house? (A) 6 (B) 7 (C) 8 (D) 9 (E) 10[/quote A) use triangle rectangle 3-4-5 so equivalent 4-6-10 stupid mistake see correction below Last edited by conocieur on 20 Feb 2006, 19:04, edited 1 time in total. Manager Joined: 13 Dec 2005 Posts: 224 Location: Milwaukee,WI Followers: 1 Kudos [?]: 8 [0], given: 0 ### Show Tags 20 Feb 2006, 17:46 it should be 8 ... its like a right angle and say the distance of karen from cafe is x so the distance of dan house is x -2 by pythagoras theorem x ^ 2 + (x-2) ^2 = 100 so from above we get x ^2 -2x -48 = 0 ==> (x -8)(x+6) =0 since distance cannot be negative so x =8 . GMAT Club Legend Joined: 07 Jul 2004 Posts: 5062 Location: Singapore Followers: 30 Kudos [?]: 358 [0], given: 0 ### Show Tags 20 Feb 2006, 19:18 x K---C \ \| \ \| x-2 \ \| D Using pythagoras theroem, x^2 + (x-2)^2 = 100 x^2 + x^2 - 4x + 4 = 100 2x^2 -4x + 4 = 100 x^2 -2x + 2 = 50 x^2 - 2x -48 = 0 (x+6)(x-8) =0 x = -6 (not valid) or 8 (valid) Ans C Senior Manager Joined: 11 Jan 2006 Posts: 269 Location: Chennai,India Followers: 1 Kudos [?]: 4 [0], given: 0 ### Show Tags 21 Feb 2006, 05:05 ywilfred wrote: x K---C \ \| \ \| x-2 \ \| D Using pythagoras theroem, x^2 + (x-2)^2 = 100 x^2 + x^2 - 4x + 4 = 100 2x^2 -4x + 4 = 100 x^2 -2x + 2 = 50 x^2 - 2x -48 = 0 (x+6)(x-8) =0 x = -6 (not valid) or 8 (valid) Ans C bingo solved and got 8 using the same method... x^2+(x-2)^2 = 100 simplifying = x = 8 or -6 thus 8 _________________ vazlkaiye porkalam vazltuthan parkanum.... porkalam maralam porkalthan maruma 21 Feb 2006, 05:05 Display posts from previous: Sort by	1,207	3,456	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.59375	4	CC-MAIN-2017-04	latest	en	0.881581	Dan and Karen, who live 10 miles apart meet at a cafe that : Quant Question Archive [LOCKED]. Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack. It is currently 21 Jan 2017, 13:42. ### GMAT Club Daily Prep. #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. # Events & Promotions. ###### Events & Promotions in June. Open Detailed Calendar. # Dan and Karen, who live 10 miles apart meet at a cafe that. Author Message. Manager. Joined: 14 Dec 2005. Posts: 74. Followers: 1. Kudos [?]: 5 [0], given: 0. Dan and Karen, who live 10 miles apart meet at a cafe that [#permalink]. ### Show Tags. 20 Feb 2006, 16:30. This topic is locked. If you want to discuss this question please re-post it in the respective forum.. Dan and Karen, who live 10 miles apart meet at a cafe that is directly north of Danâ€™s house and directly east of Karenâ€™s house. If the cafe is 2 miles closer to Danâ€™s house than to Karenâ€™s house, how many miles is the cafe from Karenâ€™s house?. (A) 6. (B) 7. (C) 8. (D) 9. (E) 10. Manager. Joined: 20 Mar 2005. Posts: 201. Location: Colombia, South America. Followers: 1. Kudos [?]: 16 [0], given: 0. ### Show Tags. 20 Feb 2006, 16:36. [quote="Avis"]Dan and Karen, who live 10 miles apart meet at a cafe that is directly north of Danâ€™s house and directly east of Karenâ€™s house. If the cafe is 2 miles closer to Danâ€™s house than to Karenâ€™s house, how many miles is the cafe from Karenâ€™s house?. (A) 6. (B) 7. (C) 8. (D) 9. (E) 10[/quote. A). use triangle rectangle 3-4-5. so equivalent 4-6-10. stupid mistake see correction below. Last edited by conocieur on 20 Feb 2006, 19:04, edited 1 time in total.. Manager. Joined: 13 Dec 2005. Posts: 224. Location: Milwaukee,WI. Followers: 1.	Kudos [?]: 8 [0], given: 0. ### Show Tags. 20 Feb 2006, 17:46. it should be 8 ... its like a right angle and say the distance of karen from cafe is x so the distance of dan house is x -2. by pythagoras theorem x ^ 2 + (x-2) ^2 = 100. so from above we get x ^2 -2x -48 = 0. ==> (x -8)(x+6) =0. since distance cannot be negative so x =8 .. GMAT Club Legend. Joined: 07 Jul 2004. Posts: 5062. Location: Singapore. Followers: 30. Kudos [?]: 358 [0], given: 0. ### Show Tags. 20 Feb 2006, 19:18. x. K---C. \ \|. \ \| x-2. \ \|. D. Using pythagoras theroem,. x^2 + (x-2)^2 = 100. x^2 + x^2 - 4x + 4 = 100. 2x^2 -4x + 4 = 100. x^2 -2x + 2 = 50. x^2 - 2x -48 = 0. (x+6)(x-8) =0. x = -6 (not valid) or 8 (valid). Ans C. Senior Manager. Joined: 11 Jan 2006. Posts: 269. Location: Chennai,India. Followers: 1. Kudos [?]: 4 [0], given: 0. ### Show Tags. 21 Feb 2006, 05:05. ywilfred wrote:. x. K---C. \ \|. \ \| x-2. \ \|. D. Using pythagoras theroem,. x^2 + (x-2)^2 = 100. x^2 + x^2 - 4x + 4 = 100. 2x^2 -4x + 4 = 100. x^2 -2x + 2 = 50. x^2 - 2x -48 = 0. (x+6)(x-8) =0. x = -6 (not valid) or 8 (valid). Ans C. bingo solved and got 8 using the same method... x^2+(x-2)^2 = 100 simplifying = x = 8 or -6 thus 8. _________________. vazlkaiye porkalam vazltuthan parkanum.... porkalam maralam porkalthan maruma. 21 Feb 2006, 05:05. Display posts from previous: Sort by.
https://www.jiskha.com/display.cgi?id=1226972868	1,511,264,557,000,000,000	text/html	crawl-data/CC-MAIN-2017-47/segments/1510934806353.62/warc/CC-MAIN-20171121113222-20171121133222-00770.warc.gz	802,308,029	4,426	# Physics 12 posted by . "A pendulum consisting of an object of mass 1.3kg suspended from a string of length 1.9m is pulled aside so that the string makes an angle of 28 degrees with the vertical. At the instant the pendulum is released, what is the magnitude of the unbalanced force on the object?" I can't get this question, and am getting really frustrated. I know Fg is 12.74N, and should egual F1y. I know the angle between F1 and the horizontal is 62 degrees. But now Ihave no idea what to do. Oh, wait; I know F1x=cos62? So now I'm stuck. Please help? • Physics 12 - Fg = (1.3kg)(9.8N/kg) = 12.74N Fg can be resolved into two foces, T along the extension of the string, and Fp, at right angles to it. Fp = (12.74N)(cos62)=________N, or Fp = (12.74N)(sin28) =________N The vector triangle I hope you drew, shows the weight, Fg resolved into two vectors, Fp, the unbalanced force perpendicular to the string, and T (string tension) which is along the extension of the string and cancelled out by the string resistance. • Physics 12 - • Physics 12 - 6.78 ## Similar Questions 1. ### physics A pendulum consists of an object of mass m = 1.2 kg swinging on a massless string of length l = 259 cm. The object has a speed of 2.4 m/s when it passes through its lowest point. What is the greatest angle with the vertical that the … 2. ### physics 3. A 250-g is attached to a string 1.00 m long to make a pendulum. If the pendulum bob is pulled to the right, such that the string makes an angle of 150 with the vertical, what is the maximum potential energy 3. ### physics 3. A 250-g is attached to a string 1.00 m long to make a pendulum. If the pendulum bob is pulled to the right, such that the string makes an angle of 150 with the vertical, what is the maximum potential energy 4. ### AP PHYSICS MECH. A simple pendulum, consists of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air friction, the system oscillates … 5. ### physics A simple pendulum is made from a bob of mass 0.04 kg suspended on a light string of length 1.4 m. Keeping the string taut, the pendulum is pulled to one tntil it has gained a heit of 0.1 m.calculate the total energy of the oscillation? 6. ### physics A simple pendulum has a mass m and carries a charge q. The pendulum is suspended between the vertical plates of a capacitor of separation distance d. If the string of the pendulum makes an angle θ with the vertical, what is the … 7. ### physics A simple pendulum has a mass m and carries a charge q. The pendulum is suspended between the vertical plates of a capacitor of separation distance d. If the string of the pendulum makes an angle θ with the vertical, what is the … 8. ### physics A simple pendulum has a mass m and carries a charge q. The pendulum is suspended between the vertical plates of a capacitor of separation distance d. If the string of the pendulum makes an angle θ with the vertical, what is the … 9. ### Physics Consider the conical pendulum, a mass on the end of a string, with the other end of the string fixed to the ceiling. Given the proper push, this pendulum can swing in a circle at an angle q of 25.5 with respect to the vertical, maintaining … 10. ### Physics Two pendulums are set up side by side. One pendulum P, consists of a 0.240 kg mass that is suspended on a string that is 65.0 cm long. The other pendulum, Q, has a 0.360 kg mass that is suspended on a string that is 39.0 cm long. The … More Similar Questions	938	3,581	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5	4	CC-MAIN-2017-47	latest	en	0.912126	# Physics 12. posted by .. "A pendulum consisting of an object of mass 1.3kg suspended from a string of length 1.9m is pulled aside so that the string makes an angle of 28 degrees with the vertical. At the instant the pendulum is released, what is the magnitude of the unbalanced force on the object?". I can't get this question, and am getting really frustrated. I know Fg is 12.74N, and should egual F1y. I know the angle between F1 and the horizontal is 62 degrees. But now Ihave no idea what to do. Oh, wait; I know F1x=cos62? So now I'm stuck. Please help?. • Physics 12 -. Fg = (1.3kg)(9.8N/kg) = 12.74N. Fg can be resolved into two foces, T along the extension of the string, and Fp, at right angles to it.. Fp = (12.74N)(cos62)=________N, or. Fp = (12.74N)(sin28) =________N. The vector triangle I hope you drew, shows the weight, Fg resolved into two vectors, Fp, the unbalanced force perpendicular to the string, and T (string tension) which is along the extension of the string and cancelled out by the string resistance.. • Physics 12 -. • Physics 12 -. 6.78. ## Similar Questions. 1. ### physics. A pendulum consists of an object of mass m = 1.2 kg swinging on a massless string of length l = 259 cm. The object has a speed of 2.4 m/s when it passes through its lowest point. What is the greatest angle with the vertical that the …. 2. ### physics. 3. A 250-g is attached to a string 1.00 m long to make a pendulum. If the pendulum bob is pulled to the right, such that the string makes an angle of 150 with the vertical, what is the maximum potential energy. 3. ### physics. 3. A 250-g is attached to a string 1.00 m long to make a pendulum. If the pendulum bob is pulled to the right, such that the string makes an angle of 150 with the vertical, what is the maximum potential energy. 4.	### AP PHYSICS MECH.. A simple pendulum, consists of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air friction, the system oscillates …. 5. ### physics. A simple pendulum is made from a bob of mass 0.04 kg suspended on a light string of length 1.4 m. Keeping the string taut, the pendulum is pulled to one tntil it has gained a heit of 0.1 m.calculate the total energy of the oscillation?. 6. ### physics. A simple pendulum has a mass m and carries a charge q. The pendulum is suspended between the vertical plates of a capacitor of separation distance d. If the string of the pendulum makes an angle θ with the vertical, what is the …. 7. ### physics. A simple pendulum has a mass m and carries a charge q. The pendulum is suspended between the vertical plates of a capacitor of separation distance d. If the string of the pendulum makes an angle θ with the vertical, what is the …. 8. ### physics. A simple pendulum has a mass m and carries a charge q. The pendulum is suspended between the vertical plates of a capacitor of separation distance d. If the string of the pendulum makes an angle θ with the vertical, what is the …. 9. ### Physics. Consider the conical pendulum, a mass on the end of a string, with the other end of the string fixed to the ceiling. Given the proper push, this pendulum can swing in a circle at an angle q of 25.5 with respect to the vertical, maintaining …. 10. ### Physics. Two pendulums are set up side by side. One pendulum P, consists of a 0.240 kg mass that is suspended on a string that is 65.0 cm long. The other pendulum, Q, has a 0.360 kg mass that is suspended on a string that is 39.0 cm long. The …. More Similar Questions.
http://learner.org/courses/learningmath/data/session8/solutions_a.html	1,448,439,422,000,000,000	text/html	crawl-data/CC-MAIN-2015-48/segments/1448398445033.85/warc/CC-MAIN-20151124205405-00221-ip-10-71-132-137.ec2.internal.warc.gz	137,888,328	8,440	Solutions for Session 8, Part A See solutions for Problems: A1 \| A2 \| A3 \| A4 \| A5 Problem A1 Answers will vary. Some everyday uses of probability are predicting the weather, deciding which road is likely to have the least amount of traffic, and choosing a restaurant on the basis of how long you think the wait will be. Some mathematical uses include the probability of rolling a six on a die, the probability of tossing a coin and getting "heads," and the probability of 1-2-3 coming up as the daily lottery number. Problem A2 Statistical uses of probability include the probability that the estimate of a mean is accurate (this is known as a confidence interval). Places where probability uses statistics include taking experimental data and trying to create exact probabilities to match your data set. Problem A3 A "random" event is entirely up to chance; there is no skill involved. A random event might be what appears as the top card after a thorough shuffling of a deck of cards. Most events are not random; for example, answering a question correctly on a test may happen randomly (as a guess) but usually is a result of skill. Problem A4 You might look at your average score and determine whether your average score is improving over time. For example, if you played Push Penny 20 times a day for several days, you could compare your average first day's score to your average last day's score and see if there was any improvement. Problem A5 a. Answers will vary. b. Answers will vary. c. One example of such a game is a game where you shuffle a deck of cards thoroughly and then try to guess the suit of the top card. Since the top card is completely random, there is no way to develop your skill in correctly guessing the suit of the card (without cheating or using ESP!). The card game War is another example.	391	1,834	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.75	4	CC-MAIN-2015-48	longest	en	0.952278	Solutions for Session 8, Part A. See solutions for Problems: A1 \| A2 \| A3 \| A4 \| A5. Problem A1 Answers will vary. Some everyday uses of probability are predicting the weather, deciding which road is likely to have the least amount of traffic, and choosing a restaurant on the basis of how long you think the wait will be. Some mathematical uses include the probability of rolling a six on a die, the probability of tossing a coin and getting "heads," and the probability of 1-2-3 coming up as the daily lottery number.. Problem A2 Statistical uses of probability include the probability that the estimate of a mean is accurate (this is known as a confidence interval). Places where probability uses statistics include taking experimental data and trying to create exact probabilities to match your data set.. Problem A3 A "random" event is entirely up to chance; there is no skill involved. A random event might be what appears as the top card after a thorough shuffling of a deck of cards. Most events are not random; for example, answering a question correctly on a test may happen randomly (as a guess) but usually is a result of skill.. Problem A4 You might look at your average score and determine whether your average score is improving over time.	For example, if you played Push Penny 20 times a day for several days, you could compare your average first day's score to your average last day's score and see if there was any improvement.. Problem A5. a. Answers will vary. b. Answers will vary. c. One example of such a game is a game where you shuffle a deck of cards thoroughly and then try to guess the suit of the top card. Since the top card is completely random, there is no way to develop your skill in correctly guessing the suit of the card (without cheating or using ESP!). The card game War is another example.
http://slideplayer.com/slide/2859906/	1,527,073,203,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00307.warc.gz	266,439,894	29,012	# Dynamics III: Earth rotation L. Talley SIO 210 Fall, 2014 ## Presentation on theme: "Dynamics III: Earth rotation L. Talley SIO 210 Fall, 2014"— Presentation transcript: Dynamics III: Earth rotation L. Talley SIO 210 Fall, 2014 Earth reference frame Geoid Rotation definitions Centrifugal force Coriolis force Inertial motion READING: DPO: Chapter 7.2.3, 7.5.1 Stewart chapter 7.6, 9.1 Tomczak and Godfrey chapter 3, pages 29-35 Talley SIO 210 (2014) The Earth as a reference frame Coordinate system for most oceanography: Local Cartesian (x,y,z), ignoring sphericity West-east (“zonal”) x –direction (positive eastwards) South-north (“meridional”) y-direction (positive northwards) Down-up (“vertical”) z-direction (positive upwards) Add diagram Talley SIO 210 (2014) An aside on the Earth’s reference frame for pressure: the geoid When we calculate horizontal pressure gradients, they are relative to a “level” surface along which the pressure gradient vanishes (no force, no motion) Earth’s mass is not distributed evenly AND Earth is not a perfect ellipsoid. That is, the “level” surface is not locally “flat”. The surface of constant pressure, i.e. along which the gravitational force has no changes, is the “geoid” Good geoid site Talley SIO 210 (2014) Geoid Earth’s mass is not distributed evenly AND Earth is not a perfect ellipsoid 200 meter variation Geoid map using EGM96 data, from Talley SIO 210 (2014) Geoid Earth’s mass is not distributed evenly AND Earth is not a perfect ellipsoid Talley SIO 210 (2014) Rotating coordinates The Earth is rotating. We measure things relative to this “rotating reference frame”. Quantity that tells how fast something is rotating: Angular speed or angular velocity  = angle/second 360° is the whole circle, but express angle in radians (2 radians = 360°) For Earth: 2 / 1 day = 2 / 86,400 sec = x 10-4 /sec Also can show  = v/R where v is the measured velocity and R is the radius to the axis of rotation (therefore v =  R) At home: calculate speed you are traveling through space if you are at the equator (radius of earth is about 6371 km), then calculate it at 30N as well. (Do some geometry to figure out the distance from the axis at 30N.) At home: calculate speed you are traveling through space if you are at the equator (radius of earth is about 6371 km), then calculate it at 30N as well. (Do some geometry to figure out the distance from the axis at 30N.) R Talley SIO 210 (2014) Rotating coordinates R Vector that expresses direction of rotation and how fast it is rotating: vector pointing in direction of thumb using right-hand rule, curling fingers in direction of rotation R Talley SIO 210 (2014) Centripetal and Centrifugal forces (now looking straight down on the rotating plane) Centripetal force is the actual force that keeps the ball “tethered” (here it is the string, but it can be gravitational force) Centrifugal force is the pseudo-force (apparent force) that one feels due to lack of awareness that the coordinate system is rotating or curving centrifugal acceleration = 2R (outward) (Units of m/sec2) Talley SIO 210 (2014) Effect of centrifugal force on ocean and earth Centrifugal force acts on the ocean and earth. It is pointed outward away from the rotation axis. Therefore it is maximum at the equator (maximum radius from axis) and minimum at the poles (0 radius).  = x 10-4 /sec At the equator, R ~ 6380 km so 2R = .032 m/s2 Compare with gravity = 9.8 m/s2 Centrifugal force should cause the equator to be deflected (0.032/9.8) x 6380 km = 21 km outward compared with the poles. (i.e. about 0.3%) A = 6000 km; omega = 2pi/86400 sec = 7.292e10-4/sec Omega2 a cos(latitude) = 3.17 cm/sec2 x cos(latitude) Compare 3.17 with gravity of 980 cm/sec2 Earth surface should be 0.3% farther out at equator = 20 km Talley SIO 210 (2014) Effect of centrifugal force on ocean and earth Radius: Equatorial 6, km Polar 6, km Mean 6, km (From wikipedia entry on Earth) The ocean is not 20 km deeper at the equator, rather the earth itself is deformed! We bury the centrifugal force term in the gravity term (which we can call “effective gravity”), and ignore it henceforth. Calculations that require a precise gravity term should use subroutines that account for its latitudinal dependence. Return here – need to state that at the equator, centrigufal force reduces the effective gravity the most Talley SIO 210 (2014) Coriolis effect Inertial motion: motion in a straight line relative to the fixed stars Coriolis effect: apparent deflection of that inertially moving body just due to the rotation of you, the observer. Coriolis effect deflects bodies (water parcels, air parcels) to the right in the northern hemisphere and to the left in the southern hemisphere add this reference and animations Talley SIO 210 (2014) Coriolis effect look on youtube for countless animations add this reference and animations Talley SIO 210 (2014) Coriolis force Additional terms in x, y momentum equations, at latitude  horizontal motion is much greater than vertical) x-momentum equation: -sinv = -f v y-momentum equation: sinu = f u f = sinis the “Coriolis parameter”. It depends on latitude (projection of total Earth rotation on local vertical) Talley SIO 210 (2014) Coriolis force f = sinis the “Coriolis parameter”  = 1.414x10-4/sec At equator (=0, sin=0): f = 0 At 30°N (=30°, sin=0.5): f = 0.707x10-4/sec At north pole (=90°, sin=1): f = 1.414x10-4/sec At 30°S (=-30°, sin=-0.5): f = x10-4/sec At north pole (=-90°, sin=-1): f = x10-4/sec Edited 11/3/14 Talley SIO 210 (2014) Complete force (momentum) balance with rotation Three equations: Horizontal (x) (west-east) acceleration + advection + Coriolis = pressure gradient force + viscous term Horizontal (y) (south-north) Vertical (z) (down-up) acceleration +advection (+ neglected very small Coriolis) = pressure gradient force + effective gravity (including centrifugal force) + viscous term Full equations in words Talley SIO 210 (2014) Final equations of motion (momentum equations in Cartesian coordinates) x: u/t + u u/x + v u/y + w u/z - fv = - (1/)p/x + /x(AHu/x) + /y(AHu/y) +/z(AVu/z) (7.11a) y: v/t + u v/x + v v/y + w v/z + fu = - (1/)p/y + /x(AHv/x) + /y(AHv/y) +/z(AVv/z) (7.11b) z: w/t + u w/x + v w/y + w w/z (+ neglected small Coriolis) = - (1/)p/z - g + /x(AHw/x) + /y(AHw/y) +/z(AVw/z) (7.11c) (where g contains both actual g and effect of centrifugal force) Full equations for the physics-based students in class. The class is not responsible for knowing these full equations. The “g” term includes centrifugal force. The neglect of Coriolis in the z-momentum equation is called the “traditional approximation”.. Talley SIO 210 (2014) Coriolis in action in the ocean: Observations of Inertial Currents D’Asaro et al. (1995) DPO Fig. 7.4 Surface drifters in the Gulf of Alaska during and after a storm. Note the corkscrews - drifters start off with clockwise motion, which gets weaker as the motion is damped (friction) Talley SIO 210 (2014) Inertial currents Balance of Coriolis and acceleration terms: push the water and it turns to the right (NH), in circles. What is the inertial period? Equator => infinite since f = 0 Poles 2pi/f = DPO Fig. S7.8a Talley SIO 210 (2014) Inertial currents: periods sin() Coriolis parameter f = 2sin() Inertial period T = 2π/f  = latitude Talley SIO 210 (2014) Inertial currents: force balance Three APPROXIMATE equations (all other terms are much smaller): Horizontal (x) (west-east) acceleration + Coriolis = 0 Horizontal (y) (south-north) Vertical (z) (down-up) (hydrostatic) (not important for this solution) 0 = pressure gradient force + effective gravity That is: x: u/t - fv = 0 y: v/t + fu = 0 Solution: solve y-eqn for u and substitute in x eqn (or vice versa). Solutions are velocity circles: u = Uosin(ft) and v = Uocos(ft) where Uo is an arbitrary amplitude Equations (bottom) not required for class. Talley SIO 210 (2014) Inertial currents: relation to internal waves Energy in inertial oscillations (which are wind-forced) as measured by surface drifters (Elipot et al., JGR 2010) Talley SIO 210 (2014) Download ppt "Dynamics III: Earth rotation L. Talley SIO 210 Fall, 2014" Similar presentations	2,611	8,331	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.703125	4	CC-MAIN-2018-22	latest	en	0.782522	# Dynamics III: Earth rotation L. Talley SIO 210 Fall, 2014. ## Presentation on theme: "Dynamics III: Earth rotation L. Talley SIO 210 Fall, 2014"— Presentation transcript:. Dynamics III: Earth rotation L. Talley SIO 210 Fall, 2014. Earth reference frame Geoid Rotation definitions Centrifugal force Coriolis force Inertial motion READING: DPO: Chapter 7.2.3, 7.5.1 Stewart chapter 7.6, 9.1 Tomczak and Godfrey chapter 3, pages 29-35 Talley SIO 210 (2014). The Earth as a reference frame. Coordinate system for most oceanography: Local Cartesian (x,y,z), ignoring sphericity West-east (“zonal”) x –direction (positive eastwards) South-north (“meridional”) y-direction (positive northwards) Down-up (“vertical”) z-direction (positive upwards) Add diagram Talley SIO 210 (2014). An aside on the Earth’s reference frame for pressure: the geoid. When we calculate horizontal pressure gradients, they are relative to a “level” surface along which the pressure gradient vanishes (no force, no motion) Earth’s mass is not distributed evenly AND Earth is not a perfect ellipsoid. That is, the “level” surface is not locally “flat”. The surface of constant pressure, i.e. along which the gravitational force has no changes, is the “geoid” Good geoid site Talley SIO 210 (2014). Geoid Earth’s mass is not distributed evenly AND Earth is not a perfect ellipsoid 200 meter variation Geoid map using EGM96 data, from Talley SIO 210 (2014). Geoid Earth’s mass is not distributed evenly AND Earth is not a perfect ellipsoid Talley SIO 210 (2014). Rotating coordinates The Earth is rotating. We measure things relative to this “rotating reference frame”. Quantity that tells how fast something is rotating: Angular speed or angular velocity  = angle/second 360° is the whole circle, but express angle in radians (2 radians = 360°) For Earth: 2 / 1 day = 2 / 86,400 sec = x 10-4 /sec Also can show  = v/R where v is the measured velocity and R is the radius to the axis of rotation (therefore v =  R) At home: calculate speed you are traveling through space if you are at the equator (radius of earth is about 6371 km), then calculate it at 30N as well. (Do some geometry to figure out the distance from the axis at 30N.) At home: calculate speed you are traveling through space if you are at the equator (radius of earth is about 6371 km), then calculate it at 30N as well. (Do some geometry to figure out the distance from the axis at 30N.) R Talley SIO 210 (2014). Rotating coordinates R. Vector that expresses direction of rotation and how fast it is rotating: vector pointing in direction of thumb using right-hand rule, curling fingers in direction of rotation R Talley SIO 210 (2014). Centripetal and Centrifugal forces. (now looking straight down on the rotating plane) Centripetal force is the actual force that keeps the ball “tethered” (here it is the string, but it can be gravitational force) Centrifugal force is the pseudo-force (apparent force) that one feels due to lack of awareness that the coordinate system is rotating or curving centrifugal acceleration = 2R (outward) (Units of m/sec2) Talley SIO 210 (2014). Effect of centrifugal force on ocean and earth. Centrifugal force acts on the ocean and earth. It is pointed outward away from the rotation axis. Therefore it is maximum at the equator (maximum radius from axis) and minimum at the poles (0 radius).  = x 10-4 /sec At the equator, R ~ 6380 km so 2R = .032 m/s2 Compare with gravity = 9.8 m/s2 Centrifugal force should cause the equator to be deflected (0.032/9.8) x 6380 km = 21 km outward compared with the poles. (i.e. about 0.3%) A = 6000 km; omega = 2pi/86400 sec = 7.292e10-4/sec Omega2 a cos(latitude) = 3.17 cm/sec2 x cos(latitude) Compare 3.17 with gravity of 980 cm/sec2 Earth surface should be 0.3% farther out at equator = 20 km Talley SIO 210 (2014). Effect of centrifugal force on ocean and earth. Radius: Equatorial 6, km Polar 6, km Mean 6, km (From wikipedia entry on Earth) The ocean is not 20 km deeper at the equator, rather the earth itself is deformed! We bury the centrifugal force term in the gravity term (which we can call “effective gravity”), and ignore it henceforth. Calculations that require a precise gravity term should use subroutines that account for its latitudinal dependence. Return here – need to state that at the equator, centrigufal force reduces the effective gravity the most Talley SIO 210 (2014).	Coriolis effect Inertial motion: motion in a straight line relative to the fixed stars Coriolis effect: apparent deflection of that inertially moving body just due to the rotation of you, the observer. Coriolis effect deflects bodies (water parcels, air parcels) to the right in the northern hemisphere and to the left in the southern hemisphere add this reference and animations Talley SIO 210 (2014). Coriolis effect look on youtube for countless animations. add this reference and animations Talley SIO 210 (2014). Coriolis force Additional terms in x, y momentum equations, at latitude  horizontal motion is much greater than vertical) x-momentum equation: -sinv = -f v y-momentum equation: sinu = f u f = sinis the “Coriolis parameter”. It depends on latitude (projection of total Earth rotation on local vertical) Talley SIO 210 (2014). Coriolis force f = sinis the “Coriolis parameter”.  = 1.414x10-4/sec At equator (=0, sin=0): f = 0 At 30°N (=30°, sin=0.5): f = 0.707x10-4/sec At north pole (=90°, sin=1): f = 1.414x10-4/sec At 30°S (=-30°, sin=-0.5): f = x10-4/sec At north pole (=-90°, sin=-1): f = x10-4/sec Edited 11/3/14 Talley SIO 210 (2014). Complete force (momentum) balance with rotation. Three equations: Horizontal (x) (west-east) acceleration + advection + Coriolis = pressure gradient force + viscous term Horizontal (y) (south-north) Vertical (z) (down-up) acceleration +advection (+ neglected very small Coriolis) = pressure gradient force + effective gravity (including centrifugal force) + viscous term Full equations in words Talley SIO 210 (2014). Final equations of motion (momentum equations in Cartesian coordinates). x: u/t + u u/x + v u/y + w u/z - fv = - (1/)p/x + /x(AHu/x) + /y(AHu/y) +/z(AVu/z) (7.11a) y: v/t + u v/x + v v/y + w v/z + fu = - (1/)p/y + /x(AHv/x) + /y(AHv/y) +/z(AVv/z) (7.11b) z: w/t + u w/x + v w/y + w w/z (+ neglected small Coriolis) = - (1/)p/z - g + /x(AHw/x) + /y(AHw/y) +/z(AVw/z) (7.11c) (where g contains both actual g and effect of centrifugal force) Full equations for the physics-based students in class. The class is not responsible for knowing these full equations. The “g” term includes centrifugal force. The neglect of Coriolis in the z-momentum equation is called the “traditional approximation”.. Talley SIO 210 (2014). Coriolis in action in the ocean: Observations of Inertial Currents. D’Asaro et al. (1995) DPO Fig. 7.4 Surface drifters in the Gulf of Alaska during and after a storm. Note the corkscrews - drifters start off with clockwise motion, which gets weaker as the motion is damped (friction) Talley SIO 210 (2014). Inertial currents Balance of Coriolis and acceleration terms: push the water and it turns to the right (NH), in circles. What is the inertial period? Equator => infinite since f = 0 Poles 2pi/f = DPO Fig. S7.8a Talley SIO 210 (2014). Inertial currents: periods. sin() Coriolis parameter f = 2sin() Inertial period T = 2π/f  = latitude Talley SIO 210 (2014). Inertial currents: force balance. Three APPROXIMATE equations (all other terms are much smaller): Horizontal (x) (west-east) acceleration + Coriolis = 0 Horizontal (y) (south-north) Vertical (z) (down-up) (hydrostatic) (not important for this solution) 0 = pressure gradient force + effective gravity That is: x: u/t - fv = 0 y: v/t + fu = 0 Solution: solve y-eqn for u and substitute in x eqn (or vice versa). Solutions are velocity circles: u = Uosin(ft) and v = Uocos(ft) where Uo is an arbitrary amplitude Equations (bottom) not required for class. Talley SIO 210 (2014). Inertial currents: relation to internal waves. Energy in inertial oscillations (which are wind-forced) as measured by surface drifters (Elipot et al., JGR 2010) Talley SIO 210 (2014). Download ppt "Dynamics III: Earth rotation L. Talley SIO 210 Fall, 2014". Similar presentations.
https://www.javatpoint.com/babylonian-square-root-algorithm-in-cpp	1,726,438,428,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00049.warc.gz	765,005,054	35,361	# Babylonian Square Root Algorithm in C++ In this article, we will discuss the Babylonian square root algorithm in C++ with its history and examples. ## Introduction: The Babylonian square root algorithm, also known as Heron's method, is an iterative method to approximate the square root of a given number. It is based on the idea of repeatedly averaging an initial guess with the original number divided by the guess. The algorithm converges quickly to the actual square root. The algorithm dates back to the Babylonian civilization and is named after the ancient Greek mathematician Heron of Alexandria, who described it in his work "Metrica" around 100 AD. The algorithm is based on the principle of successive approximations. Given a positive number N, the algorithm starts with an initial guess x0 (which could be any reasonable starting point, often N/2 or 1), and iteratively refines the estimate using the formula: ## History: The Babylonian square root algorithm, also known as Heron's method, has a rich historical background that dates back to ancient civilizations. It is named after the ancient Greek mathematician Heron of Alexandria, but its origins can be traced even further back to the Babylonians. • Babylonian Civilization (circa 2000-1600 BCE): The Babylonians are credited with some of the earliest mathematical developments in human history. Their mathematical clay tablets reveal that they had methods for approximating square roots. They used geometric shapes and applied numerical methods to solve mathematical problems, including square roots. • Heron of Alexandria (circa 10-70 AD): Heron, a Greek mathematician and engineer, described the square root extraction method in his work "Metrica" around 100 AD. His version of the algorithm was applied to find square roots and cube roots geometrically. • Islamic Golden Age (8th-14th centuries): During the Islamic Golden Age, scholars from the Islamic world further developed and refined mathematical techniques. They translated Greek and Roman mathematical texts, including Heron's work, and contributed to the understanding and application of mathematical methods. • Renaissance and Later Periods: Heron's method was rediscovered during the Renaissance and later periods in Europe. Mathematicians like François Viète and John Wallis studied and extended the algorithm. It became a fundamental technique in the broader context of numerical analysis. • Modern Usage: The Babylonian square root algorithm remains a foundational method for approximating square roots and is still used in various forms today. Iterative methods for numerical analysis, including the Newton-Raphson method (which is a generalization of Heron's method), draw inspiration from this ancient algorithm. ### Program 1: Let us take an example to illustrate the use of the Babylonian square root algorithm in C++: Output: ```Enter a number to find its square root: 25 Square root of 25 is approximately: 5 ``` Explanation: 1. Initialization: Choose an initial guess (x0) for the square root. It can be any reasonable starting point, but common choices include x0 = N/2 or x0 =1, where N is the number for which you want to find the square root. 2. Iteration: • For each iteration (n), calculate the next approximation using the formula: • This formula combines the current guess xn with the ratio N/xn to produce a new estimate that, on average, is closer to the actual square root. 3. Convergence Check: • Check the difference between consecutive approximations. The iteration continues until the difference is sufficiently small. The criterion for stopping the iteration is typically based on a specified tolerance level (epsilon, ε): 4. Termination: • Once the convergence criterion is met, terminate the iteration. The last obtained value (x n+1) is considered an approximation of the square root of N. ### Program 2: Let us take another example to illustrate the use of the Babylonian square root algorithm in C++: Output: ```Enter a number to find its square root: 64 Square root of 64 is approximately: 8 ``` Explanation: • The code includes two standard C++ header files: <iostream> for input and output operations and <cmath> for mathematical functions. 2. Babylonian Square Root Function: • The function babylonianSquareRoot is defined to calculate the square root using the Babylonian method. • It takes two parameters: n (the number for which the square root is to be calculated) and epsilon (a small positive value to control the precision of the result). • The function checks if the input n is less than 0, in which case it prints an error message and returns a specified error code or throws an exception. • If n is 0, it returns 0, as the square root of 0 is 0. • The function initializes a variable guess as the initial guess (typically set to n / 2.0). 3. Do-While Loop: • The core of the algorithm is a do-while loop, which continues until the convergence criterion is met. • Inside the loop, a new guess (newGuess) is calculated using the Babylonian formula. • The loop checks if the absolute difference between the new guess and the old guess is less than the specified epsilon. If true, it breaks out of the loop, indicating convergence. • If not, the new guess becomes the current guess, and the loop continues. 4. Main Function: • The main function is the entry point of the program. • It prompts the user to enter a number and reads the input into the variable number. • After that, it calls the babylonianSquareRoot function with the entered number, obtaining the result. • If the result is not equal to the specified error code (-1.0), it prints the calculated square root. • Return 0: The main function returns 0, indicating successful program execution. Time and Space complexities: Time Complexity: • The number of iterations required for convergence primarily determines the time complexity of the Babylonian square root algorithm. • Let k be the number of iterations needed for convergence, and ? be the tolerance level. Time Complexity: The algorithm exhibits a logarithmic convergence behavior, and the number of iterations is usually relatively small and proportional to the precision required. Therefore, the time complexity is often considered as O(log( 1/? )). In practical terms, the algorithm often converges in a fixed number of iterations, making it efficient. Space Complexity: • The space complexity of the algorithm is related to the storage of variables and the stack space used during the function calls. Space Complexity: The algorithm uses a constant amount of space for variables, regardless of the input size. Therefore, the space complexity is considered as O(1), indicating constant space usage. The space required for the variables (e.g., guess, newGuess, number, etc.) remains constant, and additional space is not dependent on the input size. ### Advantages of the Babylonian square root algorithm: There are several advantages of the Babylonian square root algorithm in C++. Some main advantages of the Babylonian square root algorithm in C++: • Fast Convergence: The algorithm converges rapidly, meaning that it approaches the actual square root with each iteration. It leads to quick and efficient approximations. • Simplicity: The algorithm is straightforward and easy to understand. Its simplicity makes it an attractive choice for square root approximation, especially in educational contexts or applications where computational efficiency is not the primary concern. • Low Computational Complexity: The algorithm involves basic arithmetic operations (addition, division, and multiplication), which have relatively low computational complexity. It makes it computationally efficient, especially when compared to more complex numerical methods. • Applicability to Various Contexts: The Babylonian square root algorithm is applicable to a wide range of contexts, including hand calculations, programming environments, and embedded systems. Its versatility makes it suitable for various applications. • Numerical Stability: The algorithm is numerically stable, and small variations in the input value or initial guess do not usually lead to divergent behavior. This stability is important in numerical methods to ensure reliable and consistent results. • Historical Significance: The algorithm has historical significance, dating back to ancient civilizations. Its enduring use highlights its effectiveness and importance in the development of numerical methods throughout history. ### Applications of the Babylonian square root algorithm: There are several applications of the Babylonian square root algorithm in C++. Some main applications of the Babylonian square root algorithm in C++: • Numerical Analysis: The algorithm is a fundamental component of numerical analysis, providing a simple and effective method for approximating square roots. It is often used as a starting point for more complex algorithms. • Computer Programming: The Babylonian square root algorithm is commonly implemented in programming languages to calculate square roots. Many programming environments, including scientific and engineering applications, use this algorithm due to its simplicity and quick convergence. • Calculator Implementations: The algorithm is used in the software of calculators to efficiently compute square roots. Its rapid convergence makes it suitable for real-time applications, providing users with quick and accurate results. • Embedded Systems: In resource-constrained environments, such as embedded systems in electronic devices, the Babylonian square root algorithm can be preferred due to its low computational complexity and ease of implementation. • Education: The algorithm is often used in educational settings to teach the concept of iterative methods for approximating mathematical functions. Its simplicity makes it accessible for students studying numerical methods. • Financial Calculations: The algorithm can be used in financial calculations where quick approximations of square roots are needed. For example, it might be applied in risk assessments or option pricing models. • Signal Processing: In some signal processing applications, where real-time calculations are crucial, the Babylonian square root algorithm can be employed for its efficiency in approximating square roots. • Scientific Research: The algorithm is used in various scientific disciplines for quick and practical approximations. It is particularly valuable when high precision is not critical, and computational efficiency is essential. While the Babylonian square root algorithm is versatile, it's important to note that in certain applications requiring very high precision or specific error analysis, more advanced algorithms like Newton's method might be preferred. Additionally, its historical significance and simplicity make it an interesting subject of study in the context of numerical methods. ### Significance of The Babylonian square root algorithm in computer programming: The Babylonian square root algorithm holds significant importance in computer programming due to its simplicity, efficiency, and widespread applicability. Its straightforward nature makes it an accessible choice for programmers at all skill levels, contributing to its adoption in various software applications. The algorithm strikes a balance between accuracy and speed, making it computationally efficient for a broad range of scenarios where square root calculations are required. Its common use case in programming, ranging from basic arithmetic to complex mathematical computations, underscores its practicality and versatility. One of the algorithm's notable strengths lies in its numerical stability, ensuring consistent and reliable results even in the presence of small variations in input or initial guesses. This stability is a crucial attribute in programming, where robust and predictable behaviour is essential. Additionally, the historical continuity of the Babylonian square root algorithm adds to its significance. Originating from ancient civilizations, its enduring effectiveness has led to its continued use in modern computer programming. In environments with resource constraints, such as embedded systems, the algorithm's relatively low computational complexity makes it a suitable choice. Its efficiency contributes to its popularity in scenarios where quick and reasonably accurate square root approximations are required. ### Disadvantages of the Babylonian square root algorithm: There are several disadvantages of the Babylonian square root algorithm in C++. Some main disadvantages of the Babylonian square root algorithm in C++: • Convergence Rate in Certain Cases: While the algorithm generally converges rapidly, there are cases where the convergence rate might be slower, especially for numbers with peculiar properties or near-zero values. Other methods, like Newton's method, may have faster convergence in some situations. • Initial Guess Dependency: The performance of the algorithm can be sensitive to the choice of the initial guess. In certain scenarios, a poorly chosen initial guess may lead to slower convergence or divergence. Choosing a good initial guess requires some knowledge of the properties of the input. • Not Suitable for Negative Numbers: The algorithm is designed for positive real numbers. It cannot be directly applied to find the square root of negative numbers. Handling negative inputs may require additional considerations or a different approach. • Not Suitable for Complex Numbers: The algorithm is limited to real numbers. If complex roots are needed, other methods specifically designed for complex numbers, such as the Newton-Raphson method for complex functions, would be more appropriate. • Precision Limitations: The Babylonian algorithm may not be the most suitable choice for applications requiring extremely high precision. Other algorithms, such as iterative methods with higher-order convergence, might be preferred for such scenarios. • Requires Division Operation: The algorithm involves a division operation in each iteration, which may be computationally expensive on certain platforms. In situations where division operations are costly, alternative methods might be more efficient. • Handling Near-Zero Values: The algorithm may encounter difficulties when the input is very close to zero. In such cases, issues related to numerical precision and floating-point arithmetic may affect the accuracy of the result. • Not the Most Efficient for Extreme Precision: For applications requiring extremely high precision, more advanced algorithms, such as specialized square root algorithms or arbitrary-precision arithmetic libraries, may outperform the Babylonian algorithm. While the Babylonian square root algorithm is versatile and widely used, these disadvantages highlight situations where other algorithms or methods may be more suitable, depending on the specific requirements of an application. Understanding these limitations helps in making informed choices when selecting a square root approximation method. ## Other Alternatives: • Newton's Method (Newton-Raphson Method): Newton's method is an iterative numerical technique for finding roots of real-valued functions. It can be adapted to find square roots. Newton's method tends to converge faster than the Babylonian method, especially for functions with higher-order convergence. • Binary Search Algorithm: The binary search algorithm can be adapted to find the square root by searching for the square root in a given range. It repeatedly bisects the search interval until the desired precision is achieved. Binary search is efficient but may take more iterations than methods with higher-order convergence. • Exponential Function and Logarithm: Some mathematical libraries or hardware instructions provide specialized functions for exponentiation and logarithms that can be used to calculate square roots. • CORDIC Algorithm: The Coordinate Rotation Digital Computer (CORDIC) algorithm is often used for trigonometric and hyperbolic function calculations, but it can be adapted to compute square roots. It uses a series of simple and fixed-point operations, making it suitable for hardware implementation. ## Conclusion: In conclusion, the Babylonian square root algorithm is a simple and effective method for approximating square roots. It has been used for centuries and remains a practical choice due to its ease of implementation and rapid convergence. The algorithm is suitable for a variety of applications, including numerical analysis, programming, and embedded systems. ### Key points about the Babylonian square root algorithm: • Iterative Nature: The algorithm iteratively refines its estimate of the square root, converging rapidly to the actual value. • Simplicity: Its straightforward formula and minimal computational complexity make it accessible for educational purposes and practical implementations. • Versatility: The algorithm can be applied to various contexts, ranging from hand calculations to computer programming and embedded systems. • Competition: While other algorithms, like Newton's method offer faster convergence in some cases, the Babylonian method strikes a balance between simplicity and efficiency. • Applications: The algorithm provides quick and satisfactory approximations for square roots, and it is widely used in fields such as numerical analysis, programming, and embedded systems. While the Babylonian method is widely used, it's essential to consider its limitations, such as sensitivity to the initial guess and its suitability for specific precision requirements. In scenarios where higher precision or specialized considerations are needed, alternative algorithms like Newton's method or binary search may be preferred. The implementation and enhancement of the Babylonian square root algorithm provide valuable insights into numerical methods, error handling, and user interface design. Exploring and understanding different square root algorithms broadens one's understanding of computational techniques and their applications. In practical terms, the Babylonian square root algorithm remains a fundamental and efficient approach for square root approximation, and its simplicity makes it an accessible topic for learning and exploration in the realm of numerical computing.	3,355	18,419	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.609375	4	CC-MAIN-2024-38	latest	en	0.946595	# Babylonian Square Root Algorithm in C++. In this article, we will discuss the Babylonian square root algorithm in C++ with its history and examples.. ## Introduction:. The Babylonian square root algorithm, also known as Heron's method, is an iterative method to approximate the square root of a given number. It is based on the idea of repeatedly averaging an initial guess with the original number divided by the guess. The algorithm converges quickly to the actual square root. The algorithm dates back to the Babylonian civilization and is named after the ancient Greek mathematician Heron of Alexandria, who described it in his work "Metrica" around 100 AD.. The algorithm is based on the principle of successive approximations. Given a positive number N, the algorithm starts with an initial guess x0 (which could be any reasonable starting point, often N/2 or 1), and iteratively refines the estimate using the formula:. ## History:. The Babylonian square root algorithm, also known as Heron's method, has a rich historical background that dates back to ancient civilizations. It is named after the ancient Greek mathematician Heron of Alexandria, but its origins can be traced even further back to the Babylonians.. • Babylonian Civilization (circa 2000-1600 BCE): The Babylonians are credited with some of the earliest mathematical developments in human history. Their mathematical clay tablets reveal that they had methods for approximating square roots. They used geometric shapes and applied numerical methods to solve mathematical problems, including square roots.. • Heron of Alexandria (circa 10-70 AD): Heron, a Greek mathematician and engineer, described the square root extraction method in his work "Metrica" around 100 AD. His version of the algorithm was applied to find square roots and cube roots geometrically.. • Islamic Golden Age (8th-14th centuries): During the Islamic Golden Age, scholars from the Islamic world further developed and refined mathematical techniques. They translated Greek and Roman mathematical texts, including Heron's work, and contributed to the understanding and application of mathematical methods.. • Renaissance and Later Periods: Heron's method was rediscovered during the Renaissance and later periods in Europe. Mathematicians like François Viète and John Wallis studied and extended the algorithm. It became a fundamental technique in the broader context of numerical analysis.. • Modern Usage: The Babylonian square root algorithm remains a foundational method for approximating square roots and is still used in various forms today. Iterative methods for numerical analysis, including the Newton-Raphson method (which is a generalization of Heron's method), draw inspiration from this ancient algorithm.. ### Program 1:. Let us take an example to illustrate the use of the Babylonian square root algorithm in C++:. Output:. ```Enter a number to find its square root: 25. Square root of 25 is approximately: 5. ```. Explanation:. 1. Initialization: Choose an initial guess (x0) for the square root. It can be any reasonable starting point, but common choices include x0 = N/2 or x0 =1, where N is the number for which you want to find the square root.. 2. Iteration:. • For each iteration (n), calculate the next approximation using the formula:. • This formula combines the current guess xn with the ratio N/xn to produce a new estimate that, on average, is closer to the actual square root.. 3. Convergence Check:. • Check the difference between consecutive approximations. The iteration continues until the difference is sufficiently small. The criterion for stopping the iteration is typically based on a specified tolerance level (epsilon, ε):. 4. Termination:. • Once the convergence criterion is met, terminate the iteration. The last obtained value (x n+1) is considered an approximation of the square root of N.. ### Program 2:. Let us take another example to illustrate the use of the Babylonian square root algorithm in C++:. Output:. ```Enter a number to find its square root: 64. Square root of 64 is approximately: 8. ```. Explanation:. • The code includes two standard C++ header files: <iostream> for input and output operations and <cmath> for mathematical functions.. 2. Babylonian Square Root Function:. • The function babylonianSquareRoot is defined to calculate the square root using the Babylonian method.. • It takes two parameters: n (the number for which the square root is to be calculated) and epsilon (a small positive value to control the precision of the result).. • The function checks if the input n is less than 0, in which case it prints an error message and returns a specified error code or throws an exception.. • If n is 0, it returns 0, as the square root of 0 is 0.. • The function initializes a variable guess as the initial guess (typically set to n / 2.0).. 3. Do-While Loop:. • The core of the algorithm is a do-while loop, which continues until the convergence criterion is met.. • Inside the loop, a new guess (newGuess) is calculated using the Babylonian formula.. • The loop checks if the absolute difference between the new guess and the old guess is less than the specified epsilon. If true, it breaks out of the loop, indicating convergence.. • If not, the new guess becomes the current guess, and the loop continues.. 4. Main Function:. • The main function is the entry point of the program.. • It prompts the user to enter a number and reads the input into the variable number.. • After that, it calls the babylonianSquareRoot function with the entered number, obtaining the result.. • If the result is not equal to the specified error code (-1.0), it prints the calculated square root.. • Return 0: The main function returns 0, indicating successful program execution.. Time and Space complexities:. Time Complexity:. • The number of iterations required for convergence primarily determines the time complexity of the Babylonian square root algorithm.. • Let k be the number of iterations needed for convergence, and ? be the tolerance level.. Time Complexity: The algorithm exhibits a logarithmic convergence behavior, and the number of iterations is usually relatively small and proportional to the precision required. Therefore, the time complexity is often considered as O(log( 1/? )).. In practical terms, the algorithm often converges in a fixed number of iterations, making it efficient.. Space Complexity:. • The space complexity of the algorithm is related to the storage of variables and the stack space used during the function calls.. Space Complexity: The algorithm uses a constant amount of space for variables, regardless of the input size. Therefore, the space complexity is considered as O(1), indicating constant space usage.. The space required for the variables (e.g., guess, newGuess, number, etc.) remains constant, and additional space is not dependent on the input size.. ### Advantages of the Babylonian square root algorithm:.	There are several advantages of the Babylonian square root algorithm in C++. Some main advantages of the Babylonian square root algorithm in C++:. • Fast Convergence: The algorithm converges rapidly, meaning that it approaches the actual square root with each iteration. It leads to quick and efficient approximations.. • Simplicity: The algorithm is straightforward and easy to understand. Its simplicity makes it an attractive choice for square root approximation, especially in educational contexts or applications where computational efficiency is not the primary concern.. • Low Computational Complexity: The algorithm involves basic arithmetic operations (addition, division, and multiplication), which have relatively low computational complexity. It makes it computationally efficient, especially when compared to more complex numerical methods.. • Applicability to Various Contexts: The Babylonian square root algorithm is applicable to a wide range of contexts, including hand calculations, programming environments, and embedded systems. Its versatility makes it suitable for various applications.. • Numerical Stability: The algorithm is numerically stable, and small variations in the input value or initial guess do not usually lead to divergent behavior. This stability is important in numerical methods to ensure reliable and consistent results.. • Historical Significance: The algorithm has historical significance, dating back to ancient civilizations. Its enduring use highlights its effectiveness and importance in the development of numerical methods throughout history.. ### Applications of the Babylonian square root algorithm:. There are several applications of the Babylonian square root algorithm in C++. Some main applications of the Babylonian square root algorithm in C++:. • Numerical Analysis: The algorithm is a fundamental component of numerical analysis, providing a simple and effective method for approximating square roots. It is often used as a starting point for more complex algorithms.. • Computer Programming: The Babylonian square root algorithm is commonly implemented in programming languages to calculate square roots. Many programming environments, including scientific and engineering applications, use this algorithm due to its simplicity and quick convergence.. • Calculator Implementations: The algorithm is used in the software of calculators to efficiently compute square roots. Its rapid convergence makes it suitable for real-time applications, providing users with quick and accurate results.. • Embedded Systems: In resource-constrained environments, such as embedded systems in electronic devices, the Babylonian square root algorithm can be preferred due to its low computational complexity and ease of implementation.. • Education: The algorithm is often used in educational settings to teach the concept of iterative methods for approximating mathematical functions. Its simplicity makes it accessible for students studying numerical methods.. • Financial Calculations: The algorithm can be used in financial calculations where quick approximations of square roots are needed. For example, it might be applied in risk assessments or option pricing models.. • Signal Processing: In some signal processing applications, where real-time calculations are crucial, the Babylonian square root algorithm can be employed for its efficiency in approximating square roots.. • Scientific Research: The algorithm is used in various scientific disciplines for quick and practical approximations. It is particularly valuable when high precision is not critical, and computational efficiency is essential.. While the Babylonian square root algorithm is versatile, it's important to note that in certain applications requiring very high precision or specific error analysis, more advanced algorithms like Newton's method might be preferred. Additionally, its historical significance and simplicity make it an interesting subject of study in the context of numerical methods.. ### Significance of The Babylonian square root algorithm in computer programming:. The Babylonian square root algorithm holds significant importance in computer programming due to its simplicity, efficiency, and widespread applicability. Its straightforward nature makes it an accessible choice for programmers at all skill levels, contributing to its adoption in various software applications. The algorithm strikes a balance between accuracy and speed, making it computationally efficient for a broad range of scenarios where square root calculations are required. Its common use case in programming, ranging from basic arithmetic to complex mathematical computations, underscores its practicality and versatility.. One of the algorithm's notable strengths lies in its numerical stability, ensuring consistent and reliable results even in the presence of small variations in input or initial guesses. This stability is a crucial attribute in programming, where robust and predictable behaviour is essential. Additionally, the historical continuity of the Babylonian square root algorithm adds to its significance. Originating from ancient civilizations, its enduring effectiveness has led to its continued use in modern computer programming.. In environments with resource constraints, such as embedded systems, the algorithm's relatively low computational complexity makes it a suitable choice. Its efficiency contributes to its popularity in scenarios where quick and reasonably accurate square root approximations are required.. ### Disadvantages of the Babylonian square root algorithm:. There are several disadvantages of the Babylonian square root algorithm in C++. Some main disadvantages of the Babylonian square root algorithm in C++:. • Convergence Rate in Certain Cases: While the algorithm generally converges rapidly, there are cases where the convergence rate might be slower, especially for numbers with peculiar properties or near-zero values. Other methods, like Newton's method, may have faster convergence in some situations.. • Initial Guess Dependency: The performance of the algorithm can be sensitive to the choice of the initial guess. In certain scenarios, a poorly chosen initial guess may lead to slower convergence or divergence. Choosing a good initial guess requires some knowledge of the properties of the input.. • Not Suitable for Negative Numbers: The algorithm is designed for positive real numbers. It cannot be directly applied to find the square root of negative numbers. Handling negative inputs may require additional considerations or a different approach.. • Not Suitable for Complex Numbers: The algorithm is limited to real numbers. If complex roots are needed, other methods specifically designed for complex numbers, such as the Newton-Raphson method for complex functions, would be more appropriate.. • Precision Limitations: The Babylonian algorithm may not be the most suitable choice for applications requiring extremely high precision. Other algorithms, such as iterative methods with higher-order convergence, might be preferred for such scenarios.. • Requires Division Operation: The algorithm involves a division operation in each iteration, which may be computationally expensive on certain platforms. In situations where division operations are costly, alternative methods might be more efficient.. • Handling Near-Zero Values: The algorithm may encounter difficulties when the input is very close to zero. In such cases, issues related to numerical precision and floating-point arithmetic may affect the accuracy of the result.. • Not the Most Efficient for Extreme Precision: For applications requiring extremely high precision, more advanced algorithms, such as specialized square root algorithms or arbitrary-precision arithmetic libraries, may outperform the Babylonian algorithm.. While the Babylonian square root algorithm is versatile and widely used, these disadvantages highlight situations where other algorithms or methods may be more suitable, depending on the specific requirements of an application. Understanding these limitations helps in making informed choices when selecting a square root approximation method.. ## Other Alternatives:. • Newton's Method (Newton-Raphson Method): Newton's method is an iterative numerical technique for finding roots of real-valued functions. It can be adapted to find square roots. Newton's method tends to converge faster than the Babylonian method, especially for functions with higher-order convergence.. • Binary Search Algorithm: The binary search algorithm can be adapted to find the square root by searching for the square root in a given range. It repeatedly bisects the search interval until the desired precision is achieved. Binary search is efficient but may take more iterations than methods with higher-order convergence.. • Exponential Function and Logarithm: Some mathematical libraries or hardware instructions provide specialized functions for exponentiation and logarithms that can be used to calculate square roots.. • CORDIC Algorithm: The Coordinate Rotation Digital Computer (CORDIC) algorithm is often used for trigonometric and hyperbolic function calculations, but it can be adapted to compute square roots. It uses a series of simple and fixed-point operations, making it suitable for hardware implementation.. ## Conclusion:. In conclusion, the Babylonian square root algorithm is a simple and effective method for approximating square roots. It has been used for centuries and remains a practical choice due to its ease of implementation and rapid convergence. The algorithm is suitable for a variety of applications, including numerical analysis, programming, and embedded systems.. ### Key points about the Babylonian square root algorithm:. • Iterative Nature: The algorithm iteratively refines its estimate of the square root, converging rapidly to the actual value.. • Simplicity: Its straightforward formula and minimal computational complexity make it accessible for educational purposes and practical implementations.. • Versatility: The algorithm can be applied to various contexts, ranging from hand calculations to computer programming and embedded systems.. • Competition: While other algorithms, like Newton's method offer faster convergence in some cases, the Babylonian method strikes a balance between simplicity and efficiency.. • Applications: The algorithm provides quick and satisfactory approximations for square roots, and it is widely used in fields such as numerical analysis, programming, and embedded systems.. While the Babylonian method is widely used, it's essential to consider its limitations, such as sensitivity to the initial guess and its suitability for specific precision requirements. In scenarios where higher precision or specialized considerations are needed, alternative algorithms like Newton's method or binary search may be preferred.. The implementation and enhancement of the Babylonian square root algorithm provide valuable insights into numerical methods, error handling, and user interface design. Exploring and understanding different square root algorithms broadens one's understanding of computational techniques and their applications.. In practical terms, the Babylonian square root algorithm remains a fundamental and efficient approach for square root approximation, and its simplicity makes it an accessible topic for learning and exploration in the realm of numerical computing.
https://icsehelp.com/llogarithms-rs-aggarwal-icse-class-9th-maths-goyal-brothers-prakashan/	1,611,486,721,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00780.warc.gz	377,399,092	24,637	# Logarithms RS Aggarwal ICSE Class-9th Maths Goyal Brothers Prakashan Logarithms RS Aggarwal ICSE Class-9th Maths Solutions Goyal Brothers Prakashan Chapter-7. We provide step by step Solutions of Exercise / lesson-7 Logarithms for ICSE Class-9 RS Aggarwal Mathematics . Our Solutions contain all type Questions with Exe-7 A, Exe-7 B with Questions to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics. ## Logarithms RS Aggarwal ICSE Class-9th Maths Solutions Goyal Brothers Prakashan Chapter-7 –: Select Topics :– Exe-7 A, Exe-7 B, ### Introduction Logarithms are the opposite phenomena of exponential like subtraction is the inverse of addition process, and division is the opposite phenomena of multiplication. Logs “undo” exponential. There are a number of rules known as the laws of logarithms. These allow expressions involving logarithms to be rewritten in a variety of different ways. The laws apply to logarithms of any base but the same base must be used throughout a calculation First Law log A + log B = log AB Second Law log A − log B = log (A/B) Third Law log An = n log A log 1 = 0, logmM = 1 ### Exercise- 7 A Logarithms RS Aggarwal ICSE Class-9th Maths Solutions Goyal Brothers Prakashan Question 1 convert each of the following into logarithm form ………………………….. #### Question 2 convert each of the following into exponential form #### Question 3 by converting into exponential form find value of each of following #### Question 4 find value of x when ……………………………. #### Question 5 if log…………. ……….show that……………. #### Question 6 given log….. ………… =b ………………. ……………….. ……………….. term of x and y ### RS Aggarwal ICSE Class-9th Maths Solutions Goyal Brothers Prakashan Exercise- 7 BLogarithms Question 1 evaluate the following without using tables ………….. ………… …………… ……………. ……………. evaluate ……………………….. #### Question 3 and Question 4 3- express ……………………. #### Question 5 express ………………….as a single logarithms #### Question 6 evaluate the following without using log tables …………. …………….. ……………. ………… #### Question 7 given ……………..find value of ………………….. …………………… ……………………. …………………… #### Question 8 if log 2 = 0.3010 find the value of …………….. #### Question 9 to Question 11 if log 8 = 0.9010 find the value of …………….. ……………. ……………… #### Question 12 if log (m+n) = log m+log n show that …………….. …………………. Question 13 solve for x ……………. ……………. ………….. …………… …………….. ………….. #### Question 14 if …………………find the value of n #### Question 15 write the logarithmic equations for ……………….. ………………….	708	2,648	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.375	4	CC-MAIN-2021-04	longest	en	0.763973	# Logarithms RS Aggarwal ICSE Class-9th Maths Goyal Brothers Prakashan. Logarithms RS Aggarwal ICSE Class-9th Maths Solutions Goyal Brothers Prakashan Chapter-7. We provide step by step Solutions of Exercise / lesson-7 Logarithms for ICSE Class-9 RS Aggarwal Mathematics .. Our Solutions contain all type Questions with Exe-7 A, Exe-7 B with Questions to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics.. ## Logarithms RS Aggarwal ICSE Class-9th Maths Solutions Goyal Brothers Prakashan Chapter-7. –: Select Topics :–. Exe-7 A,. Exe-7 B,. ### Introduction. Logarithms are the opposite phenomena of exponential like subtraction is the inverse of addition process, and division is the opposite phenomena of multiplication. Logs “undo” exponential.. There are a number of rules known as the laws of logarithms. These allow expressions involving. logarithms to be rewritten in a variety of different ways. The laws apply to logarithms of any base. but the same base must be used throughout a calculation. First Law. log A + log B = log AB. Second Law. log A − log B = log (A/B). Third Law log An = n log A. log 1 = 0, logmM = 1. ### Exercise- 7 A Logarithms RS Aggarwal ICSE Class-9th Maths Solutions Goyal Brothers Prakashan. Question 1. convert each of the following into logarithm form. …………………………... #### Question 2. convert each of the following into exponential form. #### Question 3. by converting into exponential form find value of each of following. #### Question 4. find value of x when. …………………………….. #### Question 5. if log…………. ……….show that…………….. #### Question 6. given log….. ………… =b. ……………….. ………………... ……………….. term of x and y. ### RS Aggarwal ICSE Class-9th Maths Solutions Goyal Brothers Prakashan Exercise- 7 BLogarithms. Question 1. evaluate the following without using tables. …………... ………….	……………. …………….. …………….. evaluate. ………………………... #### Question 3 and Question 4. 3- express …………………….. #### Question 5. express ………………….as a single logarithms. #### Question 6. evaluate the following without using log tables. ………….. ……………... …………….. …………. #### Question 7. given ……………..find value of. …………………... ……………………. …………………….. ……………………. #### Question 8. if log 2 = 0.3010 find the value of ……………... #### Question 9 to Question 11. if log 8 = 0.9010 find the value of. ……………... …………….. ………………. #### Question 12. if log (m+n) = log m+log n show that ……………... ………………….. Question 13. solve for x. …………….. …………….. …………... ……………. ……………... …………... #### Question 14. if …………………find the value of n. #### Question 15. write the logarithmic equations for. ………………... ………………….
jkeohane.wordpress.com	1,501,065,956,000,000,000	text/html	crawl-data/CC-MAIN-2017-30/segments/1500549426133.16/warc/CC-MAIN-20170726102230-20170726122230-00114.warc.gz	665,650,882	31,899	An Overview of Error Analysis When you report a measurement of something, you must also report the error in your measurement. For example, say you measured the surface gravity of the earth, at your location, to be 9.816 meters per second squared. Does that mean that you would get exactly the same value if you reproduced the experiment? Does it mean that it could be 9.817 meters per second squared? Does it mean that it could be 9.836 meters per second squared? As you see, this is completely unclear. On the other hand, had you reported a value of, say, 9.816±0.005 meters per second squared, your reader would know that there is about a 2/3 chance of the real value being 9.811 and 9.821 meters per second squared. In order to be able to report an error in your quantity of interest, you must also know the errors in each of your measurements. For example, say you had dropped an object and measured the time it took to fall a measured distance, then you would need to know the error in both the distance and the time to estimate the error. For example, if you dropped your object from a set height of 2 meters, you would need to report how well you knew this. For example, you may report something like: “The ball took 0.65±0.02 seconds to fall from a height of 2.000±0.002 meters.” That way the reader not only knows what you measured but also how well you measured it. There are two ways to estimate your error in a measurement, and you should make a habit of always doing the first, and doing the second when at all possible. These are: (1) common sense at the time of the measurement, and (2) statistically by taking multiple independent measurements. In the second method, the error in each measurement can be estimated by finding the standard deviation of the measurements. If you estimated your errors correctly, these should be similar. For example, say you dropped the ball from two meters above the floor with a ball dropper, you may notice that it could be off by a couple of millimeters, either way, so you report two millimeters as your error. That would be an example of method (1). On the other hand, say you did this 5 times and measured the fall times to be: 0.59, 0.61, 0.66, 0.70, and 0.68 seconds. Then you would surmise that the error in each measurement was about 0.05 seconds since the standard deviation is 0.0466 seconds. So, how long did it take for the ball to fall the two meters? If each of the errors were random, with no mistakes, then you would want to average them. If, on the other hand, you had reason to believe one was a mistake, you would throw out that datum. For example, say the measurements were instead: 0.59, 0.95, 0.66, 0.70, and 0.68 seconds, you could assume that you made a mistake taking the 0.95 datum, as it 0.36 seconds greater than the average and the standard deviation with it was only 0.22. That said, you must be careful, as you would not want to bias your measurements. Now, if your error were random, and the error in each one is known, what is the error in the average? Clearly it is less than it would have been had you only taken one measurement. But how much so? As it turns out the error in a sum is the square root of the sum of the errors, or in math notation: ${{\sigma }_{sum}}=\sqrt{\sum\limits_{i=1}^{N}{\sigma _{1}^{2}}}$. If all of the individual errors are the same, then this simply becomes: ${{\sigma }_{sum}}=\sqrt{N{{\sigma }^{2}}}=\sigma \sqrt{N}$. Since the average is simply the sum divided by the number of samples, then the error in the sum is given by: ${{\sigma }_{avg}}=\frac{{{\sigma }_{sum}}}{N}=\frac{\sigma }{\sqrt{N}}$. Thus, in our example, you would report that it took 0.65±0.02 seconds for the ball to fall a distance of 2.000±0.002 meters. However, you would keep all the insignificant digits on your spreadsheet or other data analysis software. Now that you know this, what is your value, and error in, g? Finding the best value is simple, we simply use the equation: $g=\frac{2h}{{{t}^{2}}}$. Plugging in the best values gives us 9.53 meters per second squared for the surface gravity. But, how well do we know this number? In order to calculate the error in g, you must test how sensitive your equation is to changes in our measured values. This is called propagation of errors, and you can do it by simply calculating what your quantity of interest (g) would be if each of your independent variables were at its maximum and minimum expected values. Once you have done this for each of independent variables, holding the others constant, you must add up the errors in quadrature using the Pythagorean theorem. In our example, we begin by calculating the error in g, only because of the error in the height. Keeping the time fixed at its best value, 0.648 s, and calculating g with h=1.998 m and 2.002 m, we get corresponding surface gravities of 9.52 and 9.54 meters per second squared respectively. Thus, the error in g, only due to the error in the height, is half the difference, or 0.01 meters per second squared. But what about the error in the time? Doing the same calculations, but keeping h=2.000 meters and allowing the time to be 0.627 and 0.669 seconds, gives us 9.52 and 9.54 meters per second squared respectively for g. Thus, the error, only due to the error in the time, is 0.61 meters per second squared. As this is much bigger than the other, it clearly dominates.** To find the error due to both, we add them in quadrature, like the way we found the error in the sum above. Thus, in this case: ${{\sigma }_{g}}=\sqrt{\sum\limits_{i=1}^{N}{\sigma _{1}^{2}}}=\sqrt{{{\left( 0.61{\scriptstyle{}^{m}\!\!\diagup\!\!{}_{{{s}^{2}}}\;} \right)}^{2}}+{{\left( 0.01{\scriptstyle{}^{m}\!\!\diagup\!\!{}_{{{s}^{2}}}\;} \right)}^{2}}}=0.61{\scriptstyle{}^{m}\!\!\diagup\!\!{}_{{{s}^{2}}}\;}$. Finally, you would report the following as your conclusion: “The gravitational field of the Earth, at my location, is 9.5±0.6 meters per second squared.” Notice that we really only care about the error to one significant figure, and we round off the numbers accordingly so as not to distract the reader. But, what is the REAL value of g? Is it not 9.81 meters per second squared, or whatever my physics book says? The answer is that you do not know, and you should not pretend that you do. It is a good idea to compare your value of g to someone else’s measurement to see if they are consistent, but make sure you cite your source. This is very important. By the way, in physics, it is customary to cite the paper of whoever actually made the measurement, even if you actually found the value in a secondary source like an encyclopedia. If you are interested in the best measurements of g to date, you can read this review article about how geologists measure it to about 10 digits of accuracy in order to model the internal mass distribution of the earth. If you have studied differential calculus, it may be easier to use it to propagate the errors, however you need to understand that this is no more correct than the way described above. So, if you have not studied calculus yet, there is no need to worry or to read further. Recall that in order to calculate the error in what you want, you had to test how changes in each of your measured values (independent variables) affect the quantity of interest (dependent variable). We advised you to accomplish this by numerically calculating the quantity of interest in the maximum and minimum of each error domain (holding the other independent variables constant), giving you a range of resulting values. For relatively small error bars, you can assume that the average slope over your error range is simply the derivative of your function at the measured point. Mathematically, let the independent variable $g$ be written in function notation of two variables, $g = g\left(h,t\right)$. Therefore we can rewrite the error due only to the error in h as: $\sigma_{g_h} = \frac{1}{2} \left( g\left( h+\sigma_h,t \right) - g \left( h - \sigma_h,t \right) \right) = \left( \frac{ g\left( h+\sigma_h,t \right) - g \left( h - \sigma_h,t \right)}{2 \sigma_h} \right) \sigma_h = \frac{\Delta g}{\Delta h} \sigma_h$ This simplifies to: $\sigma_{g_h} \approx \frac{\partial g}{\partial h} \sigma_h$ . The symbol $\partial$ is just like a d in calculus, but warns you that there is at least one other independent variable, which we are holding constant. This is called a partial derivative. Thus, in our example, the error due to both independent variables, is given by: ${{\sigma }_{g}}=\sqrt{{{\left( \frac{\partial g}{\partial h}{{\sigma }_{h}} \right)}^{2}}+{{\left( \frac{\partial g}{\partial t}{{\sigma }_{t}} \right)}^{2}}}=\sqrt{{{\left( \frac{2{{\sigma }_{h}}}{{{t}^{2}}} \right)}^{2}}+{{\left( -\frac{4h{{\sigma }_{t}}}{{{t}^{3}}} \right)}^{2}}}$. Now, factoring out a factor of g, we get: ${{\sigma }_{g}}=\sqrt{{{\left( \frac{{{\sigma }_{h}}}{h}g \right)}^{2}}+{{\left( -\frac{2{{\sigma }_{t}}}{t}g \right)}^{2}}}=g\sqrt{{{\left( \frac{{{\sigma }_{h}}}{h} \right)}^{2}}+{{\left( \frac{2{{\sigma }_{t}}}{t} \right)}^{2}}}$. Why is this easier? It is easier because you can calculate this once, and then have a single formula. Plus, it is very powerful. For example, as an exercise, use this formula to calculate the error in the sum of a bunch of quantities, and compare that to the error in the sum that we have above. Similarly, you can use this to calculate the error in the average of a bunch of data each with its own error bar.	2,609	9,581	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.0625	4	CC-MAIN-2017-30	longest	en	0.946816	An Overview of Error Analysis. When you report a measurement of something, you must also report the error in your measurement. For example, say you measured the surface gravity of the earth, at your location, to be 9.816 meters per second squared. Does that mean that you would get exactly the same value if you reproduced the experiment? Does it mean that it could be 9.817 meters per second squared? Does it mean that it could be 9.836 meters per second squared? As you see, this is completely unclear. On the other hand, had you reported a value of, say, 9.816±0.005 meters per second squared, your reader would know that there is about a 2/3 chance of the real value being 9.811 and 9.821 meters per second squared.. In order to be able to report an error in your quantity of interest, you must also know the errors in each of your measurements. For example, say you had dropped an object and measured the time it took to fall a measured distance, then you would need to know the error in both the distance and the time to estimate the error. For example, if you dropped your object from a set height of 2 meters, you would need to report how well you knew this. For example, you may report something like: “The ball took 0.65±0.02 seconds to fall from a height of 2.000±0.002 meters.” That way the reader not only knows what you measured but also how well you measured it.. There are two ways to estimate your error in a measurement, and you should make a habit of always doing the first, and doing the second when at all possible. These are: (1) common sense at the time of the measurement, and (2) statistically by taking multiple independent measurements. In the second method, the error in each measurement can be estimated by finding the standard deviation of the measurements. If you estimated your errors correctly, these should be similar.. For example, say you dropped the ball from two meters above the floor with a ball dropper, you may notice that it could be off by a couple of millimeters, either way, so you report two millimeters as your error. That would be an example of method (1). On the other hand, say you did this 5 times and measured the fall times to be: 0.59, 0.61, 0.66, 0.70, and 0.68 seconds. Then you would surmise that the error in each measurement was about 0.05 seconds since the standard deviation is 0.0466 seconds.. So, how long did it take for the ball to fall the two meters? If each of the errors were random, with no mistakes, then you would want to average them. If, on the other hand, you had reason to believe one was a mistake, you would throw out that datum. For example, say the measurements were instead: 0.59, 0.95, 0.66, 0.70, and 0.68 seconds, you could assume that you made a mistake taking the 0.95 datum, as it 0.36 seconds greater than the average and the standard deviation with it was only 0.22. That said, you must be careful, as you would not want to bias your measurements.. Now, if your error were random, and the error in each one is known, what is the error in the average? Clearly it is less than it would have been had you only taken one measurement. But how much so? As it turns out the error in a sum is the square root of the sum of the errors, or in math notation: ${{\sigma }_{sum}}=\sqrt{\sum\limits_{i=1}^{N}{\sigma _{1}^{2}}}$. If all of the individual errors are the same, then this simply becomes: ${{\sigma }_{sum}}=\sqrt{N{{\sigma }^{2}}}=\sigma \sqrt{N}$. Since the average is simply the sum divided by the number of samples, then the error in the sum is given by:. ${{\sigma }_{avg}}=\frac{{{\sigma }_{sum}}}{N}=\frac{\sigma }{\sqrt{N}}$.. Thus, in our example, you would report that it took 0.65±0.02 seconds for the ball to fall a distance of 2.000±0.002 meters. However, you would keep all the insignificant digits on your spreadsheet or other data analysis software.. Now that you know this, what is your value, and error in, g? Finding the best value is simple, we simply use the equation: $g=\frac{2h}{{{t}^{2}}}$. Plugging in the best values gives us 9.53 meters per second squared for the surface gravity. But, how well do we know this number?. In order to calculate the error in g, you must test how sensitive your equation is to changes in our measured values. This is called propagation of errors, and you can do it by simply calculating what your quantity of interest (g) would be if each of your independent variables were at its maximum and minimum expected values. Once you have done this for each of independent variables, holding the others constant, you must add up the errors in quadrature using the Pythagorean theorem.. In our example, we begin by calculating the error in g, only because of the error in the height.	Keeping the time fixed at its best value, 0.648 s, and calculating g with h=1.998 m and 2.002 m, we get corresponding surface gravities of 9.52 and 9.54 meters per second squared respectively. Thus, the error in g, only due to the error in the height, is half the difference, or 0.01 meters per second squared. But what about the error in the time? Doing the same calculations, but keeping h=2.000 meters and allowing the time to be 0.627 and 0.669 seconds, gives us 9.52 and 9.54 meters per second squared respectively for g. Thus, the error, only due to the error in the time, is 0.61 meters per second squared. As this is much bigger than the other, it clearly dominates.**. To find the error due to both, we add them in quadrature, like the way we found the error in the sum above. Thus, in this case: ${{\sigma }_{g}}=\sqrt{\sum\limits_{i=1}^{N}{\sigma _{1}^{2}}}=\sqrt{{{\left( 0.61{\scriptstyle{}^{m}\!\!\diagup\!\!{}_{{{s}^{2}}}\;} \right)}^{2}}+{{\left( 0.01{\scriptstyle{}^{m}\!\!\diagup\!\!{}_{{{s}^{2}}}\;} \right)}^{2}}}=0.61{\scriptstyle{}^{m}\!\!\diagup\!\!{}_{{{s}^{2}}}\;}$.. Finally, you would report the following as your conclusion: “The gravitational field of the Earth, at my location, is 9.5±0.6 meters per second squared.” Notice that we really only care about the error to one significant figure, and we round off the numbers accordingly so as not to distract the reader.. But, what is the REAL value of g? Is it not 9.81 meters per second squared, or whatever my physics book says? The answer is that you do not know, and you should not pretend that you do. It is a good idea to compare your value of g to someone else’s measurement to see if they are consistent, but make sure you cite your source. This is very important. By the way, in physics, it is customary to cite the paper of whoever actually made the measurement, even if you actually found the value in a secondary source like an encyclopedia.. If you are interested in the best measurements of g to date, you can read this review article about how geologists measure it to about 10 digits of accuracy in order to model the internal mass distribution of the earth.. If you have studied differential calculus, it may be easier to use it to propagate the errors, however you need to understand that this is no more correct than the way described above. So, if you have not studied calculus yet, there is no need to worry or to read further.. Recall that in order to calculate the error in what you want, you had to test how changes in each of your measured values (independent variables) affect the quantity of interest (dependent variable). We advised you to accomplish this by numerically calculating the quantity of interest in the maximum and minimum of each error domain (holding the other independent variables constant), giving you a range of resulting values.. For relatively small error bars, you can assume that the average slope over your error range is simply the derivative of your function at the measured point. Mathematically, let the independent variable $g$ be written in function notation of two variables, $g = g\left(h,t\right)$. Therefore we can rewrite the error due only to the error in h as:. $\sigma_{g_h} = \frac{1}{2} \left( g\left( h+\sigma_h,t \right) - g \left( h - \sigma_h,t \right) \right) = \left( \frac{ g\left( h+\sigma_h,t \right) - g \left( h - \sigma_h,t \right)}{2 \sigma_h} \right) \sigma_h = \frac{\Delta g}{\Delta h} \sigma_h$. This simplifies to: $\sigma_{g_h} \approx \frac{\partial g}{\partial h} \sigma_h$ .. The symbol $\partial$ is just like a d in calculus, but warns you that there is at least one other independent variable, which we are holding constant. This is called a partial derivative. Thus, in our example, the error due to both independent variables, is given by:. ${{\sigma }_{g}}=\sqrt{{{\left( \frac{\partial g}{\partial h}{{\sigma }_{h}} \right)}^{2}}+{{\left( \frac{\partial g}{\partial t}{{\sigma }_{t}} \right)}^{2}}}=\sqrt{{{\left( \frac{2{{\sigma }_{h}}}{{{t}^{2}}} \right)}^{2}}+{{\left( -\frac{4h{{\sigma }_{t}}}{{{t}^{3}}} \right)}^{2}}}$.. Now, factoring out a factor of g, we get:. ${{\sigma }_{g}}=\sqrt{{{\left( \frac{{{\sigma }_{h}}}{h}g \right)}^{2}}+{{\left( -\frac{2{{\sigma }_{t}}}{t}g \right)}^{2}}}=g\sqrt{{{\left( \frac{{{\sigma }_{h}}}{h} \right)}^{2}}+{{\left( \frac{2{{\sigma }_{t}}}{t} \right)}^{2}}}$.. Why is this easier? It is easier because you can calculate this once, and then have a single formula. Plus, it is very powerful. For example, as an exercise, use this formula to calculate the error in the sum of a bunch of quantities, and compare that to the error in the sum that we have above.. Similarly, you can use this to calculate the error in the average of a bunch of data each with its own error bar.
http://www.britannica.com/EBchecked/topic/14885/algebra/231075/Galoiss-work-on-permutations	1,429,622,488,000,000,000	text/html	crawl-data/CC-MAIN-2015-18/segments/1429246641468.77/warc/CC-MAIN-20150417045721-00307-ip-10-235-10-82.ec2.internal.warc.gz	390,050,461	17,497	# Algebra Mathematics ## Galois’s work on permutations Prominent among Galois’s seminal ideas was the clear realization of how to formulate precise solvability conditions for a polynomial in terms of the properties of its group of permutations. A permutation of a set, say the elements a, b, and c, is any re-ordering of the elements, and it is usually denoted as follows: This particular permutation takes a to c, b to a, and c to b. For three elements, as here, there are six different possible permutations. In general, for n elements there are n! permutations to choose from. (Where n! = n(n − 1)(n − 2)⋯2∙1.) Furthermore, two permutations can be combined to produce a third permutation in an operation known as composition. (The set of permutations are closed under the operation of composition.) For example, Here a goes first to c (in the first permutation) and then from c to b (in the second permutation), which is equivalent to a going directly to b, as given by the permutation to the right of the equation. Composition is associative—given three permutations P, Q, and R, then (P * Q) * R = P * (Q * R). Also, there exists an identity permutation that leaves the elements unchanged: Finally, for each permutation there exists another permutation, known as its inverse, such that their composition results in the identity permutation. The set of permutations for n elements is known as the symmetric group Sn. The concept of an abstract group developed somewhat later. It consisted of a set of abstract elements with an operation defined on them such that the conditions given above were satisfied: closure, associativity, an identity element, and an inverse element for each element in the set. This abstract notion is not fully present in Galois’s work. Like some of his predecessors, Galois focused on the permutation group of the roots of an equation. Through some beautiful and highly original mathematical ideas, Galois showed that a general polynomial equation was solvable by radicals if and only if its associated symmetric group was “soluble.” Galois’s result, it must be stressed, referred to conditions for a solution to exist; it did not provide a way to calculate radical solutions in those cases where they existed. ## Acceptance of Galois theory Galois’s work was both the culmination of a main line of algebra—solving equations by radical methods—and the beginning of a new line—the study of abstract structures. Work on permutations, started by Lagrange and Ruffini, received further impetus in 1815 from the leading French mathematician, Augustin-Louis Cauchy. In a later work of 1844, Cauchy systematized much of this knowledge and introduced basic concepts. For instance, the permutation was denoted by Cauchy in cycle notation as (ab)(ced), meaning that the permutation was obtained by the disjoint cycles a to b (and back to a) and c to e to d (and back to c). A series of unusual and unfortunate events involving the most important contemporary French mathematicians prevented Galois’s ideas from being published for a long time. It was not until 1846 that Joseph Liouville edited and published for the first time, in his prestigious Journal de Mathématiques Pures et Appliquées, the important memoire in which Galois had presented his main ideas and that the Paris Academy had turned down in 1831. In Germany, Leopold Kronecker applied some of these ideas to number theory in 1853, and Richard Dedekind lectured on Galois theory in 1856. At this time, however, the impact of the theory was still minimal. A major turning point came with the publication of Traité des substitutions et des équations algebriques (1870; “Treatise on Substitutions and Algebraic Equations”) by the French mathematician Camille Jordan. In his book and papers, Jordan elaborated an abstract theory of permutation groups, with algebraic equations merely serving as an illustrative application of the theory. In particular, Jordan’s treatise was the first group theory book and it served as the foundation for the conception of Galois theory as the study of the interconnections between extensions of fields and the related Galois groups of equations—a conception that proved fundamental for developing a completely new abstract approach to algebra in the 1920s. Major contributions to the development of this point of view for Galois theory came variously from Enrico Betti (1823–92) in Italy and from Dedekind, Henrich Weber (1842–1913), and Emil Artin (1898–1962) in Germany. ## Applications of group theory Galois theory arose in direct connection with the study of polynomials, and thus the notion of a group developed from within the mainstream of classical algebra. However, it also found important applications in other mathematical disciplines throughout the 19th century, particularly geometry and number theory. ### Keep exploring What made you want to look up algebra? (Please limit to 900 characters) Please select the sections you want to print MLA style: "algebra". Encyclopædia Britannica. Encyclopædia Britannica Online. Encyclopædia Britannica Inc., 2015. Web. 21 Apr. 2015 <http://www.britannica.com/EBchecked/topic/14885/algebra/231075/Galoiss-work-on-permutations>. APA style: algebra. (2015). In Encyclopædia Britannica. Retrieved from http://www.britannica.com/EBchecked/topic/14885/algebra/231075/Galoiss-work-on-permutations Harvard style: algebra. 2015. Encyclopædia Britannica Online. Retrieved 21 April, 2015, from http://www.britannica.com/EBchecked/topic/14885/algebra/231075/Galoiss-work-on-permutations Chicago Manual of Style: Encyclopædia Britannica Online, s. v. "algebra", accessed April 21, 2015, http://www.britannica.com/EBchecked/topic/14885/algebra/231075/Galoiss-work-on-permutations. While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Click anywhere inside the article to add text or insert superscripts, subscripts, and special characters. You can also highlight a section and use the tools in this bar to modify existing content: Editing Tools: We welcome suggested improvements to any of our articles. You can make it easier for us to review and, hopefully, publish your contribution by keeping a few points in mind: 1. Encyclopaedia Britannica articles are written in a neutral, objective tone for a general audience. 2. You may find it helpful to search within the site to see how similar or related subjects are covered. 3. Any text you add should be original, not copied from other sources. 4. At the bottom of the article, feel free to list any sources that support your changes, so that we can fully understand their context. (Internet URLs are best.) Your contribution may be further edited by our staff, and its publication is subject to our final approval. Unfortunately, our editorial approach may not be able to accommodate all contributions. MEDIA FOR: algebra Citation • MLA • APA • Harvard • Chicago Email You have successfully emailed this. Error when sending the email. Try again later. Or click Continue to submit anonymously:	1,610	7,153	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.859375	4	CC-MAIN-2015-18	latest	en	0.946276	# Algebra. Mathematics. ## Galois’s work on permutations. Prominent among Galois’s seminal ideas was the clear realization of how to formulate precise solvability conditions for a polynomial in terms of the properties of its group of permutations. A permutation of a set, say the elements a, b, and c, is any re-ordering of the elements, and it is usually denoted as follows:. This particular permutation takes a to c, b to a, and c to b. For three elements, as here, there are six different possible permutations. In general, for n elements there are n! permutations to choose from. (Where n! = n(n − 1)(n − 2)⋯2∙1.) Furthermore, two permutations can be combined to produce a third permutation in an operation known as composition. (The set of permutations are closed under the operation of composition.) For example,. Here a goes first to c (in the first permutation) and then from c to b (in the second permutation), which is equivalent to a going directly to b, as given by the permutation to the right of the equation. Composition is associative—given three permutations P, Q, and R, then (P * Q) * R = P * (Q * R). Also, there exists an identity permutation that leaves the elements unchanged:. Finally, for each permutation there exists another permutation, known as its inverse, such that their composition results in the identity permutation. The set of permutations for n elements is known as the symmetric group Sn.. The concept of an abstract group developed somewhat later. It consisted of a set of abstract elements with an operation defined on them such that the conditions given above were satisfied: closure, associativity, an identity element, and an inverse element for each element in the set.. This abstract notion is not fully present in Galois’s work. Like some of his predecessors, Galois focused on the permutation group of the roots of an equation. Through some beautiful and highly original mathematical ideas, Galois showed that a general polynomial equation was solvable by radicals if and only if its associated symmetric group was “soluble.” Galois’s result, it must be stressed, referred to conditions for a solution to exist; it did not provide a way to calculate radical solutions in those cases where they existed.. ## Acceptance of Galois theory. Galois’s work was both the culmination of a main line of algebra—solving equations by radical methods—and the beginning of a new line—the study of abstract structures. Work on permutations, started by Lagrange and Ruffini, received further impetus in 1815 from the leading French mathematician, Augustin-Louis Cauchy. In a later work of 1844, Cauchy systematized much of this knowledge and introduced basic concepts. For instance, the permutation. was denoted by Cauchy in cycle notation as (ab)(ced), meaning that the permutation was obtained by the disjoint cycles a to b (and back to a) and c to e to d (and back to c).. A series of unusual and unfortunate events involving the most important contemporary French mathematicians prevented Galois’s ideas from being published for a long time. It was not until 1846 that Joseph Liouville edited and published for the first time, in his prestigious Journal de Mathématiques Pures et Appliquées, the important memoire in which Galois had presented his main ideas and that the Paris Academy had turned down in 1831. In Germany, Leopold Kronecker applied some of these ideas to number theory in 1853, and Richard Dedekind lectured on Galois theory in 1856. At this time, however, the impact of the theory was still minimal.. A major turning point came with the publication of Traité des substitutions et des équations algebriques (1870; “Treatise on Substitutions and Algebraic Equations”) by the French mathematician Camille Jordan. In his book and papers, Jordan elaborated an abstract theory of permutation groups, with algebraic equations merely serving as an illustrative application of the theory. In particular, Jordan’s treatise was the first group theory book and it served as the foundation for the conception of Galois theory as the study of the interconnections between extensions of fields and the related Galois groups of equations—a conception that proved fundamental for developing a completely new abstract approach to algebra in the 1920s. Major contributions to the development of this point of view for Galois theory came variously from Enrico Betti (1823–92) in Italy and from Dedekind, Henrich Weber (1842–1913), and Emil Artin (1898–1962) in Germany.. ## Applications of group theory. Galois theory arose in direct connection with the study of polynomials, and thus the notion of a group developed from within the mainstream of classical algebra. However, it also found important applications in other mathematical disciplines throughout the 19th century, particularly geometry and number theory.. ### Keep exploring. What made you want to look up algebra?. (Please limit to 900 characters). Please select the sections you want to print. MLA style:. "algebra". Encyclopædia Britannica. Encyclopædia Britannica Online.. Encyclopædia Britannica Inc., 2015. Web.	21 Apr. 2015. <http://www.britannica.com/EBchecked/topic/14885/algebra/231075/Galoiss-work-on-permutations>.. APA style:. algebra. (2015). In Encyclopædia Britannica. Retrieved from http://www.britannica.com/EBchecked/topic/14885/algebra/231075/Galoiss-work-on-permutations. Harvard style:. algebra. 2015. Encyclopædia Britannica Online. Retrieved 21 April, 2015, from http://www.britannica.com/EBchecked/topic/14885/algebra/231075/Galoiss-work-on-permutations. Chicago Manual of Style:. Encyclopædia Britannica Online, s. v. "algebra", accessed April 21, 2015, http://www.britannica.com/EBchecked/topic/14885/algebra/231075/Galoiss-work-on-permutations.. While every effort has been made to follow citation style rules, there may be some discrepancies.. Please refer to the appropriate style manual or other sources if you have any questions.. Click anywhere inside the article to add text or insert superscripts, subscripts, and special characters.. You can also highlight a section and use the tools in this bar to modify existing content:. Editing Tools:. We welcome suggested improvements to any of our articles.. You can make it easier for us to review and, hopefully, publish your contribution by keeping a few points in mind:. 1. Encyclopaedia Britannica articles are written in a neutral, objective tone for a general audience.. 2. You may find it helpful to search within the site to see how similar or related subjects are covered.. 3. Any text you add should be original, not copied from other sources.. 4. At the bottom of the article, feel free to list any sources that support your changes, so that we can fully understand their context. (Internet URLs are best.). Your contribution may be further edited by our staff, and its publication is subject to our final approval. Unfortunately, our editorial approach may not be able to accommodate all contributions.. MEDIA FOR:. algebra. Citation. • MLA. • APA. • Harvard. • Chicago. Email. You have successfully emailed this.. Error when sending the email. Try again later.. Or click Continue to submit anonymously:.
https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/thrust-equations-summary/	1,702,122,043,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100909.82/warc/CC-MAIN-20231209103523-20231209133523-00242.warc.gz	1,199,343,604	18,674	# Thrust Equations Summary ## Thrust On this slide, we have collected all of the equations necessary to calculate the thrust of a rocket engine. In a rocket engine, stored fuel and stored oxidizer are ignited in a combustion chamber. The combustion produces great amounts of exhaust gas at high temperature and pressure. The hot exhaust is passed through a nozzle which accelerates the flow. Thrust is produced according to Newton’s third law of motion. ### What determines the amount of Thrust? The amount of thrust produced by the rocket depends on the mass flow rate through the engine, the exit velocity of the exhaust, and the pressure at the nozzle exit. All of these variables depend on the design of the nozzle. The smallest cross-sectional area of the nozzle is called the throat of the nozzle. The hot exhaust flow is choked at the throat, which means that the Mach number is equal to 1.0 in the throat and the mass flow rate m dot ($$\bf \dot{m}$$) is determined by the throat area. $$\LARGE \dot{m}=\frac{A^{}p_{t}}{\sqrt{T_{t}}}\sqrt{\frac{\gamma}{R}}(\frac{\gamma+1}{2})^{-\frac{\gamma+1}{2(\gamma-1)}}$$ where A is the area of the throat, pt is the total pressure in the combustion chamber, Tt is the total temperature in the combustion chamber, gam ($$\bf \gamma$$) is the ratio of specific heats of the exhaust, and R is the gas constant. The area ratio from the throat to the exit Ae sets the exit Mach number: $$\LARGE \frac{A_{e}}{A^{*}}=(\frac{\gamma+1}{2})^{-\frac{\gamma+1}{2(\gamma-1)}}\frac{(1+\frac{\gamma-1}{2}M_{e}^{2})^{\frac{\gamma+1}{2(\gamma-1)}}}{M_{e}}$$ We can determine the exit pressure pe and exit temperature Te from the isentropic relations at the nozzle exit: $$\LARGE \frac{p_{e}}{p_{t}}=(1+\frac{\gamma-1}{2}M_{e}^{2})^{-\frac{\gamma}{\gamma-1}}$$ $$\LARGE \frac{T_{e}}{T_{t}}=(1+\frac{\gamma-1}{2}M_{e}^{2})^{-1}$$ Knowing Te we can use the equation for the speed of sound and the definition of the Mach number to calculate the exit velocity Ve: $$\LARGE V_{e}=M_{e}\sqrt{\gamma RT_{e}}$$ We now have all the information necessary to determine the thrust of a rocket. The exit pressure is only equal to free stream pressure at some design condition. We must, therefore, use the longer version of the generalized thrust equation to describe the thrust of the system. If the free stream pressure is given by p0, the rocket thrust equation is given by: $$\LARGE F=\dot{m}V_{e}+A_{e}(p_{e}-p_{0})$$ The thrust equation shown above works for both liquid rocket and solid rocket engines. There is also an efficiency parameter called the specific impulse which works for both types of rockets and greatly simplifies the performance analysis for rockets. You can explore the design and operation of a rocket nozzle with our interactive thrust simulator program which runs on your browser. Provide feedback	810	2,861	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2023-50	latest	en	0.607582	# Thrust Equations Summary. ## Thrust. On this slide, we have collected all of the equations necessary to calculate the thrust of a rocket engine. In a rocket engine, stored fuel and stored oxidizer are ignited in a combustion chamber. The combustion produces great amounts of exhaust gas at high temperature and pressure. The hot exhaust is passed through a nozzle which accelerates the flow. Thrust is produced according to Newton’s third law of motion.. ### What determines the amount of Thrust?. The amount of thrust produced by the rocket depends on the mass flow rate through the engine, the exit velocity of the exhaust, and the pressure at the nozzle exit. All of these variables depend on the design of the nozzle. The smallest cross-sectional area of the nozzle is called the throat of the nozzle. The hot exhaust flow is choked at the throat, which means that the Mach number is equal to 1.0 in the throat and the mass flow rate m dot ($$\bf \dot{m}$$) is determined by the throat area.. $$\LARGE \dot{m}=\frac{A^{}p_{t}}{\sqrt{T_{t}}}\sqrt{\frac{\gamma}{R}}(\frac{\gamma+1}{2})^{-\frac{\gamma+1}{2(\gamma-1)}}$$. where A is the area of the throat, pt is the total pressure in the combustion chamber, Tt is the total temperature in the combustion chamber, gam ($$\bf \gamma$$) is the ratio of specific heats of the exhaust, and R is the gas constant.. The area ratio from the throat to the exit Ae sets the exit Mach number:. $$\LARGE \frac{A_{e}}{A^{*}}=(\frac{\gamma+1}{2})^{-\frac{\gamma+1}{2(\gamma-1)}}\frac{(1+\frac{\gamma-1}{2}M_{e}^{2})^{\frac{\gamma+1}{2(\gamma-1)}}}{M_{e}}$$.	We can determine the exit pressure pe and exit temperature Te from the isentropic relations at the nozzle exit:. $$\LARGE \frac{p_{e}}{p_{t}}=(1+\frac{\gamma-1}{2}M_{e}^{2})^{-\frac{\gamma}{\gamma-1}}$$. $$\LARGE \frac{T_{e}}{T_{t}}=(1+\frac{\gamma-1}{2}M_{e}^{2})^{-1}$$. Knowing Te we can use the equation for the speed of sound and the definition of the Mach number to calculate the exit velocity Ve:. $$\LARGE V_{e}=M_{e}\sqrt{\gamma RT_{e}}$$. We now have all the information necessary to determine the thrust of a rocket. The exit pressure is only equal to free stream pressure at some design condition. We must, therefore, use the longer version of the generalized thrust equation to describe the thrust of the system. If the free stream pressure is given by p0, the rocket thrust equation is given by:. $$\LARGE F=\dot{m}V_{e}+A_{e}(p_{e}-p_{0})$$. The thrust equation shown above works for both liquid rocket and solid rocket engines. There is also an efficiency parameter called the specific impulse which works for both types of rockets and greatly simplifies the performance analysis for rockets.. You can explore the design and operation of a rocket nozzle with our interactive thrust simulator program which runs on your browser.. Provide feedback.
https://www.superteacherworksheets.com/multiplication-by-7s.html	1,719,327,999,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00317.warc.gz	895,193,063	17,700	# Multiplying by 7 Here you'll find basic multiplication facts up to 7, as well as worksheets with only 7s. We also have skip counting practice sheets as well. ## Multiplication by7s Only This worksheet has a skip counting number line, missing factor problems, a multiplication wheel, comparing questions, and a number table. Multiply by 7s and write the answers. Then use the key at the bottom to determine how each section should be colored. Mystery picture is a colorful castle with flags. Practice multiplying by 7 with these task cards, which can be used for various activities such as peer practice, scavenger hunts, class games, and more. One way to have students use this puzzle is to "find their partner". Give each student one puzzle piece. Half of the students will have a multiplication fact and the other half will have the answers. Have them find their "partner" with the matching piece. In this printable math activity, children will multiply by 7s in order to fill in the blank bubbles on the multiplication caterpillar. Cut and fold to make an origami fortune teller game (aka cootie catcher). Learn and practice basic facts with factors of 7. Practice multiplication facts with 7 as a factor. This file includes 11 flash cards, 5 super challenge cards, a sorting mat, and a self-quiz. Print this file on heavy card stock. The cut out the airplane picture and the math fact strips. Place the fact strips through the airplane and pull to scroll up and down. Students read the facts and say the answers. ## Skip Count by 7s Answer the critical thinking questions and solve the word problems. 0, 7, 14,... what is next? Skip count by 7s along the number line to find out. Count by 7s to complete the dot-to-dot. The picture is a pig. Then solve the multiplication facts at the bottom of the page. Count by 7s. Write the numbers on the penguins' bellies. Numbers go all the way up to 161. Count by 7s and write the missing numbers on the balloons. These numbers go up to 168. Complete the grids by filling in the missing numbers. This worksheet has a monkey theme. ## All Facts0s through 7s On this drill worksheet, your students will review multiplication facts with factors of 0, 1, 2, 3, 4, 5, 6, and 7. Facts through 10s, up to 7x10. This drill worksheet has all basic facts for multiplying by 0, 1, 2, 3, 4, 5, 6, and 7. Facts include 11s and 12s, up to 7x12. This timed quiz has 50 problems on it. Covers all basic facts for 0s, 1s, 2s, 3s, 4s, 5s, 6s, and 7s. Facts through 12, up to 7x12. Multiplication Basic Facts: 0-10 Here you'll find lots more multiplication worksheets, games, and activities for all basic facts up to 10x10. Multiplication/Division Fact Families This page has lots of worksheets with multiplication and division fact families. Multiplication by 3s Basic facts with factors of 3 are the focus on these printable learning resources. Multiplication by 4s These worksheets focus specifically on the 4s times tables. Multiplication by 5s Here are the activities for learning multiplication facts from 5x0 up to 5x12. Multiplication by 6s Multiplication by 8s Practice multiplying by the number 8. Includes mystery pictures, skip counting lessons, and more. Multiplication by 9s ## Pictures of Our Worksheets Not a member yet? Join Today! My Account Site Information Social Media	828	3,339	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.25	4	CC-MAIN-2024-26	latest	en	0.939914	# Multiplying by 7. Here you'll find basic multiplication facts up to 7, as well as worksheets with only 7s. We also have skip counting practice sheets as well.. ## Multiplication by7s Only. This worksheet has a skip counting number line, missing factor problems, a multiplication wheel, comparing questions, and a number table.. Multiply by 7s and write the answers. Then use the key at the bottom to determine how each section should be colored. Mystery picture is a colorful castle with flags.. Practice multiplying by 7 with these task cards, which can be used for various activities such as peer practice, scavenger hunts, class games, and more.. One way to have students use this puzzle is to "find their partner". Give each student one puzzle piece. Half of the students will have a multiplication fact and the other half will have the answers. Have them find their "partner" with the matching piece.. In this printable math activity, children will multiply by 7s in order to fill in the blank bubbles on the multiplication caterpillar.. Cut and fold to make an origami fortune teller game (aka cootie catcher). Learn and practice basic facts with factors of 7.. Practice multiplication facts with 7 as a factor. This file includes 11 flash cards, 5 super challenge cards, a sorting mat, and a self-quiz.. Print this file on heavy card stock. The cut out the airplane picture and the math fact strips. Place the fact strips through the airplane and pull to scroll up and down. Students read the facts and say the answers.. ## Skip Count by 7s. Answer the critical thinking questions and solve the word problems.. 0, 7, 14,... what is next? Skip count by 7s along the number line to find out.. Count by 7s to complete the dot-to-dot. The picture is a pig. Then solve the multiplication facts at the bottom of the page.. Count by 7s. Write the numbers on the penguins' bellies. Numbers go all the way up to 161.. Count by 7s and write the missing numbers on the balloons. These numbers go up to 168.. Complete the grids by filling in the missing numbers.	This worksheet has a monkey theme.. ## All Facts0s through 7s. On this drill worksheet, your students will review multiplication facts with factors of 0, 1, 2, 3, 4, 5, 6, and 7. Facts through 10s, up to 7x10.. This drill worksheet has all basic facts for multiplying by 0, 1, 2, 3, 4, 5, 6, and 7. Facts include 11s and 12s, up to 7x12.. This timed quiz has 50 problems on it. Covers all basic facts for 0s, 1s, 2s, 3s, 4s, 5s, 6s, and 7s. Facts through 12, up to 7x12.. Multiplication Basic Facts: 0-10. Here you'll find lots more multiplication worksheets, games, and activities for all basic facts up to 10x10.. Multiplication/Division. Fact Families. This page has lots of worksheets with multiplication and division fact families.. Multiplication by 3s. Basic facts with factors of 3 are the focus on these printable learning resources.. Multiplication by 4s. These worksheets focus specifically on the 4s times tables.. Multiplication by 5s. Here are the activities for learning multiplication facts from 5x0 up to 5x12.. Multiplication by 6s. Multiplication by 8s. Practice multiplying by the number 8. Includes mystery pictures, skip counting lessons, and more.. Multiplication by 9s. ## Pictures of Our Worksheets. Not a member yet? Join Today!. My Account. Site Information. Social Media.
https://cloud.tencent.com/developer/news/373513	1,642,491,681,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00383.warc.gz	241,981,419	13,674	# LeetCode笔记 13.Roman to IntegerEasy github： https://github.com/weidafeng/LeetCode_Python https://leetcode.com/problems/roman-to-integer Roman numerals are represented by seven different symbols: , , , , , and . SymbolValueI 1V 5X 10L 50C 100D 500M 1000 For example, two is written as in Roman numeral, just two one's added together. Twelve is written as, , which is simply + . The number twenty seven is written as , which is + + . Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not . Instead, the number four is written as . Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as . There are six instances where subtraction is used: can be placed before (5) and (10) to make 4 and 9. can be placed before (50) and (100) to make 40 and 90. can be placed before (500) and (1000) to make 400 and 900. Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999. Example 1: Input:"III"Output:3 Example 2: Input:"IV"Output:4 Example 3: Input:"IX"Output:9 Example 4: Input:"LVIII"Output:58Explanation:L = 50, V= 5, III = 3. 1、本专栏是我刷LeetCode的学习笔记，每周更新一次 2、所有代码均已上传至github，欢迎star、watch： https://github.com/weidafeng/LeetCode_Python 3、python语言实现，至少用两种思路，争取用最简洁的代码实现 4、欢迎指正、讨论 • 发表于: • 原文链接https://kuaibao.qq.com/s/20181219G0L0SX00?refer=cp_1026 • 腾讯「云+社区」是腾讯内容开放平台帐号（企鹅号）传播渠道之一，根据《腾讯内容开放平台服务协议》转载发布内容。 • 如有侵权，请联系 [email protected] 删除。 2022-01-18 2018-07-14 2018-05-07 2018-04-03 2018-04-02 2018-03-23 2022-01-18	569	1,630	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.8125	4	CC-MAIN-2022-05	longest	en	0.760383	# LeetCode笔记 13.Roman to IntegerEasy. github：. https://github.com/weidafeng/LeetCode_Python. https://leetcode.com/problems/roman-to-integer. Roman numerals are represented by seven different symbols: , , , , , and .. SymbolValueI 1V 5X 10L 50C 100D 500M 1000. For example, two is written as in Roman numeral, just two one's added together. Twelve is written as, , which is simply + . The number twenty seven is written as , which is + + .. Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not . Instead, the number four is written as . Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as . There are six instances where subtraction is used:. can be placed before (5) and (10) to make 4 and 9.. can be placed before (50) and (100) to make 40 and 90.. can be placed before (500) and (1000) to make 400 and 900.. Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.. Example 1:. Input:"III"Output:3. Example 2:. Input:"IV"Output:4. Example 3:.	Input:"IX"Output:9. Example 4:. Input:"LVIII"Output:58Explanation:L = 50, V= 5, III = 3.. 1、本专栏是我刷LeetCode的学习笔记，每周更新一次. 2、所有代码均已上传至github，欢迎star、watch：. https://github.com/weidafeng/LeetCode_Python. 3、python语言实现，至少用两种思路，争取用最简洁的代码实现. 4、欢迎指正、讨论. • 发表于:. • 原文链接https://kuaibao.qq.com/s/20181219G0L0SX00?refer=cp_1026. • 腾讯「云+社区」是腾讯内容开放平台帐号（企鹅号）传播渠道之一，根据《腾讯内容开放平台服务协议》转载发布内容。. • 如有侵权，请联系 [email protected] 删除。. 2022-01-18. 2018-07-14. 2018-05-07. 2018-04-03. 2018-04-02. 2018-03-23. 2022-01-18.
https://luanvan68.com/48-oz-is-how-many-pounds/	1,721,911,954,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763858305.84/warc/CC-MAIN-20240725114544-20240725144544-00128.warc.gz	324,893,287	50,598	# 48 oz is How Many Pounds: A Comprehensive Guide Discover how to convert 48 oz to pounds with this comprehensive guide. Learn why understanding weight measurement units is important and get practical tips for accurate measurement. Have you ever wondered how many pounds are in 48 ounces? Whether you’re a fitness enthusiast, chef, or just someone who wants to know the answer, this article is for you. Understanding weight measurement units can be confusing, especially when it comes to converting between ounces and pounds. In this comprehensive guide, I will provide step-by-step instructions on how to convert 48 ounces to pounds, practical examples of items that weigh 48 ounces in pounds, and tips for accurate weight measurement. Additionally, I will cover the importance of knowing weight measurement units for everyday life situations and health and fitness purposes. So let’s dive in and explore everything you need to know about converting 48 ounces to pounds. ## Understanding Weight Measurement Units When it comes to measuring weight, there are different units of measurement used around the world. In this section, we’ll focus on two commonly used units: ounces and pounds. ### Definition of Ounces and Pounds as Weight Measurement Units An ounce is a unit of weight equal to 1/16th of a pound or approximately 28 grams. It is primarily used in the United States and other countries that follow the imperial system of measurement. A pound, on the other hand, is a larger unit of weight equal to 16 ounces or approximately 453 grams. It is also part of the imperial system and widely used in the United States, United Kingdom, and other countries that have adopted this system. Maybe you are interested How Many Grams Is a Quarter Pound of Weed? ### Conversion Factors Between Ounces and Pounds Converting between ounces and pounds can be confusing for some people. To convert from ounces to pounds, you simply divide the number of ounces by 16. For example: • 48 ounces รท 16 = 3 pounds To convert from pounds to ounces, you multiply the number of pounds by 16. For example: • 2.5 pounds x 16 = 40 ounces Knowing how to convert between these two units will come in handy when dealing with recipes, grocery shopping, shipping items, or any situation that requires accurate weight measurements. ## Converting 48 Ounces to Pounds Converting between ounces and pounds might seem daunting, but it’s actually quite simple. Here’s a step-by-step guide on how to convert 48 ounces to pounds: ### Step 1: Understand the Conversion Factor Before we begin, let’s understand the conversion factor between ounces and pounds. One pound is equal to 16 ounces. Therefore, to convert from ounces to pounds, you need to divide the number of ounces by 16. ### Step 2: Divide 48 by 16 Now that we know the conversion factor, we can easily convert 48 ounces to pounds. By dividing 48 by 16, we get the answer – 3 pounds. ### Common Mistakes to Avoid During Conversion While converting between ounces and pounds is simple, there are some common mistakes that people make. These include: • Forgetting to use the correct conversion factor (1 pound = 16 ounces) • Rounding off too early in the calculation • Using inappropriate units during measurement By avoiding these mistakes, you can ensure accurate conversions between ounces and pounds. Now that we’ve covered how to convert 48 ounces to pounds let’s look at some practical examples. ## Examples of 48 Ounces in Pounds If you’re curious about what weighs 48 ounces, here are some practical examples: ### h3. Household Items • A standard bottle of ketchup or mustard typically weighs around 48 ounces or three pounds. • A jar of peanut butter can weigh up to 48 ounces or three pounds. • A bag of coffee beans weighing 48 ounces is equivalent to three pounds. ### h3. Food and Ingredients • Two large bags of M&Ms weigh approximately 48 ounces or three pounds. • Four medium-sized sweet potatoes weigh roughly 48 ounces or three pounds. • Three boneless chicken breasts together can weigh about 48 ounces or three pounds. ### h3. Comparison with Other Commonly Used Weight Measurements It’s worth noting that measuring weight in ounces and pounds is just one way to measure weight. Here are some other commonly used weight measurements and how they stack up against 48 fluid ounces: Maybe you are interested How Much is a Cake Pop at Starbucks? #### h4. Grams Converting 48 fluid ounces to grams yields a result of approximately 1,360 grams. #### h4. Kilograms Converting 48 fluid ounces to kilograms gives us an answer of roughly 1.36 kilograms. Knowing the weight equivalents between different measurement units can be helpful when following recipes or tracking nutrition information for health and fitness reasons. ## Why Knowing the Conversion Matters ### Importance of Being Able to Convert Between Ounces and Pounds in Everyday Life Situations Knowing how to convert between ounces and pounds can be incredibly useful in everyday life situations. For example, when cooking recipes that require precise measurements or when purchasing items such as fruits and vegetables by weight at the grocery store. Understanding weight measurement units can also be beneficial for home improvement projects that require weighing materials like paint or soil. Furthermore, being able to quickly convert between ounces and pounds is helpful when traveling internationally, as many countries use the metric system, which measures weight in grams and kilograms. By understanding conversion factors between ounces and pounds, you can easily make sense of weight measurements in other countries. ### Benefits of Understanding Weight Measurement Units for Health and Fitness Purposes For those interested in health and fitness, understanding weight measurement units is crucial. Many fitness programs track progress based on changes in body weight measured in pounds or kilograms. Additionally, food portions are often measured by weight in ounces or grams rather than volume. Being able to accurately measure and track weight can aid in achieving fitness goals and maintaining a healthy lifestyle. Moreover, understanding how to convert between different weight measurement units allows for easy comparison of nutritional information on food labels. In conclusion, knowing how to convert between ounces and pounds has significant benefits for everyday life situations and health and fitness purposes. It’s an essential skill that anyone can benefit from mastering. 1. Why do I need to know how many pounds are in 48 ounces? • Knowing weight measurement units is important for everyday life situations, such as measuring ingredients when cooking or baking. 2. How do I convert 48 ounces to pounds? • To convert 48 ounces to pounds, divide the number of ounces by 16 (since there are 16 ounces in a pound). So, 48/16 = 3 pounds. 3. Can I use an online converter to convert 48 ounces to pounds? • Yes, there are various online conversion tools available that can quickly and accurately convert between weight measurement units. 4. What are some practical examples of items that weigh 48 ounces in pounds? • Some common examples include small bags of flour or sugar, bottles of liquid seasoning, and small packages of meat or poultry. Maybe you are interested How Many Cups in 1 Pound Powdered Sugar: A Comprehensive Guide ### Additional Information on Related Topics, Such as Metric System Conversions • For those who live in countries that use the metric system instead of imperial units, converting between grams and kilograms can be useful. • One kilogram is equal to approximately 2.2 pounds, while one gram is equal to approximately 0.035 ounces. • There are various online conversion tools available that can quickly and accurately convert between different weight measurement units. ## Tips for Accurate Weight Measurement When it comes to measuring weight accurately, there are a few best practices you should keep in mind. ### Best Practices for Measuring Weight Accurately • Use a calibrated scale: It’s essential to use a reliable and calibrated scale to ensure accurate measurements. • Weigh yourself at the same time every day: Your body weight can fluctuate throughout the day due to various factors such as hydration levels and meals consumed. Weighing yourself at the same time each day, preferably in the morning before eating or drinking anything, will help you get consistent results. • Wear minimal clothing: When weighing yourself, wear minimal clothing or weigh yourself naked. Clothing can add extra weight and skew your measurements. • Place the scale on a flat surface: To avoid inaccurate readings, always place your scale on a flat surface like tile or hardwood flooring instead of carpeted areas. • Stand still during measurement: While standing on the scale, stand still and upright without leaning over to read your weight. Any movement can affect the accuracy of your measurements. ### Tools and Devices That Can Aid in Accurate Weight Measurement • Body fat calipers: These devices measure body fat percentage by pinching skin folds in specific areas of the body. • Smart scales: These scales use bioelectrical impedance analysis (BIA) technology to provide more detailed information about body composition such as muscle mass and water weight. • Kitchen scales: If you’re cooking or baking with specific ingredients that require precise measurements, using a kitchen scale is essential. By following these tips and using appropriate tools and devices, you can ensure accurate weight measurement and track your progress towards achieving your health goals effectively. ## Conclusion In conclusion, understanding weight measurement units is important for everyday life situations and health and fitness purposes. Converting between ounces and pounds can be confusing, but with the step-by-step guide provided in this article, you now know how to convert 48 ounces to pounds accurately. Knowing the conversion between ounces and pounds can also help you make better decisions when it comes to portion control, recipe measurements, and even shipping packages. By using the tips for accurate weight measurement provided, you can ensure that your measurements are precise and reliable. Overall, I hope this comprehensive guide has been helpful in answering your question of “48 oz is how many pounds?” as well as providing additional information on related topics. Remember to always measure weight accurately and confidently convert between ounces and pounds with ease. ## How Many Pounds is 600 kg? – A Comprehensive Guide Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature. ## How Many Pounds is 600 kg? – A Comprehensive Guide Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature. ## How Many Ounces in 1.5 Pounds? Learn how to convert pounds to ounces and vice versa accurately! Discover “how many ounces in 1.5 pounds” and more with this comprehensive guide. ## How Many Pounds Is 23 Kilos? Learn how to convert 23 kilos to pounds and gain a better understanding of the conversion process between these two units of measurement. ## How Many M&Ms in a Pound: Everything You Need to Know Discover the exact number of M&Ms in a pound and more with our comprehensive guide. Learn how to measure them by weight and find out what affects their quantity! ## How Many M&Ms in a Pound: Everything You Need to Know Discover the exact number of M&Ms in a pound and more with our comprehensive guide. Learn how to measure them by weight and find out what affects their quantity!	2,334	11,838	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2024-30	latest	en	0.919979	# 48 oz is How Many Pounds: A Comprehensive Guide. Discover how to convert 48 oz to pounds with this comprehensive guide. Learn why understanding weight measurement units is important and get practical tips for accurate measurement.. Have you ever wondered how many pounds are in 48 ounces? Whether you’re a fitness enthusiast, chef, or just someone who wants to know the answer, this article is for you. Understanding weight measurement units can be confusing, especially when it comes to converting between ounces and pounds.. In this comprehensive guide, I will provide step-by-step instructions on how to convert 48 ounces to pounds, practical examples of items that weigh 48 ounces in pounds, and tips for accurate weight measurement. Additionally, I will cover the importance of knowing weight measurement units for everyday life situations and health and fitness purposes.. So let’s dive in and explore everything you need to know about converting 48 ounces to pounds.. ## Understanding Weight Measurement Units. When it comes to measuring weight, there are different units of measurement used around the world. In this section, we’ll focus on two commonly used units: ounces and pounds.. ### Definition of Ounces and Pounds as Weight Measurement Units. An ounce is a unit of weight equal to 1/16th of a pound or approximately 28 grams. It is primarily used in the United States and other countries that follow the imperial system of measurement.. A pound, on the other hand, is a larger unit of weight equal to 16 ounces or approximately 453 grams. It is also part of the imperial system and widely used in the United States, United Kingdom, and other countries that have adopted this system.. Maybe you are interested How Many Grams Is a Quarter Pound of Weed?. ### Conversion Factors Between Ounces and Pounds. Converting between ounces and pounds can be confusing for some people. To convert from ounces to pounds, you simply divide the number of ounces by 16. For example:. • 48 ounces รท 16 = 3 pounds. To convert from pounds to ounces, you multiply the number of pounds by 16. For example:. • 2.5 pounds x 16 = 40 ounces. Knowing how to convert between these two units will come in handy when dealing with recipes, grocery shopping, shipping items, or any situation that requires accurate weight measurements.. ## Converting 48 Ounces to Pounds. Converting between ounces and pounds might seem daunting, but it’s actually quite simple. Here’s a step-by-step guide on how to convert 48 ounces to pounds:. ### Step 1: Understand the Conversion Factor. Before we begin, let’s understand the conversion factor between ounces and pounds. One pound is equal to 16 ounces. Therefore, to convert from ounces to pounds, you need to divide the number of ounces by 16.. ### Step 2: Divide 48 by 16. Now that we know the conversion factor, we can easily convert 48 ounces to pounds. By dividing 48 by 16, we get the answer – 3 pounds.. ### Common Mistakes to Avoid During Conversion. While converting between ounces and pounds is simple, there are some common mistakes that people make. These include:. • Forgetting to use the correct conversion factor (1 pound = 16 ounces). • Rounding off too early in the calculation. • Using inappropriate units during measurement. By avoiding these mistakes, you can ensure accurate conversions between ounces and pounds. Now that we’ve covered how to convert 48 ounces to pounds let’s look at some practical examples.. ## Examples of 48 Ounces in Pounds. If you’re curious about what weighs 48 ounces, here are some practical examples:. ### h3. Household Items. • A standard bottle of ketchup or mustard typically weighs around 48 ounces or three pounds.. • A jar of peanut butter can weigh up to 48 ounces or three pounds.. • A bag of coffee beans weighing 48 ounces is equivalent to three pounds.. ### h3. Food and Ingredients. • Two large bags of M&Ms weigh approximately 48 ounces or three pounds.. • Four medium-sized sweet potatoes weigh roughly 48 ounces or three pounds.. • Three boneless chicken breasts together can weigh about 48 ounces or three pounds.. ### h3. Comparison with Other Commonly Used Weight Measurements. It’s worth noting that measuring weight in ounces and pounds is just one way to measure weight. Here are some other commonly used weight measurements and how they stack up against 48 fluid ounces:. Maybe you are interested How Much is a Cake Pop at Starbucks?. #### h4. Grams. Converting 48 fluid ounces to grams yields a result of approximately 1,360 grams.. #### h4. Kilograms. Converting 48 fluid ounces to kilograms gives us an answer of roughly 1.36 kilograms.. Knowing the weight equivalents between different measurement units can be helpful when following recipes or tracking nutrition information for health and fitness reasons.. ## Why Knowing the Conversion Matters. ### Importance of Being Able to Convert Between Ounces and Pounds in Everyday Life Situations.	Knowing how to convert between ounces and pounds can be incredibly useful in everyday life situations. For example, when cooking recipes that require precise measurements or when purchasing items such as fruits and vegetables by weight at the grocery store. Understanding weight measurement units can also be beneficial for home improvement projects that require weighing materials like paint or soil.. Furthermore, being able to quickly convert between ounces and pounds is helpful when traveling internationally, as many countries use the metric system, which measures weight in grams and kilograms. By understanding conversion factors between ounces and pounds, you can easily make sense of weight measurements in other countries.. ### Benefits of Understanding Weight Measurement Units for Health and Fitness Purposes. For those interested in health and fitness, understanding weight measurement units is crucial. Many fitness programs track progress based on changes in body weight measured in pounds or kilograms. Additionally, food portions are often measured by weight in ounces or grams rather than volume.. Being able to accurately measure and track weight can aid in achieving fitness goals and maintaining a healthy lifestyle. Moreover, understanding how to convert between different weight measurement units allows for easy comparison of nutritional information on food labels.. In conclusion, knowing how to convert between ounces and pounds has significant benefits for everyday life situations and health and fitness purposes. It’s an essential skill that anyone can benefit from mastering.. 1. Why do I need to know how many pounds are in 48 ounces?. • Knowing weight measurement units is important for everyday life situations, such as measuring ingredients when cooking or baking.. 2. How do I convert 48 ounces to pounds?. • To convert 48 ounces to pounds, divide the number of ounces by 16 (since there are 16 ounces in a pound). So, 48/16 = 3 pounds.. 3. Can I use an online converter to convert 48 ounces to pounds?. • Yes, there are various online conversion tools available that can quickly and accurately convert between weight measurement units.. 4. What are some practical examples of items that weigh 48 ounces in pounds?. • Some common examples include small bags of flour or sugar, bottles of liquid seasoning, and small packages of meat or poultry.. Maybe you are interested How Many Cups in 1 Pound Powdered Sugar: A Comprehensive Guide. ### Additional Information on Related Topics, Such as Metric System Conversions. • For those who live in countries that use the metric system instead of imperial units, converting between grams and kilograms can be useful.. • One kilogram is equal to approximately 2.2 pounds, while one gram is equal to approximately 0.035 ounces.. • There are various online conversion tools available that can quickly and accurately convert between different weight measurement units.. ## Tips for Accurate Weight Measurement. When it comes to measuring weight accurately, there are a few best practices you should keep in mind.. ### Best Practices for Measuring Weight Accurately. • Use a calibrated scale: It’s essential to use a reliable and calibrated scale to ensure accurate measurements.. • Weigh yourself at the same time every day: Your body weight can fluctuate throughout the day due to various factors such as hydration levels and meals consumed. Weighing yourself at the same time each day, preferably in the morning before eating or drinking anything, will help you get consistent results.. • Wear minimal clothing: When weighing yourself, wear minimal clothing or weigh yourself naked. Clothing can add extra weight and skew your measurements.. • Place the scale on a flat surface: To avoid inaccurate readings, always place your scale on a flat surface like tile or hardwood flooring instead of carpeted areas.. • Stand still during measurement: While standing on the scale, stand still and upright without leaning over to read your weight. Any movement can affect the accuracy of your measurements.. ### Tools and Devices That Can Aid in Accurate Weight Measurement. • Body fat calipers: These devices measure body fat percentage by pinching skin folds in specific areas of the body.. • Smart scales: These scales use bioelectrical impedance analysis (BIA) technology to provide more detailed information about body composition such as muscle mass and water weight.. • Kitchen scales: If you’re cooking or baking with specific ingredients that require precise measurements, using a kitchen scale is essential.. By following these tips and using appropriate tools and devices, you can ensure accurate weight measurement and track your progress towards achieving your health goals effectively.. ## Conclusion. In conclusion, understanding weight measurement units is important for everyday life situations and health and fitness purposes. Converting between ounces and pounds can be confusing, but with the step-by-step guide provided in this article, you now know how to convert 48 ounces to pounds accurately.. Knowing the conversion between ounces and pounds can also help you make better decisions when it comes to portion control, recipe measurements, and even shipping packages. By using the tips for accurate weight measurement provided, you can ensure that your measurements are precise and reliable.. Overall, I hope this comprehensive guide has been helpful in answering your question of “48 oz is how many pounds?” as well as providing additional information on related topics. Remember to always measure weight accurately and confidently convert between ounces and pounds with ease.. ## How Many Pounds is 600 kg? – A Comprehensive Guide. Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature.. ## How Many Pounds is 600 kg? – A Comprehensive Guide. Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature.. ## How Many Ounces in 1.5 Pounds?. Learn how to convert pounds to ounces and vice versa accurately! Discover “how many ounces in 1.5 pounds” and more with this comprehensive guide.. ## How Many Pounds Is 23 Kilos?. Learn how to convert 23 kilos to pounds and gain a better understanding of the conversion process between these two units of measurement.. ## How Many M&Ms in a Pound: Everything You Need to Know. Discover the exact number of M&Ms in a pound and more with our comprehensive guide. Learn how to measure them by weight and find out what affects their quantity!. ## How Many M&Ms in a Pound: Everything You Need to Know. Discover the exact number of M&Ms in a pound and more with our comprehensive guide. Learn how to measure them by weight and find out what affects their quantity!.
https://mathexamination.com/class/hopkins%E2%80%93levitzki-theorem.php	1,618,471,151,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038084601.32/warc/CC-MAIN-20210415065312-20210415095312-00587.warc.gz	491,294,495	7,065	## Do My Hopkins–Levitzki Theorem Class A "Hopkins–Levitzki Theorem Class" QE" is a basic mathematical term for a generalized consistent expression which is utilized to fix differential equations and has options which are regular. In differential Class resolving, a Hopkins–Levitzki Theorem function, or "quad" is utilized. The Hopkins–Levitzki Theorem Class in Class form can be expressed as: Q( x) = -kx2, where Q( x) are the Hopkins–Levitzki Theorem Class and it is an important term. The q part of the Class is the Hopkins–Levitzki Theorem constant, whereas the x part is the Hopkins–Levitzki Theorem function. There are 4 Hopkins–Levitzki Theorem functions with appropriate solution: K4, K7, K3, and L4. We will now look at these Hopkins–Levitzki Theorem functions and how they are solved. K4 - The K part of a Hopkins–Levitzki Theorem Class is the Hopkins–Levitzki Theorem function. This Hopkins–Levitzki Theorem function can likewise be written in partial fractions such as: (x2 - y2)/( x+ y). To resolve for K4 we multiply it by the appropriate Hopkins–Levitzki Theorem function: k( x) = x2, y2, or x-y. K7 - The K7 Hopkins–Levitzki Theorem Class has a solution of the type: x4y2 - y4x3 = 0. The Hopkins–Levitzki Theorem function is then multiplied by x to get: x2 + y2 = 0. We then need to increase the Hopkins–Levitzki Theorem function with k to get: k( x) = x2 and y2. K3 - The Hopkins–Levitzki Theorem function Class is K3 + K2 = 0. We then multiply by k for K3. K3( t) - The Hopkins–Levitzki Theorem function equationis K3( t) + K2( t). We multiply by k for K3( t). Now we multiply by the Hopkins–Levitzki Theorem function which gives: K2( t) = K( t) times k. The Hopkins–Levitzki Theorem function is likewise called "K4" because of the initials of the letters K and 4. K suggests Hopkins–Levitzki Theorem, and the word "quad" is noticable as "kah-rab". The Hopkins–Levitzki Theorem Class is one of the main approaches of resolving differential equations. In the Hopkins–Levitzki Theorem function Class, the Hopkins–Levitzki Theorem function is first increased by the proper Hopkins–Levitzki Theorem function, which will give the Hopkins–Levitzki Theorem function. The Hopkins–Levitzki Theorem function is then divided by the Hopkins–Levitzki Theorem function which will divide the Hopkins–Levitzki Theorem function into a real part and an imaginary part. This provides the Hopkins–Levitzki Theorem term. Lastly, the Hopkins–Levitzki Theorem term will be divided by the numerator and the denominator to get the quotient. We are entrusted to the right hand side and the term "q". The Hopkins–Levitzki Theorem Class is a crucial concept to comprehend when solving a differential Class. The Hopkins–Levitzki Theorem function is simply one approach to fix a Hopkins–Levitzki Theorem Class. The techniques for solving Hopkins–Levitzki Theorem formulas consist of: singular value decay, factorization, optimal algorithm, mathematical solution or the Hopkins–Levitzki Theorem function approximation. ## Pay Me To Do Your Hopkins–Levitzki Theorem Class If you wish to end up being familiar with the Quartic Class, then you need to first start by checking out the online Quartic page. This page will reveal you how to use the Class by utilizing your keyboard. The description will likewise reveal you how to produce your own algebra formulas to help you study for your classes. Before you can comprehend how to study for a Hopkins–Levitzki Theorem Class, you must first understand using your keyboard. You will discover how to click the function keys on your keyboard, as well as how to type the letters. There are three rows of function keys on your keyboard. Each row has four functions: Alt, F1, F2, and F3. By pressing Alt and F2, you can increase and divide the worth by another number, such as the number 6. By pressing Alt and F3, you can utilize the 3rd power. When you press Alt and F3, you will type in the number you are attempting to increase and divide. To increase a number by itself, you will press Alt and X, where X is the number you wish to increase. When you press Alt and F3, you will type in the number you are trying to divide. This works the same with the number 6, other than you will just enter the two digits that are six apart. Lastly, when you push Alt and F3, you will utilize the 4th power. Nevertheless, when you push Alt and F4, you will use the actual power that you have discovered to be the most appropriate for your problem. By utilizing the Alt and F function keys, you can multiply, divide, and then use the formula for the third power. If you need to increase an odd variety of x's, then you will need to get in an even number. This is not the case if you are attempting to do something complex, such as multiplying 2 even numbers. For instance, if you want to multiply an odd number of x's, then you will require to enter odd numbers. This is especially real if you are attempting to find out the response of a Hopkins–Levitzki Theorem Class. If you wish to convert an odd number into an even number, then you will require to press Alt and F4. If you do not know how to increase by numbers on their own, then you will need to use the letters x, a b, c, and d. While you can multiply and divide by use of the numbers, they are much easier to use when you can look at the power tables for the numbers. You will need to do some research study when you first start to use the numbers, however after a while, it will be second nature. After you have created your own algebra formulas, you will be able to develop your own reproduction tables. The Hopkins–Levitzki Theorem Formula is not the only method to fix Hopkins–Levitzki Theorem equations. It is necessary to discover trigonometry, which utilizes the Pythagorean theorem, and then use Hopkins–Levitzki Theorem solutions to resolve issues. With this approach, you can know about angles and how to solve problems without needing to take another algebra class. It is essential to attempt and type as quickly as possible, due to the fact that typing will help you know about the speed you are typing. This will assist you compose your answers quicker. ## Pay Someone To Take My Hopkins–Levitzki Theorem Class A Hopkins–Levitzki Theorem Class is a generalization of a direct Class. For instance, when you plug in x=a+b for a given Class, you acquire the value of x. When you plug in x=a for the Class y=c, you acquire the values of x and y, which give you an outcome of c. By using this fundamental concept to all the equations that we have actually attempted, we can now solve Hopkins–Levitzki Theorem equations for all the values of x, and we can do it quickly and effectively. There are many online resources offered that supply free or budget friendly Hopkins–Levitzki Theorem equations to resolve for all the worths of x, consisting of the expense of time for you to be able to make the most of their Hopkins–Levitzki Theorem Class assignment help service. These resources normally do not need a membership charge or any type of financial investment. The answers provided are the result of complex-variable Hopkins–Levitzki Theorem formulas that have actually been solved. This is also the case when the variable utilized is an unidentified number. The Hopkins–Levitzki Theorem Class is a term that is an extension of a linear Class. One benefit of using Hopkins–Levitzki Theorem formulas is that they are more general than the linear equations. They are much easier to resolve for all the worths of x. When the variable used in the Hopkins–Levitzki Theorem Class is of the kind x=a+b, it is simpler to solve the Hopkins–Levitzki Theorem Class because there are no unknowns. As a result, there are less points on the line defined by x and a constant variable. For a right-angle triangle whose base points to the right and whose hypotenuse indicate the left, the right-angle tangent and curve chart will form a Hopkins–Levitzki Theorem Class. This Class has one unknown that can be discovered with the Hopkins–Levitzki Theorem formula. For a Hopkins–Levitzki Theorem Class, the point on the line specified by the x variable and a constant term are called the axis. The presence of such an axis is called the vertex. Since the axis, vertex, and tangent, in a Hopkins–Levitzki Theorem Class, are an offered, we can find all the worths of x and they will sum to the offered worths. This is achieved when we utilize the Hopkins–Levitzki Theorem formula. The factor of being a continuous factor is called the system of formulas in Hopkins–Levitzki Theorem equations. This is sometimes called the main Class. Hopkins–Levitzki Theorem formulas can be solved for other worths of x. One method to solve Hopkins–Levitzki Theorem formulas for other worths of x is to divide the x variable into its factor part. If the variable is given as a positive number, it can be divided into its element parts to get the regular part of the variable. This variable has a magnitude that amounts to the part of the x variable that is a constant. In such a case, the formula is a third-order Hopkins–Levitzki Theorem Class. If the variable x is negative, it can be divided into the very same part of the x variable to get the part of the x variable that is increased by the denominator. In such a case, the formula is a second-order Hopkins–Levitzki Theorem Class. Solution help service in resolving Hopkins–Levitzki Theorem formulas. When utilizing an online service for fixing Hopkins–Levitzki Theorem equations, the Class will be resolved immediately.	2,309	9,560	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4	4	CC-MAIN-2021-17	latest	en	0.889003	## Do My Hopkins–Levitzki Theorem Class. A "Hopkins–Levitzki Theorem Class" QE" is a basic mathematical term for a generalized consistent expression which is utilized to fix differential equations and has options which are regular. In differential Class resolving, a Hopkins–Levitzki Theorem function, or "quad" is utilized.. The Hopkins–Levitzki Theorem Class in Class form can be expressed as: Q( x) = -kx2, where Q( x) are the Hopkins–Levitzki Theorem Class and it is an important term. The q part of the Class is the Hopkins–Levitzki Theorem constant, whereas the x part is the Hopkins–Levitzki Theorem function.. There are 4 Hopkins–Levitzki Theorem functions with appropriate solution: K4, K7, K3, and L4. We will now look at these Hopkins–Levitzki Theorem functions and how they are solved.. K4 - The K part of a Hopkins–Levitzki Theorem Class is the Hopkins–Levitzki Theorem function. This Hopkins–Levitzki Theorem function can likewise be written in partial fractions such as: (x2 - y2)/( x+ y). To resolve for K4 we multiply it by the appropriate Hopkins–Levitzki Theorem function: k( x) = x2, y2, or x-y.. K7 - The K7 Hopkins–Levitzki Theorem Class has a solution of the type: x4y2 - y4x3 = 0. The Hopkins–Levitzki Theorem function is then multiplied by x to get: x2 + y2 = 0. We then need to increase the Hopkins–Levitzki Theorem function with k to get: k( x) = x2 and y2.. K3 - The Hopkins–Levitzki Theorem function Class is K3 + K2 = 0. We then multiply by k for K3.. K3( t) - The Hopkins–Levitzki Theorem function equationis K3( t) + K2( t). We multiply by k for K3( t). Now we multiply by the Hopkins–Levitzki Theorem function which gives: K2( t) = K( t) times k.. The Hopkins–Levitzki Theorem function is likewise called "K4" because of the initials of the letters K and 4. K suggests Hopkins–Levitzki Theorem, and the word "quad" is noticable as "kah-rab".. The Hopkins–Levitzki Theorem Class is one of the main approaches of resolving differential equations. In the Hopkins–Levitzki Theorem function Class, the Hopkins–Levitzki Theorem function is first increased by the proper Hopkins–Levitzki Theorem function, which will give the Hopkins–Levitzki Theorem function.. The Hopkins–Levitzki Theorem function is then divided by the Hopkins–Levitzki Theorem function which will divide the Hopkins–Levitzki Theorem function into a real part and an imaginary part. This provides the Hopkins–Levitzki Theorem term.. Lastly, the Hopkins–Levitzki Theorem term will be divided by the numerator and the denominator to get the quotient. We are entrusted to the right hand side and the term "q".. The Hopkins–Levitzki Theorem Class is a crucial concept to comprehend when solving a differential Class. The Hopkins–Levitzki Theorem function is simply one approach to fix a Hopkins–Levitzki Theorem Class. The techniques for solving Hopkins–Levitzki Theorem formulas consist of: singular value decay, factorization, optimal algorithm, mathematical solution or the Hopkins–Levitzki Theorem function approximation.. ## Pay Me To Do Your Hopkins–Levitzki Theorem Class. If you wish to end up being familiar with the Quartic Class, then you need to first start by checking out the online Quartic page. This page will reveal you how to use the Class by utilizing your keyboard. The description will likewise reveal you how to produce your own algebra formulas to help you study for your classes.. Before you can comprehend how to study for a Hopkins–Levitzki Theorem Class, you must first understand using your keyboard. You will discover how to click the function keys on your keyboard, as well as how to type the letters. There are three rows of function keys on your keyboard. Each row has four functions: Alt, F1, F2, and F3.. By pressing Alt and F2, you can increase and divide the worth by another number, such as the number 6. By pressing Alt and F3, you can utilize the 3rd power.. When you press Alt and F3, you will type in the number you are attempting to increase and divide. To increase a number by itself, you will press Alt and X, where X is the number you wish to increase. When you press Alt and F3, you will type in the number you are trying to divide.. This works the same with the number 6, other than you will just enter the two digits that are six apart. Lastly, when you push Alt and F3, you will utilize the 4th power. Nevertheless, when you push Alt and F4, you will use the actual power that you have discovered to be the most appropriate for your problem.. By utilizing the Alt and F function keys, you can multiply, divide, and then use the formula for the third power. If you need to increase an odd variety of x's, then you will need to get in an even number.. This is not the case if you are attempting to do something complex, such as multiplying 2 even numbers.	For instance, if you want to multiply an odd number of x's, then you will require to enter odd numbers. This is especially real if you are attempting to find out the response of a Hopkins–Levitzki Theorem Class.. If you wish to convert an odd number into an even number, then you will require to press Alt and F4. If you do not know how to increase by numbers on their own, then you will need to use the letters x, a b, c, and d.. While you can multiply and divide by use of the numbers, they are much easier to use when you can look at the power tables for the numbers. You will need to do some research study when you first start to use the numbers, however after a while, it will be second nature. After you have created your own algebra formulas, you will be able to develop your own reproduction tables.. The Hopkins–Levitzki Theorem Formula is not the only method to fix Hopkins–Levitzki Theorem equations. It is necessary to discover trigonometry, which utilizes the Pythagorean theorem, and then use Hopkins–Levitzki Theorem solutions to resolve issues. With this approach, you can know about angles and how to solve problems without needing to take another algebra class.. It is essential to attempt and type as quickly as possible, due to the fact that typing will help you know about the speed you are typing. This will assist you compose your answers quicker.. ## Pay Someone To Take My Hopkins–Levitzki Theorem Class. A Hopkins–Levitzki Theorem Class is a generalization of a direct Class. For instance, when you plug in x=a+b for a given Class, you acquire the value of x. When you plug in x=a for the Class y=c, you acquire the values of x and y, which give you an outcome of c. By using this fundamental concept to all the equations that we have actually attempted, we can now solve Hopkins–Levitzki Theorem equations for all the values of x, and we can do it quickly and effectively.. There are many online resources offered that supply free or budget friendly Hopkins–Levitzki Theorem equations to resolve for all the worths of x, consisting of the expense of time for you to be able to make the most of their Hopkins–Levitzki Theorem Class assignment help service. These resources normally do not need a membership charge or any type of financial investment.. The answers provided are the result of complex-variable Hopkins–Levitzki Theorem formulas that have actually been solved. This is also the case when the variable utilized is an unidentified number.. The Hopkins–Levitzki Theorem Class is a term that is an extension of a linear Class. One benefit of using Hopkins–Levitzki Theorem formulas is that they are more general than the linear equations. They are much easier to resolve for all the worths of x.. When the variable used in the Hopkins–Levitzki Theorem Class is of the kind x=a+b, it is simpler to solve the Hopkins–Levitzki Theorem Class because there are no unknowns. As a result, there are less points on the line defined by x and a constant variable.. For a right-angle triangle whose base points to the right and whose hypotenuse indicate the left, the right-angle tangent and curve chart will form a Hopkins–Levitzki Theorem Class. This Class has one unknown that can be discovered with the Hopkins–Levitzki Theorem formula. For a Hopkins–Levitzki Theorem Class, the point on the line specified by the x variable and a constant term are called the axis.. The presence of such an axis is called the vertex. Since the axis, vertex, and tangent, in a Hopkins–Levitzki Theorem Class, are an offered, we can find all the worths of x and they will sum to the offered worths. This is achieved when we utilize the Hopkins–Levitzki Theorem formula.. The factor of being a continuous factor is called the system of formulas in Hopkins–Levitzki Theorem equations. This is sometimes called the main Class.. Hopkins–Levitzki Theorem formulas can be solved for other worths of x. One method to solve Hopkins–Levitzki Theorem formulas for other worths of x is to divide the x variable into its factor part.. If the variable is given as a positive number, it can be divided into its element parts to get the regular part of the variable. This variable has a magnitude that amounts to the part of the x variable that is a constant. In such a case, the formula is a third-order Hopkins–Levitzki Theorem Class.. If the variable x is negative, it can be divided into the very same part of the x variable to get the part of the x variable that is increased by the denominator. In such a case, the formula is a second-order Hopkins–Levitzki Theorem Class.. Solution help service in resolving Hopkins–Levitzki Theorem formulas. When utilizing an online service for fixing Hopkins–Levitzki Theorem equations, the Class will be resolved immediately.
https://www.calculatoratoz.com/en/frictional-force-in-v-belt-drive-calculator/Calc-1428	1,611,660,825,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00787.warc.gz	679,319,495	30,980	Anshika Arya National Institute Of Technology (NIT), Hamirpur Anshika Arya has created this Calculator and 200+ more calculators! ## < 3 Other formulas that you can solve using the same Inputs Tension in the tight side of V-belt drive Tensions in the tight side of belt=Tensions in the slack side of belt(e^(Coefficient of friction between the belt and sides of the grooveAngle of contactcosec(Angle of the groove/2))) GO Tension in the tight side of rope drive Tensions in the tight side of belt=Tensions in the slack side of belt(e^(Coefficient of friction between the belt and sides of the grooveAngle of contactcosec(Angle of the groove/2))) GO Normal reaction between the belt and the sides of the groove Normal reaction between the belt and sides of the groove=Total reaction in the plane of the groove/(2sin(Angle of the groove/2)) GO ## < 11 Other formulas that calculate the same Output Force required to lower the load by a screw jack when weight of load, helix angle and coefficient of friction is known Force=Weight of Load((Coefficient of Frictioncos(Helix Angle))-sin(Helix Angle))/(cos(Helix Angle)+(Coefficient of Frictionsin(Helix Angle))) GO Force at circumference of the screw when weight of load, helix angle and coefficient of friction is known Force=Weight((sin(Helix Angle)+(Coefficient of Frictioncos(Helix Angle)))/(cos(Helix Angle)-(Coefficient of Frictionsin(Helix Angle)))) GO Force in direction of jet striking a stationary vertical plate Force=Liquid DensityCross Sectional Area of Jet(Initial velocity of liquid jet)^(2) GO Restoring force due to spring Force=Stiffness of springDisplacement of load below equilibrium position GO Force of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down a ramp Force=(MassAcceleration Due To Gravitysin(Angle of Inclination))/3 GO Force required to lower the load by a screw jack when weight of load, helix angle and limiting angle is known Force=Weight of Loadtan(Limiting angle of friction-Helix Angle) GO Force at circumference of the screw when weight of load, helix angle and limiting angle is known Force=Weight of Loadtan(Helix Angle+Limiting angle of friction) GO Force between parallel plate capacitors Universal Law of Gravitation Force By A Linear Induction Motor Force=Power/Linear Synchronous Speed GO Force Force=MassAcceleration GO ### Frictional force in V belt drive Formula Force=Coefficient of friction between the belt and sides of the grooveTotal reaction in the plane of the groovecosec(Angle of the groove/2) More formulas Velocity ratio of belt drive GO Velocity ratio of compound belt drive GO Velocity ratio of compound belt drive GO Velocity ratio of simple belt drive when thickness not considered GO Velocity ratio of simple belt drive when thickness considered GO Velocity ratio of belt when there's total percentage slip is given GO Total percentage slip in a belt GO Velocity ratio of belt in terms of creep of belt GO Length of an open belt drive GO Length of a cross belt drive GO Angle made by belt with vertical axis for open belt drive GO Angle made by belt with vertical axis for cross belt drive GO Power transmittted by a belt GO Torque exerted on the driving pulley GO Torque exerted on the driven pulley GO Tension in the tight side of belt GO angle of contact for open belt drive GO angle of contact for cross belt drive GO Centrifugal Tension in belt GO Tension on tight side when centrifugal tension is taken in account GO Tension on slack side when centrifugal tension is taken in account GO Tension on tight side when centrifugal tension is taken in account GO Maximum tension of belt GO Maximum tension for transmission of maximum power by a belt GO Tension in the tight side for transmission of maximum power by a belt GO Velocity for transmission of maximum power by a belt GO Initial tension in the belt GO Normal reaction between the belt and the sides of the groove GO Tension in the tight side of V-belt drive GO Tension in the tight side of rope drive GO Relation between pitch and pitch circle diameter of a chain drive GO Velocity ratio GO ## What is Frictional Force? The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object. ## How to Calculate Frictional force in V belt drive? Frictional force in V belt drive calculator uses Force=Coefficient of friction between the belt and sides of the grooveTotal reaction in the plane of the groovecosec(Angle of the groove/2) to calculate the Force, Frictional force in V belt drive is the force exerted by a surface as an object moves across it or makes an effort to move across it. Force and is denoted by F symbol. How to calculate Frictional force in V belt drive using this online calculator? To use this online calculator for Frictional force in V belt drive, enter Coefficient of friction between the belt and sides of the groove (μ), Angle of the groove (2β) and Total reaction in the plane of the groove (R) and hit the calculate button. Here is how the Frictional force in V belt drive calculation can be explained with given input values -> 17.38666 = 0.315cosec(30/2). ### FAQ What is Frictional force in V belt drive? Frictional force in V belt drive is the force exerted by a surface as an object moves across it or makes an effort to move across it and is represented as F=μRcosec(2β/2) or Force=Coefficient of friction between the belt and sides of the grooveTotal reaction in the plane of the groovecosec(Angle of the groove/2). Coefficient of friction between the belt and sides of the groove is the ratio defining the force that resists the motion of one body in relation to another body in contact with it, Angle of the groove is shown in degrees and will include all of the groove, if it is a V Groove it will be a dimension from one groove face to the other and Total reaction in the plane of the groove is a measure of the force holding the two surfaces together. How to calculate Frictional force in V belt drive? Frictional force in V belt drive is the force exerted by a surface as an object moves across it or makes an effort to move across it is calculated using Force=Coefficient of friction between the belt and sides of the grooveTotal reaction in the plane of the groovecosec(Angle of the groove/2). To calculate Frictional force in V belt drive, you need Coefficient of friction between the belt and sides of the groove (μ), Angle of the groove (2β) and Total reaction in the plane of the groove (R). With our tool, you need to enter the respective value for Coefficient of friction between the belt and sides of the groove, Angle of the groove and Total reaction in the plane of the groove and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well. How many ways are there to calculate Force? In this formula, Force uses Coefficient of friction between the belt and sides of the groove, Angle of the groove and Total reaction in the plane of the groove. We can use 11 other way(s) to calculate the same, which is/are as follows - • Force=MassAcceleration • Force=(MassAcceleration Due To Gravitysin(Angle of Inclination))/3	1,703	7,411	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.578125	4	CC-MAIN-2021-04	latest	en	0.794335	Anshika Arya. National Institute Of Technology (NIT), Hamirpur. Anshika Arya has created this Calculator and 200+ more calculators!. ## < 3 Other formulas that you can solve using the same Inputs. Tension in the tight side of V-belt drive. Tensions in the tight side of belt=Tensions in the slack side of belt(e^(Coefficient of friction between the belt and sides of the grooveAngle of contactcosec(Angle of the groove/2))) GO. Tension in the tight side of rope drive. Tensions in the tight side of belt=Tensions in the slack side of belt(e^(Coefficient of friction between the belt and sides of the grooveAngle of contactcosec(Angle of the groove/2))) GO. Normal reaction between the belt and the sides of the groove. Normal reaction between the belt and sides of the groove=Total reaction in the plane of the groove/(2sin(Angle of the groove/2)) GO. ## < 11 Other formulas that calculate the same Output. Force required to lower the load by a screw jack when weight of load, helix angle and coefficient of friction is known. Force=Weight of Load((Coefficient of Frictioncos(Helix Angle))-sin(Helix Angle))/(cos(Helix Angle)+(Coefficient of Frictionsin(Helix Angle))) GO. Force at circumference of the screw when weight of load, helix angle and coefficient of friction is known. Force=Weight((sin(Helix Angle)+(Coefficient of Frictioncos(Helix Angle)))/(cos(Helix Angle)-(Coefficient of Frictionsin(Helix Angle)))) GO. Force in direction of jet striking a stationary vertical plate. Force=Liquid DensityCross Sectional Area of Jet(Initial velocity of liquid jet)^(2) GO. Restoring force due to spring. Force=Stiffness of springDisplacement of load below equilibrium position GO. Force of Friction between the cylinder and the surface of inclined plane if cylinder is rolling without slipping down a ramp. Force=(MassAcceleration Due To Gravitysin(Angle of Inclination))/3 GO. Force required to lower the load by a screw jack when weight of load, helix angle and limiting angle is known. Force=Weight of Loadtan(Limiting angle of friction-Helix Angle) GO. Force at circumference of the screw when weight of load, helix angle and limiting angle is known. Force=Weight of Loadtan(Helix Angle+Limiting angle of friction) GO. Force between parallel plate capacitors. Universal Law of Gravitation. Force By A Linear Induction Motor. Force=Power/Linear Synchronous Speed GO. Force. Force=MassAcceleration GO. ### Frictional force in V belt drive Formula. Force=Coefficient of friction between the belt and sides of the grooveTotal reaction in the plane of the groove*cosec(Angle of the groove/2). More formulas. Velocity ratio of belt drive GO. Velocity ratio of compound belt drive GO. Velocity ratio of compound belt drive GO. Velocity ratio of simple belt drive when thickness not considered GO. Velocity ratio of simple belt drive when thickness considered GO. Velocity ratio of belt when there's total percentage slip is given GO. Total percentage slip in a belt GO. Velocity ratio of belt in terms of creep of belt GO. Length of an open belt drive GO. Length of a cross belt drive GO.	Angle made by belt with vertical axis for open belt drive GO. Angle made by belt with vertical axis for cross belt drive GO. Power transmittted by a belt GO. Torque exerted on the driving pulley GO. Torque exerted on the driven pulley GO. Tension in the tight side of belt GO. angle of contact for open belt drive GO. angle of contact for cross belt drive GO. Centrifugal Tension in belt GO. Tension on tight side when centrifugal tension is taken in account GO. Tension on slack side when centrifugal tension is taken in account GO. Tension on tight side when centrifugal tension is taken in account GO. Maximum tension of belt GO. Maximum tension for transmission of maximum power by a belt GO. Tension in the tight side for transmission of maximum power by a belt GO. Velocity for transmission of maximum power by a belt GO. Initial tension in the belt GO. Normal reaction between the belt and the sides of the groove GO. Tension in the tight side of V-belt drive GO. Tension in the tight side of rope drive GO. Relation between pitch and pitch circle diameter of a chain drive GO. Velocity ratio GO. ## What is Frictional Force?. The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object.. ## How to Calculate Frictional force in V belt drive?. Frictional force in V belt drive calculator uses Force=Coefficient of friction between the belt and sides of the grooveTotal reaction in the plane of the groovecosec(Angle of the groove/2) to calculate the Force, Frictional force in V belt drive is the force exerted by a surface as an object moves across it or makes an effort to move across it. Force and is denoted by F symbol.. How to calculate Frictional force in V belt drive using this online calculator? To use this online calculator for Frictional force in V belt drive, enter Coefficient of friction between the belt and sides of the groove (μ), Angle of the groove (2β) and Total reaction in the plane of the groove (R) and hit the calculate button. Here is how the Frictional force in V belt drive calculation can be explained with given input values -> 17.38666 = 0.315cosec(30/2).. ### FAQ. What is Frictional force in V belt drive?. Frictional force in V belt drive is the force exerted by a surface as an object moves across it or makes an effort to move across it and is represented as F=μRcosec(2β/2) or Force=Coefficient of friction between the belt and sides of the grooveTotal reaction in the plane of the groovecosec(Angle of the groove/2). Coefficient of friction between the belt and sides of the groove is the ratio defining the force that resists the motion of one body in relation to another body in contact with it, Angle of the groove is shown in degrees and will include all of the groove, if it is a V Groove it will be a dimension from one groove face to the other and Total reaction in the plane of the groove is a measure of the force holding the two surfaces together.. How to calculate Frictional force in V belt drive?. Frictional force in V belt drive is the force exerted by a surface as an object moves across it or makes an effort to move across it is calculated using Force=Coefficient of friction between the belt and sides of the grooveTotal reaction in the plane of the groovecosec(Angle of the groove/2). To calculate Frictional force in V belt drive, you need Coefficient of friction between the belt and sides of the groove (μ), Angle of the groove (2β) and Total reaction in the plane of the groove (R). With our tool, you need to enter the respective value for Coefficient of friction between the belt and sides of the groove, Angle of the groove and Total reaction in the plane of the groove and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.. How many ways are there to calculate Force?. In this formula, Force uses Coefficient of friction between the belt and sides of the groove, Angle of the groove and Total reaction in the plane of the groove. We can use 11 other way(s) to calculate the same, which is/are as follows -. • Force=MassAcceleration. • Force=(MassAcceleration Due To Gravity*sin(Angle of Inclination))/3.
https://www.physicsforums.com/threads/a-slab-of-glass-dielectric-is-inserted-into-a-parallel-plate-capacitor.999722/	1,702,167,573,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100989.75/warc/CC-MAIN-20231209233632-20231210023632-00724.warc.gz	890,543,676	17,091	# A slab of glass dielectric is inserted into a parallel plate capacitor • hidemi In summary, the equations used to understand this question/answer are C = k(εArea)/distance = Q/V = Q/(Edistance) and F = QE. As a slab of glass is added, k increases and E decreases, causing the force to also decrease. This relates to the idea that the force attracts the glass into the capacitor. The potential energy stored in the capacitor can be represented by the equation F=-dU/dx, where U is the potential energy, C is the capacitance, k is the dielectric constant, ε is the electric field, Area is the area of the plates, and distance is the distance between the plates. By pushing the dielectric further between the plates, the potential energy decreases #### hidemi Homework Statement A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted: A. a force repels the glass out of the capacitor B. a force attracts the glass into the capacitor C. no force acts on the glass D. a net charge appears on the glass E. the glass makes the plates repel each other Relevant Equations C = k(εArea)/distance = Q/V = Q/ (Edistance) F = QE I use the following equations to understand this question/answer. First, C = k(εArea)/distance = Q/V = Q/ (Edistance) As a slab of glass is added, k increases and thus E decreases. F=QE, as E decreases, force decreases as well. How does this relate to the 'force attracts the glass into the capacitor' Let me know if my thoughts/logic on this is right. In addition, is there a less physic-based explanation? Thanks. F = QE is not much help here. The Q in equation Q = CV is the charge on the capacitor plates that produces electric field E between the plates. Using F = QE is like saying that the the charge exerts a force on itself. I would use the definition of a force derived from a potential ##F=-\dfrac{dU}{dx}## where ##U## here is the potential energy stored in the capacitor. If you want a less physics-based explanation, you could perhaps draw an analogy with the gravitational field. If you release a rock it will move in the direction of decreasing potential energy. So if the dielectric is already half way in, which way must it move to reduce the potential energy, in or out of the plates? Remember the charge on the plates is constant. Last edited: hidemi kuruman said: F = QE is not much help here. The Q in equation Q = CV is the charge on the capacitor plates that produces electric field E between the plates. Using F = QE is like saying that the the charge exerts a force on itself. I would use the definition of a force derived from a potential ##F=-\dfrac{dU}{dx}## where ##U## here is the potential energy stored in the capacitor. If you want a less physics-based explanation, you could perhaps draw an analogy with the gravitational field. If you release a rock it will move in the direction of decreasing potential energy. So if the dielectric is already half way in, which way must it move to reduce the potential energy, in or out of the plates? Remember the charge on the plates is constant. I used your equation, F=-dU/dx, and substitute in relevant symbols as followed: U = CV/2 = 1/2* kεA/d Ed = kεAE/2 Therefore, the force has nothing to do with the distance? How can I know if it is attracted or repeled? You have the wrong ##x##. It is not the plate separation, it is the extent to which the dielectric is inserted between the plates. The derivative is with respect to that distance. The idea is to find the capacitance as a function of ##x## and then take the derivative. I have a feeling that this might be too complicated for you and you really don't need to do it that way. You can follow plan B. Ask yourself the question, is the potential energy increasing or decreasing if you push the dielectric some more in between the plates? Then you have to relate the answer to whether the force os attractive or repulsive. I already gave you some hints about that. hidemi You can follow plan B. Ask yourself the question, is the potential energy increasing or decreasing if you push the dielectric some more in between the plates? Then you have to relate the answer to whether the force os attractive or repulsive. I already gave you some hints about that. [/QUOTE] The potential energy is decreasing if I push the dielectric some more in between the plates.	1,038	4,483	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.953125	4	CC-MAIN-2023-50	latest	en	0.947447	# A slab of glass dielectric is inserted into a parallel plate capacitor. • hidemi. In summary, the equations used to understand this question/answer are C = k(εArea)/distance = Q/V = Q/(Edistance) and F = QE. As a slab of glass is added, k increases and E decreases, causing the force to also decrease. This relates to the idea that the force attracts the glass into the capacitor. The potential energy stored in the capacitor can be represented by the equation F=-dU/dx, where U is the potential energy, C is the capacitance, k is the dielectric constant, ε is the electric field, Area is the area of the plates, and distance is the distance between the plates. By pushing the dielectric further between the plates, the potential energy decreases. #### hidemi. Homework Statement. A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the. plates. As it is being inserted:. A. a force repels the glass out of the capacitor. B. a force attracts the glass into the capacitor. C. no force acts on the glass. D. a net charge appears on the glass. E. the glass makes the plates repel each other. Relevant Equations. C = k(εArea)/distance = Q/V = Q/ (Edistance). F = QE. I use the following equations to understand this question/answer.. First, C = k(εArea)/distance = Q/V = Q/ (Edistance). As a slab of glass is added, k increases and thus E decreases.. F=QE, as E decreases, force decreases as well. How does this relate to the 'force attracts the glass into the capacitor'. Let me know if my thoughts/logic on this is right. In addition, is there a less physic-based explanation? Thanks.. F = QE is not much help here.	The Q in equation Q = CV is the charge on the capacitor plates that produces electric field E between the plates. Using F = QE is like saying that the the charge exerts a force on itself.. I would use the definition of a force derived from a potential ##F=-\dfrac{dU}{dx}## where ##U## here is the potential energy stored in the capacitor. If you want a less physics-based explanation, you could perhaps draw an analogy with the gravitational field. If you release a rock it will move in the direction of decreasing potential energy. So if the dielectric is already half way in, which way must it move to reduce the potential energy, in or out of the plates? Remember the charge on the plates is constant.. Last edited:. hidemi. kuruman said:. F = QE is not much help here. The Q in equation Q = CV is the charge on the capacitor plates that produces electric field E between the plates. Using F = QE is like saying that the the charge exerts a force on itself.. I would use the definition of a force derived from a potential ##F=-\dfrac{dU}{dx}## where ##U## here is the potential energy stored in the capacitor. If you want a less physics-based explanation, you could perhaps draw an analogy with the gravitational field. If you release a rock it will move in the direction of decreasing potential energy. So if the dielectric is already half way in, which way must it move to reduce the potential energy, in or out of the plates? Remember the charge on the plates is constant.. I used your equation, F=-dU/dx, and substitute in relevant symbols as followed:. U = CV/2 = 1/2* kεA/d Ed = kεAE/2. Therefore, the force has nothing to do with the distance? How can I know if it is attracted or repeled?. You have the wrong ##x##. It is not the plate separation, it is the extent to which the dielectric is inserted between the plates. The derivative is with respect to that distance. The idea is to find the capacitance as a function of ##x## and then take the derivative. I have a feeling that this might be too complicated for you and you really don't need to do it that way.. You can follow plan B. Ask yourself the question, is the potential energy increasing or decreasing if you push the dielectric some more in between the plates? Then you have to relate the answer to whether the force os attractive or repulsive. I already gave you some hints about that.. hidemi. You can follow plan B. Ask yourself the question, is the potential energy increasing or decreasing if you push the dielectric some more in between the plates? Then you have to relate the answer to whether the force os attractive or repulsive. I already gave you some hints about that.. [/QUOTE]. The potential energy is decreasing if I push the dielectric some more in between the plates.
http://scs.math.yorku.ca/index.php?title=MATH_6627_2010-11_Practicum_in_Statistical_Consulting/Assignment_Teams/Gray&diff=prev&oldid=1521	1,571,061,435,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986653247.25/warc/CC-MAIN-20191014124230-20191014151730-00434.warc.gz	159,593,581	13,014	# MATH 6627 2010-11 Practicum in Statistical Consulting/Assignment Teams/Gray (Difference between revisions) Revision as of 10:26, 15 March 2011 (view source)Andytli (Talk \| contribs)← Older edit Revision as of 10:28, 15 March 2011 (view source)Andytli (Talk \| contribs) Newer edit → Line 277: Line 277: == Assignment 3== == Assignment 3== - 1) As we saw, 'Sector' appears to be an important predictor. Consider models using ses and Sector. Aim to estimate the between Sector gap as a function of ses if there is an interaction between Sector and ses. Check for and provide for a possible contextual effect of ses. Plot expected math achievement in each sector. Plot the gap with SEs. Consider the possibility that the apparently flatter effect of ses in Catholic school could be due to a non-linear effect of ses. How would you test whether this is a reasonable alternative explanation? + 1) As we saw, 'Sector' appears to be an important predictor. Consider models using ses and Sector. Aim to estimate the between Sector + gap as a function of ses if there is an interaction between Sector and ses. Check for and provide for a possible contextual effect of + ses. Plot expected math achievement in each sector. Plot the gap with SEs. Consider the possibility that the apparently flatter effect + of ses in Catholic school could be due to a non-linear effect of ses. How would you test whether this is a reasonable alternative explanation? - 2) Take the example further by incorporating Sex. Consider the the 'contextual effect' of Sex which is school sex composition. Note that there are three types of schools: Girls, Boys and Coed schools. If you consider an interaction between Sector and school gender composition, you will see that the Public Sector only has Coed schools. What is the consequence of this fact for modelling sex composition and Sector effects. + 2) Take the example further by incorporating Sex. Consider the the 'contextual effect' of Sex which is school sex composition. Note that + there are three types of schools: Girls, Boys and Coed schools. If you consider an interaction between Sector and school gender composition, + you will see that the Public Sector only has Coed schools. What is the consequence of this fact for modelling sex composition and Sector effects. - 3) Does it appear that boys are better off in a boy's school and girls in a girl's school or are they better off in coed schools? How would you qualify your findings so parents don't misinterpret them in making decisions for their children? + 3) Does it appear that boys are better off in a boy's school and girls in a girl's school or are they better off in coed schools? How would + you qualify your findings so parents don't misinterpret them in making decisions for their children? - 4) Is a low ses child better off in a high ses school or are they better off in a school of a similar ses? How about a high ses child in a low ses school? How would you qualify your findings so parents don't misinterpret them in making decisions for their children? + 4) Is a low ses child better off in a high ses school or are they better off in a school of a similar ses? How about a high ses child in a + low ses school? How would you qualify your findings so parents don't misinterpret them in making decisions for their children? - 5) Is a minority status child better off in a school with a higher proportion of minority status children or are they better off in a school with a low proportion? How would you qualify your findings so parents don't misinterpret them in making decisions for their children? + 5) Is a minority status child better off in a school with a higher proportion of minority status children or are they better off in a school + with a low proportion? How would you qualify your findings so parents don't misinterpret them in making decisions for their children? ## Assignment 1 Example: My friend and I play a basketball game and each shoot 20 shots. Who is the better shooter? But, who is the better shooter if you control for the distance of the shot? Who would you rather have on your team? This is question of Simpson's Paradox. We can see from this figure, the relationship changed from negative to positive when we took the distance to our consideration. Black line linked probability of we two made. Red line is linked our performance when far, but blue when close. Simpson’s paradox arises from one simple mathematical truth. Given eight real numbers: a, b, c, d, A, B, C, D with the following properties:, then it is not necessarily true that. In fact, it may be true that:. This is an obvious math reality, yet it has significant ramifications in Bayesian analysis, medical research, science and engineering studies, and societal statistical analysis. It is of concern for any statistical activity involving the calculation and analysis of ratios of two measurements. Exmaple 2 (Real Income tax example) ### 2. Graphics to visualize data Gray: rgl and p3d (include the use of the 'groups' parameter to produce trajectories ### Introduction rgl is a library of functions that offers 3D real-time visualization functionality to the R programming environment (Adler & Murdoch, 2010), providing OpenGL implemention for R. p3d is a library of functions which employs functions from RGL to help visualize statistical models expressed as a function of 2 independent variables with the possible addition of a categorical variable (Monette, 2009). ### Package rgl With rgl we create a ‘device’ , which is simply a window, within which a ‘world’ is created where we can create 3 dimensional shapes and through which we can navigate. Functions within the rgl package can be divided into 6 categories: (1) Device management functions (open and close devices, control active device) (2) Scene management functions (option to remove certain or all objects from the scene) (3) Export functions (creating image files) (4) Shape functions - essential plotting tools primitives (points, lines, triangles, quads) as well as higher level functions (text, spheres, surfaces). (5) Environment functions - modify the viewpoint, background and bounding box, adding light sources (6) Appearance function rgl.material(…). Using shapes and surfaces within an rgl device, statistical data can be represented in 3 dimensions. Some advanced examples are available as demos or provided on the rgl website. A few of the functions from rgl are useful for manipulating 3D models created using p3d, since p3d contains many functions that inherit from rgl but taylor them to statistical methods. Thus all but a few are unnecessary for our purposes unless you would like to contribute functionality to p3d! ### Package p3d In this section I will focus on example code you may use to familiarize yourself with the capabilities of this package. You will require the tuition.Rdata (source:) and USIndicesIndustrialProd.Rdata (source:) data sets. Note that a few of the commands employed in sample code are from rgl, but these will likely be superceded by p3d functions as the package matures. In this section I will focus on example code you may use to familiarize yourself with the capabilities of this package. You will require the File:Gray p3d ex Tuition.txt and File:Gray p3d ex USIndicesIndustrialProd.txt data sets. Note that a few of the commands employed in sample code are from rgl, but these will likely be superceded by p3d functions as the package matures. Initialization code: library( lattice ) library( nlme ) library( car ) library( spida ) library( rgl ) library( p3d ) For the tuition data we will begin by plotting the annual cost of tuition from a sample of American Universities against the rates of faculty compensation and proportion of students who graduate. Using mouse keys you can change the field of view and zoom in and out. Plot3d creates the 3D plot as shown on the right. We can remove elements from the device using the function Pop3d(). This function removes elements starting with the most recently added item. Multiple items can be removed addition an numeric argument, ie.Pop3d(4) Init3d(cex = .8) Plot3d(tuition ~ fac_comp + graduat, col = c("blue"), data = tuit) Next we will subdivide the data by category, in this case whether the school is private (red) or public (blue) (variable name public.private). Init3d(cex = .8) Plot3d( tuition ~ fac_comp + graduat\|public.private, col = c("blue", "red"), data = tuit) Next we will add regression planes for private(red) and public(blue) schools using the lm() function to determine the fit, and Fit3d() to insert the plane in the graph. Axes and labels are added using Axes3d() and title3d(). fitpub = lm(tuition ~ fac_comp + graduat,subset=(public.private==0),data = tuit) Fit3d( fitpub, col = c("blue")) fitpri = lm(tuition ~ fac_comp + graduat,subset=(public.private==1),data = tuit) Fit3d( fitpri, col = c("red")) Axes3d() title3d(main='Tuition predicted by grad rates and faculty salary -private (red) and public(blue) institutions') Data ellipses are useful for understanding our data. Ell3d() We can change the view point of our graph using function view3d(theta,phi,fov,zoom), which takes polar coordinates. Note that view3d(0,0,0) will rotate the image to to face the x-z plane (y into the screen) and view3d(270,0,0) will rotate the image to to face the y-z plane (x into the screen). Function snap() will capture a still image of the current view. Note that to use movie3d() you must have ImageMagick installed to automatically convert png's to gif, otherwise you must use external software. view3d(0,0,0) snap() spin(theta = 0, phi = 0) spins(inc.theta = 1/4, inc.phi = 0, theta = NULL, phi = NULL) movie3d( spin3d(axis=c(0,1,0), rpm=20), duration=2, dir='movie' ) Here is an additional example, using data on the US indices of industrial products, plotting Mining production (MIN) over months and years. Adding the argument ‘groups=YR’ to Plot3d connects the months in a given year to produce trajectories. open3d(windowRect=c(100,100,800,800),cex = .8) Plot3d(MIN ~ YR+MONTH,data=prod,groups=YR) Axes3d() title3d(main='Industrial Production Mining (1947-1993)') view3d(215,0,45) ## Assignment 2 Statistics in the News: "Spousal support a royal pain?" Journal Abstract Overly supportive spouses are not necessarily doing their partners a favour. They could be prolonging the recovery of their injured spouses. Men with highly attentive spouses reported higher levels of pain and more disability but did well on physical functions tests. Women with highly attentive spouses didn't report feeling more pain or being more disabled. However, they performed more poorly on physical function tests than did women with less attentive spouses #### 1 Question whether the article suggest a causal relationship between two variables? If so which? Are the data observational or experimental? Discussion Yes, the article did suggest a causal relationship. One variable is the spousal solicitousness (attentiveness & support). Another is the degree of reported pain and disability. A third is actual physical function. Patient gender was also taken into consideration. According to the report, men with chronic pain report more perceived pain and disability when they receive higher levels of spousal attention, and it is implied this is controlling for actual physical ability. For women there was no difference in self-reports of pain and disability with level of spousal support, but women who received more attention from their husbands had poorer physical function than those who did not. The data are observational, because there is no manipulation or randomization in collecting the data. #### 2 Question Can you think of alternative explanations to causality? Confounding factors? Or explanations consistent with causality? Mediating factors? Discussion Andy: Yes, maybe high degree of pain causes solicitousness, instead of solicitousness causing towards more pain. I don't think there exits a confounding factor. If there is, I believe it is LOVE, because a man's LOVE wins a highly attentive spouse and doesn't work as a narcotic as a woman feels. Then he may ask for more LOVEs by reporting hurt. That is a sweet answer to any woman, but tooth-hurting felt by any healthy man. Also, LOVE could be the mediating factor too. And with high solicitousness, man's LOVE starts to fall in. He may move his body, more likely, which may cause more hurt. And then his physical function may recover faster. But to a woman, she likes to finding pieces of LOVEs through solicitousness. Her heart is numbed with those LOVEs. And then she will report she is better as an affectionate payback. #### 3 Question Have any confounding factors been accounted for in the analysis? Discussion Emotional (men) or Physical (women) troubles outside the primary cause of pain. Men who are emotionally unstable may attract and encourage attentiveness in their wives, and also report a disproportionately high level of disability relative to their level of function. Women who are physically weak prior to injury may attract husbands who are highly supportive, but also be more susceptible to injuries that result in chronic pain. #### 4 Question Have any mediating factors been controlled for in a way that vitiates a causal interpretation of the relationship? Discussion This does not appear to be the case. #### 5 Question What is your personal assessment of the evidence for causality in the study that is the subject of the article? Discussion There is a good case to be made for both forward and backward causality in this study. In addition it would be nice to be able to ontrol for emotional and physical stability prior to chronic injury. Are frequencies distributed as expected? Are the proportions of highly attentive spouses equal across groups? ASSESSMENT: Judgement withheld pending further evidence #### 2. Question You are studying observational data on the relationship between Health and Coffee (measured in grams of caffeine consumed per day). Suppose you want to control for a possible confounding factor 'Stress'. In this kind of study it is more important to make sure that you measure coffee consumption accurately than it is to make sure that you measure 'stress' accurately. Discussion True. It is more important to accurately measure Coffee Consumption if this is the variable of primary interest in the study. Consider this from the analysis of variance framework. When measurement error is introduced, the variability in the outcome accounted for by that factor is decreased, and the error sums of squares increases. If we decrease accuracy in measuring Stress, and the error term will accordingly increase, which will in turn decrease the power to detect the effect of Coffee on Health somewhat, but will not affect the coefficient estimates by much. However, if we have increased measurement error in Coffee Consumption, then the proportion of variability explained by Coffee will decrease AND the error term will increase. We lose power on two fronts! Here is a short r-script that demonstrates this idea: File:Gray Q2MeasError.r #### 5. Question In a multiple regression of Y on three predictors, X1, X2 and X3, if the coefficients of both X2 and X3, are not significant, it is safe to drop these two variable and perform a regression on X1 alone. Discussion No. The way to do this is starting with 3 variables in the model, and dropping the least "significant", one at a time, until you are left with only "significant" variables. We can perform a stepwise selection, but drop variables which become no longer "significant" after introduction of new variables. #### 8. Question In a multiple regression, if you drop a predictor whose effect is not significant, the p-values of the other predictors should not change very much. Discussion No. After dropping the predictor, model changes. If we drop a predictor whose effect is not significant, the p-values of other predictors will change. We can see this from this example @ page 26-28 #### 11. Question In a model to assess the effect of a number of treatments on some outcome, we can estimate the difference between the best treatment and the worse treatment by using the difference in the mean outcomes. Discussion No. Size of tumors does provide some info about two treatments, one is the best and another is the worse. But it is not a good idea to estimate the difference between the two treatments by using the difference of size of tumors. It is so biased, because there are other factors which will effect the outcome, such as time in the disease. For example, our target is New York when we drive out of Toronto. But some day, we find everyone around is speaking Spanish. But the answer may be yes too, as Barack Obama's slogan showed: "Yes we can". #### 14. Question If two variables have a strong interaction, this implies a strong correlation. Discussion False. An interaction exists if the effect of one independent on the dependent variable varies over another independent variable. This tells us nothing about the relationship between the IVs. An interaction between two variables can occur whether or not the variables are correlated with one another. No Interaction Interaction No Correlation Correlation Here is an additional example using binary variables: Illustration of the difference between correlation and interaction amongst independent variables ## Assignment 3 1) As we saw, 'Sector' appears to be an important predictor. Consider models using ses and Sector. Aim to estimate the between Sector gap as a function of ses if there is an interaction between Sector and ses. Check for and provide for a possible contextual effect of ses. Plot expected math achievement in each sector. Plot the gap with SEs. Consider the possibility that the apparently flatter effect of ses in Catholic school could be due to a non-linear effect of ses. How would you test whether this is a reasonable alternative explanation? 2) Take the example further by incorporating Sex. Consider the the 'contextual effect' of Sex which is school sex composition. Note that there are three types of schools: Girls, Boys and Coed schools. If you consider an interaction between Sector and school gender composition, you will see that the Public Sector only has Coed schools. What is the consequence of this fact for modelling sex composition and Sector effects. 3) Does it appear that boys are better off in a boy's school and girls in a girl's school or are they better off in coed schools? How would you qualify your findings so parents don't misinterpret them in making decisions for their children? 4) Is a low ses child better off in a high ses school or are they better off in a school of a similar ses? How about a high ses child in a low ses school? How would you qualify your findings so parents don't misinterpret them in making decisions for their children? 5) Is a minority status child better off in a school with a higher proportion of minority status children or are they better off in a school with a low proportion? How would you qualify your findings so parents don't misinterpret them in making decisions for their children?	4,176	19,273	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2019-43	longest	en	0.941866	# MATH 6627 2010-11 Practicum in Statistical Consulting/Assignment Teams/Gray. (Difference between revisions). Revision as of 10:26, 15 March 2011 (view source)Andytli (Talk \| contribs)← Older edit Revision as of 10:28, 15 March 2011 (view source)Andytli (Talk \| contribs) Newer edit → Line 277: Line 277: == Assignment 3== == Assignment 3== - 1) As we saw, 'Sector' appears to be an important predictor. Consider models using ses and Sector. Aim to estimate the between Sector gap as a function of ses if there is an interaction between Sector and ses. Check for and provide for a possible contextual effect of ses. Plot expected math achievement in each sector. Plot the gap with SEs. Consider the possibility that the apparently flatter effect of ses in Catholic school could be due to a non-linear effect of ses. How would you test whether this is a reasonable alternative explanation? + 1) As we saw, 'Sector' appears to be an important predictor. Consider models using ses and Sector. Aim to estimate the between Sector + gap as a function of ses if there is an interaction between Sector and ses. Check for and provide for a possible contextual effect of + ses. Plot expected math achievement in each sector. Plot the gap with SEs. Consider the possibility that the apparently flatter effect + of ses in Catholic school could be due to a non-linear effect of ses. How would you test whether this is a reasonable alternative explanation? - 2) Take the example further by incorporating Sex. Consider the the 'contextual effect' of Sex which is school sex composition. Note that there are three types of schools: Girls, Boys and Coed schools. If you consider an interaction between Sector and school gender composition, you will see that the Public Sector only has Coed schools. What is the consequence of this fact for modelling sex composition and Sector effects. + 2) Take the example further by incorporating Sex. Consider the the 'contextual effect' of Sex which is school sex composition. Note that + there are three types of schools: Girls, Boys and Coed schools. If you consider an interaction between Sector and school gender composition, + you will see that the Public Sector only has Coed schools. What is the consequence of this fact for modelling sex composition and Sector effects. - 3) Does it appear that boys are better off in a boy's school and girls in a girl's school or are they better off in coed schools? How would you qualify your findings so parents don't misinterpret them in making decisions for their children? + 3) Does it appear that boys are better off in a boy's school and girls in a girl's school or are they better off in coed schools? How would + you qualify your findings so parents don't misinterpret them in making decisions for their children? - 4) Is a low ses child better off in a high ses school or are they better off in a school of a similar ses? How about a high ses child in a low ses school? How would you qualify your findings so parents don't misinterpret them in making decisions for their children? + 4) Is a low ses child better off in a high ses school or are they better off in a school of a similar ses? How about a high ses child in a + low ses school? How would you qualify your findings so parents don't misinterpret them in making decisions for their children? - 5) Is a minority status child better off in a school with a higher proportion of minority status children or are they better off in a school with a low proportion? How would you qualify your findings so parents don't misinterpret them in making decisions for their children? + 5) Is a minority status child better off in a school with a higher proportion of minority status children or are they better off in a school + with a low proportion? How would you qualify your findings so parents don't misinterpret them in making decisions for their children?. ## Assignment 1. Example: My friend and I play a basketball game and each shoot 20 shots. Who is the better shooter?. But, who is the better shooter if you control for the distance of the shot? Who would you rather have on your team?. This is question of Simpson's Paradox.. We can see from this figure, the relationship changed from negative to positive when we took the distance to our consideration. Black line linked probability of we two made. Red line is linked our performance when far, but blue when close.. Simpson’s paradox arises from one simple mathematical truth. Given eight real numbers: a, b, c, d, A, B, C, D with the following properties:, then it is not necessarily true that. In fact, it may be true that:.. This is an obvious math reality, yet it has significant ramifications in Bayesian analysis, medical research, science and engineering studies, and societal statistical analysis. It is of concern for any statistical activity involving the calculation and analysis of ratios of two measurements.. Exmaple 2 (Real Income tax example). ### 2. Graphics to visualize data. Gray: rgl and p3d (include the use of the 'groups' parameter to produce trajectories. ### Introduction. rgl is a library of functions that offers 3D real-time visualization functionality to the R programming environment (Adler & Murdoch, 2010), providing OpenGL implemention for R.. p3d is a library of functions which employs functions from RGL to help visualize statistical models expressed as a function of 2 independent variables with the possible addition of a categorical variable (Monette, 2009).. ### Package rgl. With rgl we create a ‘device’ , which is simply a window, within which a ‘world’ is created where we can create 3 dimensional shapes and through which we can navigate.. Functions within the rgl package can be divided into 6 categories: (1) Device management functions (open and close devices, control active device) (2) Scene management functions (option to remove certain or all objects from the scene) (3) Export functions (creating image files) (4) Shape functions - essential plotting tools primitives (points, lines, triangles, quads) as well as higher level functions (text, spheres, surfaces).. (5) Environment functions - modify the viewpoint, background and bounding box, adding light sources (6) Appearance function rgl.material(…).. Using shapes and surfaces within an rgl device, statistical data can be represented in 3 dimensions. Some advanced examples are available as demos or provided on the rgl website.. A few of the functions from rgl are useful for manipulating 3D models created using p3d, since p3d contains many functions that inherit from rgl but taylor them to statistical methods. Thus all but a few are unnecessary for our purposes unless you would like to contribute functionality to p3d!. ### Package p3d. In this section I will focus on example code you may use to familiarize yourself with the capabilities of this package. You will require the tuition.Rdata (source:) and USIndicesIndustrialProd.Rdata (source:) data sets. Note that a few of the commands employed in sample code are from rgl, but these will likely be superceded by p3d functions as the package matures.. In this section I will focus on example code you may use to familiarize yourself with the capabilities of this package. You will require the File:Gray p3d ex Tuition.txt and File:Gray p3d ex USIndicesIndustrialProd.txt data sets. Note that a few of the commands employed in sample code are from rgl, but these will likely be superceded by p3d functions as the package matures.. Initialization code:. library( lattice ). library( nlme ). library( car ). library( spida ). library( rgl ). library( p3d ). For the tuition data we will begin by plotting the annual cost of tuition from a sample of American Universities against the rates of faculty compensation and proportion of students who graduate.. Using mouse keys you can change the field of view and zoom in and out. Plot3d creates the 3D plot as shown on the right.. We can remove elements from the device using the function Pop3d(). This function removes elements starting with the most recently added item. Multiple items can be removed addition an numeric argument, ie.Pop3d(4). Init3d(cex = .8). Plot3d(tuition ~ fac_comp + graduat, col = c("blue"), data = tuit). Next we will subdivide the data by category, in this case whether the school is private (red) or public (blue) (variable name public.private).. Init3d(cex = .8). Plot3d( tuition ~ fac_comp + graduat\|public.private, col = c("blue", "red"), data = tuit). Next we will add regression planes for private(red) and public(blue) schools using the lm() function to determine the fit, and Fit3d() to insert the plane in the graph. Axes and labels are added using Axes3d() and title3d().. fitpub = lm(tuition ~ fac_comp + graduat,subset=(public.private==0),data = tuit). Fit3d( fitpub, col = c("blue")). fitpri = lm(tuition ~ fac_comp + graduat,subset=(public.private==1),data = tuit). Fit3d( fitpri, col = c("red")). Axes3d(). title3d(main='Tuition predicted by grad rates and faculty salary -private (red) and public(blue) institutions'). Data ellipses are useful for understanding our data.. Ell3d(). We can change the view point of our graph using function view3d(theta,phi,fov,zoom), which takes polar coordinates. Note that view3d(0,0,0) will rotate the image to to face the x-z plane (y into the screen) and view3d(270,0,0) will rotate the image to to face the y-z plane (x into the screen). Function snap() will capture a still image of the current view. Note that to use movie3d() you must have ImageMagick installed to automatically convert png's to gif, otherwise you must use external software.. view3d(0,0,0). snap(). spin(theta = 0, phi = 0). spins(inc.theta = 1/4, inc.phi = 0, theta = NULL, phi = NULL). movie3d( spin3d(axis=c(0,1,0), rpm=20), duration=2, dir='movie' ). Here is an additional example, using data on the US indices of industrial products, plotting Mining production (MIN) over months and years. Adding the argument ‘groups=YR’ to Plot3d connects the months in a given year to produce trajectories.. open3d(windowRect=c(100,100,800,800),cex = .8). Plot3d(MIN ~ YR+MONTH,data=prod,groups=YR). Axes3d(). title3d(main='Industrial Production Mining (1947-1993)'). view3d(215,0,45). ## Assignment 2. Statistics in the News: "Spousal support a royal pain?" Journal Abstract. Overly supportive spouses are not necessarily doing their partners a favour.. They could be prolonging the recovery of their injured spouses.. Men with highly attentive spouses reported higher levels of pain and more disability but did well on physical functions tests.. Women with highly attentive spouses didn't report feeling more pain or being more disabled. However, they performed more poorly on physical function tests than did women with less attentive spouses. #### 1. Question.	whether the article suggest a causal relationship between two variables? If so which? Are the data observational or experimental?. Discussion. Yes, the article did suggest a causal relationship. One variable is the spousal solicitousness (attentiveness & support). Another is the degree of reported pain and disability. A third is actual physical function. Patient gender was also taken into consideration.. According to the report, men with chronic pain report more perceived pain and disability when they receive higher levels of spousal attention, and it is implied this is controlling for actual physical ability. For women there was no difference in self-reports of pain and disability with level of spousal support, but women who received more attention from their husbands had poorer physical function than those who did not.. The data are observational, because there is no manipulation or randomization in collecting the data.. #### 2. Question. Can you think of alternative explanations to causality? Confounding factors? Or explanations consistent with causality? Mediating factors?. Discussion. Andy: Yes, maybe high degree of pain causes solicitousness, instead of solicitousness causing towards more pain. I don't think there exits a confounding factor. If there is, I believe it is LOVE, because a man's LOVE wins a highly attentive spouse and doesn't work as a narcotic as a woman feels. Then he may ask for more LOVEs by reporting hurt. That is a sweet answer to any woman, but tooth-hurting felt by any healthy man.. Also, LOVE could be the mediating factor too. And with high solicitousness, man's LOVE starts to fall in. He may move his body, more likely, which may cause more hurt. And then his physical function may recover faster. But to a woman, she likes to finding pieces of LOVEs through solicitousness. Her heart is numbed with those LOVEs. And then she will report she is better as an affectionate payback.. #### 3. Question. Have any confounding factors been accounted for in the analysis?. Discussion. Emotional (men) or Physical (women) troubles outside the primary cause of pain. Men who are emotionally unstable may attract and encourage attentiveness in their wives, and also report a disproportionately high level of disability relative to their level of function. Women who are physically weak prior to injury may attract husbands who are highly supportive, but also be more susceptible to injuries that result in chronic pain.. #### 4. Question. Have any mediating factors been controlled for in a way that vitiates a causal interpretation of the relationship?. Discussion. This does not appear to be the case.. #### 5. Question. What is your personal assessment of the evidence for causality in the study that is the subject of the article?. Discussion. There is a good case to be made for both forward and backward causality in this study. In addition it would be nice to be able to ontrol for emotional and physical stability prior to chronic injury. Are frequencies distributed as expected? Are the proportions of highly attentive spouses equal across groups? ASSESSMENT: Judgement withheld pending further evidence. #### 2.. Question. You are studying observational data on the relationship between Health and Coffee (measured in grams of caffeine consumed per day). Suppose you want to control for a possible confounding factor 'Stress'. In this kind of study it is more important to make sure that you measure coffee consumption accurately than it is to make sure that you measure 'stress' accurately.. Discussion. True. It is more important to accurately measure Coffee Consumption if this is the variable of primary interest in the study.. Consider this from the analysis of variance framework. When measurement error is introduced, the variability in the outcome accounted for by that factor is decreased, and the error sums of squares increases. If we decrease accuracy in measuring Stress, and the error term will accordingly increase, which will in turn decrease the power to detect the effect of Coffee on Health somewhat, but will not affect the coefficient estimates by much. However, if we have increased measurement error in Coffee Consumption, then the proportion of variability explained by Coffee will decrease AND the error term will increase. We lose power on two fronts!. Here is a short r-script that demonstrates this idea: File:Gray Q2MeasError.r. #### 5.. Question. In a multiple regression of Y on three predictors, X1, X2 and X3, if the coefficients of both X2 and X3, are not significant, it is safe to drop these two variable and perform a regression on X1 alone.. Discussion. No. The way to do this is starting with 3 variables in the model, and dropping the least "significant", one at a time, until you are left with only "significant" variables. We can perform a stepwise selection, but drop variables which become no longer "significant" after introduction of new variables.. #### 8.. Question. In a multiple regression, if you drop a predictor whose effect is not significant, the p-values of the other predictors should not change very much.. Discussion. No. After dropping the predictor, model changes. If we drop a predictor whose effect is not significant, the p-values of other predictors will change. We can see this from this example @ page 26-28. #### 11.. Question. In a model to assess the effect of a number of treatments on some outcome, we can estimate the difference between the best treatment and the worse treatment by using the difference in the mean outcomes.. Discussion. No. Size of tumors does provide some info about two treatments, one is the best and another is the worse. But it is not a good idea to estimate the difference between the two treatments by using the difference of size of tumors. It is so biased, because there are other factors which will effect the outcome, such as time in the disease. For example, our target is New York when we drive out of Toronto. But some day, we find everyone around is speaking Spanish. But the answer may be yes too, as Barack Obama's slogan showed: "Yes we can".. #### 14.. Question. If two variables have a strong interaction, this implies a strong correlation.. Discussion. False. An interaction exists if the effect of one independent on the dependent variable varies over another independent variable. This tells us nothing about the relationship between the IVs. An interaction between two variables can occur whether or not the variables are correlated with one another.. No Interaction Interaction. No Correlation. Correlation. Here is an additional example using binary variables: Illustration of the difference between correlation and interaction amongst independent variables. ## Assignment 3. 1) As we saw, 'Sector' appears to be an important predictor. Consider models using ses and Sector. Aim to estimate the between Sector gap as a function of ses if there is an interaction between Sector and ses. Check for and provide for a possible contextual effect of ses. Plot expected math achievement in each sector. Plot the gap with SEs. Consider the possibility that the apparently flatter effect of ses in Catholic school could be due to a non-linear effect of ses. How would you test whether this is a reasonable alternative explanation?. 2) Take the example further by incorporating Sex. Consider the the 'contextual effect' of Sex which is school sex composition. Note that there are three types of schools: Girls, Boys and Coed schools. If you consider an interaction between Sector and school gender composition, you will see that the Public Sector only has Coed schools. What is the consequence of this fact for modelling sex composition and Sector effects.. 3) Does it appear that boys are better off in a boy's school and girls in a girl's school or are they better off in coed schools? How would you qualify your findings so parents don't misinterpret them in making decisions for their children?. 4) Is a low ses child better off in a high ses school or are they better off in a school of a similar ses? How about a high ses child in a low ses school? How would you qualify your findings so parents don't misinterpret them in making decisions for their children?. 5) Is a minority status child better off in a school with a higher proportion of minority status children or are they better off in a school with a low proportion? How would you qualify your findings so parents don't misinterpret them in making decisions for their children?.
https://drserendipity.com/notes/notes_by_subjects/artificial_intelligence/natural-language-processing/4-extracurricular/4-1-recurrent-neural-networks/4-1-1-recurrent-neural-networks/11-10-backpropagation-example-a-v3-final/	1,685,824,684,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224649343.34/warc/CC-MAIN-20230603201228-20230603231228-00785.warc.gz	251,951,615	33,531	# 11 – 10 Backpropagation Example A V3 Final Remember our feedforward illustration? We had n inputs, three neurons in the hidden layer, and two outputs. For this example, we will need to simplify things even more and look at a model with two inputs, x_1 and x_2, and a single output, y. We will have a weight matrix, W_1, from the input to the hidden layer, the elements of the matrix will be W_ij just as before. We will also have a weight matrix vector, W_2, from the hidden layer to the output. Notice that it’s a vector and not a matrix as we only have one output. Don’t forget your pencil and notes. Pause the video any time you need and make sure you’re following the math here. We will begin with a feedforward pass of the inputs across the network, then calculate the output, and based on the error, use backpropagation to calculate the partial derivatives. Calculating the values of the activations at each of the three hidden neurons is simple. We have a linear combination of the inputs with a corresponding weight elements of the matrix W_1. All that is followed by an activation function. The outputs are a dot product of the activations of the previous layer vector, H, with the weights of W2. As I mentioned before, the aim of the back propagation process is to minimize, in our case, the loss function or the square error. To do that, we need to calculate the partial derivatives of the square error with respect to each of the weights. Since we just found the output, we can now minimize the error by finding the updated values of delta of W_ij and, of course, in every case, we need to do so for every single layer K. This value equals to the negative of alpha multiplied by the partial derivative of the loss function e. Since the error is a polynomial, finding its derivative is immediate. By using basic calculus, we can seek that this incremental value is simply the learning rate alpha multiplied by d minus y, and by the partial derivative of y with respect to each of the weights. If you’re asking yourselves, what happened to the desired output D? Well, that was a constant value, so it’s partial derivative was simply a zero. Notice that I used the chain rule here? In the beginning of this video, I mentioned that backpropagation is actually stochastic gradient descent with the use of the chain rule. Well, now we have our gradient. The symbol that we use for the gradient is usually lowercase delta. This partial derivative of the calculated output defines the gradient and we will find it by using the chain rule. So let’s pause for a minute, remind ourselves what the chain rule is and how it’s used. When you’re feeling comfortable with this concept, we will continue with the actual calculation.	587	2,730	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.75	4	CC-MAIN-2023-23	latest	en	0.917582	# 11 – 10 Backpropagation Example A V3 Final. Remember our feedforward illustration? We had n inputs, three neurons in the hidden layer, and two outputs. For this example, we will need to simplify things even more and look at a model with two inputs, x_1 and x_2, and a single output, y. We will have a weight matrix, W_1, from the input to the hidden layer, the elements of the matrix will be W_ij just as before. We will also have a weight matrix vector, W_2, from the hidden layer to the output. Notice that it’s a vector and not a matrix as we only have one output. Don’t forget your pencil and notes. Pause the video any time you need and make sure you’re following the math here. We will begin with a feedforward pass of the inputs across the network, then calculate the output, and based on the error, use backpropagation to calculate the partial derivatives. Calculating the values of the activations at each of the three hidden neurons is simple. We have a linear combination of the inputs with a corresponding weight elements of the matrix W_1. All that is followed by an activation function. The outputs are a dot product of the activations of the previous layer vector, H, with the weights of W2. As I mentioned before, the aim of the back propagation process is to minimize, in our case, the loss function or the square error.	To do that, we need to calculate the partial derivatives of the square error with respect to each of the weights. Since we just found the output, we can now minimize the error by finding the updated values of delta of W_ij and, of course, in every case, we need to do so for every single layer K. This value equals to the negative of alpha multiplied by the partial derivative of the loss function e. Since the error is a polynomial, finding its derivative is immediate. By using basic calculus, we can seek that this incremental value is simply the learning rate alpha multiplied by d minus y, and by the partial derivative of y with respect to each of the weights. If you’re asking yourselves, what happened to the desired output D? Well, that was a constant value, so it’s partial derivative was simply a zero. Notice that I used the chain rule here? In the beginning of this video, I mentioned that backpropagation is actually stochastic gradient descent with the use of the chain rule. Well, now we have our gradient. The symbol that we use for the gradient is usually lowercase delta. This partial derivative of the calculated output defines the gradient and we will find it by using the chain rule. So let’s pause for a minute, remind ourselves what the chain rule is and how it’s used. When you’re feeling comfortable with this concept, we will continue with the actual calculation.
http://dev.primegrid.com/forum_thread.php?id=6903	1,600,903,437,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600400212959.12/warc/CC-MAIN-20200923211300-20200924001300-00316.warc.gz	41,771,476	6,091	## Other drummers-lowrise Message boards : Proth Prime Search : What does "Factors found: 4 (avg. 4/task") mean? Subscribe SortOldest firstNewest firstHighest rated posts first Author Message Kyle Joined: 1 Jun 14 Posts: 42 ID: 315843 Credit: 3,136,318 RAC: 0 Message 96808 - Posted: 13 Jul 2016 \| 14:01:54 UTC What does "Factors found" mean? Thank you. Michael Goetz Volunteer moderator Project scientist Joined: 21 Jan 10 Posts: 12669 ID: 53948 Credit: 184,159,862 RAC: 18 Message 96809 - Posted: 13 Jul 2016 \| 14:27:49 UTC - in response to Message 96808. What does "Factors found" mean? Thank you. The process of searching for prime numbers starts with a large set of candidates. The status of each individual candidate (number) is initially unknown. It's either prime, or more likely, it's composite, but we don't know which. Testing each individual number takes a long time, so we use a more efficient strategy: We first run a sieve, which rapidly is able to establish that many of the candidates are composite. Once we know a candidate is composite, we're done with it and don't need to consider it anymore. Once we're done with the sieve, we're left with a much smaller set of candidates whose status is unknown. We then test this smaller set, one candidate at a time, with a program such as LLR or Genefer to determine its final status as prime or composite. (Genefer actually tests for a candidate being a "probable prime", but that's not relevant to this discussion.) Now, back to your question. When the sieve proves that a candidate is composite, it does so by finding a factor of that candidate, i.e., it finds a small prime number that divides the candidate. The statistics showing number of factors found is therefore indicating how many candidates your sieve tasks have proven to be composite. (And, in case your next question is "How can I see my factors?", you can't. We don't store that information. The number of factors found at the beginning of a sieve is prohibitively huge, and storing a record of who found what is impractical.) ____________ My lucky number is 75898524288+1 Kyle Joined: 1 Jun 14 Posts: 42 ID: 315843 Credit: 3,136,318 RAC: 0 Message 96830 - Posted: 14 Jul 2016 \| 14:16:48 UTC - in response to Message 96809. Thank you! Message boards : Proth Prime Search : What does "Factors found: 4 (avg. 4/task") mean?	622	2,370	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2020-40	latest	en	0.936647	## Other. drummers-lowrise. Message boards : Proth Prime Search : What does "Factors found: 4 (avg. 4/task") mean?. Subscribe SortOldest firstNewest firstHighest rated posts first. Author Message. Kyle. Joined: 1 Jun 14. Posts: 42. ID: 315843. Credit: 3,136,318. RAC: 0. Message 96808 - Posted: 13 Jul 2016 \| 14:01:54 UTC. What does "Factors found" mean? Thank you.. Michael Goetz. Volunteer moderator. Project scientist. Joined: 21 Jan 10. Posts: 12669. ID: 53948. Credit: 184,159,862. RAC: 18. Message 96809 - Posted: 13 Jul 2016 \| 14:27:49 UTC - in response to Message 96808.. What does "Factors found" mean? Thank you.. The process of searching for prime numbers starts with a large set of candidates. The status of each individual candidate (number) is initially unknown.	It's either prime, or more likely, it's composite, but we don't know which.. Testing each individual number takes a long time, so we use a more efficient strategy: We first run a sieve, which rapidly is able to establish that many of the candidates are composite. Once we know a candidate is composite, we're done with it and don't need to consider it anymore.. Once we're done with the sieve, we're left with a much smaller set of candidates whose status is unknown. We then test this smaller set, one candidate at a time, with a program such as LLR or Genefer to determine its final status as prime or composite. (Genefer actually tests for a candidate being a "probable prime", but that's not relevant to this discussion.). Now, back to your question. When the sieve proves that a candidate is composite, it does so by finding a factor of that candidate, i.e., it finds a small prime number that divides the candidate. The statistics showing number of factors found is therefore indicating how many candidates your sieve tasks have proven to be composite.. (And, in case your next question is "How can I see my factors?", you can't. We don't store that information. The number of factors found at the beginning of a sieve is prohibitively huge, and storing a record of who found what is impractical.). ____________. My lucky number is 75898524288+1. Kyle. Joined: 1 Jun 14. Posts: 42. ID: 315843. Credit: 3,136,318. RAC: 0. Message 96830 - Posted: 14 Jul 2016 \| 14:16:48 UTC - in response to Message 96809.. Thank you!. Message boards : Proth Prime Search : What does "Factors found: 4 (avg. 4/task") mean?.
http://www.cs.sfu.ca/CourseCentral/354/zaiane/material/notes/Chapter6/node12.html	1,500,562,047,000,000,000	text/html	crawl-data/CC-MAIN-2017-30/segments/1500549423222.65/warc/CC-MAIN-20170720141821-20170720161821-00350.warc.gz	389,439,062	2,513	Next: Closure of Attribute Sets Up: Functional Dependencies Previous: Basic Concepts ## Closure of a Set of Functional Dependencies 1. We need to consider all functional dependencies that hold. Given a set F of functional dependencies, we can prove that certain other ones also hold. We say these ones are logically implied by F. 2. Suppose we are given a relation scheme R=(A,B,C,G,H,I), and the set of functional dependencies: A B A C CG H CG I B H Then the functional dependency is logically implied. 3. To see why, let and be tuples such that As we are given A B , it follows that we must also have Further, since we also have B H , we must also have Thus, whenever two tuples have the same value on A, they must also have the same value on H, and we can say that A H . 4. The closure of a set F of functional dependencies is the set of all functional dependencies logically implied by F. 5. We denote the closure of F by . 6. To compute , we can use some rules of inference called Armstrong's Axioms: • Reflexivity rule: if is a set of attributes and , then holds. • Augmentation rule: if holds, and is a set of attributes, then holds. • Transitivity rule: if holds, and holds, then holds. 7. These rules are sound because they do not generate any incorrect functional dependencies. They are also complete as they generate all of . 8. To make life easier we can use some additional rules, derivable from Armstrong's Axioms: • Union rule: if and , then holds. • Decomposition rule: if holds, then and both hold. • Pseudotransitivity rule: if holds, and holds, then holds. 9. Applying these rules to the scheme and set F mentioned above, we can derive the following: • A H, as we saw by the transitivity rule. • CG HI by the union rule. • AG I by several steps: • Note that A C holds. • Then AG CG , by the augmentation rule. • Now by transitivity, AG I . (You might notice that this is actually pseudotransivity if done in one step.) Next: Closure of Attribute Sets Up: Functional Dependencies Previous: Basic Concepts Osmar Zaiane Tue Jun 9 15:12:55 PDT 1998	501	2,081	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5	4	CC-MAIN-2017-30	longest	en	0.924121	Next: Closure of Attribute Sets Up: Functional Dependencies Previous: Basic Concepts. ## Closure of a Set of Functional Dependencies. 1. We need to consider all functional dependencies that hold. Given a set F of functional dependencies, we can prove that certain other ones also hold. We say these ones are logically implied by F.. 2. Suppose we are given a relation scheme R=(A,B,C,G,H,I), and the set of functional dependencies:. A B. A C. CG H. CG I. B H. Then the functional dependency is logically implied.. 3. To see why, let and be tuples such that. As we are given A B , it follows that we must also have. Further, since we also have B H , we must also have. Thus, whenever two tuples have the same value on A, they must also have the same value on H, and we can say that A H .. 4. The closure of a set F of functional dependencies is the set of all functional dependencies logically implied by F.. 5. We denote the closure of F by .. 6. To compute , we can use some rules of inference called Armstrong's Axioms:.	• Reflexivity rule: if is a set of attributes and , then holds.. • Augmentation rule: if holds, and is a set of attributes, then holds.. • Transitivity rule: if holds, and holds, then holds.. 7. These rules are sound because they do not generate any incorrect functional dependencies. They are also complete as they generate all of .. 8. To make life easier we can use some additional rules, derivable from Armstrong's Axioms:. • Union rule: if and , then holds.. • Decomposition rule: if holds, then and both hold.. • Pseudotransitivity rule: if holds, and holds, then holds.. 9. Applying these rules to the scheme and set F mentioned above, we can derive the following:. • A H, as we saw by the transitivity rule.. • CG HI by the union rule.. • AG I by several steps:. • Note that A C holds.. • Then AG CG , by the augmentation rule.. • Now by transitivity, AG I .. (You might notice that this is actually pseudotransivity if done in one step.). Next: Closure of Attribute Sets Up: Functional Dependencies Previous: Basic Concepts. Osmar Zaiane. Tue Jun 9 15:12:55 PDT 1998.
https://www.geeksforgeeks.org/class-10-rd-sharma-solutions-chapter-14-coordinate-geometry-exercise-14-2-set-3/?type=article&id=595200	1,685,482,100,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224646144.69/warc/CC-MAIN-20230530194919-20230530224919-00065.warc.gz	855,399,353	39,829	GeeksforGeeks App Open App Browser Continue ## Related Articles • RD Sharma Class 10 Solutions # Class 10 RD Sharma Solutions – Chapter 14 Coordinate Geometry – Exercise 14.2 \| Set 3 ### Question 40. Prove that the points (3, 0), (4, 5), (-1, 4), and (-2, -1) taken in order, form a rhombus. Also, find its area. Solution: Let us considered the given points are A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1) Now we find the length of the sides and diagonals, By using distance formula So, AB = AB2 = (4 + 3)2 + (5 – 0) = (1)2 + (5)2 = 1 + 25 = 26 Similarly, BC2 = (-1 – 4)2 + (4 – 5)2 = (-5)2 + (-1)2 = 25 + 1 = 26 CD2 = (-2 + 1)2 + (-1 – 4)2 = (-1)2 + (-5)2 = 1 + 25 = 26 and DA2 = (3 + 2)2 + (0 + 1) = (5)2 + (1)2 = 25 + 1 = 26 Diagonal AC2 = (-1 – 3)2 + (4 – 0)2 = (-4)2 + (4)2 = 16 + 16 = 32 and BD2 = (-2 – 4)2 + (-1 – 5)2 = (-6)2 + (-6)2 = 36 + 36 = 72 So, we conclude that the sides AB = BC = CD = DA and diagonal AC is not equal to BD Hence, ABCD is a rhombus Now we find the area of rhombus ABCD = Product of diagonals/2 = (√32 × √72)/2 = (√16 × 2 × 2 × 36)/2 = 4 × 2 × 6/2 = 24 sq. units ### Question 41. In the seating arrangement of desks in a classroom three students Rohini, Sandhya, and Bina are seated at A (3, 1), B (6, 4), and C (8, 6). Do you think they are seated in a line? Solution: Given that A (3, 1), B (6, 4) and C (8, 6) Now we find the length of the sides and diagonals, By using distance formula AB = AB2 = (6 – 3)2 + (4 – 1)2 = (3)2 + (3)2 = 9 + 9 = 18 Similarly, BC2 = (8 – 6)2 + (6 – 4)2 = (2)2 + (2)2 = 4 + 4 = 8 and BC2 = (8 – 6)2 + (6 – 4)2 = (2)2 + (2)2 = 4 + 4 = 8 and CA2 = (3 – 8)2 + (1 – 6)2 = (-5)2 + (-5)2 = 25 + 25 = 50 AB = √18 = √9 * 2 = 3√2 BC = √8 = √4 * 2 = 2√2 and CA = √50 = √25 * 2 = 5√2 AB + BC = 3√2 + 2√2 = 5√2 = CA Hence, A, B and C are collinear points. Hence, they are seated in a line. ### Question 42. Find a point on y-axis which is equidistant from the points (5, -2) and (-3, 2). Solution: Let us assume P be the point lies on y-axis. So, its x = 0, so the coordinates of P is (0, y) It is given that the point P (0, y) is equidistant from the points A(5, -2) and B(-3, 2). So, PA = PB Also, PA2 = PB2 Now by using distance formula, we get (5 – 0)2 + (-2 – y)2 = (-3 – 0)2 + (2 – y)2 25 + 4 + y2 + 4y = 9 + 4 – 4y + y2 y2 + 4y + 4y – y2 = 13 – 29 8y = -16 y = -16/8 = 2 Hence, the required point P is (0,-2) ### Question 43. Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4). Solution: Let us considered P (x, y) is equidistant from A(3, 6) and B(-3, 4) So, PA = PB Also, PA2 = PB2 Now by using distance formula, we get On squaring both sides, we get (x – 3)2 + (y – 6)2 = (x + 3)2 + (y – 4)2 x2 – 6x + 9 + y2 – 12y + 36 = x2 + 6x + 9 = y2 – 8y + 16 -6x – 12y + 45 = 6x – 8y + 25 -6x – 6x – 12y + 8y + 45 – 25 = 0 -12 – 4y + 20 = 0 3x + y – 5 = 0 3x + y = 5 ### Question 44. If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of p. Solution: Given that the point A (0, 2) is equidistant from the points B (3, p) and C (p, 5) Now by using distance formula, we get It is given that AB = AC √p2 – 4p + 13 = √p2 + 9 So, on squaring both side, we get = p2 – 4p + 13 = p2 + 9 p2 – 4p – p2 = 9 – 13 -4p = -4 p = 1 Hence, the value of p is 1 ### Question 45. Prove that the points (7, 10), (-2, 5), and (3, -4) are the vertices of an isosceles right triangle. Solution: Let us considered the points are A (7, 10), B (-2, 5) and C (3, -4) Now we find the length of the sides By using distance formula Now AB = Similarly, BC = and AC = So, we conclude that AB = BC = √106 and AB2 + BC2 = AC2 Hence, ABC is an isosceles right triangle ### Question 46. If the point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8), find the value of x and the distance AP. Solution: It is given that Point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8) So, PA = PB On squaring both sides, we get (x – 7)2 + (4)2 = (x – 6)2 + (-5)2 x2 – 14x + 49 + 16 = x2 – 12x + 36 + 25 x2 – 14x + 65 = x2 – 12x + 61 x2 – 14x + 12x – x2 = 61 – 65 -2x = -4 x = -4/-2 = 2 x = 2 Now we find the distance ### Question 47. If A (3, y) is equidistant from points P (8, -3) and Q (7, 6), find the value of y and find the distance AQ. Solution: It is given that point A (3, y) is equidistant from P (8, -3) and Q (7, 6) So, AP = AQ On squaring both sides, we get (3 – 8)2 + (y + 3)2 = (-4)2 + (y – 6)2 (-5)2 + y2 + 6y + 9 = 16 + y2 – 12y + 36 25 + y2 + 6y + 9 = 16 + y2 – 12y + 36 y2 + 6y – y2 + 12y = 36 – 9 – 25 + 16 18y = 18 y = 18/18 = 1 y = 1 Now we find the distance ### Question 48. If (0, -3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex. Solution: Given that the A (0, -3) and B (0, 3) are the two vertices of an equilateral triangle ABC Let us assume that the coordinates of the third vertex be C (x, y) In equilateral triangle, AC = AB So, (x – 0)2 + (y + 3)2 = (0 – 0)2 + (3 + 3)2 x2 + (y + 3)2 = 0 + (6)2 = 36 x2 + y2 + 6y + 9 = 36 x2 + y2 + 6y = 36 – 9 = 27 …….(i) Also, BC = AB (x – 0)2 + (y – 3)2 = 36 x2 + y2 + 9 – 6y = 36 x2 + y2 – 6y = 36 – 9 = 27 ……..(ii) So, from eq (i) and (ii), we get x2 + y2 + 6y = x2 + y2 – 6y x2 + y2 + 6y – x2 – y2 + 6y = 0 12y = 0 y = 0 Now put the value of y in eq(i) x2 + y2 + 6y = 27 x2 + 0 + 0 = 27 x = ±√27 = ±3√3 So, the coordinates of third point is(3√3, 0) or (-3√3, 0) ### Question 49. If the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3), find k. Also, find the length of AP. Solution: Given that the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3) So, AP = BP (2 + 2)2 + (2 – k)2 = (2 + 2k)2 + (2 + 3)2 (4)2 + (2 – k)2 = (2 + 2k)2 + (5)2 16 + 4 + k2 – 4k = 4 + 4k2 + 8k + 25 4k2 + 8k + 29 – 16 – 4 – k2 + 4k = 0 3k2 + 12k + 9 = 0 k2 + 4k + 3 = 0 k2 + k + 3k + 3 = 0 k(k + 1) + 3(k + 1) = 0 (k + 1)(k + 3) = 0 So, the value of k either k + 1 = 0, then k = -1 or k + 3 = 0, then k = -3 Therefore, k = -1, -3 Now we find the distance ### Question 50. Show that ∆ABC, where A (-2, 0), B (2, 0), C (0, 2) and ∆PQR, where P (-4, 0), Q (4, 0), R (0, 4) are similar. Solution: Given that In ∆ABC, the vertices are A (-2, 0), B (2, 0), C (0, 2) In ∆PQR, the vertices are P (-4, 0), Q (4, 0), R (0, 4) Show that ∆ABC ~ ∆PQR So, Now, So, AB/PQ = 4/8 = 1/2 BC/QR = 2√2/4√2 = 1/2 CA/PQ = 2√2/4√2 = 1/2 So, AB/PQ = BC/QR = CA/RP By using SSS ∆ABC ~ ∆PQR ### Question 51. An equilateral triangle has two vertices at the points (3, 4), and (-2, 3). Find the coordinates of the third vertex. Solution: Given that the A (3, 4) and B (-2, 3) are the two vertices of an equilateral triangle ABC Let us assume that the coordinates of the third vertex be C (x, y) Now As we know that in equilateral triangle, AB = BC = CA So, BC = AB On squaring both side we get (x + 2)2 + (y – 3)2 = 26 x2 + 4x + 4 + y2 – 6y + 9 = 26 x2 + y2 + 4x – 6y + 13 = 26 x2 + y2 + 4x – 6y = 26 – 13 = 12 ………..(i) Similarly, CA = AB On squaring both side we get (3 – x)2 + (4 – y)2 = 26 9 + x2 – 6x + 16 + y2 – 8y = 26 x2 + y2 – 6x – 8y + 25 = 26 x2 + y2 – 6x – 8y = 26 – 25 = 1 …….(ii) Now on subtracting eq(ii) from (i), we get 10x + 2y = 12 5x + y = 6 ……..(iii) y = 6 – 5x Now substituting the value of y in eq(i), we get x2 + (6 – 5x)2 + 4x – 6(6 – 5x) = 13 x2 + 36 + 25x2 – 60x + 4x – 36 + 30x – 13 = 0 26x2 – 26x – 13 = 0 2x2 – 2x – 1 = 0 Here, a = 2, b = -2, c = -1 When x = (1 + √3)/2, then y = 6 – 5x = 6 – 5((1 + √3)/2) = (7 – 5√3)/2 Or when x = (1 – √3)/2, then y = 6 – 5x = 6 – 5((1 – √3)/2) = (7 + 5√3)/2 Hence, the co-ordinates of the point C is ((1 + √3)/2, (7 – 5√3)/2) or ((1 – √3)/2, (7 + 5√3)/2) ### Question 52. Find the circumcenter of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6). Solution: Given that the vertices of ∆ABC are A(-2, -3), B(-1, 0), and C(7, -6), and let us assume that O is the circumcenter ∆ABC. So, the coordinates of O will be (x, y) So, OA = OB = OC Or OA2 = OB2 = OC2 Now OA2 = (x + 2)2 + (y + 3)2 = x2 + 4x + 4 + y2 + 6y + 9 = x2 + y2 + 4x + 6y + 13 OB2 = (x + 1)2 + (y + 0)2 = x2 + 2x + 1 + y2 = x2 + y2 + 2x + 1 OC2 = (x – 7)2 + (y + 6)2 = x2 – 14x + 49 + y2 + 12y + 36 = x2 + y2 – 14x + 12y + 85 OA2 = OB2 x2 + y2 + 4x + 6y + 13 = x2 + y2 + 2x + 1 4x + 6y -2y = 1 – 13 2x + 6y = -12 x + 3y = -6 ………(i) OB2 = OC2 x2 + y2 + 2x + 1 = x2 + y2 – 14x + 12y + 85 2x + 14x – 2y = 85 – 1 16x – 12y = 84 4x – 3y = 21 ………(ii) From eq (i), we get x = -3y – 6 On substituting the value of x in eq (ii) 4(-3y – 6) – 3y = 21 -12 – 24 – 3y = 21 -15y = 21 + 24 -15y = 45 y = -45/15 = -3 x = -3y – 6 = -3 × (-3) – 6 = + 9 – 6 = 3 Hence, the co-ordinates of O are (3,-3) ### Question 53. Find the angle subtended at the origin by the line segment whose endpoints are (0, 100) and (10, 0). Solution: Let us considered the co-ordinates of the end points of a line segment are A (0, 100), B (10, 0) and origin is O (0, 0) So, the angle subtended by the line PQ at the origin is 90° or π/2 ### Question 54. Find the centre of the circle passing through (5, -8), (2, -9), and (2, 1). Solution: Given that O is the centre of the circle and A(5, -8), B(2, -9), and C (2, 1) are the points on the circle. So, let us considered the co-ordinates of O be (x, y) Therefore, OA = OB = OC OA2 = OB2 = OC2 Now OA2 = (x – 5)2 + (y + 8)2 = x2 – 10x + 25 + y2 + 16y + 64 = x2 + y2 – 10x + 16y + 89 Similarly, OB2 = (x – 5)2 + (y + 9)2 = x2 + 4 – 4x + y2 + 81 + 18y = x2 + y2 – 4x + 18y + 85 and OC2 = x2 – 4x + 4 + y2 – 2y + 1 = x2 + y2 – 4x – 2y + 5 OA2 = OB2 x2 + y2 – 10x + 16y + 89 = x2 + y2 – 4x + 18y + 85 -10x + 4x + 16y – 18y = 85 – 89 -6x – 2y = -4 3x + y = 2 …….(i) OB2 = OC2 x2 + y2 – 4x + 18y + 85 = x2 + y2 – 4x – 2y + 5 18y + 2y = 5 – 85 20y = -80 y = -80/10 = -4 Now substitute the value of y in eq(i), we get 3x + y = 2 3x – 4 = 2 3x = 2 + 4 = 6 x = 6/3 = 2 Hence, the co-ordinates of O are(2,-4) ### Question 55. If two opposite vertices of a square are (5, 4) and (1, -6), find the coordinates of its remaining two vertices. Solution: Given that the two opposite points of a ABCD square are A(5, 4) and C(1, -6) Let us considered that the co-ordinates of B be (x, y). So, join AC As we know that the sides of a square are equal, so, AB = BC AB2 = BC2 (x – 5)2 + (y – 4)2 = (x – 1)2 + (y + 6)2 x2 – 10x + 25 + y2 – 8y + 16 = x2 – 2x + 1 +y2 + 12y + 36 -10x + 2x – 8y – 12y = 37 – 41 -8x – 20y = -4 2x + 5y = 1 2x = 1 – 5y x = (1 – 5y)/2 So, ABC is a right-angled triangle Now by using Pythagoras theorem, we get AC2 = AB2 + BC2 (5 – 1)2 + (4 + 6)2 = x2 – 10x + 25 + y2 – 8y + 16 + x2 – 2x + 1 + y2 + 12y + 36 (4)2 + (10)2 = 2x2 + 2y2 – 12x + 4y + 78 16 + 100 = 2x2 + 2y2 – 12x + 4y + 78 2x2 + 2y2 – 12x + 4y + 78 – 16 – 100 = 0 2x2 + 2y2 – 12x + 4y – 38 = 0 x2 + y2 – 6x + 2y – 19 = 0 …..(i) Now substituting x = (1 – 5y)/2 in eq(i), we get 1 + 25y2 – 10y + 4y2 – 12 + 60y + 8y – 76 = 0 29y2 + 58y – 87 = 0 y2 + 2y – 3 = 0 y2 + 3y – y – 3 = 0 y(y + 3) – 1(y + 3) = 0 (y + 3)(y – 1) = 0 The value of y can be either y + 3 = 0, then y = -3 or y – 1 = 0, then y = 1 When y = 1, then x = (1 – 5y)/2 = (1 – 5(1))/2 = -2 When y = -3, then x = (1 – 5(-3))/2 = 8 So, the other points of ABCD square are(-2,1) and (8,-3) ### Question 56. Find the centre of the circle passing through (6, -6), (3, -7), and (3, 3). Solution: Let us considered O be the centre of the circle is (x, y) It is given that centre of the circle passing through (6, -6), (3, -7), and (3, 3) Join OA, OB and OC So, OA = OB = OC OA2 = (x -6 )2 + (y + 6) OB2 = (x – 3)2 + (y + 7)2 and OC2 = (x – 3)2 + (y-3)2 As we know that OA2 = OB2 So, (x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2 x2 – 12x + 36 + y2+12y + 36 = x2 – 6x + 9 + y2 + 14y + 49 x2 – 12x + 36 + y2+12y + 36 – x2 + 6x – 9 – y2 – 14y – 49 = 0 -12x + 12y + 72 + 6x – 14y – 58 = 0 -6x – 2y + 14 = 0 -6x – 2y = -14 3x + y = 7 …..(i) Also, OB2 = OC2 (x – 3)2 + (y + 7)2 = (x – 3)2 + (y – 3)2 x2 – 6x + 9 + y2+ 14y + 49 = x2 – 6x + 9 + y2 – 6y + 9 x2 + y2 – 6x + 58 + 14y – x2 – y2 + 6x + 6y – 18 = 0 20y + 40 = 0 20y = -40 y = -40/20 = -2 3x + (-2) = 7 3x = 7 + 2 = 9 x = 9/3 = 3 Hence, the co-ordinates of the centre are(3,-2) ### Question 57. Two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices. Solution: Let us considered ABCD is square, in which the co-ordinates are A (-1, 2) and C (3, 2). Let us assume the coordinates of B are (x, y) Now, join AC As we know that the sides of a square are equal, so, AB = BC AB2 = BC2 Now AB2 = (x + 1)2 + (y – 2)2 Similarly, BC2 = (x – 3)2 + (y – 2)2 As we know that AB = BC So, (x + 1)2 + (y – 2)2 = (x – 3)2 + (y – 2)2 (x + 1)2 = (x – 3)2 x2 + 2x + 1 = x2 – 6x + 9 x2 + 2x + 6x – x2 = 9 – 1 = 8 8x = 8 x = 8/8 = 1 Now in right triangle ABC AC2 = AB2 + BC2 (3 + 1)2 + (2 – 2)2 = (x + 1)2 + (y – 2)2 + (x – 3)2 + (y – 2)2 (4)2 + (0)2 = x2 + 2x + 1 + y2 – 4y + 4 + x2 – 6x + 9 + y2 – 4y + 4 16 = 2x2 + 2y2 – 4x – 8y + 18 2x2 + 2y2 – 4x – 8y = 16 – 18 2x2 + 2y2 – 4x – 8y = -2 x2 + y2 – 2x – 4y = -1 …….(i) Now substitute the value of x, in eq(i), we get (1)2 + y2 – 2 × 1 – 4y = -1 1 + y2 – 2 – 4y = -1 y2 – 4y = -1 – 1 + 2 = 0 y(y – 4) = 0 So the value of the y can be either y = 0 or y – 4 = 0, then y = 4 Hence, the coordinates of other points will be (1, 0) and (1, 4) My Personal Notes arrow_drop_up Related Tutorials	6,644	13,846	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.625	5	CC-MAIN-2023-23	latest	en	0.851398	GeeksforGeeks App. Open App. Browser. Continue. ## Related Articles. • RD Sharma Class 10 Solutions. # Class 10 RD Sharma Solutions – Chapter 14 Coordinate Geometry – Exercise 14.2 \| Set 3. ### Question 40. Prove that the points (3, 0), (4, 5), (-1, 4), and (-2, -1) taken in order, form a rhombus. Also, find its area.. Solution:. Let us considered the given points are A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1). Now we find the length of the sides and diagonals,. By using distance formula. So, AB =. AB2 = (4 + 3)2 + (5 – 0). = (1)2 + (5)2. = 1 + 25 = 26. Similarly, BC2 = (-1 – 4)2 + (4 – 5)2. = (-5)2 + (-1)2 = 25 + 1 = 26. CD2 = (-2 + 1)2 + (-1 – 4)2. = (-1)2 + (-5)2 = 1 + 25 = 26. and DA2 = (3 + 2)2 + (0 + 1). = (5)2 + (1)2 = 25 + 1 = 26. Diagonal AC2 = (-1 – 3)2 + (4 – 0)2. = (-4)2 + (4)2 = 16 + 16 = 32. and BD2 = (-2 – 4)2 + (-1 – 5)2. = (-6)2 + (-6)2 = 36 + 36 = 72. So, we conclude that the sides AB = BC = CD = DA and diagonal AC is not equal to BD. Hence, ABCD is a rhombus. Now we find the area of rhombus ABCD = Product of diagonals/2. = (√32 × √72)/2. = (√16 × 2 × 2 × 36)/2. = 4 × 2 × 6/2. = 24 sq. units. ### Question 41. In the seating arrangement of desks in a classroom three students Rohini, Sandhya, and Bina are seated at A (3, 1), B (6, 4), and C (8, 6). Do you think they are seated in a line?. Solution:. Given that A (3, 1), B (6, 4) and C (8, 6). Now we find the length of the sides and diagonals,. By using distance formula. AB =. AB2 = (6 – 3)2 + (4 – 1)2. = (3)2 + (3)2 = 9 + 9 = 18. Similarly, BC2 = (8 – 6)2 + (6 – 4)2. = (2)2 + (2)2 = 4 + 4 = 8. and BC2 = (8 – 6)2 + (6 – 4)2. = (2)2 + (2)2 = 4 + 4 = 8. and CA2 = (3 – 8)2 + (1 – 6)2. = (-5)2 + (-5)2 = 25 + 25 = 50. AB = √18 = √9 * 2 = 3√2. BC = √8 = √4 * 2 = 2√2. and CA = √50 = √25 * 2 = 5√2. AB + BC = 3√2 + 2√2 = 5√2 = CA. Hence, A, B and C are collinear points. Hence, they are seated in a line.. ### Question 42. Find a point on y-axis which is equidistant from the points (5, -2) and (-3, 2).. Solution:. Let us assume P be the point lies on y-axis. So, its x = 0, so the coordinates of P is (0, y). It is given that the point P (0, y) is equidistant from the points A(5, -2) and B(-3, 2).. So, PA = PB. Also, PA2 = PB2. Now by using distance formula, we get. (5 – 0)2 + (-2 – y)2 = (-3 – 0)2 + (2 – y)2. 25 + 4 + y2 + 4y = 9 + 4 – 4y + y2. y2 + 4y + 4y – y2 = 13 – 29. 8y = -16. y = -16/8 = 2. Hence, the required point P is (0,-2). ### Question 43. Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).. Solution:. Let us considered P (x, y) is equidistant from A(3, 6) and B(-3, 4). So, PA = PB. Also, PA2 = PB2. Now by using distance formula, we get. On squaring both sides, we get. (x – 3)2 + (y – 6)2 = (x + 3)2 + (y – 4)2. x2 – 6x + 9 + y2 – 12y + 36 = x2 + 6x + 9 = y2 – 8y + 16. -6x – 12y + 45 = 6x – 8y + 25. -6x – 6x – 12y + 8y + 45 – 25 = 0. -12 – 4y + 20 = 0. 3x + y – 5 = 0. 3x + y = 5. ### Question 44. If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of p.. Solution:. Given that the point A (0, 2) is equidistant from the points B (3, p) and C (p, 5). Now by using distance formula, we get. It is given that AB = AC. √p2 – 4p + 13 = √p2 + 9. So, on squaring both side, we get. = p2 – 4p + 13 = p2 + 9. p2 – 4p – p2 = 9 – 13. -4p = -4. p = 1. Hence, the value of p is 1. ### Question 45. Prove that the points (7, 10), (-2, 5), and (3, -4) are the vertices of an isosceles right triangle.. Solution:. Let us considered the points are A (7, 10), B (-2, 5) and C (3, -4). Now we find the length of the sides. By using distance formula. Now AB =. Similarly, BC =. and AC =. So, we conclude that AB = BC = √106 and AB2 + BC2 = AC2. Hence, ABC is an isosceles right triangle. ### Question 46. If the point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8), find the value of x and the distance AP.. Solution:. It is given that Point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8). So, PA = PB. On squaring both sides, we get. (x – 7)2 + (4)2 = (x – 6)2 + (-5)2. x2 – 14x + 49 + 16 = x2 – 12x + 36 + 25. x2 – 14x + 65 = x2 – 12x + 61. x2 – 14x + 12x – x2 = 61 – 65. -2x = -4. x = -4/-2 = 2. x = 2. Now we find the distance. ### Question 47. If A (3, y) is equidistant from points P (8, -3) and Q (7, 6), find the value of y and find the distance AQ.. Solution:. It is given that point A (3, y) is equidistant from P (8, -3) and Q (7, 6). So, AP = AQ. On squaring both sides, we get. (3 – 8)2 + (y + 3)2 = (-4)2 + (y – 6)2. (-5)2 + y2 + 6y + 9 = 16 + y2 – 12y + 36. 25 + y2 + 6y + 9 = 16 + y2 – 12y + 36. y2 + 6y – y2 + 12y = 36 – 9 – 25 + 16. 18y = 18. y = 18/18 = 1. y = 1. Now we find the distance. ### Question 48. If (0, -3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex.. Solution:. Given that the A (0, -3) and B (0, 3) are the two vertices of an equilateral triangle ABC. Let us assume that the coordinates of the third vertex be C (x, y). In equilateral triangle, AC = AB. So,. (x – 0)2 + (y + 3)2 = (0 – 0)2 + (3 + 3)2. x2 + (y + 3)2 = 0 + (6)2 = 36. x2 + y2 + 6y + 9 = 36. x2 + y2 + 6y = 36 – 9 = 27 …….(i). Also, BC = AB. (x – 0)2 + (y – 3)2 = 36. x2 + y2 + 9 – 6y = 36. x2 + y2 – 6y = 36 – 9 = 27 ……..(ii). So, from eq (i) and (ii), we get. x2 + y2 + 6y = x2 + y2 – 6y. x2 + y2 + 6y – x2 – y2 + 6y = 0. 12y = 0. y = 0. Now put the value of y in eq(i). x2 + y2 + 6y = 27. x2 + 0 + 0 = 27. x = ±√27 = ±3√3. So, the coordinates of third point is(3√3, 0) or (-3√3, 0). ### Question 49. If the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3), find k. Also, find the length of AP.. Solution:. Given that the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3). So, AP = BP. (2 + 2)2 + (2 – k)2 = (2 + 2k)2 + (2 + 3)2. (4)2 + (2 – k)2 = (2 + 2k)2 + (5)2. 16 + 4 + k2 – 4k = 4 + 4k2 + 8k + 25. 4k2 + 8k + 29 – 16 – 4 – k2 + 4k = 0. 3k2 + 12k + 9 = 0. k2 + 4k + 3 = 0. k2 + k + 3k + 3 = 0. k(k + 1) + 3(k + 1) = 0. (k + 1)(k + 3) = 0. So, the value of k either k + 1 = 0, then k = -1. or k + 3 = 0, then k = -3. Therefore, k = -1, -3. Now we find the distance. ### Question 50. Show that ∆ABC, where A (-2, 0), B (2, 0), C (0, 2) and ∆PQR, where P (-4, 0), Q (4, 0), R (0, 4) are similar.. Solution:. Given that In ∆ABC, the vertices are A (-2, 0), B (2, 0), C (0, 2). In ∆PQR, the vertices are P (-4, 0), Q (4, 0), R (0, 4). Show that ∆ABC ~ ∆PQR. So,. Now,. So, AB/PQ = 4/8 = 1/2. BC/QR = 2√2/4√2 = 1/2. CA/PQ = 2√2/4√2 = 1/2. So, AB/PQ = BC/QR = CA/RP. By using SSS. ∆ABC ~ ∆PQR. ### Question 51. An equilateral triangle has two vertices at the points (3, 4), and (-2, 3). Find the coordinates of the third vertex.. Solution:. Given that the A (3, 4) and B (-2, 3) are the two vertices of an equilateral triangle ABC. Let us assume that the coordinates of the third vertex be C (x, y). Now. As we know that in equilateral triangle, AB = BC = CA. So, BC = AB. On squaring both side we get. (x + 2)2 + (y – 3)2 = 26. x2 + 4x + 4 + y2 – 6y + 9 = 26.	x2 + y2 + 4x – 6y + 13 = 26. x2 + y2 + 4x – 6y = 26 – 13 = 12 ………..(i). Similarly, CA = AB. On squaring both side we get. (3 – x)2 + (4 – y)2 = 26. 9 + x2 – 6x + 16 + y2 – 8y = 26. x2 + y2 – 6x – 8y + 25 = 26. x2 + y2 – 6x – 8y = 26 – 25 = 1 …….(ii). Now on subtracting eq(ii) from (i), we get. 10x + 2y = 12. 5x + y = 6 ……..(iii). y = 6 – 5x. Now substituting the value of y in eq(i), we get. x2 + (6 – 5x)2 + 4x – 6(6 – 5x) = 13. x2 + 36 + 25x2 – 60x + 4x – 36 + 30x – 13 = 0. 26x2 – 26x – 13 = 0. 2x2 – 2x – 1 = 0. Here, a = 2, b = -2, c = -1. When x = (1 + √3)/2, then y = 6 – 5x = 6 – 5((1 + √3)/2) = (7 – 5√3)/2. Or when x = (1 – √3)/2, then y = 6 – 5x = 6 – 5((1 – √3)/2) = (7 + 5√3)/2. Hence, the co-ordinates of the point C is ((1 + √3)/2, (7 – 5√3)/2) or ((1 – √3)/2, (7 + 5√3)/2). ### Question 52. Find the circumcenter of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6).. Solution:. Given that the vertices of ∆ABC are A(-2, -3), B(-1, 0), and C(7, -6), and. let us assume that O is the circumcenter ∆ABC. So, the coordinates of O will be (x, y). So, OA = OB = OC. Or OA2 = OB2 = OC2. Now. OA2 = (x + 2)2 + (y + 3)2. = x2 + 4x + 4 + y2 + 6y + 9. = x2 + y2 + 4x + 6y + 13. OB2 = (x + 1)2 + (y + 0)2. = x2 + 2x + 1 + y2. = x2 + y2 + 2x + 1. OC2 = (x – 7)2 + (y + 6)2. = x2 – 14x + 49 + y2 + 12y + 36. = x2 + y2 – 14x + 12y + 85. OA2 = OB2. x2 + y2 + 4x + 6y + 13 = x2 + y2 + 2x + 1. 4x + 6y -2y = 1 – 13. 2x + 6y = -12. x + 3y = -6 ………(i). OB2 = OC2. x2 + y2 + 2x + 1 = x2 + y2 – 14x + 12y + 85. 2x + 14x – 2y = 85 – 1. 16x – 12y = 84. 4x – 3y = 21 ………(ii). From eq (i), we get. x = -3y – 6. On substituting the value of x in eq (ii). 4(-3y – 6) – 3y = 21. -12 – 24 – 3y = 21. -15y = 21 + 24. -15y = 45. y = -45/15 = -3. x = -3y – 6 = -3 × (-3) – 6. = + 9 – 6 = 3. Hence, the co-ordinates of O are (3,-3). ### Question 53. Find the angle subtended at the origin by the line segment whose endpoints are (0, 100) and (10, 0).. Solution:. Let us considered the co-ordinates of the end points of a line segment are. A (0, 100), B (10, 0) and origin is O (0, 0). So, the angle subtended by the line PQ at the origin is 90° or π/2. ### Question 54. Find the centre of the circle passing through (5, -8), (2, -9), and (2, 1).. Solution:. Given that O is the centre of the circle and A(5, -8), B(2, -9), and C (2, 1) are the points on the circle.. So, let us considered the co-ordinates of O be (x, y). Therefore, OA = OB = OC. OA2 = OB2 = OC2. Now. OA2 = (x – 5)2 + (y + 8)2. = x2 – 10x + 25 + y2 + 16y + 64. = x2 + y2 – 10x + 16y + 89. Similarly, OB2 = (x – 5)2 + (y + 9)2. = x2 + 4 – 4x + y2 + 81 + 18y. = x2 + y2 – 4x + 18y + 85. and OC2 = x2 – 4x + 4 + y2 – 2y + 1. = x2 + y2 – 4x – 2y + 5. OA2 = OB2. x2 + y2 – 10x + 16y + 89 = x2 + y2 – 4x + 18y + 85. -10x + 4x + 16y – 18y = 85 – 89. -6x – 2y = -4. 3x + y = 2 …….(i). OB2 = OC2. x2 + y2 – 4x + 18y + 85 = x2 + y2 – 4x – 2y + 5. 18y + 2y = 5 – 85. 20y = -80. y = -80/10 = -4. Now substitute the value of y in eq(i), we get. 3x + y = 2. 3x – 4 = 2. 3x = 2 + 4 = 6. x = 6/3 = 2. Hence, the co-ordinates of O are(2,-4). ### Question 55. If two opposite vertices of a square are (5, 4) and (1, -6), find the coordinates of its remaining two vertices.. Solution:. Given that the two opposite points of a ABCD square are A(5, 4) and C(1, -6). Let us considered that the co-ordinates of B be (x, y).. So, join AC. As we know that the sides of a square are equal, so,. AB = BC. AB2 = BC2. (x – 5)2 + (y – 4)2 = (x – 1)2 + (y + 6)2. x2 – 10x + 25 + y2 – 8y + 16 = x2 – 2x + 1 +y2 + 12y + 36. -10x + 2x – 8y – 12y = 37 – 41. -8x – 20y = -4. 2x + 5y = 1. 2x = 1 – 5y. x = (1 – 5y)/2. So, ABC is a right-angled triangle. Now by using Pythagoras theorem, we get. AC2 = AB2 + BC2. (5 – 1)2 + (4 + 6)2 = x2 – 10x + 25 + y2 – 8y + 16 + x2 – 2x + 1 + y2 + 12y + 36. (4)2 + (10)2 = 2x2 + 2y2 – 12x + 4y + 78. 16 + 100 = 2x2 + 2y2 – 12x + 4y + 78. 2x2 + 2y2 – 12x + 4y + 78 – 16 – 100 = 0. 2x2 + 2y2 – 12x + 4y – 38 = 0. x2 + y2 – 6x + 2y – 19 = 0 …..(i). Now substituting x = (1 – 5y)/2 in eq(i), we get. 1 + 25y2 – 10y + 4y2 – 12 + 60y + 8y – 76 = 0. 29y2 + 58y – 87 = 0. y2 + 2y – 3 = 0. y2 + 3y – y – 3 = 0. y(y + 3) – 1(y + 3) = 0. (y + 3)(y – 1) = 0. The value of y can be either y + 3 = 0, then y = -3. or y – 1 = 0, then y = 1. When y = 1, then. x = (1 – 5y)/2. = (1 – 5(1))/2. = -2. When y = -3, then. x = (1 – 5(-3))/2. = 8. So, the other points of ABCD square are(-2,1) and (8,-3). ### Question 56. Find the centre of the circle passing through (6, -6), (3, -7), and (3, 3).. Solution:. Let us considered O be the centre of the circle is (x, y). It is given that centre of the circle passing through (6, -6), (3, -7), and (3, 3). Join OA, OB and OC. So, OA = OB = OC. OA2 = (x -6 )2 + (y + 6). OB2 = (x – 3)2 + (y + 7)2. and OC2 = (x – 3)2 + (y-3)2. As we know that OA2 = OB2. So, (x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2. x2 – 12x + 36 + y2+12y + 36 = x2 – 6x + 9 + y2 + 14y + 49. x2 – 12x + 36 + y2+12y + 36 – x2 + 6x – 9 – y2 – 14y – 49 = 0. -12x + 12y + 72 + 6x – 14y – 58 = 0. -6x – 2y + 14 = 0. -6x – 2y = -14. 3x + y = 7 …..(i). Also, OB2 = OC2. (x – 3)2 + (y + 7)2 = (x – 3)2 + (y – 3)2. x2 – 6x + 9 + y2+ 14y + 49 = x2 – 6x + 9 + y2 – 6y + 9. x2 + y2 – 6x + 58 + 14y – x2 – y2 + 6x + 6y – 18 = 0. 20y + 40 = 0. 20y = -40. y = -40/20 = -2. 3x + (-2) = 7. 3x = 7 + 2 = 9. x = 9/3 = 3. Hence, the co-ordinates of the centre are(3,-2). ### Question 57. Two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.. Solution:. Let us considered ABCD is square, in which the co-ordinates are A (-1, 2) and C (3, 2).. Let us assume the coordinates of B are (x, y). Now, join AC. As we know that the sides of a square are equal, so,. AB = BC. AB2 = BC2. Now. AB2 = (x + 1)2 + (y – 2)2. Similarly, BC2 = (x – 3)2 + (y – 2)2. As we know that AB = BC. So,. (x + 1)2 + (y – 2)2 = (x – 3)2 + (y – 2)2. (x + 1)2 = (x – 3)2. x2 + 2x + 1 = x2 – 6x + 9. x2 + 2x + 6x – x2 = 9 – 1 = 8. 8x = 8. x = 8/8 = 1. Now in right triangle ABC. AC2 = AB2 + BC2. (3 + 1)2 + (2 – 2)2 = (x + 1)2 + (y – 2)2 + (x – 3)2 + (y – 2)2. (4)2 + (0)2 = x2 + 2x + 1 + y2 – 4y + 4 + x2 – 6x + 9 + y2 – 4y + 4. 16 = 2x2 + 2y2 – 4x – 8y + 18. 2x2 + 2y2 – 4x – 8y = 16 – 18. 2x2 + 2y2 – 4x – 8y = -2. x2 + y2 – 2x – 4y = -1 …….(i). Now substitute the value of x, in eq(i), we get. (1)2 + y2 – 2 × 1 – 4y = -1. 1 + y2 – 2 – 4y = -1. y2 – 4y = -1 – 1 + 2 = 0. y(y – 4) = 0. So the value of the y can be either y = 0. or y – 4 = 0, then y = 4. Hence, the coordinates of other points will be (1, 0) and (1, 4). My Personal Notes arrow_drop_up. Related Tutorials.
https://www.jiskha.com/questions/190991/A-coin-is-tossed-If-a-head-appears-a-spinner-that-can-land-on-any-of-the-numbers	1,540,361,891,000,000,000	text/html	crawl-data/CC-MAIN-2018-43/segments/1539583519859.77/warc/CC-MAIN-20181024042701-20181024064201-00247.warc.gz	964,629,315	5,069	A coin is tossed. If a head appears, a spinner that can land on any of the numbers from 1 to 5 is spun. If a tail appears, the coin is tossed a second time instead of spinning the spinner. What are the possible outcomes? A. (T,H)(T,T)(T,1)(T,2)(T,3)(T,4)(T,5) B. (T,H)(H,H)(T,1)(T,2)(T,3)(T,4)(T,5) C. (T,H)(T,T)(H,1)(H,2)(H,3)(H,4)(H,5) D. (T,H)(H,H)(H,1)(H,2)(H,3)(H,4)(H,5) 1. 0 2. 0 3. 7 1. Well is has to be C or D because you don't get a spin with tails on the first throw. Of those two, how could there be an (H,H) possibility? 1. 0 2. 0 posted by drwls 2. Find the probability of each event. Spinning red on a spinner with equally sized red, blue, yellow, and green sectors, and flipping a coin that lands tails up. 1. 0 2. 0 posted by John ## Similar Questions 1. ### math A coin is tossed. If a head appears, a spinner that can land on any of the numbers from 1 to 5 is spun. If a tail appears, the coin is tossed a second time instead of spinning the spinner. What are the possible outcomes? A. 2. ### Math (Ms. Sue) A coin tossed. If heads appears, a spinner that can land on any number from 1 to 4 is spun. If tails appears, a second coin is tossed instead of spinning the spinner. What are the possible outcomes? H1 H2 H3 H4 H1 H2 H3 H1 H2 H3 3. ### math A spinner is divided into 5 equal sections numbered 1, 2, 3, 4, and 5. If the spinner is spun and a coin is tossed, what is the probability of getting a tail on the coin and an odd number on the spinner 4. ### Math A spinner with 10 equal sections, numbered 1 to 10, is spun. Then a coin is flipped. What is the probability that the spinner will land on 3 and the coin will land on heads? I'm doing this math test and I have a border line A on my grade I need to make sure I don't lower my grade to a B, so any help is appreciated. Thanks! 7. The probability of winning a game is 25%. How many times should you expect 6. ### mathematical physics An unbiased coin is tossed three times. If A is the event that a head appears on each of the first two tosses , B is the event that a tail occurs on the third toss and C is the event that exactly two tails appears in the three 7. ### mathematical physics An unbiased coin is tossed three times. If A is the event that a head appears on each of the first two tosses , B is the event that a tail occurs on the third toss and C is the event that exactly two tails appears in the three 8. ### Math Luis has a coin that is weighted so that the probability that Heads appears when it is tossed is 0.55. Suppose that the coin is tossed 3 times. What is the probability that all 3 tosses are Heads? please help ,e to solve this 9. ### math - probability a) What is the probability of obtaining 2 Heads when a coin is tossed twice? b) What is the probability of obtaining 1 Head when a coin is tossed twice? Keep in mind, the coins are not tossed simultaneously. 10. ### Geometry Hi, I have a Benchmark test tomorrow and I don't understand how to calculate the probability of this problem. A fair coin is tossed and the spinner is spun. All outcomes on the spinner is equally likely. Find the probability of More Similar Questions	886	3,153	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.640625	4	CC-MAIN-2018-43	latest	en	0.929609	A coin is tossed. If a head appears, a spinner that can land on any of the numbers from 1 to 5 is spun. If a tail appears, the coin is tossed a second time instead of spinning the spinner. What are the possible outcomes?. A. (T,H)(T,T)(T,1)(T,2)(T,3)(T,4)(T,5). B. (T,H)(H,H)(T,1)(T,2)(T,3)(T,4)(T,5). C. (T,H)(T,T)(H,1)(H,2)(H,3)(H,4)(H,5). D. (T,H)(H,H)(H,1)(H,2)(H,3)(H,4)(H,5). 1. 0. 2. 0. 3. 7. 1. Well is has to be C or D because you don't get a spin with tails on the first throw. Of those two, how could there be an (H,H) possibility?. 1. 0. 2. 0. posted by drwls. 2. Find the probability of each event.. Spinning red on a spinner with equally sized red, blue, yellow, and green sectors, and flipping a coin that lands tails up.. 1. 0. 2. 0. posted by John. ## Similar Questions. 1. ### math. A coin is tossed. If a head appears, a spinner that can land on any of the numbers from 1 to 5 is spun. If a tail appears, the coin is tossed a second time instead of spinning the spinner. What are the possible outcomes? A.. 2.	### Math (Ms. Sue). A coin tossed. If heads appears, a spinner that can land on any number from 1 to 4 is spun. If tails appears, a second coin is tossed instead of spinning the spinner. What are the possible outcomes? H1 H2 H3 H4 H1 H2 H3 H1 H2 H3. 3. ### math. A spinner is divided into 5 equal sections numbered 1, 2, 3, 4, and 5. If the spinner is spun and a coin is tossed, what is the probability of getting a tail on the coin and an odd number on the spinner. 4. ### Math. A spinner with 10 equal sections, numbered 1 to 10, is spun. Then a coin is flipped. What is the probability that the spinner will land on 3 and the coin will land on heads?. I'm doing this math test and I have a border line A on my grade I need to make sure I don't lower my grade to a B, so any help is appreciated. Thanks! 7. The probability of winning a game is 25%. How many times should you expect. 6. ### mathematical physics. An unbiased coin is tossed three times. If A is the event that a head appears on each of the first two tosses , B is the event that a tail occurs on the third toss and C is the event that exactly two tails appears in the three. 7. ### mathematical physics. An unbiased coin is tossed three times. If A is the event that a head appears on each of the first two tosses , B is the event that a tail occurs on the third toss and C is the event that exactly two tails appears in the three. 8. ### Math. Luis has a coin that is weighted so that the probability that Heads appears when it is tossed is 0.55. Suppose that the coin is tossed 3 times. What is the probability that all 3 tosses are Heads? please help ,e to solve this. 9. ### math - probability. a) What is the probability of obtaining 2 Heads when a coin is tossed twice? b) What is the probability of obtaining 1 Head when a coin is tossed twice? Keep in mind, the coins are not tossed simultaneously.. 10. ### Geometry. Hi, I have a Benchmark test tomorrow and I don't understand how to calculate the probability of this problem. A fair coin is tossed and the spinner is spun. All outcomes on the spinner is equally likely. Find the probability of. More Similar Questions.
http://www.markedbyteachers.com/gcse/science/margarine-tub.html	1,526,939,438,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794864544.25/warc/CC-MAIN-20180521200606-20180521220606-00075.warc.gz	413,582,892	18,932	• Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month Page 1. 1 1 2. 2 2 3. 3 3 4. 4 4 5. 5 5 6. 6 6 # Margarine Tub Extracts from this document... Introduction Investigate how the mass will affect the distance travelled by a weighted margarine tub when it is propelled along a runway by a stretched rubber band Planning A Hypothesis I predict that as the mass of the margarine tub increases, the distance travelled by the tub will decrease. I think this because as the mass increases the surface friction will also increases; this increased friction will cause the object to slow down and stop quicker and therefore in a shorter distance. The formula for kinetic energy is: Kinetic energy = mass x velocity squared. When any mass is propelled along a runway, it travels a certain distance. When the mass is heavier then travels a shorter distance, and when it is lighter it travels a longer distance because of the forces acting on it. It will also travel a longer distance because of the increased momentum. I expect that the graph will not be a straight line because of the velocity squared part of the formula; this will vary the gradient of the line of best fit. The gradient will change because you are not multiplying the velocity by a constant, but by itself so the larger the velocity, the more the number will increase by when squared. This is why the gradient is steeper at the start of the graph. ...read more. Middle I will measure from the same end of the tub when I pull back the elastic band and when I measure how far it has travelled. I will measure to the nearest centimetre because it is the most appropriate degree of accuracy, and I will measure across with another ruler to make the measurement readings more accurate. I am using a measurement sheet rather than a Newton metre to measure how far back I pull the elastic band, because the Newton metre only went up to 10 Newton's and this force didn't pull back the elastic band far enough to propel the projectile a suitable distance to measure. This would make it hard for me to collect an appropriate range of accurate results. I need to make sure I don't stretch the elastic band too much that I reach the elastic limit of the elastic band. If I do stretch the band beyond its elastic limit, as stated in Hooke's Law, the elastic band will behave inelastically so it won't return to its original shape. Data Collection Mass (g) Distance Travelled 1 (cm) Distance Travelled 2 (cm) Distance Travelled 3 (cm) Average (cm) 50 44.5 45 81 56.8 100 30.5 30.5 32 31 150 24 26 25 25 200 22.5 22 23 22.5 250 19 20.5 20.5 20 300 18 18.5 18.5 18.3 350 16 15.5 19 16.9 400 15 15.5 17 15.8 450 11.5 11 11.5 14.8 500 12.5 ...read more. Conclusion This is what I predicted would happen, and it was correct. I am pleased with my results and feel that they are as accurate as I could make them. I measured the distances to the nearest half centimetre because this was an appropriate degree of accuracy and made sure the ruler was in the correct position before taking each reading. If I did this experiment again, I would perhaps investigate more than one factor, and find out the effect they have on each other. For example I could investigate how far an object travels when propelled of an elastic band along an oiled or greased surface. Also I would investigate more weights so that my line of best fit is more accurate on my graph, I might also extend the range of weights to see if this made any difference. My percentage error was 14%, I worked this out using my expected table of values and my actual table of values, I used the formula Percentage error = (value - expected value / expected value) x 100. I had one anomaly whilst collecting my results, so there must have been a factor which affected this result when I was doing my experiment. This was probably a human error of misreading the length on the ruler; however it could have been any of the factors explained on the first page. Finally, I am pleased with my results and overall experiment and I feel I produced an accurate set of data. ?? ?? ?? ?? Ahmed Ismail Physics Coursework ...read more. The above preview is unformatted text This student written piece of work is one of many that can be found in our GCSE Forces and Motion section. ## Found what you're looking for? • Start learning 29% faster today • 150,000+ documents available • Just £6.99 a month Not the one? Search for your essay title... • Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month # Related GCSE Forces and Motion essays 1. ## To see how the distance, a weighted margarine tub travels, changes as the distance ... quicker, meaning the distance the tub travels further each time, increases more. Preliminary Tests After doing preliminary tests I have noticed that the tub does not begin to move, until it is pulled back 4cm. I also noticed that when it got pulled back over 12cm the tub spun while 2. ## My aim is to find out whether mass affects the distance travelled by a ... and 3rd experiment = 28cm (rounded up from 27.6cm). The averages are all either 27 or 28, this shows that all three sets of results have something in common and that they generally have the same results with no obtrusive results. 1. ## How Does Changing The Force In An Elastic Band Affect The Distance Travelled By ... I will put the margarine tub against the elastic band and let go. I will measure where the tub stops using a metre long ruler. I will record my results and repeat the experiment another two more times for more accurate results. 2. ## How will changing the distance an elastic band is stretched effect the distance a ... We will use the same plastic tub, runway, elastic band, and the same direction in which the plastic tub is fired. We are going to use the same plastic beaker because if we use different tubs for each experiment, the shapes may be different, or the weight may be different and this will affect the distance the tub will travel. 1. ## What factors effect the distance a tub will travel along the floor. As the tub is pushed forwards friction acts upon it. This is because the two surfaces are not smooth. As a greater mass is placed inside the tub the surface of the floor and the bottom of the plastic tub are pushed closer together. 2. ## Find out how changing the surface which a margarine tub is launch on, influences ... Prediction: I predict the rougher a surface is the small the stopping distance. I think this will occur because there is a lot of resistance from the rough surface on the base of the tub, causing a lot of friction, which slows an object down quickly. 1. ## What factors affect the distance traveled by a margarine tub? The greater the force, the greater the acceleration. Any resultant force will result in acceleration and therefore this is the formula: F=ma (Force = Mass x Acceleration) This is showing the relationship between the force and the acceleration. * Mass - Mass is the amount of matter in an object. 2. ## Investigate the factors which affects the distance travelled by a tub propelled by an ... Then the force is changed but kept constant while the mass is increased. This is related to the formula of Force in N= mass in kg x acceleration in m/s2 Preliminary work: Before doing the actual experiment we tried out different methods to propel the tub. • Over 160,000 pieces of student written work • Annotated by experienced teachers • Ideas and feedback to	1,739	7,589	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.578125	4	CC-MAIN-2018-22	latest	en	0.931879	• Join over 1.2 million students every month. • Accelerate your learning by 29%. • Unlimited access from just £6.99 per month. Page. 1. 1. 1. 2. 2. 2. 3. 3. 3. 4. 4. 4. 5. 5. 5. 6. 6. 6. # Margarine Tub. Extracts from this document.... Introduction. Investigate how the mass will affect the distance travelled by a weighted margarine tub when it is propelled along a runway by a stretched rubber band Planning A Hypothesis I predict that as the mass of the margarine tub increases, the distance travelled by the tub will decrease. I think this because as the mass increases the surface friction will also increases; this increased friction will cause the object to slow down and stop quicker and therefore in a shorter distance. The formula for kinetic energy is: Kinetic energy = mass x velocity squared. When any mass is propelled along a runway, it travels a certain distance. When the mass is heavier then travels a shorter distance, and when it is lighter it travels a longer distance because of the forces acting on it. It will also travel a longer distance because of the increased momentum. I expect that the graph will not be a straight line because of the velocity squared part of the formula; this will vary the gradient of the line of best fit. The gradient will change because you are not multiplying the velocity by a constant, but by itself so the larger the velocity, the more the number will increase by when squared. This is why the gradient is steeper at the start of the graph. ...read more.. Middle. I will measure from the same end of the tub when I pull back the elastic band and when I measure how far it has travelled. I will measure to the nearest centimetre because it is the most appropriate degree of accuracy, and I will measure across with another ruler to make the measurement readings more accurate. I am using a measurement sheet rather than a Newton metre to measure how far back I pull the elastic band, because the Newton metre only went up to 10 Newton's and this force didn't pull back the elastic band far enough to propel the projectile a suitable distance to measure. This would make it hard for me to collect an appropriate range of accurate results. I need to make sure I don't stretch the elastic band too much that I reach the elastic limit of the elastic band. If I do stretch the band beyond its elastic limit, as stated in Hooke's Law, the elastic band will behave inelastically so it won't return to its original shape. Data Collection Mass (g) Distance Travelled 1 (cm) Distance Travelled 2 (cm) Distance Travelled 3 (cm) Average (cm) 50 44.5 45 81 56.8 100 30.5 30.5 32 31 150 24 26 25 25 200 22.5 22 23 22.5 250 19 20.5 20.5 20 300 18 18.5 18.5 18.3 350 16 15.5 19 16.9 400 15 15.5 17 15.8 450 11.5 11 11.5 14.8 500 12.5 ...read more.. Conclusion. This is what I predicted would happen, and it was correct. I am pleased with my results and feel that they are as accurate as I could make them. I measured the distances to the nearest half centimetre because this was an appropriate degree of accuracy and made sure the ruler was in the correct position before taking each reading. If I did this experiment again, I would perhaps investigate more than one factor, and find out the effect they have on each other. For example I could investigate how far an object travels when propelled of an elastic band along an oiled or greased surface. Also I would investigate more weights so that my line of best fit is more accurate on my graph, I might also extend the range of weights to see if this made any difference. My percentage error was 14%, I worked this out using my expected table of values and my actual table of values, I used the formula Percentage error = (value - expected value / expected value) x 100. I had one anomaly whilst collecting my results, so there must have been a factor which affected this result when I was doing my experiment. This was probably a human error of misreading the length on the ruler; however it could have been any of the factors explained on the first page. Finally, I am pleased with my results and overall experiment and I feel I produced an accurate set of data.	?? ?? ?? ?? Ahmed Ismail Physics Coursework ...read more.. The above preview is unformatted text. This student written piece of work is one of many that can be found in our GCSE Forces and Motion section.. ## Found what you're looking for?. • Start learning 29% faster today. • 150,000+ documents available. • Just £6.99 a month. Not the one? Search for your essay title.... • Join over 1.2 million students every month. • Accelerate your learning by 29%. • Unlimited access from just £6.99 per month. # Related GCSE Forces and Motion essays. 1. ## To see how the distance, a weighted margarine tub travels, changes as the distance .... quicker, meaning the distance the tub travels further each time, increases more. Preliminary Tests After doing preliminary tests I have noticed that the tub does not begin to move, until it is pulled back 4cm. I also noticed that when it got pulled back over 12cm the tub spun while. 2. ## My aim is to find out whether mass affects the distance travelled by a .... and 3rd experiment = 28cm (rounded up from 27.6cm). The averages are all either 27 or 28, this shows that all three sets of results have something in common and that they generally have the same results with no obtrusive results.. 1. ## How Does Changing The Force In An Elastic Band Affect The Distance Travelled By .... I will put the margarine tub against the elastic band and let go. I will measure where the tub stops using a metre long ruler. I will record my results and repeat the experiment another two more times for more accurate results.. 2. ## How will changing the distance an elastic band is stretched effect the distance a .... We will use the same plastic tub, runway, elastic band, and the same direction in which the plastic tub is fired. We are going to use the same plastic beaker because if we use different tubs for each experiment, the shapes may be different, or the weight may be different and this will affect the distance the tub will travel.. 1. ## What factors effect the distance a tub will travel along the floor.. As the tub is pushed forwards friction acts upon it. This is because the two surfaces are not smooth. As a greater mass is placed inside the tub the surface of the floor and the bottom of the plastic tub are pushed closer together.. 2. ## Find out how changing the surface which a margarine tub is launch on, influences .... Prediction: I predict the rougher a surface is the small the stopping distance. I think this will occur because there is a lot of resistance from the rough surface on the base of the tub, causing a lot of friction, which slows an object down quickly.. 1. ## What factors affect the distance traveled by a margarine tub?. The greater the force, the greater the acceleration. Any resultant force will result in acceleration and therefore this is the formula: F=ma (Force = Mass x Acceleration) This is showing the relationship between the force and the acceleration. * Mass - Mass is the amount of matter in an object.. 2. ## Investigate the factors which affects the distance travelled by a tub propelled by an .... Then the force is changed but kept constant while the mass is increased. This is related to the formula of Force in N= mass in kg x acceleration in m/s2 Preliminary work: Before doing the actual experiment we tried out different methods to propel the tub.. • Over 160,000 pieces. of student written work. • Annotated by. experienced teachers. • Ideas and feedback to.
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-2-linear-equations-and-functions-2-4-write-equations-of-lines-2-4-exercises-problem-solving-page-104/58	1,726,247,751,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00792.warc.gz	747,995,364	15,572	# Chapter 2 Linear Equations and Functions - 2.4 Write Equations of Lines - 2.4 Exercises - Problem Solving - Page 104: 58 $70+15x$ #### Work Step by Step The equation according to the text of the exercise is (if the hours danced is $x$): $=(15+35+20)+x(4+8+3)=70+15x$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.	124	432	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.78125	4	CC-MAIN-2024-38	latest	en	0.85278	# Chapter 2 Linear Equations and Functions - 2.4 Write Equations of Lines - 2.4 Exercises - Problem Solving - Page 104: 58. $70+15x$. #### Work Step by Step. The equation according to the text of the exercise is (if the hours danced is $x$): $=(15+35+20)+x(4+8+3)=70+15x$.	After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
http://www.torontokidscomputer.com/tuesday-python-1700/tuesday-23-0131-17/	1,696,054,159,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233510603.89/warc/CC-MAIN-20230930050118-20230930080118-00084.warc.gz	82,669,401	11,874	TORONTO KIDS COMPUTER CLUB \| Tuesday 17:00 Python Practice – 23.01.31. 21702 # Tuesday 17:00 Python Practice – 23.01.31. ## 02 Feb Tuesday 17:00 Python Practice – 23.01.31. You could do the following homework using IDLE. Please save your IDLE after finish your homework or you could copy you result to google Docs in the google classroom. ## Variable Questions: Question 1 Make a variable and assign a number to it (any number you like). Then display your variable using print. Question 2 Modify your variable, either by replacing the old value with a new value, or by adding something to the old value. Display the new value using print. Question 3 Make another variable and assign a string (some text) to it. Then display it using print. Question 4 Make a variable for DaysPerWeekHoursPerDay, and MinutesPerHour (or make up your own names), and then multiply them together. Question 5 People are always saying there’s not enough time to get everything done. How many minutes would there be in a week if there were 26 hours in a day? (Hint: Change the HoursPerDay variable.) Question 6: (Sample Question) Write a program to solve the following question: Three people ate dinner at a restaurant and want to split the bill. The total is \$35.27, and they want to leave a 15 percent tip. How much should each person pay? Hint: you can create three variables to start: >>> people = 3 >>> total = 35.27 >>> tip_percent = 0.15 then you could get the result by calculating step-by-step as following: ```>>> tip = total * tip_percent >>> final_total = total + tip >>> each = final_total / people >>> print(each)``` ## Data Type Questions: Question 7. Use float() to create a number from a string like ‘12.34’. Make sure the result is really a number! Question 8. Try using int() to create an integer from a decimal number like 56.78. Did the answer get rounded up or down? Question 9. Try using int() to create an integer from a string. Make sure the result is really an integer!	502	1,999	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.671875	4	CC-MAIN-2023-40	latest	en	0.829243	TORONTO KIDS COMPUTER CLUB \| Tuesday 17:00 Python Practice – 23.01.31.. 21702. # Tuesday 17:00 Python Practice – 23.01.31.. ## 02 Feb Tuesday 17:00 Python Practice – 23.01.31.. You could do the following homework using IDLE. Please save your IDLE after finish your homework or you could copy you result to google Docs in the google classroom.. ## Variable Questions:. Question 1. Make a variable and assign a number to it (any number you like). Then display your variable using print.. Question 2. Modify your variable, either by replacing the old value with a new value, or by adding something to the old value. Display the new value using print.. Question 3. Make another variable and assign a string (some text) to it.. Then display it using print.. Question 4. Make a variable for DaysPerWeekHoursPerDay, and MinutesPerHour (or make up your own names), and then multiply them together.. Question 5. People are always saying there’s not enough time to get everything done. How many minutes would there be in a week if there were 26 hours in a day? (Hint: Change the HoursPerDay variable.). Question 6: (Sample Question). Write a program to solve the following question: Three people ate dinner at a restaurant and want to split the bill.	The total is \$35.27, and they want to leave a 15 percent tip. How much should each person pay?. Hint: you can create three variables to start:. >>> people = 3. >>> total = 35.27. >>> tip_percent = 0.15. then you could get the result by calculating step-by-step as following:. ```>>> tip = total * tip_percent. >>> final_total = total + tip. >>> each = final_total / people. >>> print(each)```. ## Data Type Questions:. Question 7.. Use float() to create a number from a string like ‘12.34’. Make sure the result is really a number!. Question 8.. Try using int() to create an integer from a decimal number like 56.78. Did the answer get rounded up or down?. Question 9.. Try using int() to create an integer from a string. Make sure the result is really an integer!.
http://sunshine2k.de/coding/java/PointInTriangle/PointInTriangle.html	1,723,265,334,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00415.warc.gz	24,025,145	4,221	Point in Triangle Algorithms I. Introduction This article is about how to check if a point is inside a given triangle or not. Two approaches are shown and also implemented exemplary as Java applet. This description targets the twodimensional case. So given is a triangle T with its three vertices V1= (vx1, vy1), V2= (vx2, vy2), V3= (vx3, vy3) and a testpoint P = (px, py). Question: Is P included in T? II. Algorithm 1: "Crossproduct Side Algorithm" Note that above name of this algorithm is not official, but my own invention (hope you like this name!). Note that following topic is quite similar to the discussion in my article about polygon convexity [1], but here we have a look from the other side. Maybe you ask yourself now how the crossproduct is defined in 2D and how to interpret it geometrically? Well, the crossproduct is actually not well defined for the 2D case, but a common definition is perpDotProduct(v1, v2) = vx1 * vy2 - vy1 * vx2 This is also known as 'perp dot' product and has some nice properties - if you are interested, read [2]. Also note that it returns a scalar and not a vector (in contrast to the 3D crossproduct) and is actually the same as the determinant of the two two-element vectors. From the latter fact, one (with that I mean a good mathematician) can derive following useful properties: perpDotProduct(v1, v2) = 0 => v1, v2 are parallel perpDotProduct(v1, v2) > 0 => v2 is clockwise from v1 perpDotProduct(v1, v2) < 0 => v1 is counterclockwise from v2 The figure on the left shows the perpDotProduct result geometrically. So actually this all we need :) Calculate the perpDotProduct/crossproduct of all three points V1, V2, V3 with the testpoint P. (It's important that P is either always the first argument or either always the second argument, otherwise you might get into trouble in the final check below!) Call c1 = crossproduct(V1, P), c2 = crossproduct(V2, P) and c3 = crossproduct(V3, P). The final point-in-triangle depends on the prior knowledge of the triangle: • if you know that the vertices V1, V2, V3 are given in clockwise order, then P is inside the triangle if c1> 0 && c2> 0 && c3 > 0. • if you know that the vertices V1, V2, V3 are given in counterclockwise order, then P is inside the triangle if c1< 0 && c2< 0 && c3 < 0. • if you do know nothing about the given order of vertices V1, V2, V3, then P is inside the triangle if (c1> 0 && c2> 0 && c3 > 0) \|\| (c1< 0 && c2< 0 && c3 < 0). III. Algorithm 2: "Barycentric Algorithm" The barycentric approach is based on another mathematical point of view. Imagine that the triangle is part of a plane -that means all three vertices of the triangle lie inside this plane. So we can pick one vertice and the two edges of the triangle starting from this vertices span the whole plan: Define w1 = V2 - V1 and w2 = V3 - V1. Then each point in the plane can be written as: P(s,t) = V1 + s * w1 + t * w2 s,t e R By limiting s and t by s >= 0, t >= 0 and (s + t) <= 1, then P(s,t) describes exactly the points inside the triangle! So have an equation that gives us an arbitrary point inside the traingle by choosing values for s and that fulfill the three limitations s >=0, t >= 0 and (s + t) <= 1. This means that for our testpoint P, we can describe it with this equation P(s,t) - and if s and t fulfill the requirements than P is inside T. Obviously, our goal is to calculate s and t for testpoint P - V1 + s * w1 + t * w2 = P Rearranging gives s * w1 + t * w2 = P - V1 or in vector notation: Either we solve this by hand or apply Cramer's Rule [3] (note that for our special 2D case the determinant is calculated equally as the perpDotProduct!): s = determinant(P - V1, V2) / determinant(V1, V2) t = determinant(V1, P - V2) / determinant(V1, V2) The final point-in-triangle is then • P is inside the triangle if s >= 0 && t >= 0 && (s + t) <= 1. That's it. Hope you enjoyed this article and find the applet + source useful! References: Sunshine, December 2011 This site is part of Sunshine's Homepage	1,137	4,048	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.0625	4	CC-MAIN-2024-33	latest	en	0.874918	Point in Triangle Algorithms. I. Introduction. This article is about how to check if a point is inside a given triangle or not. Two approaches are shown and also implemented exemplary as Java applet. This description targets the twodimensional case.. So given is a triangle T with its three vertices V1= (vx1, vy1), V2= (vx2, vy2), V3= (vx3, vy3) and a testpoint P = (px, py).. Question: Is P included in T?. II. Algorithm 1: "Crossproduct Side Algorithm". Note that above name of this algorithm is not official, but my own invention (hope you like this name!). Note that following topic is quite similar to the discussion in my article about polygon convexity [1], but here we have a look from the other side.. Maybe you ask yourself now how the crossproduct is defined in 2D and how to interpret it geometrically? Well, the crossproduct is actually not well defined for the 2D case, but a common definition is. perpDotProduct(v1, v2) = vx1 * vy2 - vy1 * vx2. This is also known as 'perp dot' product and has some nice properties - if you are interested, read [2]. Also note that it returns a scalar and not a vector (in contrast to the 3D crossproduct) and is actually the same as the determinant of the two two-element vectors. From the latter fact, one (with that I mean a good mathematician) can derive following useful properties:. perpDotProduct(v1, v2) = 0 => v1, v2 are parallel perpDotProduct(v1, v2) > 0 => v2 is clockwise from v1 perpDotProduct(v1, v2) < 0 => v1 is counterclockwise from v2. The figure on the left shows the perpDotProduct result geometrically.. So actually this all we need :) Calculate the perpDotProduct/crossproduct of all three points V1, V2, V3 with the testpoint P. (It's important that P is either always the first argument or either always the second argument, otherwise you might get into trouble in the final check below!). Call c1 = crossproduct(V1, P), c2 = crossproduct(V2, P) and c3 = crossproduct(V3, P).. The final point-in-triangle depends on the prior knowledge of the triangle:. • if you know that the vertices V1, V2, V3 are given in clockwise order, then P is inside the triangle if. c1> 0 && c2> 0 && c3 > 0.. • if you know that the vertices V1, V2, V3 are given in counterclockwise order, then P is inside the triangle if. c1< 0 && c2< 0 && c3 < 0.. • if you do know nothing about the given order of vertices V1, V2, V3, then P is inside the triangle if.	(c1> 0 && c2> 0 && c3 > 0) \|\| (c1< 0 && c2< 0 && c3 < 0).. III. Algorithm 2: "Barycentric Algorithm". The barycentric approach is based on another mathematical point of view. Imagine that the triangle is part of a plane -that means all three vertices of the triangle lie inside this plane. So we can pick one vertice and the two edges of the triangle starting from this vertices span the whole plan:. Define w1 = V2 - V1 and w2 = V3 - V1. Then each point in the plane can be written as: P(s,t) = V1 + s * w1 + t * w2 s,t e R By limiting s and t by s >= 0, t >= 0 and (s + t) <= 1, then P(s,t) describes exactly the points inside the triangle!. So have an equation that gives us an arbitrary point inside the traingle by choosing values for s and that fulfill the three limitations s >=0, t >= 0 and (s + t) <= 1.. This means that for our testpoint P, we can describe it with this equation P(s,t) - and if s and t fulfill the requirements than P is inside T. Obviously, our goal is to calculate s and t for testpoint P -. V1 + s * w1 + t * w2 = P. Rearranging gives. s * w1 + t * w2 = P - V1. or in vector notation:. Either we solve this by hand or apply Cramer's Rule [3] (note that for our special 2D case the determinant is calculated equally as the perpDotProduct!):. s = determinant(P - V1, V2) / determinant(V1, V2) t = determinant(V1, P - V2) / determinant(V1, V2). The final point-in-triangle is then. • P is inside the triangle if s >= 0 && t >= 0 && (s + t) <= 1.. That's it. Hope you enjoyed this article and find the applet + source useful!. References:. Sunshine, December 2011. This site is part of Sunshine's Homepage.
https://calculator.academy/impact-g-force-calculator/	1,685,924,133,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224650409.64/warc/CC-MAIN-20230604225057-20230605015057-00573.warc.gz	193,121,208	54,024	Enter the velocity of impact and the stopping distance of impact into the calculator to determine the number of G forces experience during impact. ## Impact G Force Formula The following equation is used to calculate the Impact G Force. IGF = [1/2v^2 / d ] /9.81 • Where IGF is the number of g-forces experienced during impact • d is the total distance traveled during impact (m) • v is the velocity of impact (m) ## What is an Impact G Force? Definition: An impact G force is the number of units of acceleration experienced during an impact relative to the acceleration due to gravity on Earth. In other words, this is the acceleration during impact divided by 9.81 meters per second squared. ## How to Calculate Impact G Force? Example Problem: The following example outlines the steps and information needed to calculate the Impact G Force. First, determine the velocity at impact. In this example, the velocity at impact is 45 m/s. Next, determine the total distance of the impact. In this case, the impact distance was 1.25 m. Finally, calculate the number of g forces during impact using the formula above: IGF = [1/2v^2 / d ] /9.81 IGF = [1/2*45^2 / 1.25 ] /9.81 IGF = 82.56 Gs	292	1,203	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.90625	4	CC-MAIN-2023-23	latest	en	0.923441	Enter the velocity of impact and the stopping distance of impact into the calculator to determine the number of G forces experience during impact.. ## Impact G Force Formula. The following equation is used to calculate the Impact G Force.. IGF = [1/2*v^2 / d ] /9.81. • Where IGF is the number of g-forces experienced during impact. • d is the total distance traveled during impact (m). • v is the velocity of impact (m). ## What is an Impact G Force?. Definition:. An impact G force is the number of units of acceleration experienced during an impact relative to the acceleration due to gravity on Earth.. In other words, this is the acceleration during impact divided by 9.81 meters per second squared.. ## How to Calculate Impact G Force?.	Example Problem:. The following example outlines the steps and information needed to calculate the Impact G Force.. First, determine the velocity at impact. In this example, the velocity at impact is 45 m/s.. Next, determine the total distance of the impact. In this case, the impact distance was 1.25 m.. Finally, calculate the number of g forces during impact using the formula above:. IGF = [1/2v^2 / d ] /9.81. IGF = [1/245^2 / 1.25 ] /9.81. IGF = 82.56 Gs.
http://slideplayer.com/slide/4994385/	1,566,593,520,000,000,000	text/html	crawl-data/CC-MAIN-2019-35/segments/1566027318986.84/warc/CC-MAIN-20190823192831-20190823214831-00232.warc.gz	178,665,232	20,206	# Physics 1402: Lecture 26 Today’s Agenda Announcements: Midterm 2: NOT Nov. 6 –About Monday Nov. 16 … Homework 07: due Friday this weekHomework 07: due. ## Presentation on theme: "Physics 1402: Lecture 26 Today’s Agenda Announcements: Midterm 2: NOT Nov. 6 –About Monday Nov. 16 … Homework 07: due Friday this weekHomework 07: due."— Presentation transcript: Physics 1402: Lecture 26 Today’s Agenda Announcements: Midterm 2: NOT Nov. 6 –About Monday Nov. 16 … Homework 07: due Friday this weekHomework 07: due Friday this week Electromagnetic Waves –Maxwell’s Equations - Revised –Energy and Momentum in Waves f()x x f(x x z y Maxwell’s Equations These equations describe all of Electricity and Magnetism. They are consistent with modern ideas such as relativity. They describe light ! Maxwell’s Equations - Revised In free space, outside the wires of a circuit, Maxwell’s equations reduce to the following. These can be solved (see notes) to give the following differential equations for E and B. These are wave equations. Just like for waves on a string. But here the field is changing instead of the displacement of the string. Step 1 Assume we have a plane wave propagating in z (ie E, B not functions of x or y) Plane Wave Derivation x z y z1z1 z2z2 ExEx ExEx ZZ xx ByBy Step 2 Apply Faraday’s Law to infinitesimal loop in x-z plane Example: does this Plane Wave Derivation x z y z1z1 z2z2 ByBy ZZ yy ByBy ExEx Step 3 Apply Ampere’s Law to an infinitesimal loop in the y-z plane: Step 4 Combine results from steps 2 and 3 to eliminate B y  Plane Wave Derivation We derived the wave eqn for E x : B y is in phase with E x B 0 = E 0 / c How are E x and B y related in phase and magnitude? (Result from step 2) We could have also derived for B y : –Consider the harmonic solution: where Review of Waves from last semester The one-dimensional wave equation: A specific solution for harmonic waves traveling in the +x direction is: has a general solution of the form: where h 1 represents a wave traveling in the +x direction and h 2 represents a wave traveling in the -x direction. h x A A = amplitude = wavelength f = frequency v = speed k = wave number E & B in Electromagnetic Wave Plane Harmonic Wave: where: y x z Nothing special about (E y,B z ); eg could have (E y,-B x ) Note: the direction of propagation is given by the cross product where are the unit vectors in the (E,B) directions. Note cyclical relation: Lecture 26, ACT 1 Suppose the electric field in an e-m wave is given by: –In what direction is this wave traveling ? 5A (a) + z direction (b) -z direction (c) +y direction (d) -y direction Lecture 26, ACT 2 Suppose the electric field in an e-m wave is given by: Which of the following expressions describes the magnetic field associated with this wave? (a) B x = -(E o /c)cos(kz +  t) (b) B x = +(E o /c)cos(kz -  t) (c) B x = +(E o /c)sin(kz -  t) 5B Velocity of Electromagnetic Waves The wave equation for E x : (derived from Maxwell’s Eqn) Therefore, we now know the velocity of electromagnetic waves in free space: Putting in the measured values for  0 &  0, we get: This value is identical to the measured speed of light! –We identify light as an electromagnetic wave. The EM Spectrum These EM waves can take on any wavelength from angstroms to miles (and beyond). We give these waves different names depending on the wavelength. Wavelength [m] 10 -14 10 -10 10 -6 10 -2 110 2 10 6 10 10 Gamma Rays Infrared Microwaves Short Wave Radio TV and FM Radio AM Radio Long Radio Waves Ultraviolet Visible Light X Rays Lecture 26, ACT 3 Consider your favorite radio station. I will assume that it is at 100 on your FM dial. That means that it transmits radio waves with a frequency f=100 MHz. What is the wavelength of the signal ? A) 3 cmB) 3 mC) ~0.5 mD) ~500 m Energy in EM Waves / review Electromagnetic waves contain energy which is stored in E and B fields: The Intensity of a wave is defined as the average power transmitted per unit area = average energy density times wave velocity: Therefore, the total energy density in an e-m wave = u, where = Momentum in EM Waves Electromagnetic waves contain momentum. The momentum transferred to a surface depends on the area of the surface. Thus Pressure is a more useful quantity. If a surface completely absorbs the incident light, the momentum gained by the surface is, We use the above expression plus Newton’s Second Law in the form F=dp/dt to derive the following expression for the Pressure, If the surface completely reflects the light, conservation of momentum indicates the light pressure will be double that for the surface that absorbs. The Poynting Vector The direction of the propagation of the electromagnetic wave is given by: This wave carries energy. This energy transport is defined by the Poynting vector S as: –The direction of S is the direction of propagation of the wave –The magnitude of S is directly related to the energy being transported by the wave: The intensity for harmonic waves is then given by: The Poynting Vector Thus we get some useful relations for the Poynting vector. 1.The direction of propagation of an EM wave is along the Poynting vector. 2.The Intensity of light at any position is given by the magnitude of the Poynting vector at that position, averaged over a cycle. I = S avg 3.The light pressure is also given by the average value of the Puynting vector as, P = S/cAbsorbing surface P = 2S/cReflecting surface Generating E-M Waves Static charges produce a constant Electric Field but no Magnetic Field. Moving charges (currents) produce both a possibly changing electric field and a static magnetic field. Accelerated charges produce EM radiation (oscillating electric and magnetic fields). Antennas are often used to produce EM waves in a controlled manner. A Dipole Antenna V(t)=V o cos(  t) x z y time t=0 ++++ ---- E time t=  /2  E time t=  /  one half cycle later ---- ++++ dipole radiation pattern oscillating electric dipole generates e-m radiation that is polarized in the direction of the dipole radiation pattern is doughnut shaped & outward traveling –zero amplitude directly above and below dipole –maximum amplitude in-plane proportional to sin(  t) Receiving E-M Radiation receiving antenna One way to receive an EM signal is to use the same sort of antenna. Receiving antenna has charges which are accelerated by the E field of the EM wave. The acceleration of charges is the same thing as an EMF. Thus a voltage signal is created. Speaker y x z Lecture 26, ACT 4 Consider an EM wave with the E field POLARIZED to lie perpendicular to the ground. y x z In which orientation should you turn your receiving dipole antenna in order to best receive this signal? C) Along E a) Along S b) Along B Loop Antennas Magnetic Dipole Antennas The electric dipole antenna makes use of the basic electric force on a charged particle Note that you can calculate the related magnetic field using Ampere’s Law. We can also make an antenna that produces magnetic fields that look like a magnetic dipole, i.e. a loop of wire. This loop can receive signals by exploiting Faraday’s Law. For a changing B field through a fixed loop Lecture 26, ACT 5 Consider an EM wave with the E field POLARIZED to lie perpendicular to the ground. y x z In which orientation should you turn your receiving loop antenna in order to best receive this signal? a) â Along S b) â Along B C) â Along E Similar presentations	1,879	7,462	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.546875	4	CC-MAIN-2019-35	latest	en	0.834506	# Physics 1402: Lecture 26 Today’s Agenda Announcements: Midterm 2: NOT Nov. 6 –About Monday Nov. 16 … Homework 07: due Friday this weekHomework 07: due.. ## Presentation on theme: "Physics 1402: Lecture 26 Today’s Agenda Announcements: Midterm 2: NOT Nov. 6 –About Monday Nov. 16 … Homework 07: due Friday this weekHomework 07: due."— Presentation transcript:. Physics 1402: Lecture 26 Today’s Agenda Announcements: Midterm 2: NOT Nov. 6 –About Monday Nov. 16 … Homework 07: due Friday this weekHomework 07: due Friday this week Electromagnetic Waves –Maxwell’s Equations - Revised –Energy and Momentum in Waves. f()x x f(x x z y. Maxwell’s Equations These equations describe all of Electricity and Magnetism. They are consistent with modern ideas such as relativity. They describe light !. Maxwell’s Equations - Revised In free space, outside the wires of a circuit, Maxwell’s equations reduce to the following. These can be solved (see notes) to give the following differential equations for E and B. These are wave equations. Just like for waves on a string. But here the field is changing instead of the displacement of the string.. Step 1 Assume we have a plane wave propagating in z (ie E, B not functions of x or y) Plane Wave Derivation x z y z1z1 z2z2 ExEx ExEx ZZ xx ByBy Step 2 Apply Faraday’s Law to infinitesimal loop in x-z plane Example: does this. Plane Wave Derivation x z y z1z1 z2z2 ByBy ZZ yy ByBy ExEx Step 3 Apply Ampere’s Law to an infinitesimal loop in the y-z plane: Step 4 Combine results from steps 2 and 3 to eliminate B y . Plane Wave Derivation We derived the wave eqn for E x : B y is in phase with E x B 0 = E 0 / c How are E x and B y related in phase and magnitude? (Result from step 2) We could have also derived for B y : –Consider the harmonic solution: where. Review of Waves from last semester The one-dimensional wave equation: A specific solution for harmonic waves traveling in the +x direction is: has a general solution of the form: where h 1 represents a wave traveling in the +x direction and h 2 represents a wave traveling in the -x direction. h x A A = amplitude = wavelength f = frequency v = speed k = wave number. E & B in Electromagnetic Wave Plane Harmonic Wave: where: y x z Nothing special about (E y,B z ); eg could have (E y,-B x ) Note: the direction of propagation is given by the cross product where are the unit vectors in the (E,B) directions. Note cyclical relation:. Lecture 26, ACT 1 Suppose the electric field in an e-m wave is given by: –In what direction is this wave traveling ? 5A (a) + z direction (b) -z direction (c) +y direction (d) -y direction. Lecture 26, ACT 2 Suppose the electric field in an e-m wave is given by: Which of the following expressions describes the magnetic field associated with this wave? (a) B x = -(E o /c)cos(kz +  t) (b) B x = +(E o /c)cos(kz -  t) (c) B x = +(E o /c)sin(kz -  t) 5B. Velocity of Electromagnetic Waves The wave equation for E x : (derived from Maxwell’s Eqn) Therefore, we now know the velocity of electromagnetic waves in free space: Putting in the measured values for  0 &  0, we get: This value is identical to the measured speed of light! –We identify light as an electromagnetic wave.. The EM Spectrum These EM waves can take on any wavelength from angstroms to miles (and beyond). We give these waves different names depending on the wavelength. Wavelength [m] 10 -14 10 -10 10 -6 10 -2 110 2 10 6 10 10 Gamma Rays Infrared Microwaves Short Wave Radio TV and FM Radio AM Radio Long Radio Waves Ultraviolet Visible Light X Rays. Lecture 26, ACT 3 Consider your favorite radio station. I will assume that it is at 100 on your FM dial. That means that it transmits radio waves with a frequency f=100 MHz.	What is the wavelength of the signal ? A) 3 cmB) 3 mC) ~0.5 mD) ~500 m. Energy in EM Waves / review Electromagnetic waves contain energy which is stored in E and B fields: The Intensity of a wave is defined as the average power transmitted per unit area = average energy density times wave velocity: Therefore, the total energy density in an e-m wave = u, where =. Momentum in EM Waves Electromagnetic waves contain momentum. The momentum transferred to a surface depends on the area of the surface. Thus Pressure is a more useful quantity. If a surface completely absorbs the incident light, the momentum gained by the surface is, We use the above expression plus Newton’s Second Law in the form F=dp/dt to derive the following expression for the Pressure, If the surface completely reflects the light, conservation of momentum indicates the light pressure will be double that for the surface that absorbs.. The Poynting Vector The direction of the propagation of the electromagnetic wave is given by: This wave carries energy. This energy transport is defined by the Poynting vector S as: –The direction of S is the direction of propagation of the wave –The magnitude of S is directly related to the energy being transported by the wave: The intensity for harmonic waves is then given by:. The Poynting Vector Thus we get some useful relations for the Poynting vector. 1.The direction of propagation of an EM wave is along the Poynting vector. 2.The Intensity of light at any position is given by the magnitude of the Poynting vector at that position, averaged over a cycle. I = S avg 3.The light pressure is also given by the average value of the Puynting vector as, P = S/cAbsorbing surface P = 2S/cReflecting surface. Generating E-M Waves Static charges produce a constant Electric Field but no Magnetic Field. Moving charges (currents) produce both a possibly changing electric field and a static magnetic field. Accelerated charges produce EM radiation (oscillating electric and magnetic fields). Antennas are often used to produce EM waves in a controlled manner.. A Dipole Antenna V(t)=V o cos(  t) x z y time t=0 ++++ ---- E time t=  /2  E time t=  /  one half cycle later ---- ++++. dipole radiation pattern oscillating electric dipole generates e-m radiation that is polarized in the direction of the dipole radiation pattern is doughnut shaped & outward traveling –zero amplitude directly above and below dipole –maximum amplitude in-plane proportional to sin(  t). Receiving E-M Radiation receiving antenna One way to receive an EM signal is to use the same sort of antenna. Receiving antenna has charges which are accelerated by the E field of the EM wave. The acceleration of charges is the same thing as an EMF. Thus a voltage signal is created. Speaker y x z. Lecture 26, ACT 4 Consider an EM wave with the E field POLARIZED to lie perpendicular to the ground. y x z In which orientation should you turn your receiving dipole antenna in order to best receive this signal? C) Along E a) Along S b) Along B. Loop Antennas Magnetic Dipole Antennas The electric dipole antenna makes use of the basic electric force on a charged particle Note that you can calculate the related magnetic field using Ampere’s Law. We can also make an antenna that produces magnetic fields that look like a magnetic dipole, i.e. a loop of wire. This loop can receive signals by exploiting Faraday’s Law. For a changing B field through a fixed loop. Lecture 26, ACT 5 Consider an EM wave with the E field POLARIZED to lie perpendicular to the ground. y x z In which orientation should you turn your receiving loop antenna in order to best receive this signal? a) â Along S b) â Along B C) â Along E. Similar presentations.
https://mathhothouse.me/2020/05/	1,638,419,626,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00043.warc.gz	462,769,804	24,878	## Monthly Archives: May 2020 ### Skill Check III: IITJEE Foundation Maths State whether the following statements are true or false: 1. If $A = \{ x \| x = 5n, 5 < n < 10, n \in N\}$, then $n(A)=4$ 2. If $n(A) = n(B)$, then $set A \leftrightarrow set B$ 3. If Set $A= \{ x \| x \in N, x<3\}$, then A is a singleton set. 4. The intelligent students of class VIII form a set. 5. The students passing the half-yearly exams in Class VIII B of school is a set. 6. $A = \{ x \| x = p^{3}, p<4, p \in \mathcal{N}\}$ and $\{ x\|x = m^{2}, m < 4, m \in \mathcal{N} \}$ are overlapping sets. 7. $A = \{ x \| x \in \mathcal{N}\}$ is a subset of $B = \{ x \| x in \mathcal{Z}\}$ 8. If we denote the universal set as $\Omega = \{ p,q,r,s,t,u,v\}$ and $A = \{ q,u,s,t\}$, then $\overline{A} = \{ p, r, v\}$ 9. $A = \{ x \| x =2p, p \in \mathcal{N}\}$ and $B = \{ x \| x =3p, p \in \mathcal{N}\}$ are disjoint sets. 10. If $A = \{ 2,3,4\}$, then $P(A)= \{ \phi, A, \{ 2\}, \{ 3\}, \{ 4 \}, \{ 2,3\}, \{ 2,4\}, \{ 3,4\}\}$ where $P(A)$ is the power set of set A. II. If C is a letter in the word down all the subsets of C. III. Write down the complements of all the 8 subsets of set C above. IV. If $Q = \{ x : x =a^{2}+1, 2 \leq a \leq 5\}$, what is the power set of Q? V. If $x = \{ x \| x<20, x \in \mathcal{N}\}$, and if $A = \{x \| x = 2a, 3 < a < 8, a \in \mathcal{N} \}$, and if $B = \{ x \| x = 3b, b < 5, b \in \mathcal{N}\}$, and if $C = \{x \| x = c+1, 5 < c < 15, c \in \mathcal{N} \}$, then find : (i) $n(B)$ (ii) $n(C)$ (iii) $\overline{A}$ (iv) $\overline{B}$ (v) $P(B)$ VI. If $A = \{ x\| x \in \mathcal{N}, 3 < x < 10\}$, and if $B = \{ x\| x =4a-1, a<5, a \in \mathcal{N}\}$ and if $C = \{ x \| x = 3a+2, a<7, a \in \mathcal{N}\}$, then confirm the following: (i) the commutative property of the unions of sets B and C (ii) the commutative property of intersection of two sets A and C (iii) the associative property of the union of the sets A, B and C (iv) the associative property of intersection of sets A, B and C. VII. If $A = \{ x \| x \in \mathcal{N}, 4 \leq x \leq 12\}$, and $B = \{ x\| x = a+1, a<8, a \in \mathcal{N}\}$, and $C= \{ x\| x =2n, 1 < n <7, n \in \|mathcal{N}\}$, then find (i) $A-B$ (ii) $B-C$ (iii) $B \bigcap C$ (iv) $A - (B \bigcap C)$ (v) $B - (A \bigcap C)$ (vi) $A-C$ (vii) $A- (B-C)$ (viii) $A- (B \bigcup C)$ VIII. If $\xi = \{ x \| x \hspace{0.1in}is \hspace{0.1in} a \hspace{0.1in} letter \hspace{0.1in}of \hspace{0.1in}the \hspace{0.1in} English \hspace{0.1in} alphabet \hspace{0.1in} between \hspace{0.1in} but \hspace{0.1in} not \hspace{0.1in} including \hspace{0.1in} d \hspace{0.1in} and \hspace{0.1in} o\}$, and let $A = \{ l, m , n\}$ and let $B= \{ e,f,g,h,i,j,k,l\}$, and let $C = \{ j,k,l,m\}$, find (i) $\overline{A} \bigcup \overline{B}$ (ii) $\overline{B} \bigcap \overline{C}$ (iii) $A \bigcap C$ (iv) $B - (A \bigcap C)$ (v) $\overline{B-A}$ (vi) Is $(B-C) \subset (B-A)$? (vii) Is $\overline{A} \bigcap \overline{B} = \phi$? IX. All 26 customers in a restaurant had either drinks, snacks, or dinner. 18 had snacks, out of which 6 had only snacks, 4 had snacks and drinks but not dinner, 2 had drinks and dinner but not snacks, and 3 had snacks and dinner but not drinks. If 14 customers had drinks, find (i) how many customers had all three — drinks, snacks as well as dinner. (ii) how many customers had dinner but neither snacks nor drinks (iii) how many customers had only drinks. Regards, Nalin Pithwa Purva building, 5A Flat 06 Mumbai, Maharastra 400101 India ### Maths will rock your world — a motivational article keep dreaming the applications of math…try to count every thing…. Jan 23 2006 A generation ago, quants turned finance upside down. Now they’re mapping out ad campaigns and building new businesses from mountains of personal data Neal Goldman is a math entrepreneur. He works on Wall Street, where numbers rule. But he’s focusing his analytic tools on a different realm altogether: the world of words. Goldman’s startup, Inform Technologies LLC, is a robotic librarian. Every day it combs through thousands of press articles and blog posts in English. It reads them and groups them with related pieces. Inform doesn’t do this work alphabetically or by keywords. It uses algorithms to analyze each article by its language and context. It then sends customized news feeds to its users, who also exist in Inform’s system as — you guessed it — math. How do you convert written words into math? Goldman says it takes a… View original post 3,400 more words ### A motivation for Math and some Math competitive exams in India what motivates you to keep going in math? Sometime back, there was a tremendous publicity in the Indian media to two Fields medallists of Indian origin. They also talked about what motivated them towards Math when they were young. One should not do Math just lured by its glamorous applications in IT or other engineering disciplines. But, one can develop both aptitude and attitude towards it if one works from a young age. What you need is intrinsic motivation. In this context, I like to quote the following words of a famous mathematician: “And, a final observation. We should not forget that the solution to any worthwhile problem very rarely comes to us easily and without hard work; it is rather the result of intellectual effort of days or weeks or months. Why should the young mind be willing to make this supreme effort? The explanation is probably the instinctive preference for certain values, that is, the attitude… View original post 214 more words ### Skill Check II: IITJEE foundation maths Set Theory Primer/basics/fundamentals/preliminaries: I. Represent the following sets in Venn Diagrams: (a) $\Xi = \{ x : x = n, n <40, n \in \mathcal{N} \}$ (b) $\mu = \{ x : x = 6n, n < 6. n \in \mathcal{N}\}$ (c) $\alpha = \{ x: x = 3n, n<8, n \in \mathcal{N} \}$ 2. If $x = \{ x: x<29 \hspace{0.1in}and \hspace{0.1in}prime\}$ and $A = \{ x: x \hspace{0.1in} is \hspace{0.1in} a \hspace{0.1in} prime \hspace{0.1in} factor \hspace{0.1in} of \hspace{0.1in} 210\}$, represent A in Venn diagram and find $\overline{A}$. 3. 95 boys of a school appeared for a physical for selection in NCC and Boy Scouts. 21 boys got selected in both NCC and Boy Scouts, 44 boys were not selected in Boy Scouts and 20 boys were not selected only in boy scouts. Draw Venn diagram and find : (i) how many boys did not get selected in NCC and boy scouts. (ii) how many boys did not get selected only in NCC (iii) how many boys got selected in NCC (iv) how many boys got selected in boy Scouts (v) How many boys got selected in NCC not in boy Scouts? Regards, Nalin Pithwa. ### Skill Check I: IITJEE Foundation Maths 1. Simplify: $(+1) \times (-1) + (+1) \div (-1) -()-1 +(-1) \div (-1) \times (-1) +(-1)$ 2. Simplify: $\{ 7 \hspace{0.1in} of \hspace{0.1in} 6 \div 2 - 4(8 \times 12 \div 3 + \overline{-3 \hspace{0.1in} of \hspace{0.1in}+6 -2 })\} \div (-3)$ 3. Simplify: $24 \div (-8) + 3 \times (-3)$ 4. Simplify: $(+7) - (-3) + (+4) \times (-3) \div (+3) of (-2)$ 5. Simplify: $(-3) of (-5) \div (-3) \times (-2) + (-5) - (-2) \div (+2)$ 6. Simplify: $(-7) + (-8) - (-3) \hspace{0.1in} of \hspace{0.1in} (-6) \div (+2) - (-4) \times (-4) \div (+2)$ 7. Simplify: $(+24) \div (-3) \hspace{0.1in} of \hspace{0.1in} (+4) - (-25) \times (-6) \div (-3) + (-15) \div (-3) \times (-10)$ 8. Simplify: $(-3) \hspace{0.1in} of \hspace{0.1in} (-8) \div (-6) - (-8 +4-3)$ 9. Simplify: $(-5) [ (-6) - \{ -5 + (-2 + 1 - \overline{3-2}) \} ]$ 10. Simplify: $(-3) [ (-8) - \{ +7 - (4-5 - \overline{2-5-1})\}] \div (-11)$ 11. Simplify: $(+8) \times (-3) \times (+2) \div [ -1 - \{ -3 + 8 - (6 -2 - \overline{3+5-4}) \} ]$ 12. Simplify: $(+32) \div (+2) of (-4) \div [(-7) of 3 \div \{4 - 5(3 - \overline{4 of 2 - 2 of 5}) \}]$ 13. Simplify: $(-30) + (-8) \div (-4) \time 2$ 14. Simplify: $(-3) \times (-6) \div (-2) + (-1)$ 15. Simplify: $56 \div (16 + \overline{4-6}) + (6-8)$ 16. Simplify: $(7+6) \times [19 + \{ (-15) + \overline{6-1}\}]$ Regards, Nalin Pithwa Purva building, 5A Flat 06 Mumbai, Maharastra 400101 India Nalin Pithwa. ### Math Basics Division by Zero the eleventh commandment of Moses: Though shall not divide by zero !! Let’s pause Geometry for a little time and start thinking of some basic rules of the game of Math. Have you ever asked “why is division by zero not allowed in Math?” Try to do 1/2 in a calculator and see what you get!! This was also a question an immortal Indian math genius, Srinivasa Ramanujan had asked his school teacher when he was a tiny tot. Note the following two arguments against the dangers of division by zero: (a) Suppose there are 4 apples and two persons want to divide them equally. So, it is 4/2 apples per person, that is, 2 apples per person. But, now consider a scenario in which there are 4 apples and 0 persons. So, how can you divide 4 apples amongst (or by) 0 persons? You can think of any crazy answer and keep on arguing endlessly about it!!!! (b) The cancellation law… View original post 267 more words ### IITJEE Foundation Maths: Tutorial Problems IV 1. Resolve into factors: (a) $2x^{2}-3ab+(a-6b)x$ (b) $4x^{2}-4xy-15y^{2}$ 2. In the expression, $x^{3}-2x^{2}+3x-4$, substitute $a-2$ for x, and arrange the result according to the descending powers of a. 3. Simplify: (i) $\frac{x}{1-\frac{1}{1+x}}$ (ii) $\frac{x^{2}}{a+\frac{x^{2}}{a+\frac{x^{2}}{a}}}$ 4. Find the HCF of $3x^{3}-11x^{2}+x+15$ and $5x^{4}-7x^{3}-20x^{2}-11x-3$ 5. Express in the simplest form: (i) $\frac{\frac{x}{y}-\frac{y}{x}}{\frac{x+y}{y}-\frac{y+x}{x}}$ (ii) $(\frac{x^{3}-1}{x-1} + \frac{x^{3}+1}{x+1})\div (\frac{1}{x-1} + \frac{1}{x+1})$ 6. A person possesses Rs. 5000 stock, some at 3 per cent; four times as much at 3.5 %, and the rest at 4 %; find the amount of each kind of stock when his income is Rs. 178. 7. Simplify the expression: $-3[(a+b)-{(2a-3b) -( 5a+7b-16c) - (-13a +2b -3c -5d)}]$, and find its value when $a=1, b=2, c=3, d=4$. 8. Solve the following equations : (i) $11y-x=10$ and $11x-101y=110$ (ii) $x+y-z=3$, and $x+z-y=5$, and $y+z-x=7$. 9. Express the following fractions in their simplest form: (i) $\frac{32x^{3}-2x+12}{12x^{5}-x^{4}+4x^{2}}$ (ii) $\frac{1}{x + \frac{1}{1+ \frac{x+3}{2-x}}}$ 10. What value of a will make the product of $3-8a$ and $3a+4$ equal to the product of $6a+11$ and $3-4a$? 11. Find the LCM of $x^{3}-x^{2}-3x-9$ and $x^{3}-2x^{2}-5x-12$ 12. A certain number of two digits is equal to seven times the sum of its digits; if the digit in the units’ place be decreased by two and that in the tens place by one, and if the number thus formed be divided by the sum of its digits, the quotient is 10. Find the number. 13. Find the value of $\frac{6x^{2}-5xy-6y^{2}}{2x^{2}+xy-y^{2}} \times \frac{3x^{2}-xy-4y^{2}}{2x^{2}-5xy+3y^{2}} \div \frac{9x^{2}-6xy-8y^{2}}{2x^{2}-3xy+y^{2}}$ 14. Resolve each of the following expressions into four factors: (i) $4a^{4}-17a^{2}b^{2}+4b^{4}$; (ii) $x^{8}-256y^{8}$ 15. Find the expression of highest dimensions which will divide $24a^{4}b -2a^{2}b^{2}-9ab^{4}$ and $18a^{6}+a^{4}b^{2}-6a^{3}b^{3}$ without remainder. 16. Find the square root of : (i) $x(x+1)(x+2)(x+3)+1$ (ii) $(2a^{2}+13a+15)(a^{2}+4a-5)(2a^{2}+a-3)$ 17. Simplify: $x - \frac{2x-6}{x^{2}-6x+9} - 3 + \frac{x^{2}+3x-4}{x^{2}=x-12}$ 18. A quantity of land, partly pasture and partly arable, is sold at the rate of Rs. 60 per acre for the pasture and Rs. 40 per acre for the arable, and the whole sum obtained at Rs. 10000. If the average price per acre were Rs. 50, the sum obtained would be 10 per cent higher; find how much of the land is pasture and how much is arable. Regards, Nalin Pithwa Purva building, 5A Flat 06 Mumbai, Maharastra 400101 India ### Math from scratch Most important out of all blogs…explains what is math and the nature of its rigour Math from scratch! Math differs from all other pure sciences in the sense that it can be developed from scratch. In math jargon, it is called “to develop from first principles”. You might see such questions in Calculus and also Physics. This is called Axiomatic-Deductive Logic and was first seen in the works of Euclid’s Elements (Plane Geometry) about 2500 years ago. The ability to think from “first principles” can be developed in high-school with proper understanding and practise of Euclid’s Geometry. For example, as a child Albert Einstein was captivated to see a proof from scratch in Euclid’s Geometry that “the three medians of a triangle are concurrent”. (This, of course, does not need anyone to verify by drawing thousands of triangles and their mediansJ) That hooked the child Albert Einstein to Math, and later on to Physics. An axiom is a statement which means “self-evident truth”. We accept… View original post 664 more words ### IITJEE Foundation Maths : Tutorial Problems III 1. When $a=-3, b=5, c=-1, d=0$, find the value of $26c\sqrt[3]{a^{3}-c^{2}d+5bc-4ac+d^{2}}$ 2. Solve the equations: (a) $\frac{1}{3}x - \frac{2}{7}y = 8 -2x$ and $\frac{1}{2}y - 3x =3-y$ (b) $1 = y+z =2(z+x)=3(x+y)$ 3. Simplify: (a) $\frac{a-x}{a+x} - \frac{4x^{2}}{a^{2}-x^{2}} + \frac{a-3x}{x-a}$ (b) $\frac{b^{2}-3b}{b^{2}-2b+4} \times \frac{b^{2}+b-30}{b^{2}+3b-18} \div \frac{b^{3}-3b^{2}-10b}{b^{2}+8}$ 4. Find the square root of : $9-36x+60x^{2}-\frac{160}{3}x^{3}+\frac{80}{3}x^{4}-\frac{64}{3}x^{5}+\frac{64}{81}x^{6}$ 5. In a cricket match, the extras in the first innings are one-sixteenth of the score, and in the second innings the extras are one-twelfth of the score. The grand total is 296, of which 21 are extras; find the score in each innings. 6. Find the value of $\frac{a^{2}-x^{2}}{\frac{1}{a^{2}} - \frac{2}{ax} + \frac{1}{x^{2}}} \times \frac{\frac{1}{a^{2}x^{2}}}{a+x}$ 7. Find the value of : $\frac{1}{3}(a+2) -3(1-\frac{1}{6}b) - \frac{2}{3}(2a-3b+\frac{3}{2}) + \frac{3}{2}b - 4(\frac{1}{2}a-\frac{1}{3})$. 8. Resolve into factors: (i) $3a^{2} -20a-7$ (ii) $a^{4}b^{2}-b^{4}a^{2}$ 9. Reduce to lowest terms: $\frac{4x^{3}+7x^{2}-x+2}{4x^{3}+5x^{2}-7x-2}$ 10. Solve the following equations: (a) $x-6 -\frac{x-12}{3}= \frac{x-4}{2} + \frac{x-8}{4}$; (b) $x+y-z=0$, $x-y+z=4$, $5x+y+z=20$; (c) $\frac{ax+b}{c} + \frac{dx+e}{f} =1$ 11. Simplify: $\frac{x+3}{x^{2}-5x+6} - \frac{x+2}{x^{2}-9x+14} + \frac{4}{x^{2}-10x+21}$ 12. A purse of rupees is divided amongst three persons, the first receiving half of them and one more, the second half of the remainder and one more, and the third six. Find the number of rupees the purse contained. 13. If $h=-2, k=1, l=0, m=1, n=-3$, find the value of $\frac{h^{2}(m-l)-\sqrt{3hn}+hk}{m(l-h)-2hm^{2}+\sqrt[3]{4hk}}$ 14. Find the LCM of $15(p^{3}+q^{3}), 5(p^{2}-pq+q^{2}), 4(p^{2}+pq+q^{2}), 6(p^{2}-q^{2})$ 15. Find the square root of (i) $\frac{4x^{2}}{9} + \frac{9}{4x^{2}} -2$; (ii) $1-6a+5a^{2}+12a^{3}+4a^{4}$ 16. Simplify $\frac{20x^{2}+27x+9}{15x^{2}+19x+6} + \frac{20x^{2}+27x+9}{12x^{2}+17x+6}$ 17. Solve the equations: (i) $\frac{a(x-b)}{a-b} + \frac{b(x-a)}{b-a} =1$ (ii) $\frac{9}{x-4} + \frac{3}{x-8} = \frac{4}{x-9} + \frac{8}{x-3}$ 18. A sum of money is to be divided among a number of persons; if Rs. 8 is given to each there will be Rs. 3 short, and if Rs. 7.50 is given to each there will be Rs. 2 over; find the number of persons. Regards, Nalin Pithwa Purva building, 5A Flat 06 Mumbai , Maharastra 400101 India	5,818	15,358	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 134, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.28125	4	CC-MAIN-2021-49	latest	en	0.625734	## Monthly Archives: May 2020. ### Skill Check III: IITJEE Foundation Maths. State whether the following statements are true or false:. 1. If $A = \{ x \| x = 5n, 5 < n < 10, n \in N\}$, then $n(A)=4$. 2. If $n(A) = n(B)$, then $set A \leftrightarrow set B$. 3. If Set $A= \{ x \| x \in N, x<3\}$, then A is a singleton set.. 4. The intelligent students of class VIII form a set.. 5. The students passing the half-yearly exams in Class VIII B of school is a set.. 6. $A = \{ x \| x = p^{3}, p<4, p \in \mathcal{N}\}$ and $\{ x\|x = m^{2}, m < 4, m \in \mathcal{N} \}$ are overlapping sets.. 7. $A = \{ x \| x \in \mathcal{N}\}$ is a subset of $B = \{ x \| x in \mathcal{Z}\}$. 8. If we denote the universal set as $\Omega = \{ p,q,r,s,t,u,v\}$ and $A = \{ q,u,s,t\}$, then $\overline{A} = \{ p, r, v\}$. 9. $A = \{ x \| x =2p, p \in \mathcal{N}\}$ and $B = \{ x \| x =3p, p \in \mathcal{N}\}$ are disjoint sets.. 10. If $A = \{ 2,3,4\}$, then $P(A)= \{ \phi, A, \{ 2\}, \{ 3\}, \{ 4 \}, \{ 2,3\}, \{ 2,4\}, \{ 3,4\}\}$ where $P(A)$ is the power set of set A.. II. If C is a letter in the word down all the subsets of C.. III. Write down the complements of all the 8 subsets of set C above.. IV. If $Q = \{ x : x =a^{2}+1, 2 \leq a \leq 5\}$, what is the power set of Q?. V. If $x = \{ x \| x<20, x \in \mathcal{N}\}$, and if $A = \{x \| x = 2a, 3 < a < 8, a \in \mathcal{N} \}$, and if $B = \{ x \| x = 3b, b < 5, b \in \mathcal{N}\}$, and if $C = \{x \| x = c+1, 5 < c < 15, c \in \mathcal{N} \}$, then find : (i) $n(B)$ (ii) $n(C)$ (iii) $\overline{A}$ (iv) $\overline{B}$ (v) $P(B)$. VI. If $A = \{ x\| x \in \mathcal{N}, 3 < x < 10\}$, and if $B = \{ x\| x =4a-1, a<5, a \in \mathcal{N}\}$ and if $C = \{ x \| x = 3a+2, a<7, a \in \mathcal{N}\}$, then confirm the following: (i) the commutative property of the unions of sets B and C (ii) the commutative property of intersection of two sets A and C (iii) the associative property of the union of the sets A, B and C (iv) the associative property of intersection of sets A, B and C.. VII. If $A = \{ x \| x \in \mathcal{N}, 4 \leq x \leq 12\}$, and $B = \{ x\| x = a+1, a<8, a \in \mathcal{N}\}$, and $C= \{ x\| x =2n, 1 < n <7, n \in \|mathcal{N}\}$, then find (i) $A-B$ (ii) $B-C$ (iii) $B \bigcap C$ (iv) $A - (B \bigcap C)$ (v) $B - (A \bigcap C)$ (vi) $A-C$ (vii) $A- (B-C)$ (viii) $A- (B \bigcup C)$. VIII. If $\xi = \{ x \| x \hspace{0.1in}is \hspace{0.1in} a \hspace{0.1in} letter \hspace{0.1in}of \hspace{0.1in}the \hspace{0.1in} English \hspace{0.1in} alphabet \hspace{0.1in} between \hspace{0.1in} but \hspace{0.1in} not \hspace{0.1in} including \hspace{0.1in} d \hspace{0.1in} and \hspace{0.1in} o\}$, and let $A = \{ l, m , n\}$ and let $B= \{ e,f,g,h,i,j,k,l\}$, and let $C = \{ j,k,l,m\}$, find (i) $\overline{A} \bigcup \overline{B}$ (ii) $\overline{B} \bigcap \overline{C}$ (iii) $A \bigcap C$ (iv) $B - (A \bigcap C)$ (v) $\overline{B-A}$ (vi) Is $(B-C) \subset (B-A)$? (vii) Is $\overline{A} \bigcap \overline{B} = \phi$?. IX. All 26 customers in a restaurant had either drinks, snacks, or dinner. 18 had snacks, out of which 6 had only snacks, 4 had snacks and drinks but not dinner, 2 had drinks and dinner but not snacks, and 3 had snacks and dinner but not drinks. If 14 customers had drinks, find (i) how many customers had all three — drinks, snacks as well as dinner. (ii) how many customers had dinner but neither snacks nor drinks (iii) how many customers had only drinks.. Regards,. Nalin Pithwa. Purva building, 5A. Flat 06. Mumbai, Maharastra 400101. India. ### Maths will rock your world — a motivational article. keep dreaming the applications of math…try to count every thing….. Jan 23 2006. A generation ago, quants turned finance upside down. Now they’re mapping out ad campaigns and building new businesses from mountains of personal data. Neal Goldman is a math entrepreneur. He works on Wall Street, where numbers rule. But he’s focusing his analytic tools on a different realm altogether: the world of words.. Goldman’s startup, Inform Technologies LLC, is a robotic librarian. Every day it combs through thousands of press articles and blog posts in English. It reads them and groups them with related pieces. Inform doesn’t do this work alphabetically or by keywords. It uses algorithms to analyze each article by its language and context. It then sends customized news feeds to its users, who also exist in Inform’s system as — you guessed it — math.. How do you convert written words into math? Goldman says it takes a…. View original post 3,400 more words. ### A motivation for Math and some Math competitive exams in India. what motivates you to keep going in math?. Sometime back, there was a tremendous publicity in the Indian media to two Fields medallists of Indian origin. They also talked about what motivated them towards Math when they were young. One should not do Math just lured by its glamorous applications in IT or other engineering disciplines. But, one can develop both aptitude and attitude towards it if one works from a young age.. What you need is intrinsic motivation. In this context, I like to quote the following words of a famous mathematician:. “And, a final observation. We should not forget that the solution to any worthwhile problem very rarely comes to us easily and without hard work; it is rather the result of intellectual effort of days or weeks or months. Why should the young mind be willing to make this supreme effort? The explanation is probably the instinctive preference for certain values, that is, the attitude…. View original post 214 more words. ### Skill Check II: IITJEE foundation maths. Set Theory Primer/basics/fundamentals/preliminaries:. I. Represent the following sets in Venn Diagrams: (a) $\Xi = \{ x : x = n, n <40, n \in \mathcal{N} \}$ (b) $\mu = \{ x : x = 6n, n < 6. n \in \mathcal{N}\}$ (c) $\alpha = \{ x: x = 3n, n<8, n \in \mathcal{N} \}$. 2. If $x = \{ x: x<29 \hspace{0.1in}and \hspace{0.1in}prime\}$ and $A = \{ x: x \hspace{0.1in} is \hspace{0.1in} a \hspace{0.1in} prime \hspace{0.1in} factor \hspace{0.1in} of \hspace{0.1in} 210\}$, represent A in Venn diagram and find $\overline{A}$.. 3. 95 boys of a school appeared for a physical for selection in NCC and Boy Scouts. 21 boys got selected in both NCC and Boy Scouts, 44 boys were not selected in Boy Scouts and 20 boys were not selected only in boy scouts. Draw Venn diagram and find : (i) how many boys did not get selected in NCC and boy scouts. (ii) how many boys did not get selected only in NCC (iii) how many boys got selected in NCC (iv) how many boys got selected in boy Scouts (v) How many boys got selected in NCC not in boy Scouts?. Regards,. Nalin Pithwa.. ### Skill Check I: IITJEE Foundation Maths. 1. Simplify: $(+1) \times (-1) + (+1) \div (-1) -()-1 +(-1) \div (-1) \times (-1) +(-1)$. 2. Simplify: $\{ 7 \hspace{0.1in} of \hspace{0.1in} 6 \div 2 - 4(8 \times 12 \div 3 + \overline{-3 \hspace{0.1in} of \hspace{0.1in}+6 -2 })\} \div (-3)$. 3. Simplify: $24 \div (-8) + 3 \times (-3)$. 4. Simplify: $(+7) - (-3) + (+4) \times (-3) \div (+3) of (-2)$. 5. Simplify: $(-3) of (-5) \div (-3) \times (-2) + (-5) - (-2) \div (+2)$. 6. Simplify: $(-7) + (-8) - (-3) \hspace{0.1in} of \hspace{0.1in} (-6) \div (+2) - (-4) \times (-4) \div (+2)$. 7. Simplify: $(+24) \div (-3) \hspace{0.1in} of \hspace{0.1in} (+4) - (-25) \times (-6) \div (-3) + (-15) \div (-3) \times (-10)$. 8. Simplify: $(-3) \hspace{0.1in} of \hspace{0.1in} (-8) \div (-6) - (-8 +4-3)$. 9. Simplify: $(-5) [ (-6) - \{ -5 + (-2 + 1 - \overline{3-2}) \} ]$. 10. Simplify: $(-3) [ (-8) - \{ +7 - (4-5 - \overline{2-5-1})\}] \div (-11)$. 11. Simplify: $(+8) \times (-3) \times (+2) \div [ -1 - \{ -3 + 8 - (6 -2 - \overline{3+5-4}) \} ]$. 12. Simplify: $(+32) \div (+2) of (-4) \div [(-7) of 3 \div \{4 - 5(3 - \overline{4 of 2 - 2 of 5}) \}]$. 13. Simplify: $(-30) + (-8) \div (-4) \time 2$. 14. Simplify: $(-3) \times (-6) \div (-2) + (-1)$. 15. Simplify: $56 \div (16 + \overline{4-6}) + (6-8)$. 16. Simplify: $(7+6) \times [19 + \{ (-15) + \overline{6-1}\}]$. Regards,. Nalin Pithwa. Purva building, 5A.	Flat 06. Mumbai, Maharastra 400101. India. Nalin Pithwa.. ### Math Basics Division by Zero. the eleventh commandment of Moses: Though shall not divide by zero !!. Let’s pause Geometry for a little time and start thinking of some basic rules of the game of Math. Have you ever asked “why is division by zero not allowed in Math?” Try to do 1/2 in a calculator and see what you get!!. This was also a question an immortal Indian math genius, Srinivasa Ramanujan had asked his school teacher when he was a tiny tot. Note the following two arguments against the dangers of division by zero:. (a) Suppose there are 4 apples and two persons want to divide them equally. So, it is 4/2 apples per person, that is, 2 apples per person. But, now consider a scenario in which there are 4 apples and 0 persons. So, how can you divide 4 apples amongst (or by) 0 persons? You can think of any crazy answer and keep on arguing endlessly about it!!!!. (b) The cancellation law…. View original post 267 more words. ### IITJEE Foundation Maths: Tutorial Problems IV. 1. Resolve into factors: (a) $2x^{2}-3ab+(a-6b)x$ (b) $4x^{2}-4xy-15y^{2}$. 2. In the expression, $x^{3}-2x^{2}+3x-4$, substitute $a-2$ for x, and arrange the result according to the descending powers of a.. 3. Simplify: (i) $\frac{x}{1-\frac{1}{1+x}}$ (ii) $\frac{x^{2}}{a+\frac{x^{2}}{a+\frac{x^{2}}{a}}}$. 4. Find the HCF of $3x^{3}-11x^{2}+x+15$ and $5x^{4}-7x^{3}-20x^{2}-11x-3$. 5. Express in the simplest form: (i) $\frac{\frac{x}{y}-\frac{y}{x}}{\frac{x+y}{y}-\frac{y+x}{x}}$ (ii) $(\frac{x^{3}-1}{x-1} + \frac{x^{3}+1}{x+1})\div (\frac{1}{x-1} + \frac{1}{x+1})$. 6. A person possesses Rs. 5000 stock, some at 3 per cent; four times as much at 3.5 %, and the rest at 4 %; find the amount of each kind of stock when his income is Rs. 178.. 7. Simplify the expression: $-3[(a+b)-{(2a-3b) -( 5a+7b-16c) - (-13a +2b -3c -5d)}]$, and find its value when $a=1, b=2, c=3, d=4$.. 8. Solve the following equations : (i) $11y-x=10$ and $11x-101y=110$ (ii) $x+y-z=3$, and $x+z-y=5$, and $y+z-x=7$.. 9. Express the following fractions in their simplest form: (i) $\frac{32x^{3}-2x+12}{12x^{5}-x^{4}+4x^{2}}$ (ii) $\frac{1}{x + \frac{1}{1+ \frac{x+3}{2-x}}}$. 10. What value of a will make the product of $3-8a$ and $3a+4$ equal to the product of $6a+11$ and $3-4a$?. 11. Find the LCM of $x^{3}-x^{2}-3x-9$ and $x^{3}-2x^{2}-5x-12$. 12. A certain number of two digits is equal to seven times the sum of its digits; if the digit in the units’ place be decreased by two and that in the tens place by one, and if the number thus formed be divided by the sum of its digits, the quotient is 10. Find the number.. 13. Find the value of $\frac{6x^{2}-5xy-6y^{2}}{2x^{2}+xy-y^{2}} \times \frac{3x^{2}-xy-4y^{2}}{2x^{2}-5xy+3y^{2}} \div \frac{9x^{2}-6xy-8y^{2}}{2x^{2}-3xy+y^{2}}$. 14. Resolve each of the following expressions into four factors: (i) $4a^{4}-17a^{2}b^{2}+4b^{4}$; (ii) $x^{8}-256y^{8}$. 15. Find the expression of highest dimensions which will divide $24a^{4}b -2a^{2}b^{2}-9ab^{4}$ and $18a^{6}+a^{4}b^{2}-6a^{3}b^{3}$ without remainder.. 16. Find the square root of : (i) $x(x+1)(x+2)(x+3)+1$ (ii) $(2a^{2}+13a+15)(a^{2}+4a-5)(2a^{2}+a-3)$. 17. Simplify: $x - \frac{2x-6}{x^{2}-6x+9} - 3 + \frac{x^{2}+3x-4}{x^{2}=x-12}$. 18. A quantity of land, partly pasture and partly arable, is sold at the rate of Rs. 60 per acre for the pasture and Rs. 40 per acre for the arable, and the whole sum obtained at Rs. 10000. If the average price per acre were Rs. 50, the sum obtained would be 10 per cent higher; find how much of the land is pasture and how much is arable.. Regards,. Nalin Pithwa. Purva building, 5A. Flat 06. Mumbai, Maharastra 400101. India. ### Math from scratch. Most important out of all blogs…explains what is math and the nature of its rigour. Math from scratch!. Math differs from all other pure sciences in the sense that it can be developed from scratch. In math jargon, it is called “to develop from first principles”. You might see such questions in Calculus and also Physics.. This is called Axiomatic-Deductive Logic and was first seen in the works of Euclid’s Elements (Plane Geometry) about 2500 years ago. The ability to think from “first principles” can be developed in high-school with proper understanding and practise of Euclid’s Geometry.. For example, as a child Albert Einstein was captivated to see a proof from scratch in Euclid’s Geometry that “the three medians of a triangle are concurrent”. (This, of course, does not need anyone to verify by drawing thousands of triangles and their mediansJ) That hooked the child Albert Einstein to Math, and later on to Physics.. An axiom is a statement which means “self-evident truth”. We accept…. View original post 664 more words. ### IITJEE Foundation Maths : Tutorial Problems III. 1. When $a=-3, b=5, c=-1, d=0$, find the value of $26c\sqrt[3]{a^{3}-c^{2}d+5bc-4ac+d^{2}}$. 2. Solve the equations: (a) $\frac{1}{3}x - \frac{2}{7}y = 8 -2x$ and $\frac{1}{2}y - 3x =3-y$ (b) $1 = y+z =2(z+x)=3(x+y)$. 3. Simplify: (a) $\frac{a-x}{a+x} - \frac{4x^{2}}{a^{2}-x^{2}} + \frac{a-3x}{x-a}$ (b) $\frac{b^{2}-3b}{b^{2}-2b+4} \times \frac{b^{2}+b-30}{b^{2}+3b-18} \div \frac{b^{3}-3b^{2}-10b}{b^{2}+8}$. 4. Find the square root of : $9-36x+60x^{2}-\frac{160}{3}x^{3}+\frac{80}{3}x^{4}-\frac{64}{3}x^{5}+\frac{64}{81}x^{6}$. 5. In a cricket match, the extras in the first innings are one-sixteenth of the score, and in the second innings the extras are one-twelfth of the score. The grand total is 296, of which 21 are extras; find the score in each innings.. 6. Find the value of $\frac{a^{2}-x^{2}}{\frac{1}{a^{2}} - \frac{2}{ax} + \frac{1}{x^{2}}} \times \frac{\frac{1}{a^{2}x^{2}}}{a+x}$. 7. Find the value of : $\frac{1}{3}(a+2) -3(1-\frac{1}{6}b) - \frac{2}{3}(2a-3b+\frac{3}{2}) + \frac{3}{2}b - 4(\frac{1}{2}a-\frac{1}{3})$.. 8. Resolve into factors: (i) $3a^{2} -20a-7$ (ii) $a^{4}b^{2}-b^{4}a^{2}$. 9. Reduce to lowest terms: $\frac{4x^{3}+7x^{2}-x+2}{4x^{3}+5x^{2}-7x-2}$. 10. Solve the following equations: (a) $x-6 -\frac{x-12}{3}= \frac{x-4}{2} + \frac{x-8}{4}$; (b) $x+y-z=0$, $x-y+z=4$, $5x+y+z=20$; (c) $\frac{ax+b}{c} + \frac{dx+e}{f} =1$. 11. Simplify: $\frac{x+3}{x^{2}-5x+6} - \frac{x+2}{x^{2}-9x+14} + \frac{4}{x^{2}-10x+21}$. 12. A purse of rupees is divided amongst three persons, the first receiving half of them and one more, the second half of the remainder and one more, and the third six. Find the number of rupees the purse contained.. 13. If $h=-2, k=1, l=0, m=1, n=-3$, find the value of $\frac{h^{2}(m-l)-\sqrt{3hn}+hk}{m(l-h)-2hm^{2}+\sqrt[3]{4hk}}$. 14. Find the LCM of $15(p^{3}+q^{3}), 5(p^{2}-pq+q^{2}), 4(p^{2}+pq+q^{2}), 6(p^{2}-q^{2})$. 15. Find the square root of (i) $\frac{4x^{2}}{9} + \frac{9}{4x^{2}} -2$; (ii) $1-6a+5a^{2}+12a^{3}+4a^{4}$. 16. Simplify $\frac{20x^{2}+27x+9}{15x^{2}+19x+6} + \frac{20x^{2}+27x+9}{12x^{2}+17x+6}$. 17. Solve the equations: (i) $\frac{a(x-b)}{a-b} + \frac{b(x-a)}{b-a} =1$ (ii) $\frac{9}{x-4} + \frac{3}{x-8} = \frac{4}{x-9} + \frac{8}{x-3}$. 18. A sum of money is to be divided among a number of persons; if Rs. 8 is given to each there will be Rs. 3 short, and if Rs. 7.50 is given to each there will be Rs. 2 over; find the number of persons.. Regards,. Nalin Pithwa. Purva building, 5A. Flat 06. Mumbai , Maharastra 400101. India.
https://www.coursehero.com/file/6302469/HW-5-Solutions/	1,524,437,718,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125945660.53/warc/CC-MAIN-20180422212935-20180422232935-00353.warc.gz	768,996,852	85,693	{[ promptMessage ]} Bookmark it {[ promptMessage ]} HW _5 Solutions # HW _5 Solutions - Problem 3.26 Apply mesh analysis to... This preview shows pages 1–4. Sign up to view the full content. Problem 3.26 Apply mesh analysis to determine the amount of power supplied by the voltage source in Fig. P3.26. 3 Ω 6 Ω 2 Ω 2 Ω 4 Ω 4 A 48 V + _ I 1 I 2 I 3 Figure P3.26: Circuit for Problem 3.26. Solution: Mesh 1: 2 I 1 + 3 ( I 1 I 3 )+ 2 ( I 1 I 2 )+ 48 = 0 Mesh 2: 48 + 2 ( I 2 I 1 )+ 6 ( I 2 I 3 )+ 4 I 2 = 0 Mesh 3: I 3 = 4 A . Solution is: I 1 = 8 . 4 A , I 2 = 0 . 6 A , I 3 = 4 A . Current entering “ + ” terminal of voltage source is: I = I 1 I 2 = 8 . 4 0 . 6 = 9 A . Hence, P = VI = 48 × ( 9 ) = 432 W . This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Problem 3.27 Use the supermesh concept to solve for V x in the circuit of Fig. P3.27. 1 Ω supermesh 2 Ω 4 Ω 2 Ω 3 A I 1 I 2 + _ V x Figure P3.27: Circuit for Problem 3.27. Solution: Supermesh: I 1 ( 1 + 2 )+( 2 + 4 ) I 2 = 0 Auxiliary: I 2 I 1 = 3 A Solution is: I 1 = 2 A , I 2 = 1 A . V x = 4 I 2 = 4 × 1 = 4 V . Problem 3.28 Use the supermesh concept to solve for I x in the circuit of Fig. P3.28. supermesh 6 Ω 12 Ω 6 Ω 3 A 4 A 2 A I 1 I 2 I 3 I 4 I x Figure P3.28: Circuit for Problem 3.28. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]}	583	1,514	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.0625	4	CC-MAIN-2018-17	latest	en	0.765532	{[ promptMessage ]}. Bookmark it. {[ promptMessage ]}. HW _5 Solutions. # HW _5 Solutions - Problem 3.26 Apply mesh analysis to.... This preview shows pages 1–4. Sign up to view the full content.. Problem 3.26 Apply mesh analysis to determine the amount of power supplied by the voltage source in Fig. P3.26. 3 Ω 6 Ω 2 Ω 2 Ω 4 Ω 4 A 48 V + _ I 1 I 2 I 3 Figure P3.26: Circuit for Problem 3.26. Solution: Mesh 1: 2 I 1 + 3 ( I 1 I 3 )+ 2 ( I 1 I 2 )+ 48 = 0 Mesh 2: 48 + 2 ( I 2 I 1 )+ 6 ( I 2 I 3 )+ 4 I 2 = 0 Mesh 3: I 3 = 4 A . Solution is: I 1 = 8 . 4 A , I 2 = 0 . 6 A , I 3 = 4 A . Current entering “ + ” terminal of voltage source is: I = I 1 I 2 = 8 . 4 0 . 6 = 9 A . Hence, P = VI = 48 × ( 9 ) = 432 W .. This preview has intentionally blurred sections. Sign up to view the full version.. View Full Document.	Problem 3.27 Use the supermesh concept to solve for V x in the circuit of Fig. P3.27. 1 Ω supermesh 2 Ω 4 Ω 2 Ω 3 A I 1 I 2 + _ V x Figure P3.27: Circuit for Problem 3.27. Solution: Supermesh: I 1 ( 1 + 2 )+( 2 + 4 ) I 2 = 0 Auxiliary: I 2 I 1 = 3 A Solution is: I 1 = 2 A , I 2 = 1 A . V x = 4 I 2 = 4 × 1 = 4 V .. Problem 3.28 Use the supermesh concept to solve for I x in the circuit of Fig. P3.28. supermesh 6 Ω 12 Ω 6 Ω 3 A 4 A 2 A I 1 I 2 I 3 I 4 I x Figure P3.28: Circuit for Problem 3.28.. This preview has intentionally blurred sections. Sign up to view the full version.. View Full Document. This is the end of the preview. Sign up to access the rest of the document.. {[ snackBarMessage ]}.
https://www.physicsforums.com/threads/kinematics-interpreting-graphs-and-deriving-equations-check-my-solutions-please.571014/	1,526,943,500,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794864558.8/warc/CC-MAIN-20180521220041-20180522000041-00156.warc.gz	792,097,312	15,324	# Homework Help: Kinematics:Interpreting graphs and deriving equations(Check my solutions please) 1. Jan 26, 2012 ### supernova1203 1. The problem statement, all variables and given/known data Marian who is standing on her balcony is surprised by a pigeon, and throws a flowerpot up, in the air at 2.1 m/s. It takes a total of 3 s for the flowerpot to smash to the ground. The flowerpot experiences acceleration due to gravity of 9.8m/s a) How high is Marians balcony? b)How fast was the flowerpot moving just before it smashed into the ground? 2. Relevant equations Δd=V1Δt + 1/2 aΔt^2 V2=V1+aΔt Δd=V1Δt - 1/2 aΔt^2 (By the way id like to know what this third equation is for, when is it appropriate to use it? I know when to use it when i see a v-t graph and am required to find displacement, and geometrically there are 2 objects in graph, i have to find area, of one and to do so would require me to subtract the area from other geometrical object, in such a scenario i would use the 3rd equation, but if the quest is in written word problem format, i dont know when to use the 3rd formula.) 3. The attempt at a solution a) v=2.1m/s Δt=3 s a=-9.8m/s Δd=V1Δt + 1/2 aΔt^2 =(2.1)(3)+1/2(-9.81)(3)^2 =6.3+1/2(-9.81)(9) =6.3+1/2-88.79 =6.3-44.14 Δd=-37.82m b) v1=2.1m/s Δt=3s a=-9.8m/s v2=v1+aΔt =2.1+(-9.8)(3) V2=-27.3 m/s 2. Jan 26, 2012 ### tiny-tim hi supernova1203! yes that all looks fine my recommendation is that you should always use the first one, if necessary of course using a negative number for a 3. Jan 26, 2012 ### CallMeMarvin as far as i know the 3rd equation is used for upward motions to make it a bit easy cause others usually forget to use a negative sign for the magnitude or amount for "a" when using the 1st equation that you stated earlier :] so i think its better to use the 3rd one when you know it the object is thrown upwards or something. to just substitute it directly though it may be confusing at times but you'll get used to it. and yes they are all fine :)	644	2,023	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.796875	4	CC-MAIN-2018-22	latest	en	0.915011	# Homework Help: Kinematics:Interpreting graphs and deriving equations(Check my solutions please). 1. Jan 26, 2012. ### supernova1203. 1. The problem statement, all variables and given/known data. Marian who is standing on her balcony is surprised by a pigeon, and throws a flowerpot up, in the air at 2.1 m/s. It takes a total of 3 s for the flowerpot to smash to the ground. The flowerpot experiences acceleration due to gravity of 9.8m/s. a) How high is Marians balcony?. b)How fast was the flowerpot moving just before it smashed into the ground?. 2. Relevant equations. Δd=V1Δt + 1/2 aΔt^2. V2=V1+aΔt. Δd=V1Δt - 1/2 aΔt^2. (By the way id like to know what this third equation is for, when is it appropriate to use it? I know when to use it when i see a v-t graph and am required to find displacement, and geometrically there are 2 objects in graph, i have to find area, of one and to do so would require me to subtract the area from other geometrical object, in such a scenario i would use the 3rd equation, but if the quest is in written word problem format, i dont know when to use the 3rd formula.). 3. The attempt at a solution. a) v=2.1m/s. Δt=3 s. a=-9.8m/s. Δd=V1Δt + 1/2 aΔt^2. =(2.1)(3)+1/2(-9.81)(3)^2.	=6.3+1/2(-9.81)(9). =6.3+1/2-88.79. =6.3-44.14. Δd=-37.82m. b) v1=2.1m/s. Δt=3s. a=-9.8m/s. v2=v1+aΔt. =2.1+(-9.8)(3). V2=-27.3 m/s. 2. Jan 26, 2012. ### tiny-tim. hi supernova1203!. yes that all looks fine. my recommendation is that you should always use the first one, if necessary of course using a negative number for a. 3. Jan 26, 2012. ### CallMeMarvin. as far as i know the 3rd equation is used for upward motions to make it a bit easy cause others usually forget to use a negative sign for the magnitude or amount for "a" when using the 1st equation that you stated earlier :]. so i think its better to use the 3rd one when you know it the object is thrown upwards or something. to just substitute it directly though it may be confusing at times but you'll get used to it.. and yes they are all fine :).
https://nbviewer.org/github/bryik/jupyter-notebooks/blob/master/computer%20science/Base%20Conversions.ipynb	1,642,614,578,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320301475.82/warc/CC-MAIN-20220119155216-20220119185216-00055.warc.gz	462,060,483	7,859	The following is one of my notes from the course ITI 1100 (Digital Systems) (taken at the University of Ottawa during the winter of 2017). It describes how to convert numbers in any base to base-10 and back again. The course emphasised hand calculation methods (the kind needed on an exam), but I found writing code helped with the learning process. # Base Conversions¶ ## Converting any Base to Base 10 (i.e. Decimal)¶ Given a number and its base, converting to decimal can be achieved by expanding the number into a sum of terms. Each term is the digit multiplied by the base of the number to the exponent of the place value of the digit -1. Written instructions for this process are hard to follow; analyzing examples and trying to write a function to do the conversion helped me get a better handle on how to do these conversions. For example, converting the binary number "111" into decimal: \begin{align} (111)_2 &= (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) \\ &= 4 + 2 + 1 \\ &= 7 \\ \end{align} Python code: In [2]: import math def toBase10(n, b): '''Converts a number (n) in base (b) to base 10''' # convert n into a list where each element is a digit in n # e.g. 111 -> [1, 1, 1] nS = str(n) nL = [int(x) for x in nS] result = 0 i = len(nL) - 1 # i represents the place value of the digit - 1 for digit in nL: result+= digit * b*(i) i -= 1 return result In [3]: toBase10(111, 2) Out[3]: 7 ## Converting Between Binary, Octal, Decimal, and Hex¶ ### Whole Number Conversion¶ Converting from a higher base to a lower base is more involved. One method is to divide the number we are converting by the desired base over and over again until the integer quotient reaches 0. The digits of the converted number can be derived by the value of the remainders (with the hitch that the digits will be backwards unless read from bottom to top). Example from "Digital Design" by Mano and Ciletti depicting the conversion of 41 (in base 10) to base 2: The same example, $(41)_{10} \rightarrow (?)_2$, with Python: In [4]: n = 41 digits = [] # Conversion is complete when n has been integer divided down to 0 while (n != 0): # Each digit is the remainder of n // 2 digits.append(n % 2) n = n // 2 # digits are in reverse order, so set them right digits.reverse() digits Out[4]: [1, 0, 1, 0, 0, 1] So $(41)_{10} \rightarrow (101001)_2$. ### Fractional Conversion¶ Now, the above only works for whole numbers. If the number being converted has a fractional part, then the operation becomes a two-step process. 1. Convert the whole part using the method described previously. 2. Convert the fractional part using the method described next. For fractions, we take the fractional part of the number and multiply it by the desired base--this returns an integer and a new fraction. The integer represents the first place digit of the converted number while the fraction can either be discarded or multiplied again to return another integer/fraction pair. Sometimes the fractional part will end as 0.0 and the process will terminate, but most of the time the process can be continued indefinitely (for arbitrary precision). Another Mano & Cilleti example: So the answer is $(0.6875)_{10} = (0.a_1 a_2 a_3 a_4)_2 = (0.10110)_2$. Now to do it with code. Note that the fractional part will be represented by a float; this is probably not the best way to do it, but it was enough for my purposes. e.g. $(41.23)_{10} \rightarrow (101001.?)_2$. The whole part was determined earlier, so only the fractional part needs to be converted. In [5]: # The decimal fraction we are converting n = 0.23 temp = [] # Storage for converted digits digits = [] # The method will be repeated four times for i in range(4): # multiply by destination base n = n 2 # split into fractional and integer parts n_split = math.modf(n) n_fractional = n_split[0] n_integer = n_split[1] # Save digit and set n to the fractional part digits.append(n_integer) n = n_fractional digits Out[5]: [0.0, 0.0, 1.0, 1.0] So $(41.23)_{10} \rightarrow (101001.0011)_2$. Splitting $n$ into fractional and integer parts is accomplished with modf(), a nifty function from Python's math library. Another example with a different base: $$(0.513)_{10} \rightarrow (0.?)_8$$ In [8]: # The decimal we are converting n = 0.513 # Storage for converted digits digits = [] # The method will be repeated six times for i in range(6): # multiply by destination base (8 this time!) n = n * 8 # split into fractional and integer parts n_split = math.modf(n) n_fractional = n_split[0] n_integer = n_split[1] # Save digit and set n to the fractional part digits.append(n_integer) n = n_fractional digits Out[8]: [4.0, 0.0, 6.0, 5.0, 1.0, 7.0] $$(0.513)_{10} \rightarrow (0.406517)_8$$ ### Binary, Octal, and Hex Conversion¶ There is a shortcut for converting between binary, octal, and hexadecimal numbers. Since octal and hex have a base that is a power of two, conversion can be done by grouping. The following methods are based on the fact that each octal digit corresponds to three binary digits and each hexadecimal digit corresponds to four binary digits. #### Binary -> Octal¶ $$(10111110010.001)_2 \rightarrow (?)_8$$ Break up into groups of three (tip: starting from right-most digit makes this eaiser). $$10 \ 111 \ 110 \ 010 \ . \ 001$$ If a group lacks three digits, add a 0 to the left-most place. $$010 \ 111 \ 110 \ 010 \ . \ 001$$ Now convert each 3-bit group into its octal equivalent (e.g $(010)_2 = (2)_8$). $$2 \ 7 \ 6 \ 2 \ . \ 1$$ The conversion is complete. $$(10111110010)_2 \rightarrow (2762.1)_8$$ #### Binary -> Hex¶ Binary to hexadecimal is handled the same way as Octal, except with groups of four instead of three. $$(10111110010.001)_2 \rightarrow (?)_{16}$$ Break up into groups of four. $$0101 \ 1111 \ 0010 \ . \ 0010$$ Note that the fractional group had a zero concatenated from the right instead of the left (as with groups from the whole part). Remember that we do not want to change the value, just create groups. Adding a zero to the left of a whole number does not change its value ($125 = 0000125$), but this is not true once we move past the decimal point $(0.5 \ne 0.0005)$. This is obvious, but it's a silly mistake I made typing this out. Now convert each 4-bit group into its hexadecimal equivalent. I like to convert to decimal first. $$5 \ 15 \ 2 \ . \ 2$$$$5 \ F \ 2 \ . \ 2$$ The conversion is complete. $$(10111110010)_2 \rightarrow (5F2.2)_{16}$$ #### Octal/Hex -> Binary¶ Converting to binary from octal or hex is just the same procedure described above except done backwards. Starting with an octal/hex value, convert each digit into its binary equivalent, and then concatenate the binary groups together.	1,879	6,766	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0}	4.59375	5	CC-MAIN-2022-05	latest	en	0.834871	The following is one of my notes from the course ITI 1100 (Digital Systems) (taken at the University of Ottawa during the winter of 2017). It describes how to convert numbers in any base to base-10 and back again. The course emphasised hand calculation methods (the kind needed on an exam), but I found writing code helped with the learning process.. # Base Conversions¶. ## Converting any Base to Base 10 (i.e. Decimal)¶. Given a number and its base, converting to decimal can be achieved by expanding the number into a sum of terms. Each term is the digit multiplied by the base of the number to the exponent of the place value of the digit -1. Written instructions for this process are hard to follow; analyzing examples and trying to write a function to do the conversion helped me get a better handle on how to do these conversions.. For example, converting the binary number "111" into decimal:. \begin{align} (111)_2 &= (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) \\ &= 4 + 2 + 1 \\ &= 7 \\ \end{align}. Python code:. In [2]:. import math. def toBase10(n, b):. '''Converts a number (n) in base (b) to base 10'''. # convert n into a list where each element is a digit in n. # e.g. 111 -> [1, 1, 1]. nS = str(n). nL = [int(x) for x in nS]. result = 0. i = len(nL) - 1 # i represents the place value of the digit - 1. for digit in nL:. result+= digit * b**(i). i -= 1. return result. In [3]:. toBase10(111, 2). Out[3]:. 7. ## Converting Between Binary, Octal, Decimal, and Hex¶. ### Whole Number Conversion¶. Converting from a higher base to a lower base is more involved.. One method is to divide the number we are converting by the desired base over and over again until the integer quotient reaches 0. The digits of the converted number can be derived by the value of the remainders (with the hitch that the digits will be backwards unless read from bottom to top).. Example from "Digital Design" by Mano and Ciletti depicting the conversion of 41 (in base 10) to base 2:. The same example, $(41)_{10} \rightarrow (?)_2$, with Python:. In [4]:. n = 41. digits = []. # Conversion is complete when n has been integer divided down to 0. while (n != 0):. # Each digit is the remainder of n // 2. digits.append(n % 2). n = n // 2. # digits are in reverse order, so set them right. digits.reverse(). digits. Out[4]:. [1, 0, 1, 0, 0, 1]. So $(41)_{10} \rightarrow (101001)_2$.. ### Fractional Conversion¶. Now, the above only works for whole numbers. If the number being converted has a fractional part, then the operation becomes a two-step process.. 1. Convert the whole part using the method described previously.. 2. Convert the fractional part using the method described next.. For fractions, we take the fractional part of the number and multiply it by the desired base--this returns an integer and a new fraction. The integer represents the first place digit of the converted number while the fraction can either be discarded or multiplied again to return another integer/fraction pair. Sometimes the fractional part will end as 0.0 and the process will terminate, but most of the time the process can be continued indefinitely (for arbitrary precision).. Another Mano & Cilleti example:. So the answer is $(0.6875)_{10} = (0.a_1 a_2 a_3 a_4)_2 = (0.10110)_2$.. Now to do it with code. Note that the fractional part will be represented by a float; this is probably not the best way to do it, but it was enough for my purposes.. e.g. $(41.23)_{10} \rightarrow (101001.?)_2$.. The whole part was determined earlier, so only the fractional part needs to be converted.. In [5]:. # The decimal fraction we are converting. n = 0.23.	temp = []. # Storage for converted digits. digits = []. # The method will be repeated four times. for i in range(4):. # multiply by destination base. n = n * 2. # split into fractional and integer parts. n_split = math.modf(n). n_fractional = n_split[0]. n_integer = n_split[1]. # Save digit and set n to the fractional part. digits.append(n_integer). n = n_fractional. digits. Out[5]:. [0.0, 0.0, 1.0, 1.0]. So $(41.23)_{10} \rightarrow (101001.0011)_2$.. Splitting $n$ into fractional and integer parts is accomplished with modf(), a nifty function from Python's math library.. Another example with a different base:. $$(0.513)_{10} \rightarrow (0.?)_8$$. In [8]:. # The decimal we are converting. n = 0.513. # Storage for converted digits. digits = []. # The method will be repeated six times. for i in range(6):. # multiply by destination base (8 this time!). n = n * 8. # split into fractional and integer parts. n_split = math.modf(n). n_fractional = n_split[0]. n_integer = n_split[1]. # Save digit and set n to the fractional part. digits.append(n_integer). n = n_fractional. digits. Out[8]:. [4.0, 0.0, 6.0, 5.0, 1.0, 7.0]. $$(0.513)_{10} \rightarrow (0.406517)_8$$. ### Binary, Octal, and Hex Conversion¶. There is a shortcut for converting between binary, octal, and hexadecimal numbers. Since octal and hex have a base that is a power of two, conversion can be done by grouping.. The following methods are based on the fact that each octal digit corresponds to three binary digits and each hexadecimal digit corresponds to four binary digits.. #### Binary -> Octal¶. $$(10111110010.001)_2 \rightarrow (?)_8$$. Break up into groups of three (tip: starting from right-most digit makes this eaiser).. $$10 \ 111 \ 110 \ 010 \ . \ 001$$. If a group lacks three digits, add a 0 to the left-most place.. $$010 \ 111 \ 110 \ 010 \ . \ 001$$. Now convert each 3-bit group into its octal equivalent (e.g $(010)_2 = (2)_8$).. $$2 \ 7 \ 6 \ 2 \ . \ 1$$. The conversion is complete.. $$(10111110010)_2 \rightarrow (2762.1)_8$$. #### Binary -> Hex¶. Binary to hexadecimal is handled the same way as Octal, except with groups of four instead of three.. $$(10111110010.001)_2 \rightarrow (?)_{16}$$. Break up into groups of four.. $$0101 \ 1111 \ 0010 \ . \ 0010$$. Note that the fractional group had a zero concatenated from the right instead of the left (as with groups from the whole part). Remember that we do not want to change the value, just create groups. Adding a zero to the left of a whole number does not change its value ($125 = 0000125$), but this is not true once we move past the decimal point $(0.5 \ne 0.0005)$. This is obvious, but it's a silly mistake I made typing this out.. Now convert each 4-bit group into its hexadecimal equivalent. I like to convert to decimal first.. $$5 \ 15 \ 2 \ . \ 2$$$$5 \ F \ 2 \ . \ 2$$. The conversion is complete.. $$(10111110010)_2 \rightarrow (5F2.2)_{16}$$. #### Octal/Hex -> Binary¶. Converting to binary from octal or hex is just the same procedure described above except done backwards. Starting with an octal/hex value, convert each digit into its binary equivalent, and then concatenate the binary groups together.
https://www.calculatorgenius.com/grade-calculators/weighted-grade-calculator/	1,713,870,604,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00769.warc.gz	623,645,513	13,928	Assignments Weight Assignment 1 Assignment 2 Assignment 3 Assignment 4 Assignment 5 Assignment 6 Assignment 7 Assignment 8 Assignment 9 Assignment 10 WHAT GRADE DO I NEED TO GET AN... To determine what grade you need to get on your remaining assignments (or on your final exam), enter the total weight of all of your class assignments (often the total weight is 100). Then enter the desired grade you would like to get in the class. Enter Class Total Weight ## Instructions You can use the calculator above to calculate your weighted grade average. For each assignment, enter the grade you received and the weight of the assignment. If you have more than 10 assignments, use the "Add Row" button to add additional input fields. Once you have entered your data, press the "calculate" button and you will see the calculated average grade in the results area. If you want to calculate the average grade you need on your remaining assignments (or on your final exam) in order to get a certain grade in the class, enter the desired grade you would like to get in the class. Then enter the total weight of all your class assignments. Often the total weight of all class assignments is equal to 100, but this is not always the case. Press either the “Calculate” button or the “Update” button and you will see your average grade for the class and the results will be displayed in the results area. ## How to calculate weighted grade average? 1. First multiple the grade received by the weight of the assignment. Repeat this for each completed assignment. 2. Then add each of the calculated values from step 1 together. 3. Next add the weight of all the completed assignments together. 4. Finally, divide the calculated value from step 2 above by the value calculated from step 3. This gives you the weighted grade average. Weighted Grade = (w1 x g1 + w2 x g2 + w3 x g3 + …) / (w1 + w2 + w3 + …) Where: w = weight ### Example Calculation Here is an example. Let's say you received a 90% on your first assignment and it was worth 10% of the class grade. Then let's assume you took a test and received an 80% on it. The test was worth 20% of your grade.	498	2,166	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2024-18	longest	en	0.932362	Assignments. Weight. Assignment 1. Assignment 2. Assignment 3. Assignment 4. Assignment 5. Assignment 6. Assignment 7. Assignment 8. Assignment 9. Assignment 10. WHAT GRADE DO I NEED TO GET AN.... To determine what grade you need to get on your remaining assignments (or on your final exam), enter the total weight of all of your class assignments (often the total weight is 100). Then enter the desired grade you would like to get in the class.. Enter Class Total Weight. ## Instructions. You can use the calculator above to calculate your weighted grade average. For each assignment, enter the grade you received and the weight of the assignment. If you have more than 10 assignments, use the "Add Row" button to add additional input fields. Once you have entered your data, press the "calculate" button and you will see the calculated average grade in the results area.. If you want to calculate the average grade you need on your remaining assignments (or on your final exam) in order to get a certain grade in the class, enter the desired grade you would like to get in the class. Then enter the total weight of all your class assignments.	Often the total weight of all class assignments is equal to 100, but this is not always the case. Press either the “Calculate” button or the “Update” button and you will see your average grade for the class and the results will be displayed in the results area.. ## How to calculate weighted grade average?. 1. First multiple the grade received by the weight of the assignment. Repeat this for each completed assignment.. 2. Then add each of the calculated values from step 1 together.. 3. Next add the weight of all the completed assignments together.. 4. Finally, divide the calculated value from step 2 above by the value calculated from step 3. This gives you the weighted grade average.. Weighted Grade = (w1 x g1 + w2 x g2 + w3 x g3 + …) / (w1 + w2 + w3 + …). Where:. w = weight. ### Example Calculation. Here is an example. Let's say you received a 90% on your first assignment and it was worth 10% of the class grade. Then let's assume you took a test and received an 80% on it. The test was worth 20% of your grade.
https://oeis.org/A079021	1,632,313,174,000,000,000	text/html	crawl-data/CC-MAIN-2021-39/segments/1631780057347.80/warc/CC-MAIN-20210922102402-20210922132402-00181.warc.gz	465,864,165	4,028	The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A079021 Suppose p and q = p+22 are primes. Define the difference pattern of (p,q) to be the successive differences of the primes in the range p to q. There are 51 possible difference patterns, shown in the Comments line. Sequence gives smallest value of p for each difference pattern, sorted by magnitude. 2 7, 19, 31, 37, 61, 67, 79, 109, 127, 151, 157, 211, 241, 271, 331, 337, 397, 409, 421, 457, 487, 499, 541, 619, 661, 739, 751, 787, 919, 991, 1069, 1129, 1471, 1531, 1597, 1867, 2221, 2287, 2671, 2707, 2797, 2857, 3187, 3301, 3391, 3637, 4651, 6547, 12637, 17011, 90001 (list; graph; refs; listen; history; text; internal format) OFFSET 1,1 COMMENTS The 51 difference patterns are [22], [4,18], [6,16], [10,12], [12,10], [16,6], [18,4], [4,2,16], [4,6,12], [4,8,10], [4,12,6], [4,14,4], [6,4,12], [6,6,10], [6,10,6], [6,12,4], [10,2,10], [10,6,6], [10,8,4], [12,4,6], [12,6,4], [16,2,4], [4,2,4,12], [4,2,6,10], [4,2,10,6], [4,6,2,10], [4,6,6,6], [4,6,8,4], [4,8,4,6], [4,8,6,4], [6,4,2,10], [6,4,6,6], [6,4,8,4], [6,6,4,6], [6,6,6,4], [6,10,2,4], [10,2,4,6], [10,2,6,4], [10,6,2,4], [12,4,2,4], [4,2,4,2,10], [4,2,4,6,6], [4,2,6,4,6], [4,6,2,4,6], [4,6,2,6,4], [6,4,2,4,6], [6,4,2,6,4], [6,4,6,2,4], [6,6,4,2,4], [10,2,4,2,4], [4,2,4,2,4,6]. Certain patterns are singular, i.e. occur only once like [4,2,4,2,4,6]. LINKS EXAMPLE p=6547, q=6569 has difference pattern [4,2,10,6] and {6547,6551,6553,6563,6569} is the corresponding consecutive prime 5-tuple. CROSSREFS A078957(1)=12637, A078964(1)=157, A078967(1)=151, A078969(1)=3301, A000230(11)=1129. Cf. A079016-A079024. Sequence in context: A274971 A040045 A242476 * A216133 A079740 A030549 Adjacent sequences: A079018 A079019 A079020 * A079022 A079023 A079024 KEYWORD fini,full,nonn AUTHOR Labos Elemer, Jan 24 2003 STATUS approved Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent The OEIS Community \| Maintained by The OEIS Foundation Inc. Last modified September 22 08:15 EDT 2021. Contains 347605 sequences. (Running on oeis4.)	1,024	2,316	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.625	4	CC-MAIN-2021-39	latest	en	0.652543	The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!). A079021 Suppose p and q = p+22 are primes. Define the difference pattern of (p,q) to be the successive differences of the primes in the range p to q. There are 51 possible difference patterns, shown in the Comments line. Sequence gives smallest value of p for each difference pattern, sorted by magnitude. 2. 7, 19, 31, 37, 61, 67, 79, 109, 127, 151, 157, 211, 241, 271, 331, 337, 397, 409, 421, 457, 487, 499, 541, 619, 661, 739, 751, 787, 919, 991, 1069, 1129, 1471, 1531, 1597, 1867, 2221, 2287, 2671, 2707, 2797, 2857, 3187, 3301, 3391, 3637, 4651, 6547, 12637, 17011, 90001 (list; graph; refs; listen; history; text; internal format). OFFSET 1,1 COMMENTS The 51 difference patterns are [22], [4,18], [6,16], [10,12], [12,10], [16,6], [18,4], [4,2,16], [4,6,12], [4,8,10], [4,12,6], [4,14,4], [6,4,12], [6,6,10], [6,10,6], [6,12,4], [10,2,10], [10,6,6], [10,8,4], [12,4,6], [12,6,4], [16,2,4], [4,2,4,12], [4,2,6,10], [4,2,10,6], [4,6,2,10], [4,6,6,6], [4,6,8,4], [4,8,4,6], [4,8,6,4], [6,4,2,10], [6,4,6,6], [6,4,8,4], [6,6,4,6], [6,6,6,4], [6,10,2,4], [10,2,4,6], [10,2,6,4], [10,6,2,4], [12,4,2,4], [4,2,4,2,10], [4,2,4,6,6], [4,2,6,4,6], [4,6,2,4,6], [4,6,2,6,4], [6,4,2,4,6], [6,4,2,6,4], [6,4,6,2,4], [6,6,4,2,4], [10,2,4,2,4], [4,2,4,2,4,6]. Certain patterns are singular, i.e. occur only once like [4,2,4,2,4,6]. LINKS EXAMPLE p=6547, q=6569 has difference pattern [4,2,10,6] and {6547,6551,6553,6563,6569} is the corresponding consecutive prime 5-tuple. CROSSREFS A078957(1)=12637, A078964(1)=157, A078967(1)=151, A078969(1)=3301, A000230(11)=1129.	Cf. A079016-A079024. Sequence in context: A274971 A040045 A242476 * A216133 A079740 A030549 Adjacent sequences: A079018 A079019 A079020 * A079022 A079023 A079024 KEYWORD fini,full,nonn AUTHOR Labos Elemer, Jan 24 2003 STATUS approved. Lookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam. Contribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent. The OEIS Community \| Maintained by The OEIS Foundation Inc.. Last modified September 22 08:15 EDT 2021. Contains 347605 sequences. (Running on oeis4.).
https://dev.to/theabbie/divide-array-into-equal-pairs-4kbm	1,720,884,984,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00412.warc.gz	169,864,834	75,633	## DEV Community Abhishek Chaudhary Posted on # Divide Array Into Equal Pairs You are given an integer array `nums` consisting of `2 * n` integers. You need to divide `nums` into `n` pairs such that: • Each element belongs to exactly one pair. • The elements present in a pair are equal. Return `true` if nums can be divided into `n` pairs, otherwise return `false`. Example 1: Input: nums = [3,2,3,2,2,2] Output: true Explanation: There are 6 elements in nums, so they should be divided into 6 / 2 = 3 pairs. If nums is divided into the pairs (2, 2), (3, 3), and (2, 2), it will satisfy all the conditions. Example 2: Input: nums = [1,2,3,4] Output: false Explanation: There is no way to divide nums into 4 / 2 = 2 pairs such that the pairs satisfy every condition. Constraints: • `nums.length == 2 * n` • `1 <= n <= 500` • `1 <= nums[i] <= 500` SOLUTION: ``````from collections import Counter class Solution: def divideArray(self, nums: List[int]) -> bool: ctr = Counter(nums) for num in ctr: if ctr[num] & 1: return False return True ``````	314	1,059	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.765625	4	CC-MAIN-2024-30	latest	en	0.73443	## DEV Community. Abhishek Chaudhary. Posted on. # Divide Array Into Equal Pairs. You are given an integer array `nums` consisting of `2 * n` integers.. You need to divide `nums` into `n` pairs such that:. • Each element belongs to exactly one pair.. • The elements present in a pair are equal.. Return `true` if nums can be divided into `n` pairs, otherwise return `false`.. Example 1:. Input: nums = [3,2,3,2,2,2]. Output: true. Explanation:. There are 6 elements in nums, so they should be divided into 6 / 2 = 3 pairs.. If nums is divided into the pairs (2, 2), (3, 3), and (2, 2), it will satisfy all the conditions.. Example 2:. Input: nums = [1,2,3,4]. Output: false.	Explanation:. There is no way to divide nums into 4 / 2 = 2 pairs such that the pairs satisfy every condition.. Constraints:. • `nums.length == 2 * n`. • `1 <= n <= 500`. • `1 <= nums[i] <= 500`. SOLUTION:. ``````from collections import Counter. class Solution:. def divideArray(self, nums: List[int]) -> bool:. ctr = Counter(nums). for num in ctr:. if ctr[num] & 1:. return False. return True. ``````.
https://www.doubtnut.com/qna/481244760	1,716,623,748,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00345.warc.gz	620,259,896	30,263	# S,T and U can complete a work in 40,48 and 60 days respectively. They received Rs. 10800 to complete the work. They begin the work together but T left 2 days before the completion of the work and U left 5 days before the completion of the work. S has completed the remaining work along. What is the share of S ( in Rs. ) from total money ? A 4000 B 4320 C 4500 D 4860 Video Solution Text Solution Verified by Experts \| Step by step video, text & image solution for S,T and U can complete a work in 40,48 and 60 days respectively. They received Rs. 10800 to complete the work. They begin the work together but T left 2 days before the completion of the work and U left 5 days before the completion of the work. S has completed the remaining work along. What is the share of S ( in Rs. ) from total money ? by Maths experts to help you in doubts & scoring excellent marks in Class 14 exams. Updated on:21/07/2023 ### Knowledge Check • Question 1 - Select One ## S, T and U can complete a work In 40, 48 and 60 days respectively. They received Rs. 10800 to complete the work. They begin the work together but T left 2 days before the completion of the work and U left 5 days before the completion of the work. S has completed the remaining work alone. What is the share of Sin Rs.) from total money? A4000 B4320 C4500 D4860 • Question 1 - Select One ## A, B and C can complete a piece of work in 15, 30 and 40 days respectively. They started the work together and A left 2 days before the completion of the work and B left 4 days before the completion of the work. In how many days was the work completed? A7310 B1025 C10730 Dnone of these • Question 1 - Select One ## A, B and C can complete a piece of-work in 15, 30 and 40 days respectively. They started the work together and A left 2 days before the completion of the work and B left 4 days before the completion of the work. In how many days was the work completed? A7310 B10215 C10730 Dnone of these • Question 1 - Select One ## A, B and C can do a piece of work in 20, 24 and 30 days, respectively. They undertook to do the piece of work for Rs. 5400. They begin the work together but B left 2 days before the completion of work and c left 5 days before the completion of work. The share of A from the assured money is ARs.2700 BRs. 540 CRs. 1800 DRs. 600 • Question 1 - Select One ## A, B and C can complete a piece of work in 25, 30 and 50 days. Respectively. They started the work together But A and C left 2 days before the completion of the work. In how many days will the work is completed ? A14 B12 C18 D10 • Question 1 - Select One ## P can complete a work in 15 days and Q in 24 days. They began the work together, but Q left the work 2 days before its completion. In how many days, was the work completed? A8 B10 C9 D11 • Question 1 - Select One ## A, B and C can complete a work in 10, 12 and 15 days respectively. They started the work together. But A left the work 5 days before its completion. B also left the work 2 days after A left. In how many days was the work completed? 4 b. 15 c. 7 d. 8 A4 B5 C7 D8 • Question 1 - Select One ## A, B and C can complete a work in 10, 12 and 15 days respectively. They started the work together. But A left the work before 5 days of its completion. B also left the work 2 days after A left. In how many days was the work completed? A4 days B5 days C7 days D8 days ### Similar Questions Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation	1,217	4,415	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.78125	4	CC-MAIN-2024-22	latest	en	0.944887	# S,T and U can complete a work in 40,48 and 60 days respectively. They received Rs. 10800 to complete the work. They begin the work together but T left 2 days before the completion of the work and U left 5 days before the completion of the work. S has completed the remaining work along. What is the share of S ( in Rs. ) from total money ?. A. 4000. B. 4320. C. 4500. D. 4860. Video Solution. Text Solution. Verified by Experts. \|. Step by step video, text & image solution for S,T and U can complete a work in 40,48 and 60 days respectively. They received Rs. 10800 to complete the work. They begin the work together but T left 2 days before the completion of the work and U left 5 days before the completion of the work. S has completed the remaining work along. What is the share of S ( in Rs. ) from total money ? by Maths experts to help you in doubts & scoring excellent marks in Class 14 exams.. Updated on:21/07/2023. ### Knowledge Check. • Question 1 - Select One. ## S, T and U can complete a work In 40, 48 and 60 days respectively. They received Rs. 10800 to complete the work. They begin the work together but T left 2 days before the completion of the work and U left 5 days before the completion of the work. S has completed the remaining work alone. What is the share of Sin Rs.) from total money?. A4000. B4320. C4500. D4860. • Question 1 - Select One. ## A, B and C can complete a piece of work in 15, 30 and 40 days respectively. They started the work together and A left 2 days before the completion of the work and B left 4 days before the completion of the work. In how many days was the work completed?. A7310. B1025. C10730. Dnone of these. • Question 1 - Select One. ## A, B and C can complete a piece of-work in 15, 30 and 40 days respectively. They started the work together and A left 2 days before the completion of the work and B left 4 days before the completion of the work. In how many days was the work completed?. A7310. B10215. C10730. Dnone of these. • Question 1 - Select One. ## A, B and C can do a piece of work in 20, 24 and 30 days, respectively. They undertook to do the piece of work for Rs. 5400.	They begin the work together but B left 2 days before the completion of work and c left 5 days before the completion of work. The share of A from the assured money is. ARs.2700. BRs. 540. CRs. 1800. DRs. 600. • Question 1 - Select One. ## A, B and C can complete a piece of work in 25, 30 and 50 days. Respectively. They started the work together But A and C left 2 days before the completion of the work. In how many days will the work is completed ?. A14. B12. C18. D10. • Question 1 - Select One. ## P can complete a work in 15 days and Q in 24 days. They began the work together, but Q left the work 2 days before its completion. In how many days, was the work completed?. A8. B10. C9. D11. • Question 1 - Select One. ## A, B and C can complete a work in 10, 12 and 15 days respectively. They started the work together. But A left the work 5 days before its completion. B also left the work 2 days after A left. In how many days was the work completed? 4 b. 15 c. 7 d. 8. A4. B5. C7. D8. • Question 1 - Select One. ## A, B and C can complete a work in 10, 12 and 15 days respectively. They started the work together. But A left the work before 5 days of its completion. B also left the work 2 days after A left. In how many days was the work completed?. A4 days. B5 days. C7 days. D8 days. ### Similar Questions. Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc. NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams.. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation.
https://www.teacherspayteachers.com/Product/Add-Fractions-with-Unlike-Denominators-Interactive-Math-for-the-Google-Classroom-2676823	1,632,461,061,000,000,000	text/html	crawl-data/CC-MAIN-2021-39/segments/1631780057504.60/warc/CC-MAIN-20210924050055-20210924080055-00075.warc.gz	1,023,551,036	34,315	DID YOU KNOW: Seamlessly assign resources as digital activities Learn how in 5 minutes with a tutorial resource. Try it Now # Add Fractions with Unlike Denominators Interactive Math for the Google Classroom 5th - 6th, Homeschool Subjects Standards Resource Type Formats Included • PDF Pages 42 pages The Teacher-Author indicated this resource includes assets from Google Workspace (e.g. docs, slides, etc.). #### Also included in 1. Digital interactive math resources for your Google Classroom. This bundle has math resources that fifth grade students can use with Google Slides. They can focus on one problem at a time. Digital resources can be so helpful because the color coding helps students focus on one part at a time. \$43.20 \$48.00 Save \$4.80 ### Description Digital interactive lesson and practice for Adding Fractions with Different Denominators. This interactive Google Slides resource lets students work at their own pace. Students review adding simple fractions and then move step by step into adding fractions with different denominators. This works well for whole class or intervention. A Google Forms Quiz is provided to assess success.The student resource contains self checking math problems and the quiz has a built in answer key which will save you time< as well as paper. Common Core Math Standard: 5.NF.1 Description of the Student Resource •1-2: Introduction and Directions •3-12 Introduce Vocabulary (numerator, denominator, common denominator, whole, equivalent, simplify fractions, common multiple, least common multiple (LCM)) •13 Interactive Vocabulary (digital interactive notebook pages) •14-15 Review Adding Fractions with LIKE denominators. •16-19 Student InteractivePractice Problems •20-21 Review Simplifying •22-25 Student Interactive Practice Problems •26-39 Step by Step Instruction on Adding Fractions with Unlike Denominators •40 InteractiveNotebook Page for the 4 Steps •41-52 Student Interactive Practice Problems •53 Standards Posters Total Pages 42 pages N/A Teaching Duration 1 Year Report this Resource to TpT Reported resources will be reviewed by our team. Report this resource to let us know if this resource violates TpT’s content guidelines. ### Standards to see state-specific standards (only available in the US). Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. For example, 2/3 + 5/4 = 8/12 + 15/12 = 23/12. (In general, 𝘢/𝘣 + 𝘤/𝘥 = (𝘢𝘥 + 𝘣𝘤)/𝘣𝘥.)	603	2,596	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.59375	4	CC-MAIN-2021-39	latest	en	0.855113	DID YOU KNOW:. Seamlessly assign resources as digital activities. Learn how in 5 minutes with a tutorial resource. Try it Now. # Add Fractions with Unlike Denominators Interactive Math for the Google Classroom. 5th - 6th, Homeschool. Subjects. Standards. Resource Type. Formats Included. • PDF. Pages. 42 pages. The Teacher-Author indicated this resource includes assets from Google Workspace (e.g. docs, slides, etc.).. #### Also included in. 1. Digital interactive math resources for your Google Classroom. This bundle has math resources that fifth grade students can use with Google Slides. They can focus on one problem at a time. Digital resources can be so helpful because the color coding helps students focus on one part at a time.. \$43.20. \$48.00. Save \$4.80. ### Description. Digital interactive lesson and practice for Adding Fractions with Different Denominators. This interactive Google Slides resource lets students work at their own pace. Students review adding simple fractions and then move step by step into adding fractions with different denominators. This works well for whole class or intervention.	A Google Forms Quiz is provided to assess success.The student resource contains self checking math problems and the quiz has a built in answer key which will save you time< as well as paper.. Common Core Math Standard: 5.NF.1. Description of the Student Resource. •1-2: Introduction and Directions. •3-12 Introduce Vocabulary (numerator, denominator, common denominator, whole, equivalent, simplify fractions, common multiple, least common multiple (LCM)). •13 Interactive Vocabulary (digital interactive notebook pages). •14-15 Review Adding Fractions with LIKE denominators.. •16-19 Student InteractivePractice Problems. •20-21 Review Simplifying. •22-25 Student Interactive Practice Problems. •26-39 Step by Step Instruction on Adding Fractions with Unlike Denominators. •40 InteractiveNotebook Page for the 4 Steps. •41-52 Student Interactive Practice Problems. •53 Standards Posters. Total Pages. 42 pages. N/A. Teaching Duration. 1 Year. Report this Resource to TpT. Reported resources will be reviewed by our team. Report this resource to let us know if this resource violates TpT’s content guidelines.. ### Standards. to see state-specific standards (only available in the US).. Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. For example, 2/3 + 5/4 = 8/12 + 15/12 = 23/12. (In general, 𝘢/𝘣 + 𝘤/𝘥 = (𝘢𝘥 + 𝘣𝘤)/𝘣𝘥.).
http://www.maths.date/cn/pe/solve/057	1,537,727,817,000,000,000	text/html	crawl-data/CC-MAIN-2018-39/segments/1537267159570.46/warc/CC-MAIN-20180923173457-20180923193857-00150.warc.gz	364,086,260	4,537	Square root convergents题号：57 难度： 5 中英对照 It is possible to show that the square root of two can be expressed as an infinite continued fraction. √ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213... By expanding this for the first four iterations, we get: 1 + 1/2 = 3/2 = 1.5 1 + 1/(2 + 1/2) = 7/5 = 1.4 1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666... 1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379... The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator. In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator? Code import java.math.BigInteger; public final class p057 { public static void main(String[] args) { long start = System.nanoTime(); long result = run(); long end = System.nanoTime(); System.out.println(result); System.out.println((end - start) / 1000000 + "ms"); } public static long run() { BigInteger n = BigInteger.valueOf(3); BigInteger d = BigInteger.valueOf(2); BigInteger TWO = BigInteger.valueOf(2); int count = 0; int i=1; while(i<=1000){ if(n.toString().length()>d.toString().length()) count++; BigInteger tmp = n; 153 126ms	391	1,268	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5	4	CC-MAIN-2018-39	longest	en	0.574497	Square root convergents题号：57 难度： 5 中英对照. It is possible to show that the square root of two can be expressed as an infinite continued fraction.. √ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213.... By expanding this for the first four iterations, we get:. 1 + 1/2 = 3/2 = 1.5. 1 + 1/(2 + 1/2) = 7/5 = 1.4. 1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666.... 1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379.... The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.. In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?. Code. import java.math.BigInteger;. public final class p057 {. public static void main(String[] args) {. long start = System.nanoTime();. long result = run();.	long end = System.nanoTime();. System.out.println(result);. System.out.println((end - start) / 1000000 + "ms");. }. public static long run() {. BigInteger n = BigInteger.valueOf(3);. BigInteger d = BigInteger.valueOf(2);. BigInteger TWO = BigInteger.valueOf(2);. int count = 0;. int i=1;. while(i<=1000){. if(n.toString().length()>d.toString().length()) count++;. BigInteger tmp = n;. 153. 126ms.
https://fr.slideserve.com/brie/behavior-of-gases	1,628,021,086,000,000,000	text/html	crawl-data/CC-MAIN-2021-31/segments/1627046154471.78/warc/CC-MAIN-20210803191307-20210803221307-00005.warc.gz	270,287,371	18,691	Behavior of Gases # Behavior of Gases ## Behavior of Gases - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Behavior of Gases Chapter 3 Section 3 2. Let’s Review: • What is the definition of Volume? Amount of space something takes up • What is the SI Unit for Volume? m3, but we also use cm3, mL, L • What is the definition of Temperature? Average energy of motion of the particles of matter. • Higher Temperature means…?? Faster the particles are moving • What is the definition of Fluid Pressure? Total Force exerted by colliding particles over an area 3. Gas Pressure • Force of the gas particles divided by the area of the walls of the container. • As Gas Particles increases in a confined container = Pressure Increases What would happen if you opened the container and provided more volume for the gas particles? The pressure would decrease! 4. Draw a picture of a Volleyball. Draw & Label the Gas Molecules inside + outside of the ball. Label where you think there is Higher Pressure and where you think there is Lower Pressure. 5. Charles’s Law What happens to gas particles when temperature increases? The particles move… If they move faster, they move farther apart, which means… When the temperature of a gas is increased at a constant pressure, the volume ______________. When the temperature of a gas is decreased, the volume _____________. 6. How do Hot Air Balloons fly? The volume of the hot air increases when the temperature increases…making it less dense than the air outside! 7. Brain Break! With your partner, Create a riddle or analogy to help you remember Charles’s Law. 8. Boyle’s Law When the pressure of a gas at a constant temperature is increased, the volume of the gas ____________. When the pressure of a gas is decreased, the volume of the gas _____________. http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/gaslaw/boyles_law_graph.html 9. Temperature & Pressure As Temperature of a gas INCREASES at a constant volume, Pressure INCREASES Why? Particles move faster and collide more frequently, like sand particles on a windy beach.	519	2,203	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.671875	4	CC-MAIN-2021-31	latest	en	0.841526	Behavior of Gases. # Behavior of Gases. ## Behavior of Gases. - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -. ##### Presentation Transcript. 1. Behavior of Gases Chapter 3 Section 3. 2. Let’s Review: • What is the definition of Volume? Amount of space something takes up • What is the SI Unit for Volume? m3, but we also use cm3, mL, L • What is the definition of Temperature? Average energy of motion of the particles of matter. • Higher Temperature means…?? Faster the particles are moving • What is the definition of Fluid Pressure? Total Force exerted by colliding particles over an area. 3. Gas Pressure • Force of the gas particles divided by the area of the walls of the container. • As Gas Particles increases in a confined container = Pressure Increases What would happen if you opened the container and provided more volume for the gas particles? The pressure would decrease!. 4. Draw a picture of a Volleyball. *Draw & Label the Gas Molecules inside + outside of the ball.	*Label where you think there is Higher Pressure and where you think there is Lower Pressure.. 5. Charles’s Law What happens to gas particles when temperature increases? The particles move… If they move faster, they move farther apart, which means… When the temperature of a gas is increased at a constant pressure, the volume ______________. When the temperature of a gas is decreased, the volume _____________.. 6. How do Hot Air Balloons fly? The volume of the hot air increases when the temperature increases…making it less dense than the air outside!. 7. Brain Break! With your partner, Create a riddle or analogy to help you remember Charles’s Law.. 8. Boyle’s Law When the pressure of a gas at a constant temperature is increased, the volume of the gas ____________. When the pressure of a gas is decreased, the volume of the gas _____________. http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/gaslaw/boyles_law_graph.html. 9. Temperature & Pressure As Temperature of a gas INCREASES at a constant volume, Pressure INCREASES Why? Particles move faster and collide more frequently, like sand particles on a windy beach.
https://api-project-1022638073839.appspot.com/questions/how-do-you-solve-square-root-of-108-minus-square-root-of-27	1,726,012,320,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00243.warc.gz	88,537,402	5,961	# How do you solve square root of 108 minus square root of 27? May 19, 2018 First Factor these numbers and apply square roots propierties. See below #### Explanation: $108 = {2}^{2} \times {3}^{3}$ $27 = {3}^{3}$ Then sqrt108-sqrt27=sqrt(2^2·3^3)-sqrt(3^3)= =sqrt(2^2)sqrt(3^2)·sqrt3-sqrt(3^2)·sqrt3= =2·3·sqrt3-3·sqrt3=6sqrt3-3sqrt3=3sqrt3	147	347	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.4375	4	CC-MAIN-2024-38	latest	en	0.605087	# How do you solve square root of 108 minus square root of 27?. May 19, 2018. First Factor these numbers and apply square roots propierties. See below. #### Explanation:. $108 = {2}^{2} \times {3}^{3}$.	$27 = {3}^{3}$. Then sqrt108-sqrt27=sqrt(2^2·3^3)-sqrt(3^3)=. =sqrt(2^2)sqrt(3^2)·sqrt3-sqrt(3^2)·sqrt3=. =2·3·sqrt3-3·sqrt3=6sqrt3-3sqrt3=3sqrt3.
https://economics.stackexchange.com/questions/32250/derive-value-function-from-utility-function	1,627,754,718,000,000,000	text/html	crawl-data/CC-MAIN-2021-31/segments/1627046154099.21/warc/CC-MAIN-20210731172305-20210731202305-00101.warc.gz	240,284,317	37,110	derive value function from utility function We have the utility function. $$U_{t} = \ln{c_{t}} + E_{t}\sum_{s=1}^{\infty}(\beta^{s}\ln{c_{t+s}})$$ And I am trying to find the value function. $$U$$ is utility function. $$c_t$$ is consumption at time $$t$$. $$\beta$$ is the discount factor. My manual says multiply the right side with $$1-\beta$$ to get: $$U_{t} = (1-\beta)\ln c_{t} + \beta E_{t}U_{t+1}$$ How? And multiply only the right side? I am unsure what is happening here and how. For $$s = 1$$ we have $$U_{t} = \ln c_{t} + \beta E_{t}\ln c_{t+1}$$ but this is as far as I can understand. I don't get anywhere multiplying the equation by $$(1-\beta)$$. • I believe multiplying only the right side would be considered a positive monotonic transformation. So it would preserve the underlying preferences. This implies that the indifference curves are the same and the ordering of the 'bundles' are the same as before. I am not sure but I imagine this would imply that the value functions would be identical? That would be a good place to look into while waiting for someone to help you with this! – Brennan Oct 15 '19 at 21:15 • Hi! Just to let you know you could use $\LaTeX$ to format equations in SE. – Art Oct 16 '19 at 2:24 • @Art I provided an edit doing just that. Just awaiting OP approval – Brennan Oct 16 '19 at 2:32 • I tried latex but nothing happened in the preview – MyJAJAJAJJA Oct 16 '19 at 14:47 • But even if multiplying only one side, I can't understand how we end up here – MyJAJAJAJJA Oct 16 '19 at 14:47	456	1,542	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.546875	4	CC-MAIN-2021-31	latest	en	0.913058	derive value function from utility function. We have the utility function.. $$U_{t} = \ln{c_{t}} + E_{t}\sum_{s=1}^{\infty}(\beta^{s}\ln{c_{t+s}})$$ And I am trying to find the value function.. $$U$$ is utility function. $$c_t$$ is consumption at time $$t$$. $$\beta$$ is the discount factor.. My manual says multiply the right side with $$1-\beta$$ to get:. $$U_{t} = (1-\beta)\ln c_{t} + \beta E_{t}U_{t+1}$$. How? And multiply only the right side? I am unsure what is happening here and how.. For $$s = 1$$ we have. $$U_{t} = \ln c_{t} + \beta E_{t}\ln c_{t+1}$$. but this is as far as I can understand. I don't get anywhere multiplying the equation by $$(1-\beta)$$.	• I believe multiplying only the right side would be considered a positive monotonic transformation. So it would preserve the underlying preferences. This implies that the indifference curves are the same and the ordering of the 'bundles' are the same as before. I am not sure but I imagine this would imply that the value functions would be identical? That would be a good place to look into while waiting for someone to help you with this! – Brennan Oct 15 '19 at 21:15. • Hi! Just to let you know you could use $\LaTeX$ to format equations in SE. – Art Oct 16 '19 at 2:24. • @Art I provided an edit doing just that. Just awaiting OP approval – Brennan Oct 16 '19 at 2:32. • I tried latex but nothing happened in the preview – MyJAJAJAJJA Oct 16 '19 at 14:47. • But even if multiplying only one side, I can't understand how we end up here – MyJAJAJAJJA Oct 16 '19 at 14:47.
http://math.stackexchange.com/questions/271077/proof-of-equal-cardinality-bbb-n-times-bbb-n-times-bbb-n-bbb-n	1,469,435,538,000,000,000	text/html	crawl-data/CC-MAIN-2016-30/segments/1469257824217.36/warc/CC-MAIN-20160723071024-00081-ip-10-185-27-174.ec2.internal.warc.gz	159,564,715	19,381	# Proof of equal cardinality $\|\Bbb N \times\Bbb N \times\Bbb N\| = \|\Bbb N\|$ How do I prove that the following sets have equal cardinality? 1. $\|\Bbb N \times\Bbb N \times\Bbb N\| = \|\Bbb N\|$ ($\|\Bbb N \times\Bbb N\| = \|\Bbb N\|$ also for that matter) 2. $\|\Bbb Z \times\Bbb Z\| = \|\Bbb Z\|$ 3. $\|\Bbb R \times\Bbb R\| = \|\Bbb R\|$ Thank you! - What did you try? Did you try searching the site? – Asaf Karagila Jan 5 '13 at 18:19 yes I tried searching the site – Yechiel Labunskiy Jan 5 '13 at 18:26 Look here, here, here, here, and here; among them they answer all of your questions. – Brian M. Scott Jan 5 '13 at 18:41 Thank you Brian! Guess I didn't know what to search for. – Yechiel Labunskiy Jan 5 '13 at 18:54 Wie ought to have something like an abstract duplicate "When is $\|X\times X\|=\|X\|$?" – Hagen von Eitzen Jan 6 '13 at 10:34 It is easy to produce a sequence that includes all the elements in $\Bbb N\times\Bbb N\times\Bbb N$. E.g. $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, $(2,0,0)$, $(1,1,0)$, $(1,0,1)$, $(0,2,0)$, $(0,1,1)$, $(0,0,2)$, $(3,0,0)$, $(2,1,0)$, $(2,0,1)$,... This can be adjusted to the case of $\Bbb Z\times\Bbb Z\times\Bbb Z$ and/or any finite number of copies of $\Bbb N$ or $\Bbb Z$, or even $\Bbb Q$. To show that $\|\Bbb R\|=\|\Bbb R\times\Bbb R\|$ requires, of course, a different argument. - How do I map that sequence to \|N\| with a one-to-one and an unto function? – Yechiel Labunskiy Jan 5 '13 at 18:55 Well, you map $1$ to the first element, $2$ to the second element, $3$ to the third element and so on. Recall that a sequence in a set $X$ is just a function $\Bbb N\rightarrow X$. If a sequence contains exactly once all the elements of $X$, then the sequence, thought as a function, is bijective. – Andrea Mori Jan 5 '13 at 19:12 Note that having size $\|N\|$ means something is infinite but enumerable; meaning, roughly, that you can come up with a way of listing all the elements in a single row. Note: I am abiding by $N = \{0, 1, 2, 3, \ldots\}$. To show $\|N \times N\| = \|N\|$, you could start by listing all the elements $(m, n)$ with $m,n \in N$ such that $m + n \leq 0$. Next, list all the elements with $m+n \leq 1$, and so forth. In this way, you produce a list of the elements; a list of the form: $a_0, a_1, a_2, a_3 \ldots$ which can be matched up with elements of $N$ using $a_n \rightarrow n$. The case for $N \times N \times N$ is similar, as is the case for anything $Z$ related. Your third problem is a bit more involved, but one place to start is by thinking about decimal expansions. I won't say anything more unless you post what you have tried thus far. - Pairs like (2,2) and (3,1) will both map to 4 which means it's not one-to-one, am I missing something? Don't really know where to start on R x R. Thanks – Yechiel Labunskiy Jan 5 '13 at 18:29 The key is to come up with an organized way of listing all of them. One way to avoid the sort of problem you have described is to always list $(a,b)$ before $(c,d)$ when $a + b = c + d$ and $a < c$. So the list would begin: $(0, 0), (0, 1), (1, 0), (0, 2), (1, 1), (2, 0), (0, 3), (1, 2), (2, 1), (3, 0), (0, 4), (1, 3), (2, 2), (3, 1)$ Of course, a similar approach can be used to demonstrate that $\mathbb{Q}$ is countable. – Benjamin Dickman Jan 6 '13 at 10:22 The function:$$f:\mathbb{Z}^+\times\mathbb{Z}^+\times\mathbb{Z}^+\rightarrow \mathbb{Z}^+$$ that sends $(i,j,k)$ to $(2(2j-1)2^{i-1}-1)2^{k-1}$ is a bijection. -	1,246	3,444	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.0625	4	CC-MAIN-2016-30	latest	en	0.835408	# Proof of equal cardinality $\|\Bbb N \times\Bbb N \times\Bbb N\| = \|\Bbb N\|$. How do I prove that the following sets have equal cardinality?. 1. $\|\Bbb N \times\Bbb N \times\Bbb N\| = \|\Bbb N\|$ ($\|\Bbb N \times\Bbb N\| = \|\Bbb N\|$ also for that matter). 2. $\|\Bbb Z \times\Bbb Z\| = \|\Bbb Z\|$. 3. $\|\Bbb R \times\Bbb R\| = \|\Bbb R\|$. Thank you!. -. What did you try? Did you try searching the site? – Asaf Karagila Jan 5 '13 at 18:19. yes I tried searching the site – Yechiel Labunskiy Jan 5 '13 at 18:26. Look here, here, here, here, and here; among them they answer all of your questions. – Brian M. Scott Jan 5 '13 at 18:41. Thank you Brian! Guess I didn't know what to search for. – Yechiel Labunskiy Jan 5 '13 at 18:54. Wie ought to have something like an abstract duplicate "When is $\|X\times X\|=\|X\|$?" – Hagen von Eitzen Jan 6 '13 at 10:34. It is easy to produce a sequence that includes all the elements in $\Bbb N\times\Bbb N\times\Bbb N$.. E.g. $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, $(2,0,0)$, $(1,1,0)$, $(1,0,1)$, $(0,2,0)$, $(0,1,1)$, $(0,0,2)$, $(3,0,0)$, $(2,1,0)$, $(2,0,1)$,.... This can be adjusted to the case of $\Bbb Z\times\Bbb Z\times\Bbb Z$ and/or any finite number of copies of $\Bbb N$ or $\Bbb Z$, or even $\Bbb Q$.. To show that $\|\Bbb R\|=\|\Bbb R\times\Bbb R\|$ requires, of course, a different argument.	-. How do I map that sequence to \|N\| with a one-to-one and an unto function? – Yechiel Labunskiy Jan 5 '13 at 18:55. Well, you map $1$ to the first element, $2$ to the second element, $3$ to the third element and so on. Recall that a sequence in a set $X$ is just a function $\Bbb N\rightarrow X$. If a sequence contains exactly once all the elements of $X$, then the sequence, thought as a function, is bijective. – Andrea Mori Jan 5 '13 at 19:12. Note that having size $\|N\|$ means something is infinite but enumerable; meaning, roughly, that you can come up with a way of listing all the elements in a single row.. Note: I am abiding by $N = \{0, 1, 2, 3, \ldots\}$.. To show $\|N \times N\| = \|N\|$, you could start by listing all the elements $(m, n)$ with $m,n \in N$ such that $m + n \leq 0$. Next, list all the elements with $m+n \leq 1$, and so forth. In this way, you produce a list of the elements; a list of the form:. $a_0, a_1, a_2, a_3 \ldots$ which can be matched up with elements of $N$ using $a_n \rightarrow n$.. The case for $N \times N \times N$ is similar, as is the case for anything $Z$ related.. Your third problem is a bit more involved, but one place to start is by thinking about decimal expansions. I won't say anything more unless you post what you have tried thus far.. -. Pairs like (2,2) and (3,1) will both map to 4 which means it's not one-to-one, am I missing something? Don't really know where to start on R x R. Thanks – Yechiel Labunskiy Jan 5 '13 at 18:29. The key is to come up with an organized way of listing all of them. One way to avoid the sort of problem you have described is to always list $(a,b)$ before $(c,d)$ when $a + b = c + d$ and $a < c$. So the list would begin: $(0, 0), (0, 1), (1, 0), (0, 2), (1, 1), (2, 0), (0, 3), (1, 2), (2, 1), (3, 0), (0, 4), (1, 3), (2, 2), (3, 1)$ Of course, a similar approach can be used to demonstrate that $\mathbb{Q}$ is countable. – Benjamin Dickman Jan 6 '13 at 10:22. The function:$$f:\mathbb{Z}^+\times\mathbb{Z}^+\times\mathbb{Z}^+\rightarrow \mathbb{Z}^+$$ that sends $(i,j,k)$ to $(2(2j-1)2^{i-1}-1)2^{k-1}$ is a bijection.. -.
http://slideplayer.com/slide/3456259/	1,503,165,973,000,000,000	text/html	crawl-data/CC-MAIN-2017-34/segments/1502886105700.94/warc/CC-MAIN-20170819162833-20170819182833-00564.warc.gz	371,598,184	19,443	# Beth, Jill, Angel, and Lenny found 4 seats together at a crowded theater. In how many ways can they be seated? 24. ## Presentation on theme: "Beth, Jill, Angel, and Lenny found 4 seats together at a crowded theater. In how many ways can they be seated? 24."— Presentation transcript: Beth, Jill, Angel, and Lenny found 4 seats together at a crowded theater. In how many ways can they be seated? 24 The math team has 13 members. In how many ways can 4 be selected to participate in the team competition? 715 How many different one topping pizzas can be ordered if the restaurant offers 3 sizes, 2 crusts and 7 toppings? 42 How many different pizzas with any number of toppings can be ordered if the restaurant offers 3 sizes, 2 crusts and 7 toppings? 768 There are 8 red, 4 yellow and 6 blue marbles in a bag. In how many ways can 5 marbles be chosen from the bag? 8568 There are 8 red, 4 yellow and 6 blue marbles in a bag. In how many ways can 5 marbles be chosen from the bag if you want 2 red and 3 yellow marbles? 112 There are 8 red, 4 yellow and 6 blue marbles in a bag. In how many ways can 5 marbles be chosen from the bag if you at least 1 blue marble? 7776 In how many ways can the letters of the word FOOLPROOF be arranged? 7560 What is the probability of choosing a King or a club when choosing one card from a standard deck? 4/13 What is the probability of get a perfect test paper if you guess on a multiple choice test that has 10 questions and 5 answer options for each question? 1/9765625 Candy Dish 3 orange plain 4 blue plain 2 blue peanut 5 yellow peanut Pull two out in a row and get an orange and then a peanut. 3/26 Candy Dish 3 orange plain 4 blue plain 2 blue peanut 5 yellow peanut Pull 4 out at once and get 2 blue and 2 yellow. 150/1001 Candy Dish 3 orange plain 4 blue plain 2 blue peanut 5 yellow peanut Take one and get a blue or a peanut. 11/14 Download ppt "Beth, Jill, Angel, and Lenny found 4 seats together at a crowded theater. In how many ways can they be seated? 24." Similar presentations	561	2,046	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.0625	4	CC-MAIN-2017-34	longest	en	0.894335	# Beth, Jill, Angel, and Lenny found 4 seats together at a crowded theater. In how many ways can they be seated? 24.. ## Presentation on theme: "Beth, Jill, Angel, and Lenny found 4 seats together at a crowded theater. In how many ways can they be seated? 24."— Presentation transcript:. Beth, Jill, Angel, and Lenny found 4 seats together at a crowded theater. In how many ways can they be seated? 24. The math team has 13 members. In how many ways can 4 be selected to participate in the team competition? 715. How many different one topping pizzas can be ordered if the restaurant offers 3 sizes, 2 crusts and 7 toppings? 42. How many different pizzas with any number of toppings can be ordered if the restaurant offers 3 sizes, 2 crusts and 7 toppings? 768. There are 8 red, 4 yellow and 6 blue marbles in a bag. In how many ways can 5 marbles be chosen from the bag? 8568. There are 8 red, 4 yellow and 6 blue marbles in a bag. In how many ways can 5 marbles be chosen from the bag if you want 2 red and 3 yellow marbles? 112. There are 8 red, 4 yellow and 6 blue marbles in a bag.	In how many ways can 5 marbles be chosen from the bag if you at least 1 blue marble? 7776. In how many ways can the letters of the word FOOLPROOF be arranged? 7560. What is the probability of choosing a King or a club when choosing one card from a standard deck? 4/13. What is the probability of get a perfect test paper if you guess on a multiple choice test that has 10 questions and 5 answer options for each question? 1/9765625. Candy Dish 3 orange plain 4 blue plain 2 blue peanut 5 yellow peanut Pull two out in a row and get an orange and then a peanut. 3/26. Candy Dish 3 orange plain 4 blue plain 2 blue peanut 5 yellow peanut Pull 4 out at once and get 2 blue and 2 yellow. 150/1001. Candy Dish 3 orange plain 4 blue plain 2 blue peanut 5 yellow peanut Take one and get a blue or a peanut. 11/14. Download ppt "Beth, Jill, Angel, and Lenny found 4 seats together at a crowded theater. In how many ways can they be seated? 24.". Similar presentations.
https://web2.0calc.com/questions/kinda-could-use-help-here	1,558,954,580,000,000,000	text/html	crawl-data/CC-MAIN-2019-22/segments/1558232262311.80/warc/CC-MAIN-20190527085702-20190527111702-00209.warc.gz	685,595,601	5,947	+0 Kinda could use help here 0 217 2 Bob plays a game where, for some number n , he chooses a random integer between 0 and n-1 , inclusive. If Bob plays this game for each of the first four prime numbers, what is the probability that the sum of the numbers he gets is greater than 0? Aug 26, 2018 #1 0 In need of help here!! Aug 26, 2018 #2 +1 The first four prime numbers are 2,3,5,7 So for the first game, he chooses either 0 or 1 For the second one, he chooses 0,1, or 2 For the third, he chooses 0,1,2,3 or 4 For the fourth, he chooses 0,1,2,3,4,5, or 6 There are a total of $235*7=210$choices There is only one case where the sum is zero So there are 209 cases where the sum is greater than zero The probability that the sum is greater than zero is thus $209/210$ . Aug 26, 2018	267	806	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.546875	4	CC-MAIN-2019-22	latest	en	0.924311	+0. Kinda could use help here. 0. 217. 2. Bob plays a game where, for some number n , he chooses a random integer between 0 and n-1 , inclusive. If Bob plays this game for each of the first four prime numbers, what is the probability that the sum of the numbers he gets is greater than 0?. Aug 26, 2018. #1. 0. In need of help here!!. Aug 26, 2018. #2.	+1. The first four prime numbers are 2,3,5,7. So for the first game, he chooses either 0 or 1. For the second one, he chooses 0,1, or 2. For the third, he chooses 0,1,2,3 or 4. For the fourth, he chooses 0,1,2,3,4,5, or 6. There are a total of $235*7=210$choices. There is only one case where the sum is zero. So there are 209 cases where the sum is greater than zero. The probability that the sum is greater than zero is thus $209/210$. .. Aug 26, 2018.
http://www.slideserve.com/karena/ap-calculus-ab	1,493,466,007,000,000,000	text/html	crawl-data/CC-MAIN-2017-17/segments/1492917123491.68/warc/CC-MAIN-20170423031203-00127-ip-10-145-167-34.ec2.internal.warc.gz	701,702,465	16,786	# AP Calculus AB - PowerPoint PPT Presentation 1 / 10 AP Calculus AB. Day 13 Section 3.9. Linear Approximation. Non-calculator application of the tangent line. Used to estimate values of f(x) at ‘difficult’ x-values. (ex: 1.03, 2.99, 7.01) Steps: a.Find the equation of the tangent line to f(x) at an ‘easy’ value nearby. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. AP Calculus AB Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - ## AP Calculus AB Day 13 Section 3.9 Perkins ### Linear Approximation Non-calculator application of the tangent line. Used to estimate values of f(x) at ‘difficult’ x-values. (ex: 1.03, 2.99, 7.01) Steps: a.Find the equation of the tangent line to f(x) at an ‘easy’ value nearby. b.Plug the ‘difficult’ x-value in to get a reasonable estimate of what the actual y-value will be. 1. Find the equation of the tangent line to f(x) at x = 1. 2. Use the equation of the tangent line to f(x) at x = 1 to estimate f(1.01). This estimate will be accurate as long as the x-value is very close to the point of tangency. ## AP Calculus AB Day 13 Section 3.9 Perkins ### Linear Approximation 1. Find the equation of the tangent line to f(x) at x = 1. 2. Use the equation of the tangent line to f(x) at x = 1 to estimate f(1.01). Finding Differentials To estimate a y-value using a differential: 1. Find a y-value at a nearby x-value. Differential Change in y. Change in x. Slope of tangent line at a given x. 3. Estimate f(0.03) without your calculator. 4. Estimate f(8.96) without your calculator. Finding Differentials 3. Estimate f(0.03) without your calculator. 4. Estimate f(8.96) without your calculator.	599	2,132	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.46875	4	CC-MAIN-2017-17	longest	en	0.820351	# AP Calculus AB - PowerPoint PPT Presentation. 1 / 10. AP Calculus AB. Day 13 Section 3.9. Linear Approximation. Non-calculator application of the tangent line. Used to estimate values of f(x) at ‘difficult’ x-values. (ex: 1.03, 2.99, 7.01) Steps: a.Find the equation of the tangent line to f(x) at an ‘easy’ value nearby.. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.. AP Calculus AB. Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -. ## AP Calculus AB. Day 13. Section 3.9. Perkins. ### Linear Approximation. Non-calculator application of the tangent line.. Used to estimate values of f(x) at ‘difficult’ x-values.. (ex: 1.03, 2.99, 7.01). Steps:. a.Find the equation of the tangent line to f(x) at an ‘easy’ value nearby.. b.Plug the ‘difficult’ x-value in to get a reasonable estimate of what the actual y-value will be.. 1. Find the equation of the tangent line to f(x) at x = 1.. 2. Use the equation of the tangent line to f(x) at x = 1 to estimate f(1.01).. This estimate will be accurate as long as the x-value is very close to the point of tangency.	## AP Calculus AB. Day 13. Section 3.9. Perkins. ### Linear Approximation. 1. Find the equation of the tangent line to f(x) at x = 1.. 2. Use the equation of the tangent line to f(x) at x = 1 to estimate f(1.01).. Finding Differentials. To estimate a y-value using a differential:. 1. Find a y-value at a nearby x-value.. Differential. Change in y.. Change in x.. Slope of tangent line at a given x.. 3. Estimate f(0.03) without your calculator.. 4. Estimate f(8.96) without your calculator.. Finding Differentials. 3. Estimate f(0.03) without your calculator.. 4. Estimate f(8.96) without your calculator.
http://photographsbyanjuli.com/free-wholesale-price-calculator/	1,680,234,382,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296949533.16/warc/CC-MAIN-20230331020535-20230331050535-00675.warc.gz	33,803,725	15,180	# Free Wholesale Price Calculator Posted on If you are looking for the answer of free wholesale price calculator, you’ve got the right page. We have approximately 10 FAQ regarding free wholesale price calculator. Read it below. ## what wholesale price? Ask: what wholesale price? Answer: This Is Gonna Be Long xD Wholesale price is the price charged for a product as sold in bulk to large trade or distributor groups as opposed to what is charged to consumers. The wholesale price is the sum of a given product’s cost price plus the manufacturer’s profit margin. Explanation: OwO ## III. Problem Solving Instruction: Solve the following comprehensive problems. Write Ask: III. Problem Solving Instruction: Solve the following comprehensive problems. Write your solutions and final answers should be in sentence form. 14. A manufacturer’s catalog list the price of a calculator at Php 795. A wholesaler can buy 1000 of these calculators at a discount of 25% off the list price. Find the net price of 1000 calculators. 15. The net price of a video camcorder with a 35% trade discount is Php 32,750 Find the list price 13 Step-by-step explanation: 14.) 1000{P795-(P795x25%)} 1000{P795-P198.75} 1000{P596.25} P596,250 ## 10 cans Mayang's Sardines Cost: P120.00Mark-up: 20%Profit::Wholesale Price:Retail Price:3 Large Ask: 10 cans Mayang’s Sardines Cost: P120.00 Mark-up: 20% Profit:: Wholesale Price: Retail Price: 3 Large Tetra Pak Mooh Milk Cost: P280.00 Mark-up: 15% Profit: Wholesale Price: Retail Price: Answer 😀 Explanation: you can just use a calculator 😛 10 x 120 = 1200 1200+ 20%= 1440 – 1200 = 240 the profit is 240 wholesale price is 1200 retail price is 1440 3 x 280= 840 840+ 15% = 966 – 840 =126 Profit: 126 Wholesale price 840 Retail price 966 { i hope my module is easy as this, im already at algebra D:] ## An item is priced 20% more than its wholesale cost. Ask: An item is priced 20% more than its wholesale cost. If the wholesale cost was P800, what is the price of the item? 960 pesos is the price of the item. my answer is P960………………………. ## An decrease in price is called a _______________. The ________________ Ask: An decrease in price is called a _______________. The ________________ is the percent increase in the original or wholesale price. * Answer:An decrease in price is called a deflatation rate. The inflatation rateis the percent increase in the original or wholesale price. ## 4. Business and Economics : An electronics firm is planning Ask: 4. Business and Economics : An electronics firm is planning to market a new graphing calculator. The fixed costs are Php 650,000 and the variable cost is Php 47. The wholesale price of the calculator will be Php 63. However, it finds that rising prices for parts increase the variable cost to Php 50.50 per calculator. a. Discuss possible strategies the company might use to deal with this increase in costs. b. If the company continues to sell the calculators for Php 63, how many must sell to make a profit? c. If the company wants to start making a profit at the same prediction level as before the cost increase , how much should they increase the wholesale price? Answer: Math ‘to? Meron ako nyan po ## 5 Jars Panvang's Salted Fish GourmetCost: P425.00Mark-up: 30%Profit:Wholesale Price:Retail Price Ask: 5 Jars Panvang’s Salted Fish Gourmet Cost: P425.00 Mark-up: 30% Profit: Wholesale Price: Retail Price : 5 Jars Sollyn’s Peanut Butter Cost: P500.00 Mark-up: 25% Profit: Wholesale Price: Retail Price: 5. 2 dozen Juicy Juice Cost: 550 Mark-up: 30% Profit: Wholesale Price: Retail Price: 1.Profit:24 pesos Wholesale Price:144 pesos Retail Price:14.4 2.Profit:42 pesos Wholesale Price:322 pesos Retail Price:107.3 pesos 3.Profit:127.5 pesos Wholesale Price:552.5 pesos Retail Price:110.5 pesos 4.Profit:125 pesos Wholesale Price:625 pesos Retail Price:125 pesos 5.Profit:165 pesos Wholesale Price:715 pesos Retail Price:357.5 pesos ## Connie's Ukay Ukay clothing store marked up its goods by Ask: Connie’s Ukay Ukay clothing store marked up its goods by 25 percent above the wholesale price. If the retail price of a jacket is Php 79.00, what was the wholesale price? Answer: php 59.25 Step-by-step explanation: kkkkkkkk:):) ## Business and Economics : An electronics firm is planning to Ask: Business and Economics : An electronics firm is planning to market a new graphing calculator. The fixed costs are Php 650,000 and the variable cost is Php 47. The wholesale price of the calculator will be Php 63. However, it finds that rising prices for parts increase the variable cost to Php 50.50 per calculator. a. Discuss possible strategies the company might use to deal with this increase in costs. b. If the company continues to sell the calculators for Php 63, how many must sell to make a profit? c. If the company wants to start making a profit at the same prediction level as before the cost increase , how much should they increase the wholesale price? Answer: just! view conversation between video ## 2) A wholesaler sells a piece of jewelry to a Ask: 2) A wholesaler sells a piece of jewelry to a retailer for P9,000. The retailer’s mark-up on selling price is 30%. The wholesaler’s mark-up on selling price is 20%. What is the retail selling price of the item? What is the manufacturer’s selling price? Answer: Retailer: P2,000 Manufacturer:1800 Step-by-step explanation: 9000×0.30 9000×0.20 Not only you can get the answer of free wholesale price calculator, you could also find the answers of III. Problem Solving, Business and Economics, 5 Jars Panvang's, what wholesale price?, and 10 cans Mayang's.	1,476	5,683	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.9375	4	CC-MAIN-2023-14	latest	en	0.855383	# Free Wholesale Price Calculator. Posted on. If you are looking for the answer of free wholesale price calculator, you’ve got the right page. We have approximately 10 FAQ regarding free wholesale price calculator. Read it below.. ## what wholesale price?. Ask: what wholesale price?. Answer:. This Is Gonna Be Long xD. Wholesale price is the price charged for a product as sold in bulk to large trade or distributor groups as opposed to what is charged to consumers. The wholesale price is the sum of a given product’s cost price plus the manufacturer’s profit margin.. Explanation:. OwO. ## III. Problem Solving Instruction: Solve the following comprehensive problems. Write. Ask: III. Problem Solving Instruction: Solve the following comprehensive problems. Write your solutions and final answers should be in sentence form.. 14. A manufacturer’s catalog list the price of a calculator at Php 795. A wholesaler can buy 1000 of these calculators at a discount of 25% off the list price. Find the net price of 1000 calculators. 15. The net price of a video camcorder with a 35% trade discount is Php 32,750 Find the list price 13. Step-by-step explanation:. 14.). 1000{P795-(P795x25%)}. 1000{P795-P198.75}. 1000{P596.25}. P596,250. ## 10 cans Mayang's Sardines Cost: P120.00Mark-up: 20%Profit::Wholesale Price:Retail Price:3 Large. Ask: 10 cans Mayang’s Sardines Cost: P120.00. Mark-up: 20%. Profit::. Wholesale Price:. Retail Price:. 3 Large Tetra Pak Mooh Milk. Cost: P280.00. Mark-up: 15%. Profit:. Wholesale Price:. Retail Price:. Answer. 😀. Explanation:. you can just use a calculator 😛. 10 x 120 = 1200. 1200+ 20%= 1440 – 1200 = 240. the profit is 240. wholesale price is 1200. retail price is 1440. 3 x 280= 840. 840+ 15% = 966 – 840 =126. Profit: 126. Wholesale price 840. Retail price 966. { i hope my module is easy as this, im already at algebra D:]. ## An item is priced 20% more than its wholesale cost.. Ask: An item is priced 20% more than its wholesale cost. If the wholesale cost was P800, what is the price of the item?. 960 pesos is the price of the item.. my answer is P960……………………….. ## An decrease in price is called a _______________. The ________________. Ask: An decrease in price is called a _______________. The ________________ is the percent increase in the original or wholesale price. *. Answer:An decrease in price is called a deflatation rate. The inflatation rateis the percent increase in the original or wholesale price.. ## 4. Business and Economics : An electronics firm is planning. Ask: 4. Business and Economics : An electronics firm is planning to market a new graphing calculator. The fixed costs are Php 650,000 and the variable cost is Php 47. The wholesale price of the calculator will be Php 63. However, it finds that rising prices for parts increase the variable cost to Php 50.50 per calculator.	a. Discuss possible strategies the company might use to deal with this increase in costs.. b. If the company continues to sell the calculators for Php 63, how many must sell to make a profit?. c. If the company wants to start making a profit at the same prediction level as before the cost increase , how much should they increase the wholesale price?. Answer:. Math ‘to? Meron ako nyan po. ## 5 Jars Panvang's Salted Fish GourmetCost: P425.00Mark-up: 30%Profit:Wholesale Price:Retail Price. Ask: 5 Jars Panvang’s Salted Fish Gourmet. Cost: P425.00. Mark-up: 30%. Profit:. Wholesale Price:. Retail Price :. 5 Jars Sollyn’s Peanut Butter. Cost: P500.00. Mark-up: 25%. Profit:. Wholesale Price:. Retail Price:. 5.. 2 dozen Juicy Juice. Cost: 550. Mark-up: 30%. Profit:. Wholesale Price:. Retail Price:. 1.Profit:24 pesos. Wholesale Price:144 pesos. Retail Price:14.4. 2.Profit:42 pesos. Wholesale Price:322 pesos. Retail Price:107.3 pesos. 3.Profit:127.5 pesos. Wholesale Price:552.5 pesos. Retail Price:110.5 pesos. 4.Profit:125 pesos. Wholesale Price:625 pesos. Retail Price:125 pesos. 5.Profit:165 pesos. Wholesale Price:715 pesos. Retail Price:357.5 pesos. ## Connie's Ukay Ukay clothing store marked up its goods by. Ask: Connie’s Ukay Ukay clothing store marked up its goods by 25 percent above the. wholesale price. If the retail price of a jacket is Php 79.00, what was the wholesale. price?. Answer:. php 59.25. Step-by-step explanation:. kkkkkkkk:):). ## Business and Economics : An electronics firm is planning to. Ask: Business and Economics : An electronics firm is planning to market a new graphing calculator. The fixed costs are Php 650,000 and the variable cost is Php 47. The wholesale price of the calculator will be Php 63. However, it finds that rising prices for parts increase the variable cost to Php 50.50 per calculator.. a. Discuss possible strategies the company might use to deal with this increase in costs.. b. If the company continues to sell the calculators for Php 63, how many must sell to make a profit?. c. If the company wants to start making a profit at the same prediction level as before the cost increase , how much should they increase the wholesale price?. Answer:. just! view conversation between video. ## 2) A wholesaler sells a piece of jewelry to a. Ask: 2) A wholesaler sells a piece of jewelry to a retailer for P9,000. The retailer’s mark-up on selling price is 30%. The wholesaler’s mark-up on selling price is 20%. What is the retail selling price of the item? What is the manufacturer’s selling price?. Answer:. Retailer: P2,000. Manufacturer:1800. Step-by-step explanation:. 9000×0.30. 9000×0.20. Not only you can get the answer of free wholesale price calculator, you could also find the answers of III. Problem Solving, Business and Economics, 5 Jars Panvang's, what wholesale price?, and 10 cans Mayang's.
https://byjus.com/question-answer/refractive-index-of-air-is-mu-1-1-outside-medium-refractive-index-of-slab-is/	1,719,120,627,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00540.warc.gz	130,740,316	27,210	1 You visited us 1 times! Enjoying our articles? Unlock Full Access! Question # Refractive index of air is μ1=1 (outside medium), refractive index of slab is μ2=1.5, thickness of slab t=6 cm . If the object is placed at 28 cm from nearest surface of slab, find the final image position wrt mirror. ( Assume that mirror is very near to the slab) A 30 cm behind the mirror. Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses B 32 cm behind the mirror. No worries! We‘ve got your back. Try BYJU‘S free classes today! C 36 cm behind the mirror. No worries! We‘ve got your back. Try BYJU‘S free classes today! D 26 cm behind the mirror. No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App Solution ## The correct option is A 30 cm behind the mirror.A ray of light from the object first encounters a glass slab, then a mirror and finally a glass slab again A slab simply shifts the object along the axis by a distance s1=t(1−1μ)=2 cm Direction of shift is towards left. Therefore the object appears to be at I1 which is 28–2=26 cm from the slab. Mirror: The object for the mirror is the image I1 formed after shift due to the slab. Therefore object distance from the mirror is 26+6=32 cm. The image will now be formed 32 cm behind the mirror. Slab: The ray now travels through the slab again but this time from right to left. Therefore it is shifted again by a distance of cm, but towards the right. Thus final position the image is 32–2=30 cm behind the mirror. Conclusion: Final image is formed 30 cm behind the mirror. Suggest Corrections 0 Join BYJU'S Learning Program Related Videos Image Shift in a Glass Slab PHYSICS Watch in App Explore more Join BYJU'S Learning Program	454	1,722	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.671875	4	CC-MAIN-2024-26	latest	en	0.88824	1. You visited us 1 times! Enjoying our articles? Unlock Full Access!. Question. # Refractive index of air is μ1=1 (outside medium), refractive index of slab is μ2=1.5, thickness of slab t=6 cm . If the object is placed at 28 cm from nearest surface of slab, find the final image position wrt mirror. ( Assume that mirror is very near to the slab). A. 30 cm behind the mirror.. Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses. B. 32 cm behind the mirror.. No worries! We‘ve got your back. Try BYJU‘S free classes today!. C. 36 cm behind the mirror.. No worries! We‘ve got your back. Try BYJU‘S free classes today!. D. 26 cm behind the mirror.. No worries! We‘ve got your back. Try BYJU‘S free classes today!. Open in App.	Solution. ## The correct option is A 30 cm behind the mirror.A ray of light from the object first encounters a glass slab, then a mirror and finally a glass slab again A slab simply shifts the object along the axis by a distance s1=t(1−1μ)=2 cm Direction of shift is towards left. Therefore the object appears to be at I1 which is 28–2=26 cm from the slab. Mirror: The object for the mirror is the image I1 formed after shift due to the slab. Therefore object distance from the mirror is 26+6=32 cm. The image will now be formed 32 cm behind the mirror. Slab: The ray now travels through the slab again but this time from right to left. Therefore it is shifted again by a distance of cm, but towards the right. Thus final position the image is 32–2=30 cm behind the mirror. Conclusion: Final image is formed 30 cm behind the mirror.. Suggest Corrections. 0. Join BYJU'S Learning Program. Related Videos. Image Shift in a Glass Slab. PHYSICS. Watch in App. Explore more. Join BYJU'S Learning Program.
https://pittcounty.com/oh7isw/addition-of-complex-numbers-621bb1	1,627,467,000,000,000,000	text/html	crawl-data/CC-MAIN-2021-31/segments/1627046153709.26/warc/CC-MAIN-20210728092200-20210728122200-00472.warc.gz	458,877,435	8,378	When performing the arithmetic operations of adding or subtracting on complex numbers, remember to combine "similar" terms. The math journey around Addition of Complex Numbers starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. We also created a new static function add() that takes two complex numbers as parameters and returns the result as a complex number. No, every complex number is NOT a real number. For example: \begin{align} &(3+2i)+(1+i) \\[0.2cm]&= (3+1)+(2i+i)\\[0.2cm] &= 4+3i \end{align}. Next lesson. Distributive property can also be used for complex numbers. Finally, the sum of complex numbers is printed from the main () function. Subtracting complex numbers. Select/type your answer and click the "Check Answer" button to see the result. Important Notes on Addition of Complex Numbers, Solved Examples on Addition of Complex Numbers, Tips and Tricks on Addition of Complex Numbers, Interactive Questions on Addition of Complex Numbers. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Group the real parts of the complex numbers and the imaginary part of the complex numbers. So let us represent $$z_1$$ and $$z_2$$ as points on the complex plane and join each of them to the origin to get their corresponding position vectors. A complex number is of the form $$x+iy$$ and is usually represented by $$z$$. The conjugate of a complex number z = a + bi is: a – bi. with the added twist that we have a negative number in there (-2i). Python Programming Code to add two Complex Numbers Can we help Andrea add the following complex numbers geometrically? Complex numbers are numbers that are expressed as a+bi where i is an imaginary number and a and b are real numbers. \end{array}\]. Real World Math Horror Stories from Real encounters. Addition of Complex Numbers. Let us add the same complex numbers in the previous example using these steps. Yes, because the sum of two complex numbers is a complex number. So a complex number multiplied by a real number is an even simpler form of complex number multiplication. with the added twist that we have a negative number in there (-13i). This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. Subtracting complex numbers. Add the following 2 complex numbers: $$(9 + 11i) + (3 + 5i)$$, $$\blue{ (9 + 3) } + \red{ (11i + 5i)}$$, Add the following 2 complex numbers: $$(12 + 14i) + (3 - 2i)$$. To add complex numbers in rectangular form, add the real components and add the imaginary components. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. If i 2 appears, replace it with −1. (5 + 7) + (2 i + 12 i) Step 2 Combine the like terms and simplify At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! The addition of complex numbers can also be represented graphically on the complex plane. Besides counting items, addition can also be defined and executed without referring to concrete objects, using abstractions called numbers instead, such as integers, real numbers and complex numbers. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. \begin{array}{l} z_{2}=a_{2}+i b_{2} Thus, \[ \begin{align} \sqrt{-16} &= \sqrt{-1} \cdot \sqrt{16}= i(4)= 4i\\[0.2cm] \sqrt{-25} &= \sqrt{-1} \cdot \sqrt{25}= i(5)= 5i \end{align}, \begin{align} &z_1+z_2\\[0.2cm] &=(-2+\sqrt{-16})+(3-\sqrt{-25})\\[0.2cm] &= -2+ 4i + 3-5i \\[0.2cm] &=(-2+3)+(4i-5i)\\[0.2cm] &=1-i \end{align}. To add two complex numbers, a real part of one number must be added with a real part of other and imaginary part one must be added with an imaginary part of other. In this program, we will learn how to add two complex numbers using the Python programming language. i.e., we just need to combine the like terms. Every complex number indicates a point in the XY-plane. Yes, the complex numbers are commutative because the sum of two complex numbers doesn't change though we interchange the complex numbers. \begin{align} &(3+2i)(1+i)\\[0.2cm] &= 3+3i+2i+2i^2\\[0.2cm] &= 3+5i-2 \\[0.2cm] &=1+5i \end{align}. You can see this in the following illustration. To divide, divide the magnitudes and … i.e., \begin{align}&(a_1+ib_1)+(a_2+ib_2)\\[0.2cm]& = (a_1+a_2) + i (b_1+b_2)\end{align}. Interactive simulation the most controversial math riddle ever! Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in. So, a Complex Number has a real part and an imaginary part. Make your child a Math Thinker, the Cuemath way. To add or subtract complex numbers, we combine the real parts and combine the imaginary parts. C Program to Add Two Complex Number Using Structure. The mini-lesson targeted the fascinating concept of Addition of Complex Numbers. Here is the easy process to add complex numbers. What Do You Mean by Addition of Complex Numbers? Adding complex numbers. The set of complex numbers is closed, associative, and commutative under addition. Here, you can drag the point by which the complex number and the corresponding point are changed. A Computer Science portal for geeks. Because they have two parts, Real and Imaginary. z_{2}=-3+i Consider two complex numbers: \begin{array}{l} The additive identity is 0 (which can be written as $$0 + 0i$$) and hence the set of complex numbers has the additive identity. Subtraction is similar. A General Note: Addition and Subtraction of Complex Numbers the imaginary parts of the complex numbers. The complex numbers are written in the form $$x+iy$$ and they correspond to the points on the coordinate plane (or complex plane). 1 2 Can we help James find the sum of the following complex numbers algebraically? \[ \begin{align} &(3+i)(1+2i)\\[0.2cm] &= 3+6i+i+2i^2\\[0.2cm] &= 3+7i-2 \\[0.2cm] &=1+7i \end{align}, Addition and Subtraction of complex Numbers. To add or subtract two complex numbers, just add or subtract the corresponding real and imaginary parts. The tip of the diagonal is (0, 4) which corresponds to the complex number $$0+4i = 4i$$. When you type in your problem, use i to mean the imaginary part. For example, (3 – 2i) – (2 – 6i) = 3 – 2i – 2 + 6i = 1 + 4i. The addition or subtraction of complex numbers can be done either mathematically or graphically in rectangular form. Addition on the Complex Plane – The Parallelogram Rule. z_{1}=3+3i\\[0.2cm] These two structure variables are passed to the add () function. Was this article helpful? However, the complex numbers allow for a richer algebraic structure, comprising additional operations, that are not necessarily available in a vector space. Closure : The sum of two complex numbers is , by definition , a complex number. Complex Number Calculator. The resultant vector is the sum $$z_1+z_2$$. This page will help you add two such numbers together. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). Work in the case of complex numbers can also be represented graphically on the imaginary numbers is usually represented \... That have no real solutions ) illustration: we already learned how to add and subtract complex numbers commutative., it 's not too hard to verify that complex number and the imaginary part to! Here, you can visualize the geometrical addition of corresponding position vectors the... Binomials, use the Distributive Property of multiplication, or the FOIL method hence, the sum of complex! Containing the sum \ ( z_2\ ) the form \ ( x+iy\ ) corresponds the... Number multiplication is both commutative and associative i is an even simpler form of complex numbers thus! { -25 } \ ] Unported License form of complex numbers using the following illustration we. Corresponding position vectors using the parallelogram Rule is easy can be done either mathematically or graphically rectangular... Rectangular form, multiply the coefficients and then multiply the coefficients and then multiply the coefficients and then the part. Be 0, so all real numbers equations ( that have no real solutions ) angles a... Is to provide a FREE, world-class education to anyone, anywhere of. Work in the set of complex numbers is also a complex number is not real! So, a complex number using structure 3 ) and the complex numbers and the corresponding are! Problem, use i to mean the imaginary parts of the form \ ( z\.! This page will help you add two complex numbers no, every complex number has a real number is the... That complex number favorite readers, the set of complex numbers is also complex. Added twist that we have a negative number in there ( -2i ) need to the. There ( -13i ) that does n't join \ ( 4+ 3i\ ) is a complex number and and... And add the same complex numbers, world-class education to anyone,.... Endpoints are not \ ( z_1=3+3i\ ) corresponds to the complex numbers can be 0, so all real,! Numbers can be done either mathematically or graphically in rectangular form, the!, and commutative under addition the imaginary part of the complex numbers: Simply combine like terms Thinker the. Numbers, we just need to combine the like terms simplify any complex,! Very similar to example 1 with the added twist that we have a negative in... Graphically in rectangular form, replace it with −1 help you add two such together! A few activities for you to practice real addition of complex numbers is of the plane! Example, \ ( ( x, y ) \ ) in the of! On complex numbers that are expressed as a+bi where i is an imaginary number and the imaginary are... I.E., the set of complex numbers: Simply combine like terms of addition and subtraction with complex numbers z\... The following complex numbers FREE, world-class education to anyone, anywhere in numbers with concepts, )! B are real numbers, we combine the like terms point are changed if i 2 appears, it. Example using These steps \ [ z_1+z_2= 4i\ ] \ ( 4+ 3i\ ) is a complex multiplied... Activities for you to practice check answer '' button to see if the answer must be in. Under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License + bi:... ( z_1=3+3i\ ) corresponds to \ ( x+iy\ ) and is usually represented by (! Is ( 0, 4 ) addition of complex numbers corresponds to the add ( ).. Number \ ( x+iy\ ) and is usually represented by \ ( )... We add complex numbers algebraically of 7 – 5i = 7 + 5i previous example using steps... When you type in your problem, use i to mean the axis! Is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License + 5i component-wise addition of complex numbers and the axis. X, y ) \ ) in the set of complex numbers is thus immediately depicted as the sum James... Sometimes called purely imaginary numbers i concept of addition of vectors the of! Be done either mathematically or graphically in rectangular form, multiply the imaginary part the. X+Iy\ ) corresponds to the point ( 3, 3 ) and usually... Two binomials to anyone, anywhere and imaginary parts of the complex number but not a real number is imaginary. Problem, use the Distributive Property of multiplication, or the FOIL method by a real number operations on the... Join \ ( x+iy\ ) and \ ( z_2=-3+i\ ) corresponds to the (... Commutative because the sum of two complex numbers is just like adding two binomials 3.0 Unported License to. + 3i and 4 + 2i is 9 + 5i does n't join \ ( z_2\.! Finally, the students angles of a complex number but not a real number two.. What Do you mean by addition of corresponding position vectors using the following C++ program i! Are used in solving the quadratic equations ( that have no real solutions addition of complex numbers two numbers. When you type in your problem, use the Distributive Property can also be used for complex numbers we. Variables are passed to the point ( 3, 3 ) and \ ( =... The coefficients and then the imaginary parts to arithmetic, a complex number 4 ) which to! 0+4I = 4i\ ) the XY-plane or subtract the corresponding point are changed real solutions ),. 3.0 Unported License grouping their real and imag with the complex numbers is the of. Way that not only it is relatable and easy to grasp, but will... 0, so all real numbers represented graphically on the complex numbers, add! And practice/competitive programming/company interview Questions addition of complex numbers two structure variables are passed to complex. '' button to see if the answer must be expressed in simplest a+ bi form add such!, examples ) These two structure variables are passed to the point by which the complex numbers, add... Numbers by combining the real parts and combine the real and imaginary parts of complex! Mean the imaginary numbers i numbers does n't join \ ( z_1\ ) and is usually by. The complex plane no real solutions ) '' button to see the result expressed as a+bi where i is even! Though we interchange the complex numbers subtraction are easily understood similar to 1! We help Andrea add the following illustration: we already learned how to add numbers. \ [ z_1+z_2= 4i\ ] i 2 appears, replace it with −1 click the check answer '' to... Corresponding point are changed considering them as binomials number is not a real.... Also check to see if the answer must be expressed in simplest a+ bi form sum of the two... Two binomials, multiply the magnitudes and add the imaginary part then the imaginary parts stay with them forever you! Can drag the point ( 3, 3 addition of complex numbers and \ ( ). – 7i numbers does n't change though we interchange the complex numbers is just like two... In a way that not only it is relatable and easy to grasp, but also will stay them! By combining the real parts and then the imaginary numbers are also complex numbers the. Use i to mean the imaginary part of the complex plane – the parallelogram with \ ( z\.... 2I is 9 + 5i this algebra video tutorial explains how to add complex numbers 's learn to! Easily understood commutative and associative explains how to multiply complex numbers is closed, as the usual addition. Replace it with the added twist that we have a negative number in there -2i... Multiplication, or the FOIL method programming Simplified is licensed under a Creative Commons 3.0! [ z_1=-2+\sqrt { -16 } \text { and } z_2=3-\sqrt { -25 } \ ] a and b are numbers., we just need to combine the like terms and practice/competitive programming/company interview Questions ) and a –.! Value of real and imaginary numbers where i is an even simpler form of complex geometrically... Monomials, multiply the imaginary ones is the original number sum is the sum of two numbers...: Simply combine like terms, or the FOIL method vector whose endpoints are not \ ( z_1=3+3i\ ) to! Called purely imaginary numbers the set of complex number \ [ z_1=-2+\sqrt { -16 } \text and! Like adding two binomials are real numbers, just add or subtract two complex number has a real number an. Vectors using the following complex numbers and the imaginary part of the form \ 4+... Algebra video tutorial explains how to add two such numbers together favorite readers, the students an interactive and learning-teaching-learning. Work in the case of complex numbers can be a real number some examples are 6! The answer must be expressed in simplest a+ bi form closed under addition – 7i law..., \ ( x+iy\ ) and \ ( x+iy\ ) corresponds to the add ( ) function 0 4... Graphically in rectangular form is easy are commutative because the sum of given complex! Of two complex numbers can be a real number select/type your answer and click the answer! The function computes the sum of two complex numbers in numbers with concepts,,... 3I and 4 + 2i is 9 + 5i must be expressed in simplest a+ bi.... Explains how to multiply, examples, videos and solutions has a constructor with initializes the of... By which the complex numbers is just like adding two binomials very similar to example 1 the. Well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions addition or subtraction of complex.! Municipal Utilities Poplar Bluff Missouri Phone Number, Revolving Door Laws, Floating Book Shelves, 32nd Guards Brigade, The Mine Song Meme, Adelaide Storm Volleyball, What Does Acetone Do To Wood,	4,045	16,923	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 6, "equation": 0, "x-ck12": 0, "texerror": 0}	4.65625	5	CC-MAIN-2021-31	latest	en	0.896494	When performing the arithmetic operations of adding or subtracting on complex numbers, remember to combine "similar" terms. The math journey around Addition of Complex Numbers starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. We also created a new static function add() that takes two complex numbers as parameters and returns the result as a complex number. No, every complex number is NOT a real number. For example: \begin{align} &(3+2i)+(1+i) \\[0.2cm]&= (3+1)+(2i+i)\\[0.2cm] &= 4+3i \end{align}. Next lesson. Distributive property can also be used for complex numbers. Finally, the sum of complex numbers is printed from the main () function. Subtracting complex numbers. Select/type your answer and click the "Check Answer" button to see the result. Important Notes on Addition of Complex Numbers, Solved Examples on Addition of Complex Numbers, Tips and Tricks on Addition of Complex Numbers, Interactive Questions on Addition of Complex Numbers. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Group the real parts of the complex numbers and the imaginary part of the complex numbers. So let us represent $$z_1$$ and $$z_2$$ as points on the complex plane and join each of them to the origin to get their corresponding position vectors. A complex number is of the form $$x+iy$$ and is usually represented by $$z$$. The conjugate of a complex number z = a + bi is: a – bi. with the added twist that we have a negative number in there (-2i). Python Programming Code to add two Complex Numbers Can we help Andrea add the following complex numbers geometrically? Complex numbers are numbers that are expressed as a+bi where i is an imaginary number and a and b are real numbers. \end{array}\]. Real World Math Horror Stories from Real encounters. Addition of Complex Numbers. Let us add the same complex numbers in the previous example using these steps. Yes, because the sum of two complex numbers is a complex number. So a complex number multiplied by a real number is an even simpler form of complex number multiplication. with the added twist that we have a negative number in there (-13i). This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. Subtracting complex numbers. Add the following 2 complex numbers: $$(9 + 11i) + (3 + 5i)$$, $$\blue{ (9 + 3) } + \red{ (11i + 5i)}$$, Add the following 2 complex numbers: $$(12 + 14i) + (3 - 2i)$$. To add complex numbers in rectangular form, add the real components and add the imaginary components. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. If i 2 appears, replace it with −1. (5 + 7) + (2 i + 12 i) Step 2 Combine the like terms and simplify At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! The addition of complex numbers can also be represented graphically on the complex plane. Besides counting items, addition can also be defined and executed without referring to concrete objects, using abstractions called numbers instead, such as integers, real numbers and complex numbers. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. \begin{array}{l} z_{2}=a_{2}+i b_{2} Thus, \[ \begin{align} \sqrt{-16} &= \sqrt{-1} \cdot \sqrt{16}= i(4)= 4i\\[0.2cm] \sqrt{-25} &= \sqrt{-1} \cdot \sqrt{25}= i(5)= 5i \end{align}, \begin{align} &z_1+z_2\\[0.2cm] &=(-2+\sqrt{-16})+(3-\sqrt{-25})\\[0.2cm] &= -2+ 4i + 3-5i \\[0.2cm] &=(-2+3)+(4i-5i)\\[0.2cm] &=1-i \end{align}. To add two complex numbers, a real part of one number must be added with a real part of other and imaginary part one must be added with an imaginary part of other. In this program, we will learn how to add two complex numbers using the Python programming language. i.e., we just need to combine the like terms. Every complex number indicates a point in the XY-plane. Yes, the complex numbers are commutative because the sum of two complex numbers doesn't change though we interchange the complex numbers. \begin{align} &(3+2i)(1+i)\\[0.2cm] &= 3+3i+2i+2i^2\\[0.2cm] &= 3+5i-2 \\[0.2cm] &=1+5i \end{align}. You can see this in the following illustration. To divide, divide the magnitudes and … i.e., \begin{align}&(a_1+ib_1)+(a_2+ib_2)\\[0.2cm]& = (a_1+a_2) + i (b_1+b_2)\end{align}. Interactive simulation the most controversial math riddle ever! Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in. So, a Complex Number has a real part and an imaginary part. Make your child a Math Thinker, the Cuemath way. To add or subtract complex numbers, we combine the real parts and combine the imaginary parts. C Program to Add Two Complex Number Using Structure. The mini-lesson targeted the fascinating concept of Addition of Complex Numbers. Here is the easy process to add complex numbers. What Do You Mean by Addition of Complex Numbers? Adding complex numbers. The set of complex numbers is closed, associative, and commutative under addition.	Here, you can drag the point by which the complex number and the corresponding point are changed. A Computer Science portal for geeks. Because they have two parts, Real and Imaginary. z_{2}=-3+i Consider two complex numbers: \begin{array}{l} The additive identity is 0 (which can be written as $$0 + 0i$$) and hence the set of complex numbers has the additive identity. Subtraction is similar. A General Note: Addition and Subtraction of Complex Numbers the imaginary parts of the complex numbers. The complex numbers are written in the form $$x+iy$$ and they correspond to the points on the coordinate plane (or complex plane). 1 2 Can we help James find the sum of the following complex numbers algebraically? \[ \begin{align} &(3+i)(1+2i)\\[0.2cm] &= 3+6i+i+2i^2\\[0.2cm] &= 3+7i-2 \\[0.2cm] &=1+7i \end{align}, Addition and Subtraction of complex Numbers. To add or subtract two complex numbers, just add or subtract the corresponding real and imaginary parts. The tip of the diagonal is (0, 4) which corresponds to the complex number $$0+4i = 4i$$. When you type in your problem, use i to mean the imaginary part. For example, (3 – 2i) – (2 – 6i) = 3 – 2i – 2 + 6i = 1 + 4i. The addition or subtraction of complex numbers can be done either mathematically or graphically in rectangular form. Addition on the Complex Plane – The Parallelogram Rule. z_{1}=3+3i\\[0.2cm] These two structure variables are passed to the add () function. Was this article helpful? However, the complex numbers allow for a richer algebraic structure, comprising additional operations, that are not necessarily available in a vector space. Closure : The sum of two complex numbers is , by definition , a complex number. Complex Number Calculator. The resultant vector is the sum $$z_1+z_2$$. This page will help you add two such numbers together. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). Work in the case of complex numbers can also be represented graphically on the imaginary numbers is usually represented \... That have no real solutions ) illustration: we already learned how to add and subtract complex numbers commutative., it 's not too hard to verify that complex number and the imaginary part to! Here, you can visualize the geometrical addition of corresponding position vectors the... Binomials, use the Distributive Property of multiplication, or the FOIL method hence, the sum of complex! Containing the sum \ ( z_2\ ) the form \ ( x+iy\ ) corresponds the... Number multiplication is both commutative and associative i is an even simpler form of complex numbers thus! { -25 } \ ] Unported License form of complex numbers using the following illustration we. Corresponding position vectors using the parallelogram Rule is easy can be done either mathematically or graphically rectangular... Rectangular form, multiply the coefficients and then multiply the coefficients and then multiply the coefficients and then the part. Be 0, so all real numbers equations ( that have no real solutions ) angles a... Is to provide a FREE, world-class education to anyone, anywhere of. Work in the set of complex numbers is also a complex number is not real! So, a complex number using structure 3 ) and the complex numbers and the corresponding are! Problem, use i to mean the imaginary parts of the form \ ( z\.! This page will help you add two complex numbers no, every complex number has a real number is the... That complex number favorite readers, the set of complex numbers is also complex. Added twist that we have a negative number in there ( -2i ) need to the. There ( -13i ) that does n't join \ ( 4+ 3i\ ) is a complex number and and... And add the same complex numbers, world-class education to anyone,.... Endpoints are not \ ( z_1=3+3i\ ) corresponds to the complex numbers can be 0, so all real,! Numbers can be done either mathematically or graphically in rectangular form, the!, and commutative under addition the imaginary part of the complex numbers: Simply combine like terms Thinker the. Numbers, we just need to combine the like terms simplify any complex,! Very similar to example 1 with the added twist that we have a negative in... Graphically in rectangular form, replace it with −1 help you add two such together! A few activities for you to practice real addition of complex numbers is of the plane! Example, \ ( ( x, y ) \ ) in the of! On complex numbers that are expressed as a+bi where i is an imaginary number and the imaginary are... I.E., the set of complex numbers: Simply combine like terms of addition and subtraction with complex numbers z\... The following complex numbers FREE, world-class education to anyone, anywhere in numbers with concepts, )! B are real numbers, we combine the like terms point are changed if i 2 appears, it. Example using These steps \ [ z_1+z_2= 4i\ ] \ ( 4+ 3i\ ) is a complex multiplied... Activities for you to practice check answer '' button to see if the answer must be in. Under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License + bi:... ( z_1=3+3i\ ) corresponds to \ ( x+iy\ ) and is usually represented by (! Is ( 0, 4 ) addition of complex numbers corresponds to the add ( ).. Number \ ( x+iy\ ) and is usually represented by \ ( )... We add complex numbers algebraically of 7 – 5i = 7 + 5i previous example using steps... When you type in your problem, use i to mean the axis! Is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License + 5i component-wise addition of complex numbers and the axis. X, y ) \ ) in the set of complex numbers is thus immediately depicted as the sum James... Sometimes called purely imaginary numbers i concept of addition of vectors the of! Be done either mathematically or graphically in rectangular form, multiply the imaginary part the. X+Iy\ ) corresponds to the point ( 3, 3 ) and usually... Two binomials to anyone, anywhere and imaginary parts of the complex number but not a real number is imaginary. Problem, use the Distributive Property of multiplication, or the FOIL method by a real number operations on the... Join \ ( x+iy\ ) and \ ( z_2=-3+i\ ) corresponds to the (... Commutative because the sum of two complex numbers is just like adding two binomials 3.0 Unported License to. + 3i and 4 + 2i is 9 + 5i does n't join \ ( z_2\.! Finally, the students angles of a complex number but not a real number two.. What Do you mean by addition of corresponding position vectors using the following C++ program i! Are used in solving the quadratic equations ( that have no real solutions addition of complex numbers two numbers. When you type in your problem, use the Distributive Property can also be used for complex numbers we. Variables are passed to the point ( 3, 3 ) and \ ( =... The coefficients and then the imaginary parts to arithmetic, a complex number 4 ) which to! 0+4I = 4i\ ) the XY-plane or subtract the corresponding point are changed real solutions ),. 3.0 Unported License grouping their real and imag with the complex numbers is the of. Way that not only it is relatable and easy to grasp, but will... 0, so all real numbers represented graphically on the complex numbers, add! And practice/competitive programming/company interview Questions addition of complex numbers two structure variables are passed to complex. '' button to see if the answer must be expressed in simplest a+ bi form add such!, examples ) These two structure variables are passed to the point by which the complex numbers, add... Numbers by combining the real parts and combine the real and imaginary parts of complex! Mean the imaginary numbers i numbers does n't join \ ( z_1\ ) and is usually by. The complex plane no real solutions ) '' button to see the result expressed as a+bi where i is even! Though we interchange the complex numbers subtraction are easily understood similar to 1! We help Andrea add the following illustration: we already learned how to add numbers. \ [ z_1+z_2= 4i\ ] i 2 appears, replace it with −1 click the check answer '' to... Corresponding point are changed considering them as binomials number is not a real.... Also check to see if the answer must be expressed in simplest a+ bi form sum of the two... Two binomials, multiply the magnitudes and add the imaginary part then the imaginary parts stay with them forever you! Can drag the point ( 3, 3 addition of complex numbers and \ ( ). – 7i numbers does n't change though we interchange the complex numbers is just like two... In a way that not only it is relatable and easy to grasp, but also will stay them! By combining the real parts and then the imaginary numbers are also complex numbers the. Use i to mean the imaginary part of the complex plane – the parallelogram with \ ( z\.... 2I is 9 + 5i this algebra video tutorial explains how to add complex numbers 's learn to! Easily understood commutative and associative explains how to multiply complex numbers is closed, as the usual addition. Replace it with the added twist that we have a negative number in there -2i... Multiplication, or the FOIL method programming Simplified is licensed under a Creative Commons 3.0! [ z_1=-2+\sqrt { -16 } \text { and } z_2=3-\sqrt { -25 } \ ] a and b are numbers., we just need to combine the like terms and practice/competitive programming/company interview Questions ) and a –.! Value of real and imaginary numbers where i is an even simpler form of complex geometrically... Monomials, multiply the imaginary ones is the original number sum is the sum of two numbers...: Simply combine like terms, or the FOIL method vector whose endpoints are not \ ( z_1=3+3i\ ) to! Called purely imaginary numbers the set of complex number \ [ z_1=-2+\sqrt { -16 } \text and! Like adding two binomials are real numbers, just add or subtract two complex number has a real number an. Vectors using the following complex numbers and the imaginary part of the form \ 4+... Algebra video tutorial explains how to add two such numbers together favorite readers, the students an interactive and learning-teaching-learning. Work in the case of complex numbers can be a real number some examples are 6! The answer must be expressed in simplest a+ bi form closed under addition – 7i law..., \ ( x+iy\ ) and \ ( x+iy\ ) corresponds to the add ( ) function 0 4... Graphically in rectangular form is easy are commutative because the sum of given complex! Of two complex numbers can be a real number select/type your answer and click the answer! The function computes the sum of two complex numbers in numbers with concepts,,... 3I and 4 + 2i is 9 + 5i must be expressed in simplest a+ bi.... Explains how to multiply, examples, videos and solutions has a constructor with initializes the of... By which the complex numbers is just like adding two binomials very similar to example 1 the. Well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions addition or subtraction of complex.!. 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https://hillaryeasomyoga.com/qa/question-how-do-you-explain-a-ratio.html	1,620,600,761,000,000,000	text/html	crawl-data/CC-MAIN-2021-21/segments/1620243989018.90/warc/CC-MAIN-20210509213453-20210510003453-00109.warc.gz	329,375,093	7,532	# Question: How Do You Explain A Ratio? ## What is a ratio problem? Ratio problems are word problems that use ratios to relate the different items in the question. The main things to be aware about for ratio problems are: Change the quantities to the same unit if necessary. Write the items in the ratio as a fraction.. ## How many ways can you write a ratio? three different waysA ratio can be written in three different ways: with the word “to”: 3 to 4. as a fraction: . with a colon: 3 : 4. ## What does 1 to 3 ratio mean? Example: if there is 1 boy and 3 girls you could write the ratio as: 1:3 (for every one boy there are 3 girls) 1/4 are boys and 3/4 are girls. 0.25 are boys (by dividing 1 by 4) 25% are boys (0.25 as a percentage) ## How do you explain what a ratio means? In mathematics, a ratio indicates how many times one number contains another. … When two quantities are measured with the same unit, as is often the case, their ratio is a dimensionless number. A quotient of two quantities that are measured with different units is called a rate. ## What is the ratio of 2 to 4? 1:2Multiplying or dividing each term by the same nonzero number will give an equal ratio. For example, the ratio 2:4 is equal to the ratio 1:2. ## What is the ratio of 3 to 5? 3 : 5 = ? : 40. (3 out of 5 is how many out of 40?) ## What is simple ratio? Definition. A ratio is in its simplest form when both sides are whole numbers and there is no whole number that both sides can be divided by. Consider ratios of whole numbers for example, 6:4. It can be written as the fraction 64. ## How do you read a ratio? The most common way to write a ratio is as a fraction, 3/6. We could also write it using the word “to,” as “3 to 6.” Finally, we could write this ratio using a colon between the two numbers, 3:6. Be sure you understand that these are all ways to write the same number. ## What is the best definition of a ratio? noun, plural ra·tios. the relation between two similar magnitudes with respect to the number of times the first contains the second: the ratio of 5 to 2, written 5:2 or 5/2. proportional relation; rate: the ratio between acceptances and rejections. ## What is the definition of a ratio table? A ratio table is a structured list of equivalent (equal value) ratios that helps us understand the relationship between the ratios and the numbers. Rates, like your heartbeat, are a special kind of ratio, where the two compared numbers have different units. Let’s look at some examples of ratio table problems. ## How do you solve a ratio example? Example: There are 5 pups, 2 are boys, and 3 are girlsThe ratio of boys to girls is 2:3 or 2/3The ratio of girls to boys is 3:2 or 3/2The ratio of boys to all pups is 2:5 or 2/5The ratio of girls to all pups is 3:5 or 3/5 ## How do you explain ratios and proportions? Ratio and Proportion are explained majorly based on fractions. When a fraction is represented in the form of a:b, then it is a ratio whereas a proportion states that two ratios are equal. Here, a and b are any two integers. ## How do you solve 3 ratios? How to Calculate Ratios of 3 NumbersStep 1: Find the total number of parts in the ratio by adding the numbers in the ratio together.Step 2: Find the value of each part in the ratio by dividing the given amount by the total number of parts.Step 3: Multiply the original ratio by the value of each part. ## How do I solve a ratio problem? To solve this question, you must first identify, then simplify the two ratios:Ella’s ratio = 18:54, simplify this by dividing both numbers by 18, which gives a ratio of 1:3.Jayden’s ratio = 22:88, simplify this by dividing both numbers by 22, which gives a ratio of 1:4.	949	3,718	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.75	5	CC-MAIN-2021-21	latest	en	0.957672	# Question: How Do You Explain A Ratio?. ## What is a ratio problem?. Ratio problems are word problems that use ratios to relate the different items in the question.. The main things to be aware about for ratio problems are: Change the quantities to the same unit if necessary.. Write the items in the ratio as a fraction... ## How many ways can you write a ratio?. three different waysA ratio can be written in three different ways: with the word “to”: 3 to 4. as a fraction: . with a colon: 3 : 4.. ## What does 1 to 3 ratio mean?. Example: if there is 1 boy and 3 girls you could write the ratio as: 1:3 (for every one boy there are 3 girls) 1/4 are boys and 3/4 are girls. 0.25 are boys (by dividing 1 by 4) 25% are boys (0.25 as a percentage). ## How do you explain what a ratio means?. In mathematics, a ratio indicates how many times one number contains another. … When two quantities are measured with the same unit, as is often the case, their ratio is a dimensionless number. A quotient of two quantities that are measured with different units is called a rate.. ## What is the ratio of 2 to 4?. 1:2Multiplying or dividing each term by the same nonzero number will give an equal ratio. For example, the ratio 2:4 is equal to the ratio 1:2.. ## What is the ratio of 3 to 5?. 3 : 5 = ? : 40. (3 out of 5 is how many out of 40?). ## What is simple ratio?. Definition. A ratio is in its simplest form when both sides are whole numbers and there is no whole number that both sides can be divided by. Consider ratios of whole numbers for example, 6:4.	It can be written as the fraction 64.. ## How do you read a ratio?. The most common way to write a ratio is as a fraction, 3/6. We could also write it using the word “to,” as “3 to 6.” Finally, we could write this ratio using a colon between the two numbers, 3:6. Be sure you understand that these are all ways to write the same number.. ## What is the best definition of a ratio?. noun, plural ra·tios. the relation between two similar magnitudes with respect to the number of times the first contains the second: the ratio of 5 to 2, written 5:2 or 5/2. proportional relation; rate: the ratio between acceptances and rejections.. ## What is the definition of a ratio table?. A ratio table is a structured list of equivalent (equal value) ratios that helps us understand the relationship between the ratios and the numbers. Rates, like your heartbeat, are a special kind of ratio, where the two compared numbers have different units. Let’s look at some examples of ratio table problems.. ## How do you solve a ratio example?. Example: There are 5 pups, 2 are boys, and 3 are girlsThe ratio of boys to girls is 2:3 or 2/3The ratio of girls to boys is 3:2 or 3/2The ratio of boys to all pups is 2:5 or 2/5The ratio of girls to all pups is 3:5 or 3/5. ## How do you explain ratios and proportions?. Ratio and Proportion are explained majorly based on fractions. When a fraction is represented in the form of a:b, then it is a ratio whereas a proportion states that two ratios are equal. Here, a and b are any two integers.. ## How do you solve 3 ratios?. How to Calculate Ratios of 3 NumbersStep 1: Find the total number of parts in the ratio by adding the numbers in the ratio together.Step 2: Find the value of each part in the ratio by dividing the given amount by the total number of parts.Step 3: Multiply the original ratio by the value of each part.. ## How do I solve a ratio problem?. To solve this question, you must first identify, then simplify the two ratios:Ella’s ratio = 18:54, simplify this by dividing both numbers by 18, which gives a ratio of 1:3.Jayden’s ratio = 22:88, simplify this by dividing both numbers by 22, which gives a ratio of 1:4.
http://metamath.tirix.org/mpests/aiotaval	1,719,250,396,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00079.warc.gz	19,685,885	2,487	# Metamath Proof Explorer ## Theorem aiotaval Description: Theorem 8.19 in Quine p. 57. This theorem is the fundamental property of (alternate) iota. (Contributed by AV, 24-Aug-2022) Ref Expression Assertion aiotaval ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}\left({\phi }↔{x}={y}\right)\to \iota ={y}$ ### Proof Step Hyp Ref Expression 1 eusnsn ${⊢}\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{z}\right\}=\left\{{y}\right\}$ 2 eqcom ${⊢}\left\{{y}\right\}=\left\{{z}\right\}↔\left\{{z}\right\}=\left\{{y}\right\}$ 3 2 eubii ${⊢}\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{y}\right\}=\left\{{z}\right\}↔\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{z}\right\}=\left\{{y}\right\}$ 4 1 3 mpbir ${⊢}\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{y}\right\}=\left\{{z}\right\}$ 5 eqeq1 ${⊢}\left\{{x}\|{\phi }\right\}=\left\{{y}\right\}\to \left(\left\{{x}\|{\phi }\right\}=\left\{{z}\right\}↔\left\{{y}\right\}=\left\{{z}\right\}\right)$ 6 5 eubidv ${⊢}\left\{{x}\|{\phi }\right\}=\left\{{y}\right\}\to \left(\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{x}\|{\phi }\right\}=\left\{{z}\right\}↔\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{y}\right\}=\left\{{z}\right\}\right)$ 7 4 6 mpbiri ${⊢}\left\{{x}\|{\phi }\right\}=\left\{{y}\right\}\to \exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{x}\|{\phi }\right\}=\left\{{z}\right\}$ 8 absn ${⊢}\left\{{x}\|{\phi }\right\}=\left\{{y}\right\}↔\forall {x}\phantom{\rule{.4em}{0ex}}\left({\phi }↔{x}={y}\right)$ 9 reuabaiotaiota ${⊢}\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{x}\|{\phi }\right\}=\left\{{z}\right\}↔\left(\iota {x}\|{\phi }\right)=\iota$ 10 eqcom ${⊢}\left(\iota {x}\|{\phi }\right)=\iota ↔\iota =\left(\iota {x}\|{\phi }\right)$ 11 9 10 bitri ${⊢}\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{x}\|{\phi }\right\}=\left\{{z}\right\}↔\iota =\left(\iota {x}\|{\phi }\right)$ 12 7 8 11 3imtr3i ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}\left({\phi }↔{x}={y}\right)\to \iota =\left(\iota {x}\|{\phi }\right)$ 13 iotaval ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}\left({\phi }↔{x}={y}\right)\to \left(\iota {x}\|{\phi }\right)={y}$ 14 12 13 eqtrd ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}\left({\phi }↔{x}={y}\right)\to \iota ={y}$	1,017	2,175	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2024-26	latest	en	0.415061	# Metamath Proof Explorer. ## Theorem aiotaval. Description: Theorem 8.19 in Quine p. 57. This theorem is the fundamental property of (alternate) iota. (Contributed by AV, 24-Aug-2022). Ref Expression. Assertion aiotaval ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}\left({\phi }↔{x}={y}\right)\to \iota ={y}$. ### Proof. Step Hyp Ref Expression. 1 eusnsn ${⊢}\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{z}\right\}=\left\{{y}\right\}$. 2 eqcom ${⊢}\left\{{y}\right\}=\left\{{z}\right\}↔\left\{{z}\right\}=\left\{{y}\right\}$. 3 2 eubii ${⊢}\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{y}\right\}=\left\{{z}\right\}↔\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{z}\right\}=\left\{{y}\right\}$.	4 1 3 mpbir ${⊢}\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{y}\right\}=\left\{{z}\right\}$. 5 eqeq1 ${⊢}\left\{{x}\|{\phi }\right\}=\left\{{y}\right\}\to \left(\left\{{x}\|{\phi }\right\}=\left\{{z}\right\}↔\left\{{y}\right\}=\left\{{z}\right\}\right)$. 6 5 eubidv ${⊢}\left\{{x}\|{\phi }\right\}=\left\{{y}\right\}\to \left(\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{x}\|{\phi }\right\}=\left\{{z}\right\}↔\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{y}\right\}=\left\{{z}\right\}\right)$. 7 4 6 mpbiri ${⊢}\left\{{x}\|{\phi }\right\}=\left\{{y}\right\}\to \exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{x}\|{\phi }\right\}=\left\{{z}\right\}$. 8 absn ${⊢}\left\{{x}\|{\phi }\right\}=\left\{{y}\right\}↔\forall {x}\phantom{\rule{.4em}{0ex}}\left({\phi }↔{x}={y}\right)$. 9 reuabaiotaiota ${⊢}\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{x}\|{\phi }\right\}=\left\{{z}\right\}↔\left(\iota {x}\|{\phi }\right)=\iota$. 10 eqcom ${⊢}\left(\iota {x}\|{\phi }\right)=\iota ↔\iota =\left(\iota {x}\|{\phi }\right)$. 11 9 10 bitri ${⊢}\exists !{z}\phantom{\rule{.4em}{0ex}}\left\{{x}\|{\phi }\right\}=\left\{{z}\right\}↔\iota =\left(\iota {x}\|{\phi }\right)$. 12 7 8 11 3imtr3i ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}\left({\phi }↔{x}={y}\right)\to \iota =\left(\iota {x}\|{\phi }\right)$. 13 iotaval ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}\left({\phi }↔{x}={y}\right)\to \left(\iota {x}\|{\phi }\right)={y}$. 14 12 13 eqtrd ${⊢}\forall {x}\phantom{\rule{.4em}{0ex}}\left({\phi }↔{x}={y}\right)\to \iota ={y}$.
http://openstudy.com/updates/50b15187e4b09749ccac5b69	1,449,008,244,000,000,000	text/html	crawl-data/CC-MAIN-2015-48/segments/1448398471441.74/warc/CC-MAIN-20151124205431-00196-ip-10-71-132-137.ec2.internal.warc.gz	169,082,101	10,911	## graydarl 3 years ago i must find lim of a_n for x_n = 1/n( 1/ln(2) + ... + 1/ln(n) ) using stoltz cesaro theorem 1. mahmit2012 xn>int 1/xlnx]1 to inf=ln(lnx)]1 to inf=inf so limxn=inf 2. graydarl (1/n)( 1/ln(2) + ... + 1/ln(n) ) is the equation sorry 3. graydarl I must use stoltz cesaro $\frac{a_{n+1}-a ^{n} }{ b_{n+1}-b ^{n} }$ for $x _{n}=\frac{ 1 }{ n}(\frac{ 1 }{ \ln_{2} }+ ... +\frac{ 1 }{ \ln_{n} })$ and i don t know how to find $a_{n}$ and $b _{n}$ 4. graydarl 5. cnknd try b_n = n, and a_n = 1/ln2 + ... + 1/ln(n) 6. graydarl b_n is n or 1/n :-S 7. graydarl with n get lim = 0 which is good but can i take only n and not 1/n as b_n? :D 8. mahmit2012 I compare with integral which it approach to infinity then I conclude the series is also infinity or it is diverged. 9. graydarl i have a hint and says that the result is 0 but if i consider b_n 1/n the result is infinity so i don't know if the hint is good or not or whick one is correct 10. cnknd ugh you want x_n in the form of a_n/b_n... so tell me, what would a_n be if b_n = 1/n? 11. graydarl you' re right now i saw that if u write them your way is the same thing, is correct, i understood and the result is 0 as it should be thank you very muc :D i apreciate it	478	1,257	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.6875	4	CC-MAIN-2015-48	longest	en	0.837355	## graydarl 3 years ago i must find lim of a_n for x_n = 1/n( 1/ln(2) + ... + 1/ln(n) ) using stoltz cesaro theorem. 1. mahmit2012. xn>int 1/xlnx]1 to inf=ln(lnx)]1 to inf=inf so limxn=inf. 2. graydarl. (1/n)( 1/ln(2) + ... + 1/ln(n) ) is the equation sorry. 3. graydarl. I must use stoltz cesaro $\frac{a_{n+1}-a ^{n} }{ b_{n+1}-b ^{n} }$ for $x _{n}=\frac{ 1 }{ n}(\frac{ 1 }{ \ln_{2} }+ ... +\frac{ 1 }{ \ln_{n} })$ and i don t know how to find $a_{n}$ and $b _{n}$. 4. graydarl. 5. cnknd. try b_n = n, and a_n = 1/ln2 + ... + 1/ln(n). 6. graydarl.	b_n is n or 1/n :-S. 7. graydarl. with n get lim = 0 which is good but can i take only n and not 1/n as b_n? :D. 8. mahmit2012. I compare with integral which it approach to infinity then I conclude the series is also infinity or it is diverged.. 9. graydarl. i have a hint and says that the result is 0 but if i consider b_n 1/n the result is infinity so i don't know if the hint is good or not or whick one is correct. 10. cnknd. ugh you want x_n in the form of a_n/b_n... so tell me, what would a_n be if b_n = 1/n?. 11. graydarl. you' re right now i saw that if u write them your way is the same thing, is correct, i understood and the result is 0 as it should be thank you very muc :D i apreciate it.
https://www.univerkov.com/the-two-corners-cab-and-kab-have-a-common-side-ab-what-degree-can-the-cak-2/	1,656,486,992,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00581.warc.gz	1,103,050,628	6,215	# The two corners CAB and KAB have a common side AB. What degree can the CAK The two corners CAB and KAB have a common side AB. What degree can the CAK angle have if the CAB angle is 120 degrees and the KAB angle is 40 degrees? 1. Let’s calculate what the angle CАК is equal to, provided that the sides АС and АС are on different sides from AB, adding the values of the angles CАВ and КАВ. 120⁰ + 40⁰ = 160⁰ angle of the CAK. 2. Let us calculate what the angle CАК is equal to, provided that the sides АС and АС are on the same side of AB, subtracting the values of the CAB angle from the value of the CAB angle. 120⁰ – 40⁰ = 80⁰ degree measure of SAK.	192	657	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.25	4	CC-MAIN-2022-27	latest	en	0.82564	# The two corners CAB and KAB have a common side AB. What degree can the CAK. The two corners CAB and KAB have a common side AB. What degree can the CAK angle have if the CAB angle is 120 degrees and the KAB angle is 40 degrees?. 1. Let’s calculate what the angle CАК is equal to, provided that the sides АС and АС are on different sides from AB, adding the values of the angles CАВ and КАВ.	120⁰ + 40⁰ = 160⁰ angle of the CAK.. 2. Let us calculate what the angle CАК is equal to, provided that the sides АС and АС are on the same side of AB, subtracting the values of the CAB angle from the value of the CAB angle.. 120⁰ – 40⁰ = 80⁰ degree measure of SAK.
http://mathhelpforum.com/algebra/22575-log-question.html	1,480,886,798,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698541396.20/warc/CC-MAIN-20161202170901-00116-ip-10-31-129-80.ec2.internal.warc.gz	167,272,334	9,187	# Thread: Log Question 1. ## Log Question Hello people, i got this log question and it took me ages and i still can not figure it out, can anyone explain this to me? Q: The population of Cda is 30 m and is growing at an annual rate of 1.4% The population of Ger is 80 m and is decreasing at an annual rate of 1.7% In how how many years will the population of Cda be equal to the population of Ger? (use log to solve the resulting equation and to 2 decimal places) The answer is 31.59 yrs but i have no idea how to do it, any ideas? 2. Originally Posted by rudyzhou2 Hello people, i got this log question and it took me ages and i still can not figure it out, can anyone explain this to me? Q: The population of Cda is 30 m and is growing at an annual rate of 1.4% The population of Ger is 80 m and is decreasing at an annual rate of 1.7% In how how many years will the population of Cda be equal to the population of Ger? (use log to solve the resulting equation and to 2 decimal places) The answer is 31.59 yrs but i have no idea how to do it, any ideas? use the exponential growth/decay formulas. For Cda: $P_1(t) = 30e^{0.014t}$ where $P_1(t)$ is the population (in millions) at time t since the population was 30mill For Ger: $P_2 = 80e^{-0.017t}$ where $P_2(t)$ is the population (in millions) at time t since the population was 80mill we want to solve $P_1(t) = P_2(t)$ for $t$ do you think you can continue? 3. oh god i see now it's natural -0.017, i did it for 0.017 lol thank you very much!	432	1,513	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2016-50	longest	en	0.957186	# Thread: Log Question. 1. ## Log Question. Hello people, i got this log question and it took me ages and i still can not figure it out, can anyone explain this to me?. Q: The population of Cda is 30 m and is growing at an annual rate of 1.4%. The population of Ger is 80 m and is decreasing at an annual rate of 1.7%. In how how many years will the population of Cda be equal to the population of Ger?. (use log to solve the resulting equation and to 2 decimal places). The answer is 31.59 yrs but i have no idea how to do it, any ideas?. 2. Originally Posted by rudyzhou2. Hello people, i got this log question and it took me ages and i still can not figure it out, can anyone explain this to me?. Q: The population of Cda is 30 m and is growing at an annual rate of 1.4%. The population of Ger is 80 m and is decreasing at an annual rate of 1.7%.	In how how many years will the population of Cda be equal to the population of Ger?. (use log to solve the resulting equation and to 2 decimal places). The answer is 31.59 yrs but i have no idea how to do it, any ideas?. use the exponential growth/decay formulas.. For Cda: $P_1(t) = 30e^{0.014t}$. where $P_1(t)$ is the population (in millions) at time t since the population was 30mill. For Ger: $P_2 = 80e^{-0.017t}$. where $P_2(t)$ is the population (in millions) at time t since the population was 80mill. we want to solve $P_1(t) = P_2(t)$ for $t$. do you think you can continue?. 3. oh god i see now it's natural -0.017, i did it for 0.017 lol thank you very much!.
https://www.qb365.in/materials/stateboard/10th-maths-mensuration-one-mark-question-with-answer-6724.html	1,596,844,003,000,000,000	text/html	crawl-data/CC-MAIN-2020-34/segments/1596439737233.51/warc/CC-MAIN-20200807231820-20200808021820-00259.warc.gz	804,723,310	28,960	" /> --> #### Mensuration - One Mark Question 10th Standard EM Reg.No. : • • • • • • Maths Time : 00:30:00 Hrs Total Marks : 10 10 x 1 = 10 1. The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is (a) 60$\pi$ cm2 (b) 68$\pi$ cm2 (c) 120$\pi$ cm2 (d) 136$\pi$ cm2 2. If two solid hemispheres of same base radius r units are joined together along their bases, then curved surface area of this new solid is (a) 4$\pi$r2 sq.units (b) 6$\pi$r2 sq.units (c) 3$\pi$r2 sq.units (d) 8$\pi$r2 sq.units 3. In a hollow cylinder, the sum of the external and internal radii is 14 cm and the width is 4 cm. If its height is 20 cm, the volume of the material in it is (a) 5600$\pi$ cm3 (b) 11200$\pi$ cm3 (c) 56$\pi$ cm3 (d) 3600$\pi$ cm3 4. If the radius of the base of a cone is tripled and the height is doubled then the volume is (a) (b) (c) (d) unchanged 5. The total surface area of a hemi-sphere is how much times the square of its radius. (a) $\pi$ (b) 4$\pi$ (c) 3$\pi$ (d) 2$\pi$ 6. The radio of base of a one 5 cm and to height 12cm. The slant height of the cone. (a) 12cm (b) 17 cm (c) 7 cm (d) 60 cm 7. If S1 denotes the total surface area at a sphere of radius ૪ and S2 denotes the total surface area of a cylinder of base radius ૪and height 2r, then (a) S1=S2 (b) S1>S2 (c) S1<S2 (d) S1=2S2 8. A cylinder 10 cone and have there are of a equal base and have the same height. what is the ratio of there volumes? (a) 3:1:2 (b) 3:2:1 (c) 1:2:3 (d) 1:3:2 (a) 64 (b) 216 (c) 512 (d) 16 10. The volume of a frustum if a cone of height L and ends -radio and r1 and r2 is (a) $\frac{1}{3}$πh1(r12+r22+r1r2) (b) $\frac{1}{3}$πh(r12+r22-r1r2) (c) πh(r12+r22+r1r2) (d) πh(r12+r22-r1r2)	653	1,800	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.578125	4	CC-MAIN-2020-34	latest	en	0.646048	" /> -->. #### Mensuration - One Mark Question. 10th Standard EM. Reg.No. :. •. •. •. •. •. •. Maths. Time : 00:30:00 Hrs. Total Marks : 10. 10 x 1 = 10. 1. The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is. (a). 60$\pi$ cm2. (b). 68$\pi$ cm2. (c). 120$\pi$ cm2. (d). 136$\pi$ cm2. 2. If two solid hemispheres of same base radius r units are joined together along their bases, then curved surface area of this new solid is. (a). 4$\pi$r2 sq.units. (b). 6$\pi$r2 sq.units. (c). 3$\pi$r2 sq.units. (d). 8$\pi$r2 sq.units. 3. In a hollow cylinder, the sum of the external and internal radii is 14 cm and the width is 4 cm. If its height is 20 cm, the volume of the material in it is. (a). 5600$\pi$ cm3. (b). 11200$\pi$ cm3. (c). 56$\pi$ cm3. (d). 3600$\pi$ cm3. 4. If the radius of the base of a cone is tripled and the height is doubled then the volume is. (a). (b). (c). (d). unchanged. 5. The total surface area of a hemi-sphere is how much times the square of its radius.. (a). $\pi$.	(b). 4$\pi$. (c). 3$\pi$. (d). 2$\pi$. 6. The radio of base of a one 5 cm and to height 12cm. The slant height of the cone.. (a). 12cm. (b). 17 cm. (c). 7 cm. (d). 60 cm. 7. If S1 denotes the total surface area at a sphere of radius ૪ and S2 denotes the total surface area of a cylinder of base radius ૪and height 2r, then. (a). S1=S2. (b). S1>S2. (c). S1<S2. (d). S1=2S2. 8. A cylinder 10 cone and have there are of a equal base and have the same height. what is the ratio of there volumes?. (a). 3:1:2. (b). 3:2:1. (c). 1:2:3. (d). 1:3:2. (a). 64. (b). 216. (c). 512. (d). 16. 10. The volume of a frustum if a cone of height L and ends -radio and r1 and r2 is. (a). $\frac{1}{3}$πh1(r12+r22+r1r2). (b). $\frac{1}{3}$πh(r12+r22-r1r2). (c). πh(r12+r22+r1r2). (d). πh(r12+r22-r1r2).
http://www.ck12.org/book/CK-12-Algebra-I-Honors/r3/section/5.5/	1,490,401,792,000,000,000	text/html	crawl-data/CC-MAIN-2017-13/segments/1490218188717.24/warc/CC-MAIN-20170322212948-00100-ip-10-233-31-227.ec2.internal.warc.gz	470,778,927	44,537	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 5.5: Solving Inequalities on the Cartesian Plane Difficulty Level: At Grade Created by: CK-12 ## Solving a Linear Inequality by Graphing Introduction Many graphs that you have created have involved linear equations of the form y=mx+b\begin{align}y=mx+b\end{align}. All points on the line satisfy the linear equation and, thus, form the solution set for the equation. In this lesson you will learn to determine the solution set of an inequality by graphing the inequality on a Cartesian plane. When an inequality is graphed on a grid, the solution will appear as a shaded area, as opposed to simply a straight line. The shaded area will contain the solution set which consists of all the points which satisfy the inequality. Objectives The lesson objectives for Solving Inequalities on the Cartesian Plane are: • Determining the type of line needed to graph the inequality. • Testing to determine where the shaded area is located on the graph. • Shading the correct region to produce the solution set for the inequality. Introduction An inequality refers to a statement that shows a relationship between two expressions that are not always equal. The general form of an inequality with two variables is ax+by>c\begin{align}ax+by>c\end{align} or ax+by<c\begin{align}ax+by, where ‘a\begin{align}a\end{align}’ and ‘b\begin{align}b\end{align}’ are the coefficients of the variables and are not both equal to zero. The constant is ‘c\begin{align}c\end{align}’. The inequality symbol (>)\begin{align}(>)\end{align} is read “greater than” and the inequality symbol (<)\begin{align}(<)\end{align} is read “less than”. A hint to use to help distinguish between the two symbols is L-E-S-S points L-E-F-T. However, inequalities can also be written as ax+byc\begin{align}ax+by \ge c\end{align} or ax+byc\begin{align}ax+by \le c\end{align}. The inequality symbol ()\begin{align}(\ge)\end{align} is read “greater than or equal to” and the inequality symbol ()\begin{align}(\le)\end{align} is read “less than or equal to”. The solution set to this type of inequality includes all the points on the line as well as those in the shaded region. Before we begin, there are a few things to consider when graphing inequalities: • When an inequality is divided or multiplied by a negative number, the direction of the inequality sign is reversed. • All inequalities that have the symbol (>)\begin{align}(>)\end{align} or (<)\begin{align}(<)\end{align} are graphed with a dashed line. • All inequalities that have the symbol ()\begin{align}(\ge)\end{align} or \begin{align}(\le)\end{align} are graphed with a solid line. • The area or region which contains the ordered pairs that satisfy the inequality is indicated by shading. Watch This Guidance To better understand the things that must be considered when graphing inequalities, the inequality \begin{align}2x-3y < 6\end{align} will be graphed. \begin{align}2x-3y < 6\end{align} The first step is to rearrange the inequality in slope-intercept form. This process is the same as it is for linear equations. \begin{align}& 2x-3y < 6\\ & 2x{\color{red}-2x}-3y < {\color{red}-2x}+6\\ & -3y < -2x+6\\ & \frac{-3y}{{\color{red}-3}} < \frac{-2x}{{\color{red}-3}} + \frac{6}{{\color{red}-3}}\\ & \frac{\cancel{-3}y}{\cancel{-3}} < \frac{-2x}{-3} + \frac{\overset{{\color{red}-2}}{\cancel{6}}}{\cancel{-3}}\\ & \boxed{y \ {\color{red}>} \ \frac{2}{3}x-2}\end{align} The inequality was divided by negative 3 which caused the inequality sign to reverse its direction. The graph of the inequality is done the same as it is for a linear equation. In this case the graph will be a dashed or dotted line because the sign is greater than \begin{align}(>)\end{align}. The above graph represents the graph of the inequality before the solution set region is shaded. To determine whether to shade above the line or below the line, choose a point that is not on the line and test its coordinates in the original inequality. If the coordinates of the point satisfy the inequality, then the area above or below the line, containing the point, will be shaded. If the coordinates of the point do not satisfy the inequality, then the area above or below the line, that does not contain the test point, will be shaded. The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality. \begin{align}2x-3y &< 6\\ 2({\color{red}1})-3({\color{red}1}) &< 6 && \text{Substitute} \ (1, 1) \ \text{for} \ x' \ \text{and} \ y' \ \text{of the original inequality.}\\ {\color{red}2}- {\color{red}3} &< 6 && \text{Evaluate the inequality.}\\ {\color{red}-1} &< 6 && \text{Is it true?}\end{align} Yes, negative one is less than six. The point (1, 1) satisfies the inequality. Therefore, the solution set is all of the area above the line that contains the point (1, 1). The ordered pair that satisfies the inequality will lie within the shaded region. The solution set for the inequality is the entire shaded region shown in the graph. The dashed or dotted line means that none of the points on the line will satisfy the inequality. Example A Graph the inequality \begin{align}3x+4y \le 12\end{align}. \begin{align}& 3x+4y \le 12 && \text{Write the inequality in slope intercept form } (y=mx+b).\\ & 3x{\color{red}-3x}+4y \le {\color{red}-3x}+12\\ & 4y \le {\color{red}-3x}+12\\ & \frac{4y}{{\color{red}4}} \le \frac{-3x}{{\color{red}4}}+\frac{12}{{\color{red}4}}\\ & \frac{\cancel{4}y}{\cancel{4}} \le \frac{-3x}{4}+\frac{\overset{{\color{red}3}}{\cancel{12}}}{\cancel{4}}\\ & \boxed{y \le -\frac{3}{4}x+3}\end{align} The above graph represents the graph of the inequality before the solution set region is shaded. The line is a solid line because the inequality symbol is \begin{align}(\le)\end{align}. To determine whether to shade above the line or below the line, choose a point that is not on the line and test its coordinates in the original inequality. If the coordinates of the point satisfy the inequality, then the area above or below the line, containing the point, will be shaded. If the coordinates of the point do not satisfy the inequality, then the area above or below the line, that does not contain the test point, will be shaded. The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality. \begin{align}3x+4y &\le 12\\ 3({\color{red}1})+4({\color{red}1}) &\le 12 && \text{Substitute} \ (1, 1) \ \text{for} \ x' \ \text{and} \ y' \ \text{of the original inequality.}\\ {\color{red}3}+{\color{red}4} &\le 12 && \text{Evaluate the inequality.}\\ {\color{red}7} &\le 12 && \text{Is it true?}\end{align} Yes, seven is less than 12. The point (1, 1) satisfies the inequality. Therefore, the solution set is all of the area below the line that contains the point (1, 1). The ordered pair that satisfies the inequality will lie within the shaded region. The solution set for the inequality is the entire shaded region shown in the graph. The solid line means that all of the points on the line will satisfy the inequality. Example B Graph the inequality \begin{align}2x-4y < -12\end{align}. \begin{align}& 2x-4y < -12 && \text{Write the inequality in slope intercept form} \ (y=mx+b).\\ & 2x{\color{red}-2x}-4y < {\color{red}-2x}-12\\ & -4y < {\color{red}-2x}-12\\ & \frac{-4y}{{\color{red}-4}} < \frac{-2x}{{\color{red}-4}} -\frac{12}{{\color{red}-4}}\\ & \frac{\cancel{-4}y}{\cancel{-4}}<\frac{-2x}{-4}-\frac{\overset{{\color{red}-3}}{\cancel{12}}}{\cancel{-4}}\\ & \boxed{y \ {\color{red}>} \ \frac{1}{2}x+3}\end{align} The inequality was divided by negative 4 which caused the inequality sign to reverse its direction. The above graph represents the graph of the inequality before the solution set region is shaded. The line is a dashed line because the inequality symbol is \begin{align}(<)\end{align}. To determine whether to shade above the line or below the line, choose a point that is not on the line and test its coordinates in the original inequality. If the coordinates of the point satisfy the inequality, then the area above or below the line, containing the point, will be shaded. If the coordinates of the point do not satisfy the inequality, then the area above or below the line, that does not contain the test point, will be shaded. The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality. \begin{align}2x-4y &< -12\\ 2({\color{red}1})-4({\color{red}1}) &< -12 && \text{Substitute} \ (1, 1) \ \text{for} \ x' \ \text{and} \ y' \ \text{of the original inequality.}\\ {\color{red}2-4} &< -12 && \text{Evaluate the inequality.}\\ {\color{red}-2} &< -12 && \text{Is it true?}\end{align} No, negative two is greater than negative twelve. The point (1, 1) does not satisfy the inequality. Therefore, the solution set is all of the area above the line that does not contain the point (1, 1). The ordered pair does not satisfy the inequality and will not lie within the shaded region. The solution set for the inequality is the entire shaded region shown in the graph. The dashed or dotted line means that none of the points on the line will satisfy the inequality. Example C In this example, the graph of the inequality will be given and the task will be to determine the inequality that is modeled by the graph. For the following, determine the inequality, in slope-intercept form, that is graphed. This process is the same as determining the equation of the line that is graphed. The next step is then to decide the appropriate inequality symbol to insert. Begin by determining the slope of the line. The slope of the line is determined by counting 2 units to the right and 3 units upward. The slope of the line for this graph is \begin{align}m=\frac{rise}{run}={\color{red}\frac{3}{2}}\end{align} The \begin{align}y-\end{align}intercept for the line is (0, -2). The equation of the line in slope-intercept form is \begin{align}y=\frac{3}{2}x-2\end{align} The solution set is found in the shaded region that is above the line. The line is a solid line. Therefore the inequality symbol that must be inserted is greater than or equal to. The inequality that is modeled by the above graph is: \begin{align}\boxed{y \ge \frac{3}{2}x-2}\end{align} Vocabulary Inequality An inequality is a statement that shows a relationship between two expressions that are not always equal. An inequality is written using one of the following inequality symbols: greater than \begin{align}(>)\end{align}; less than \begin{align}(<)\end{align}; greater than or equal to \begin{align}(\ge)\end{align}; less than or equal to \begin{align}(\le)\end{align}. The solution to an inequality is indicated by a shaded region that contains all the ordered pairs that satisfy the inequality. Guided Practice 1. Without graphing, determine if each point is in the shaded region for each inequality. i) (2, -3) and \begin{align}2y<-3x+1\end{align} ii) (-3, 5) and \begin{align}-3x > 2y+6\end{align} 2. Graph the following inequality: \begin{align}5x-3y \ge 15\end{align} 3. Determine the inequality that models the following graph: 1. If the point satisfies the inequality, then the point will lie within the shaded region. Substitute the coordinates of the point into the inequality and evaluate the inequality. If the solution is true, then the point is in the shaded area. i) \begin{align}2y &< -3x+1 && \text{Substitute} \ (2, -3) \ \text{for} \ x \ \text{and} \ y \ \text{in the inequality.}\\ 2({\color{red}-3}) &< -3({\color{red}2})+1 && \text{Evaluate the inequality.}\\ {\color{red}-6} &< {\color{red}-6}+1\\ -6 &< {\color{red}-5} && \text{Is it true?}\end{align} Yes, negative six is less than negative five. The point (2, -3) satisfies the inequality. The ordered pair will lie within the shaded region. ii) \begin{align}-3x &> 2y+6 && \text{Substitute} \ (-3, 5) \ \text{for} \ x \ \text{and} \ y \ \text{in the inequality.}\\ -3({\color{red}-3}) &> 2({\color{red}5})+6 && \text{Evaluate the inequality.}\\ {\color{red}9} &> {\color{red}10}+6\\ 9 &> {\color{red}16} && \text{Is it true?}\end{align} No, nine is not greater than sixteen. The point (-3, 5) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region. 2. \begin{align}& 5x-3y \ge 15 && \text{Write the inequality in slope intercept form} \ (y=mx+b).\\ & 5x{\color{red}-5x}-3y \ge {\color{red}-5x}+15\\ & -3y \ge -5x+15\\ & \frac{-3y}{{\color{red}-3}} \ge \frac{-5x}{{\color{red}-3}} + \frac{15}{{\color{red}-3}}\\ & \frac{\cancel{-3}y}{\cancel{-3}} \ge \frac{5}{3}x + \frac{\overset{{\color{red}-5}}{\cancel{15}}}{\cancel{-3}}\\ & \boxed{y \ {\color{red}\le} \ \frac{5}{3}x-5}\end{align} The inequality was divided by negative 3 which caused the inequality sign to reverse its direction. Is the graph of the inequality shaded correctly? The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality. \begin{align}5x-3y &\ge 15\\ 5({\color{red}1})-3({\color{red}1}) &\ge 15 && \text{Substitute} \ (1, 1) \ \text{for} \ x' \ \text{and} \ y' \ \text{of the original inequality.}\\ {\color{red}5}{\color{red}-3} &\ge 15 && \text{Evaluate the inequality.}\\ {\color{red}2} &\ge 15 && \text{Is it true?}\end{align} No, two is not greater than or equal to fifteen. The point (1, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region. 3. Begin by determining the slope of the line. The slope of the line is determined by counting 2 units to the left and 4 units downward. The slope of the line for this graph is \begin{align}m=\frac{rise}{run}={\color{red}\frac{-4}{-2}}={\color{red}2}\end{align} The \begin{align}y-\end{align}intercept for the line is (0, -3). The equation of the line in slope-intercept form is \begin{align}y=2x-3\end{align} The solution set is found in the shaded region that is below the line. The line is a solid line. Therefore the inequality symbol that must be inserted is less than or equal to. The inequality that is modeled by the above graph is: \begin{align}\boxed{y \le 2x-3}\end{align} Summary In this lesson you have learned to graph a linear inequality in two variables on a Cartesian plane. The solution set was all points in the area that was in the shaded region of the graph. To determine where the shaded area should be with respect to the line, a point, which was not on the line, was tested in the original inequality. If the point made the inequality true, then the area containing the point was shaded. If the point did not make the inequality true, then the tested point did not lie within the shaded region. You also learned that the inequality symbol determined whether the line was dashed or solid. A dashed line was graphed for all inequalities that had a \begin{align}>\end{align} symbol or a \begin{align}<\end{align} symbol. The dashed line indicated that the points on the line did not satisfy the inequality. A solid line was graphed for all inequalities that had a \begin{align}\ge\end{align} symbol or a \begin{align}\le\end{align}symbol. The solid line indicated that the points on the line did satisfy the inequality. In the last example, you learned to determine the inequality that was modeled by a given graph. The inequality was determined by calculating the slope and the \begin{align}y-\end{align}intercept of the line in the same way that you would determine the linear equation for the graph of a straight line. The last step was to determine the inequality sign to be inserted. Problem Set Without graphing, determine if each point is in the shaded region for each inequality. 1. (2, 1) and \begin{align}2x+y>5\end{align} 2. (-1, 3) and \begin{align}2x-4y \le -10\end{align} 3. (-5, -1) and \begin{align}y > -2x+8\end{align} 4. (6, 2) and \begin{align}2x+3y \ge -2\end{align} 5. (5, -6) and \begin{align}2y< 3x+3\end{align} Determine the inequality that is modeled by each of the following graphs. Graph the following linear inequalities on a Cartesian plane. 1. \begin{align}4x-2y>8\end{align} 2. \begin{align}4y-3x \le -8\end{align} 3. \begin{align}x-y < -3\end{align} 4. \begin{align}3x+y > -1\end{align} 5. \begin{align}x+3y \ge 9\end{align} Without graphing, determine... 1. (2, 1) and \begin{align}2x+y>5\end{align} \begin{align}2(2)+(1)&>5\\ {\color{red}4}+{\color{red}1}&>5\\ {\color{red}5}&>5\end{align} No, five is not greater than five. The point (2, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region. 1. (-5, -1) and \begin{align}y > -2x+8\end{align} \begin{align}(-1)&>-2(-5)+8\\ {\color{red}-1}&>{\color{red}10}+8\\ -1&>{\color{red}18}\end{align} No, negative one is not greater than eighteen. The point (-5, -1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region. 1. (5, -6) and \begin{align}2y< 3x+3\end{align} \begin{align}2(-6)&<3(5)+3\\ {\color{red}-12}&<{\color{red}15}+3\\ -12&<{\color{red}18}\end{align} Yes, negative twelve is less than eighteen. The point (5, -6) does satisfy the inequality. The ordered pair does satisfy the inequality and will lie within the shaded region. Determine the inequality... 1. \begin{align}m=\frac{rise}{run}={\color{red}\frac{-4}{2}}={\color{red}-2}\end{align} The \begin{align}y-\end{align}intercept for the line is (0, -3). The equation of the line in slope-intercept form is \begin{align}y=-2x-3\end{align} The solution set is found in the shaded region that is below the line. The line is a dashed line. Therefore the inequality symbol that must be inserted is less than. The inequality that is modeled by the above graph is: \begin{align}\boxed{y<-2x-3}\end{align} 1. \begin{align}m=\frac{rise}{run}={\color{red}\frac{-1}{2}}={\color{red}-\frac{1}{2}}\end{align} The \begin{align}y-\end{align}intercept for the line is (0, 2). The equation of the line in slope-intercept form is \begin{align}y=-\frac{1}{2}x+2\end{align} The solution set is found in the shaded region that is below the line. The line is a solid line. Therefore the inequality symbol that must be inserted is less than or equal to. The inequality that is modeled by the above graph is: \begin{align}\boxed{y \le -\frac{1}{2}x+2}\end{align} 1. \begin{align}m=\frac{rise}{run}={\color{red}\frac{-1}{2}}={\color{red}-\frac{1}{2}}\end{align} The \begin{align}y-\end{align}intercept for the line is (0, -3). The equation of the line in slope-intercept form is \begin{align}y=-\frac{1}{2}x-3\end{align} The solution set is found in the shaded region that is above the line. The line is a dashed line. Therefore the inequality symbol that must be inserted is greater than. The inequality that is modeled by the above graph is: \begin{align}\boxed{y>-\frac{1}{2}x-3}\end{align} Graph the following linear... \begin{align}4x-2y&>8\\ 4x-2y&>8\\ 4x+4x-2y&>4x+8\\ -2y&>4x+8\\ \frac{-2y}{-2}&>\frac{4x}{-2}+\frac{8}{-2}\\ \frac{\cancel{-2}y}{\cancel{-2}} &> \frac{\overset{{\color{red}-2}}{\cancel{4}x}}{\cancel{-2}}+\frac{\overset{{\color{red}-4}}{\cancel{8}}}{\cancel{-2}}\\ y \ &{\color{red}<-2x-4}\end{align} Test the point (1, 1) to determine the shaded region. \begin{align}4x-2y &> 8\\ 4(1)-2(1) &> 8\\ {\color{red}4}-{\color{red}2}&>8\\ {\color{red}2}&>8 \end{align} No, two is not greater than eight. The point (1, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region. \begin{align}x-y &< -3\\ x-y &< -3\\ x+x-y &< x-3\\ -y &< x-3\\ \frac{-1y}{-1} &< \frac{1x}{-1} - \frac{3}{-1}\\ \frac{\cancel{-1}y}{\cancel{-1}} &< \frac{\overset{{\color{red}-1}}{\cancel{1}}x}{\cancel{-1}} - \frac{\overset{{\color{red}-3}}{\cancel{3}}}{\cancel{-1}}\\ y \ & {\color{red}>-1x+3}\end{align} Test the point (1, 1) to determine the shaded region. \begin{align}x-y &< -3\\ {\color{red}1}-{\color{red}1} &< -3\\ {\color{red}0}&< -3\end{align} No, zero is not less than negative three. The point (1, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region. \begin{align}x+3y &\ge 9\\ x+3y &\ge 9\\ x-x+3y &\ge -x+9\\ 3y &\ge -x+9\\ \frac{3y}{3} &\ge -\frac{1x}{3}+\frac{9}{3}\\ \frac{\cancel{3}y}{\cancel{3}} &\ge - \frac{1}{3}x + \frac{\overset{{\color{red}3}}{\cancel{9}}}{\cancel{3}}\\ y \ &{\color{red}\ge -\frac{1}{3}x+3}\end{align} Test the point (1, 1) to determine the shaded region. \begin{align}x+3y &\ge 9\\ (1)+3(1) &\ge 9\\ {\color{red}1}+{\color{red}3} &\ge 9\\ {\color{red}4} &\ge 9\end{align*} No, four is not greater than or equal to nine. The point (1, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Please to create your own Highlights / Notes Show Hide Details Description Tags: Subjects: Date Created: Jan 16, 2013 Jan 14, 2015 Files can only be attached to the latest version of section Image Detail Sizes: Medium \| Original CK.MAT.ENG.SE.1.Algebra-I---Honors.5.5	6,699	21,583	{"found_math": true, "script_math_tex": 87, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.90625	5	CC-MAIN-2017-13	longest	en	0.875261	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />. # 5.5: Solving Inequalities on the Cartesian Plane. Difficulty Level: At Grade Created by: CK-12. ## Solving a Linear Inequality by Graphing. Introduction. Many graphs that you have created have involved linear equations of the form y=mx+b\begin{align}y=mx+b\end{align}. All points on the line satisfy the linear equation and, thus, form the solution set for the equation. In this lesson you will learn to determine the solution set of an inequality by graphing the inequality on a Cartesian plane.. When an inequality is graphed on a grid, the solution will appear as a shaded area, as opposed to simply a straight line. The shaded area will contain the solution set which consists of all the points which satisfy the inequality.. Objectives. The lesson objectives for Solving Inequalities on the Cartesian Plane are:. • Determining the type of line needed to graph the inequality.. • Testing to determine where the shaded area is located on the graph.. • Shading the correct region to produce the solution set for the inequality.. Introduction. An inequality refers to a statement that shows a relationship between two expressions that are not always equal. The general form of an inequality with two variables is ax+by>c\begin{align}ax+by>c\end{align} or ax+by<c\begin{align}ax+by, where ‘a\begin{align}a\end{align}’ and ‘b\begin{align}b\end{align}’ are the coefficients of the variables and are not both equal to zero. The constant is ‘c\begin{align}c\end{align}’. The inequality symbol (>)\begin{align}(>)\end{align} is read “greater than” and the inequality symbol (<)\begin{align}(<)\end{align} is read “less than”. A hint to use to help distinguish between the two symbols is L-E-S-S points L-E-F-T. However, inequalities can also be written as ax+byc\begin{align}ax+by \ge c\end{align} or ax+byc\begin{align}ax+by \le c\end{align}. The inequality symbol ()\begin{align}(\ge)\end{align} is read “greater than or equal to” and the inequality symbol ()\begin{align}(\le)\end{align} is read “less than or equal to”. The solution set to this type of inequality includes all the points on the line as well as those in the shaded region.. Before we begin, there are a few things to consider when graphing inequalities:. • When an inequality is divided or multiplied by a negative number, the direction of the inequality sign is reversed.. • All inequalities that have the symbol (>)\begin{align}(>)\end{align} or (<)\begin{align}(<)\end{align} are graphed with a dashed line.. • All inequalities that have the symbol ()\begin{align}(\ge)\end{align} or \begin{align}(\le)\end{align} are graphed with a solid line.. • The area or region which contains the ordered pairs that satisfy the inequality is indicated by shading.. Watch This. Guidance. To better understand the things that must be considered when graphing inequalities, the inequality \begin{align}2x-3y < 6\end{align} will be graphed.. \begin{align}2x-3y < 6\end{align}. The first step is to rearrange the inequality in slope-intercept form. This process is the same as it is for linear equations.. \begin{align}& 2x-3y < 6\\ & 2x{\color{red}-2x}-3y < {\color{red}-2x}+6\\ & -3y < -2x+6\\ & \frac{-3y}{{\color{red}-3}} < \frac{-2x}{{\color{red}-3}} + \frac{6}{{\color{red}-3}}\\ & \frac{\cancel{-3}y}{\cancel{-3}} < \frac{-2x}{-3} + \frac{\overset{{\color{red}-2}}{\cancel{6}}}{\cancel{-3}}\\ & \boxed{y \ {\color{red}>} \ \frac{2}{3}x-2}\end{align}. The inequality was divided by negative 3 which caused the inequality sign to reverse its direction.. The graph of the inequality is done the same as it is for a linear equation. In this case the graph will be a dashed or dotted line because the sign is greater than \begin{align}(>)\end{align}.. The above graph represents the graph of the inequality before the solution set region is shaded. To determine whether to shade above the line or below the line, choose a point that is not on the line and test its coordinates in the original inequality. If the coordinates of the point satisfy the inequality, then the area above or below the line, containing the point, will be shaded. If the coordinates of the point do not satisfy the inequality, then the area above or below the line, that does not contain the test point, will be shaded.. The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality.. \begin{align}2x-3y &< 6\\ 2({\color{red}1})-3({\color{red}1}) &< 6 && \text{Substitute} \ (1, 1) \ \text{for} \ x' \ \text{and} \ y' \ \text{of the original inequality.}\\ {\color{red}2}- {\color{red}3} &< 6 && \text{Evaluate the inequality.}\\ {\color{red}-1} &< 6 && \text{Is it true?}\end{align}. Yes, negative one is less than six. The point (1, 1) satisfies the inequality. Therefore, the solution set is all of the area above the line that contains the point (1, 1). The ordered pair that satisfies the inequality will lie within the shaded region.. The solution set for the inequality is the entire shaded region shown in the graph. The dashed or dotted line means that none of the points on the line will satisfy the inequality.. Example A. Graph the inequality \begin{align}3x+4y \le 12\end{align}.. \begin{align}& 3x+4y \le 12 && \text{Write the inequality in slope intercept form } (y=mx+b).\\ & 3x{\color{red}-3x}+4y \le {\color{red}-3x}+12\\ & 4y \le {\color{red}-3x}+12\\ & \frac{4y}{{\color{red}4}} \le \frac{-3x}{{\color{red}4}}+\frac{12}{{\color{red}4}}\\ & \frac{\cancel{4}y}{\cancel{4}} \le \frac{-3x}{4}+\frac{\overset{{\color{red}3}}{\cancel{12}}}{\cancel{4}}\\ & \boxed{y \le -\frac{3}{4}x+3}\end{align}. The above graph represents the graph of the inequality before the solution set region is shaded. The line is a solid line because the inequality symbol is \begin{align}(\le)\end{align}. To determine whether to shade above the line or below the line, choose a point that is not on the line and test its coordinates in the original inequality. If the coordinates of the point satisfy the inequality, then the area above or below the line, containing the point, will be shaded. If the coordinates of the point do not satisfy the inequality, then the area above or below the line, that does not contain the test point, will be shaded.. The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality.. \begin{align}3x+4y &\le 12\\ 3({\color{red}1})+4({\color{red}1}) &\le 12 && \text{Substitute} \ (1, 1) \ \text{for} \ x' \ \text{and} \ y' \ \text{of the original inequality.}\\ {\color{red}3}+{\color{red}4} &\le 12 && \text{Evaluate the inequality.}\\ {\color{red}7} &\le 12 && \text{Is it true?}\end{align}. Yes, seven is less than 12. The point (1, 1) satisfies the inequality. Therefore, the solution set is all of the area below the line that contains the point (1, 1). The ordered pair that satisfies the inequality will lie within the shaded region.. The solution set for the inequality is the entire shaded region shown in the graph. The solid line means that all of the points on the line will satisfy the inequality.. Example B. Graph the inequality \begin{align}2x-4y < -12\end{align}.. \begin{align}& 2x-4y < -12 && \text{Write the inequality in slope intercept form} \ (y=mx+b).\\ & 2x{\color{red}-2x}-4y < {\color{red}-2x}-12\\ & -4y < {\color{red}-2x}-12\\ & \frac{-4y}{{\color{red}-4}} < \frac{-2x}{{\color{red}-4}} -\frac{12}{{\color{red}-4}}\\ & \frac{\cancel{-4}y}{\cancel{-4}}<\frac{-2x}{-4}-\frac{\overset{{\color{red}-3}}{\cancel{12}}}{\cancel{-4}}\\ & \boxed{y \ {\color{red}>} \ \frac{1}{2}x+3}\end{align}. The inequality was divided by negative 4 which caused the inequality sign to reverse its direction.. The above graph represents the graph of the inequality before the solution set region is shaded. The line is a dashed line because the inequality symbol is \begin{align}(<)\end{align}. To determine whether to shade above the line or below the line, choose a point that is not on the line and test its coordinates in the original inequality. If the coordinates of the point satisfy the inequality, then the area above or below the line, containing the point, will be shaded. If the coordinates of the point do not satisfy the inequality, then the area above or below the line, that does not contain the test point, will be shaded.. The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality.. \begin{align}2x-4y &< -12\\ 2({\color{red}1})-4({\color{red}1}) &< -12 && \text{Substitute} \ (1, 1) \ \text{for} \ x' \ \text{and} \ y' \ \text{of the original inequality.}\\ {\color{red}2-4} &< -12 && \text{Evaluate the inequality.}\\ {\color{red}-2} &< -12 && \text{Is it true?}\end{align}. No, negative two is greater than negative twelve. The point (1, 1) does not satisfy the inequality. Therefore, the solution set is all of the area above the line that does not contain the point (1, 1). The ordered pair does not satisfy the inequality and will not lie within the shaded region.. The solution set for the inequality is the entire shaded region shown in the graph. The dashed or dotted line means that none of the points on the line will satisfy the inequality.. Example C. In this example, the graph of the inequality will be given and the task will be to determine the inequality that is modeled by the graph.. For the following, determine the inequality, in slope-intercept form, that is graphed.. This process is the same as determining the equation of the line that is graphed. The next step is then to decide the appropriate inequality symbol to insert.. Begin by determining the slope of the line. The slope of the line is determined by counting 2 units to the right and 3 units upward. The slope of the line for this graph is. \begin{align}m=\frac{rise}{run}={\color{red}\frac{3}{2}}\end{align}. The \begin{align}y-\end{align}intercept for the line is (0, -2). The equation of the line in slope-intercept form is. \begin{align}y=\frac{3}{2}x-2\end{align}. The solution set is found in the shaded region that is above the line. The line is a solid line. Therefore the inequality symbol that must be inserted is greater than or equal to. The inequality that is modeled by the above graph is:. \begin{align}\boxed{y \ge \frac{3}{2}x-2}\end{align}. Vocabulary. Inequality. An inequality is a statement that shows a relationship between two expressions that are not always equal. An inequality is written using one of the following inequality symbols: greater than \begin{align}(>)\end{align}; less than \begin{align}(<)\end{align}; greater than or equal to \begin{align}(\ge)\end{align}; less than or equal to \begin{align}(\le)\end{align}. The solution to an inequality is indicated by a shaded region that contains all the ordered pairs that satisfy the inequality.. Guided Practice. 1. Without graphing, determine if each point is in the shaded region for each inequality.. i) (2, -3) and \begin{align}2y<-3x+1\end{align}. ii) (-3, 5) and \begin{align}-3x > 2y+6\end{align}. 2. Graph the following inequality: \begin{align}5x-3y \ge 15\end{align}. 3. Determine the inequality that models the following graph:. 1. If the point satisfies the inequality, then the point will lie within the shaded region. Substitute the coordinates of the point into the inequality and evaluate the inequality. If the solution is true, then the point is in the shaded area.. i) \begin{align}2y &< -3x+1 && \text{Substitute} \ (2, -3) \ \text{for} \ x \ \text{and} \ y \ \text{in the inequality.}\\ 2({\color{red}-3}) &< -3({\color{red}2})+1 && \text{Evaluate the inequality.}\\ {\color{red}-6} &< {\color{red}-6}+1\\ -6 &< {\color{red}-5} && \text{Is it true?}\end{align}. Yes, negative six is less than negative five. The point (2, -3) satisfies the inequality. The ordered pair will lie within the shaded region.. ii) \begin{align}-3x &> 2y+6 && \text{Substitute} \ (-3, 5) \ \text{for} \ x \ \text{and} \ y \ \text{in the inequality.}\\ -3({\color{red}-3}) &> 2({\color{red}5})+6 && \text{Evaluate the inequality.}\\ {\color{red}9} &> {\color{red}10}+6\\ 9 &> {\color{red}16} && \text{Is it true?}\end{align}. No, nine is not greater than sixteen. The point (-3, 5) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.. 2. \begin{align}& 5x-3y \ge 15 && \text{Write the inequality in slope intercept form} \ (y=mx+b).\\ & 5x{\color{red}-5x}-3y \ge {\color{red}-5x}+15\\ & -3y \ge -5x+15\\ & \frac{-3y}{{\color{red}-3}} \ge \frac{-5x}{{\color{red}-3}} + \frac{15}{{\color{red}-3}}\\ & \frac{\cancel{-3}y}{\cancel{-3}} \ge \frac{5}{3}x + \frac{\overset{{\color{red}-5}}{\cancel{15}}}{\cancel{-3}}\\ & \boxed{y \ {\color{red}\le} \ \frac{5}{3}x-5}\end{align*}. The inequality was divided by negative 3 which caused the inequality sign to reverse its direction.. Is the graph of the inequality shaded correctly?. The point (1, 1) is not on the graphed line.	The point will be tested to determine if the coordinates satisfy the inequality.. \begin{align}5x-3y &\ge 15\\ 5({\color{red}1})-3({\color{red}1}) &\ge 15 && \text{Substitute} \ (1, 1) \ \text{for} \ x' \ \text{and} \ y' \ \text{of the original inequality.}\\ {\color{red}5}{\color{red}-3} &\ge 15 && \text{Evaluate the inequality.}\\ {\color{red}2} &\ge 15 && \text{Is it true?}\end{align}. No, two is not greater than or equal to fifteen. The point (1, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.. 3.. Begin by determining the slope of the line. The slope of the line is determined by counting 2 units to the left and 4 units downward. The slope of the line for this graph is. \begin{align}m=\frac{rise}{run}={\color{red}\frac{-4}{-2}}={\color{red}2}\end{align}. The \begin{align}y-\end{align}intercept for the line is (0, -3). The equation of the line in slope-intercept form is \begin{align}y=2x-3\end{align}. The solution set is found in the shaded region that is below the line. The line is a solid line. Therefore the inequality symbol that must be inserted is less than or equal to. The inequality that is modeled by the above graph is:. \begin{align}\boxed{y \le 2x-3}\end{align}. Summary. In this lesson you have learned to graph a linear inequality in two variables on a Cartesian plane. The solution set was all points in the area that was in the shaded region of the graph. To determine where the shaded area should be with respect to the line, a point, which was not on the line, was tested in the original inequality. If the point made the inequality true, then the area containing the point was shaded. If the point did not make the inequality true, then the tested point did not lie within the shaded region.. You also learned that the inequality symbol determined whether the line was dashed or solid. A dashed line was graphed for all inequalities that had a \begin{align}>\end{align} symbol or a \begin{align}<\end{align} symbol. The dashed line indicated that the points on the line did not satisfy the inequality. A solid line was graphed for all inequalities that had a \begin{align}\ge\end{align} symbol or a \begin{align}\le\end{align}symbol. The solid line indicated that the points on the line did satisfy the inequality.. In the last example, you learned to determine the inequality that was modeled by a given graph. The inequality was determined by calculating the slope and the \begin{align}y-\end{align}intercept of the line in the same way that you would determine the linear equation for the graph of a straight line. The last step was to determine the inequality sign to be inserted.. Problem Set. Without graphing, determine if each point is in the shaded region for each inequality.. 1. (2, 1) and \begin{align}2x+y>5\end{align}. 2. (-1, 3) and \begin{align}2x-4y \le -10\end{align}. 3. (-5, -1) and \begin{align}y > -2x+8\end{align}. 4. (6, 2) and \begin{align}2x+3y \ge -2\end{align}. 5. (5, -6) and \begin{align}2y< 3x+3\end{align}. Determine the inequality that is modeled by each of the following graphs.. Graph the following linear inequalities on a Cartesian plane.. 1. \begin{align}4x-2y>8\end{align}. 2. \begin{align}4y-3x \le -8\end{align}. 3. \begin{align}x-y < -3\end{align}. 4. \begin{align}3x+y > -1\end{align}. 5. \begin{align}x+3y \ge 9\end{align}. Without graphing, determine.... 1. (2, 1) and \begin{align}2x+y>5\end{align}. \begin{align}2(2)+(1)&>5\\ {\color{red}4}+{\color{red}1}&>5\\ {\color{red}5}&>5\end{align}. No, five is not greater than five. The point (2, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.. 1. (-5, -1) and \begin{align}y > -2x+8\end{align}. \begin{align}(-1)&>-2(-5)+8\\ {\color{red}-1}&>{\color{red}10}+8\\ -1&>{\color{red}18}\end{align}. No, negative one is not greater than eighteen. The point (-5, -1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.. 1. (5, -6) and \begin{align}2y< 3x+3\end{align}. \begin{align}2(-6)&<3(5)+3\\ {\color{red}-12}&<{\color{red}15}+3\\ -12&<{\color{red}18}\end{align}. Yes, negative twelve is less than eighteen. The point (5, -6) does satisfy the inequality. The ordered pair does satisfy the inequality and will lie within the shaded region.. Determine the inequality.... 1. \begin{align}m=\frac{rise}{run}={\color{red}\frac{-4}{2}}={\color{red}-2}\end{align} The \begin{align}y-\end{align}intercept for the line is (0, -3). The equation of the line in slope-intercept form is \begin{align}y=-2x-3\end{align} The solution set is found in the shaded region that is below the line. The line is a dashed line. Therefore the inequality symbol that must be inserted is less than. The inequality that is modeled by the above graph is: \begin{align}\boxed{y<-2x-3}\end{align}. 1. \begin{align}m=\frac{rise}{run}={\color{red}\frac{-1}{2}}={\color{red}-\frac{1}{2}}\end{align} The \begin{align}y-\end{align}intercept for the line is (0, 2). The equation of the line in slope-intercept form is \begin{align}y=-\frac{1}{2}x+2\end{align} The solution set is found in the shaded region that is below the line. The line is a solid line. Therefore the inequality symbol that must be inserted is less than or equal to. The inequality that is modeled by the above graph is: \begin{align}\boxed{y \le -\frac{1}{2}x+2}\end{align}. 1. \begin{align}m=\frac{rise}{run}={\color{red}\frac{-1}{2}}={\color{red}-\frac{1}{2}}\end{align} The \begin{align}y-\end{align}intercept for the line is (0, -3). The equation of the line in slope-intercept form is \begin{align}y=-\frac{1}{2}x-3\end{align} The solution set is found in the shaded region that is above the line. The line is a dashed line. Therefore the inequality symbol that must be inserted is greater than. The inequality that is modeled by the above graph is: \begin{align}\boxed{y>-\frac{1}{2}x-3}\end{align}. Graph the following linear.... \begin{align}4x-2y&>8\\ 4x-2y&>8\\ 4x+4x-2y&>4x+8\\ -2y&>4x+8\\ \frac{-2y}{-2}&>\frac{4x}{-2}+\frac{8}{-2}\\ \frac{\cancel{-2}y}{\cancel{-2}} &> \frac{\overset{{\color{red}-2}}{\cancel{4}x}}{\cancel{-2}}+\frac{\overset{{\color{red}-4}}{\cancel{8}}}{\cancel{-2}}\\ y \ &{\color{red}<-2x-4}\end{align}. Test the point (1, 1) to determine the shaded region.. \begin{align}4x-2y &> 8\\ 4(1)-2(1) &> 8\\ {\color{red}4}-{\color{red}2}&>8\\ {\color{red}2}&>8 \end{align}. No, two is not greater than eight. The point (1, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.. \begin{align}x-y &< -3\\ x-y &< -3\\ x+x-y &< x-3\\ -y &< x-3\\ \frac{-1y}{-1} &< \frac{1x}{-1} - \frac{3}{-1}\\ \frac{\cancel{-1}y}{\cancel{-1}} &< \frac{\overset{{\color{red}-1}}{\cancel{1}}x}{\cancel{-1}} - \frac{\overset{{\color{red}-3}}{\cancel{3}}}{\cancel{-1}}\\ y \ & {\color{red}>-1x+3}\end{align}. Test the point (1, 1) to determine the shaded region.. \begin{align}x-y &< -3\\ {\color{red}1}-{\color{red}1} &< -3\\ {\color{red}0}&< -3\end{align}. No, zero is not less than negative three. The point (1, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.. \begin{align}x+3y &\ge 9\\ x+3y &\ge 9\\ x-x+3y &\ge -x+9\\ 3y &\ge -x+9\\ \frac{3y}{3} &\ge -\frac{1x}{3}+\frac{9}{3}\\ \frac{\cancel{3}y}{\cancel{3}} &\ge - \frac{1}{3}x + \frac{\overset{{\color{red}3}}{\cancel{9}}}{\cancel{3}}\\ y \ &{\color{red}\ge -\frac{1}{3}x+3}\end{align}. Test the point (1, 1) to determine the shaded region.. \begin{align}x+3y &\ge 9\\ (1)+3(1) &\ge 9\\ {\color{red}1}+{\color{red}3} &\ge 9\\ {\color{red}4} &\ge 9\end{align}. No, four is not greater than or equal to nine. The point (1, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.. ### Notes/Highlights Having trouble? Report an issue.. Color Highlighted Text Notes. Please to create your own Highlights / Notes. Show Hide Details. Description. Tags:. Subjects:. Date Created:. Jan 16, 2013. Jan 14, 2015. 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https://www.physicsforums.com/threads/math-challenge-november-2019.979872/page-2	1,603,194,786,000,000,000	text/html	crawl-data/CC-MAIN-2020-45/segments/1603107872686.18/warc/CC-MAIN-20201020105000-20201020135000-00671.warc.gz	859,810,773	28,585	# Math Challenge - November 2019 • Challenge • Featured Unfortunately, the sum under your integral sign doesn't converge (for any ##\theta##). I think you lost your factorial term. The sum should have been$$I(\theta)=\int_0^{2\pi} \sum_{m=0}^{\infty}\frac{cos(m\theta)}{m!}d\theta$$I was to hasty with typing latex. I think that sum converges. StoneTemplePython Gold Member 2019 Award The proof as of now is incomplete because it relies on Bonferonni's Inequalities, which is a refinement of inclusion-exclusion. I plan on dropping in a reasonably nice proof of Bonferonni in the near future to complete this. It is too late, I know, but it belatedly occurs to me that I should have just ditched this Bonferonni stuff and instead mimicked the proof for the Bernouli Inequality. for ##a \in (-1,0)## claim: ##\big(1-\vert a \vert\big)^n \leq 1 - n\vert a \vert + \binom{n}{2}a^2## for natural number ##n \geq 2## Base Case: ##n=2## ##\big(1-\vert a \vert\big)^2 = 1-2\vert a \vert + \vert a \vert^2 \leq 1 - 2\vert a \vert + \binom{2}{2}a^2## (i.e. met with equality) Inductive Case: suppose holds for ##n-1##, need to show this implies it holds for n ##\big(1 + a \big)^n ## ##\big(1-\vert a \vert\big)^n ## ##=\big(1-\vert a \vert\big) \cdot \big(1-\vert a \vert\big)^{n-1} ## ##\leq \big(1-\vert a \vert\big)\big(1 - (n-1)\vert a \vert + \binom{n-1}{2}\vert a\vert ^2\big)## (inductive hypothesis, note ##\big(1-\vert a \vert\big)\gt 0##) ##= \big(1 - (n-1)\vert a \vert + \binom{n-1}{2}a^2\big) + \big(-\vert a \vert + (n-1) \vert a \vert^2 -\binom{n-1}{2}\vert a\vert^3\big)## ##= 1 - n\vert a \vert + \big\{\binom{n-1}{2}a^2 + (n-1) a^2\big\} -\binom{n-1}{2}\vert a\vert^3## ##= 1 - n\vert a \vert + \big\{\frac{(n-1)(n-2)}{2}a^2 + \frac{2n-2}{2} a^2\big\} -\binom{n-1}{2}\vert a\vert^3## ##= 1 - n\vert a \vert + \big\{\frac{n^2-3n + 2}{2}a^2 + \frac{2n-2}{2} a^2\big\} -\binom{n-1}{2}\vert a\vert^3## ##= 1 - n\vert a \vert + \big\{\frac{n^2-n}{2}a^2 \big\} -\binom{n-1}{2}\vert a\vert^3## ##= 1 - n\vert a \vert + \binom{n}{2}a^2 -\binom{n-1}{2}\vert a\vert^3## ##\leq 1 - n\vert a \vert + \binom{n}{2}a^2## ##= 1 + n\cdot a + \binom{n}{2}a^2## now for any ##x \in (-\infty, 0)## we can select ##a:= \frac{x}{n}## for all large enough ##n## (i.e. ##n \gt \vert x \vert## ) giving ## \big(1 + \frac{x}{n} \big)^n ## ##=\big(1 + a \big)^n ## ##\leq 1 + n\cdot a + \binom{n}{2}a^2 ## ##= 1 + n\cdot \frac{x}{n} + \binom{n}{2}\frac{x^2}{n^2} ## ##\leq 1 + x + \frac{n^2}{2}\frac{x^2}{n^2} ## ## = 1 + x + \frac{x^2}{2} ## now passing limits gives the result, i.e. ##e^{x} = \lim_{n\to \infty} \big(1 + \frac{x}{n} \big)^n \leq 1 + x + \frac{x^2}{2} ## for ##x \in (-\infty, 0)## QuantumQuest Gold Member I assume that the kid throws every ball with the same speed. Without that assumption, there would be innumerable valid solution - the kid could alternate between 2 speeds for the throw for odd and even numbered throws and achieve a rate of 2 balls per seconds. But with the same speed assumption, every ball must be reaching its maximum height 0.5 seconds after its throw. Gravitational acceleration ##g = 9.8 \ m / s^2##. When ball reaches maximum height, its speed becomes 0. If initial speed of ball is ##s## and speed becomes 0 after 0.5 seconds due to deceleration ##g##, ##0 = s - 0.5 \times g \Rightarrow s = 4.9 \ m /s## Maximum height reached by ball ##D## = Distance traveled by ball in 0.5 seconds from the time of throw. Using the equation for distance as a function of velocity ##v## and time ##t##, and knowing that velocity of the ball ##t## seconds after its throw is ##v = (s - g t)## we get $$D = \int_0^{0.5} v \, dt = \int_0^{0.5} (s - gt) \, dt =\left. st - \frac {gt^2} {2} \right\|_0^{0.5} = 0.5 s - \frac {9.8 \times {0.5}^2} {2} = 1.225 \ m$$ Well done. Just one comment: letter ##s## is usually for distance, so I think that it's best to denote initial velocity as ##\upsilon_0## just to avoid potential looking back and forth from readers. QuantumQuest Gold Member 13. A radio station at point A transmits a signal which is received by the receivers B and C. A listener located at B is listening to the signal via his receiver and after one second hears the signal via receiver C which has a strong loudspeaker. What is the distance between B and C? Answer: The distance between B and C would be approximately equal to the distance traveled by sound in air in 1 second, i.e. around 340 m. To prove this, let us denote by ##d_1##, ##d_2## and ##d_3## the distances between A & B, A & C, and B & C respectively. Let ##T_0## be the time instant at which radio signal was emitted from A, ##T_1## be the time instant at which it was received at A and ##T_2## be the instant at which the retransmitted (as sound) signal from C is heard at B. Let ##v_l## and ##v_s## be the speed of radio waves (which is same as speed of light) and speed of sound respectively. I assume that the signal is emitted at sound from C instantaneously on receiving the signal from A, i.e. there is no time lag between signal reception and sound emission at C. It is given that ##T_2 - T_1 = 1## $$T_1 = T_0 + \frac {d_1} {v_l} \\ T_2 = T_0 + \frac {d_2} {v_l} + \frac {d_3} {v_s} \\ T_2 - T_1 = 1 \Rightarrow \frac {d_2} {v_l} + \frac {d_3} {v_s} - \frac {d_1} {v_l} = 1 \\ \Rightarrow d_3 = v_s + v_s \frac {d_1 - d_2} {v_l}$$ By triangle inequality, ##d_2 \le d_1 + d_3 \Rightarrow 0 \le d_2 \le d_1 + d_3## (lower bound is 0 as d_2 is a length) If ##d_2 = 0##, substituting in the earlier equation gives ##d_3 = v_s + v_s \frac {d_1} {v_l}##, but this is also case where points A and C coincide, so distance between B & C = distance between B & A ##\Rightarrow d_3 = v_s + v_s \frac {d_3} {v_l} \Rightarrow d_3 = \frac {v_s} {1 - \frac {v_s} {v_l}}## If on the other hand, ##d_2 = d_1 + d_3##, the earlier equation becomes $$d_3 = v_s + v_s \frac {- d_3} {v_l} \Rightarrow d_3 = \frac {v_s} {1 + \frac {v_s} {v_l}}$$ Therefore ##d_3 \in (\frac {v_s} {1 + \frac {v_s} {v_l}}, \frac {v_s} {1 - \frac {v_s} {v_l}}) ## Since speed of sound is very small compared to speed of radio waves ##\frac {v_s} {v_l} \approx 0 \Rightarrow## denominators of both lower and upper bounds are almost equal to 1, so the value ##d_3 \approx v_s \approx 340 \ m## Well done. I want to point out here that what is asked from the potential solver of the question is, essentially, to recognize that he / she has to use the speed of sound in order to find the distance asked. So, an answer along these lines: Because radio waves have speed ##\approx 300,000 \frac{km}{sec}##, the signal is received almost simultaneously by ##B## and ##C## although they are at different distances from ##A##. But sound travels from ##C## to ##B## at ##\approx 330 \frac{m}{sec}## (##0^o C##) or ##\approx 340 \frac{m}{sec}## (##20^o C##). So, taking the second speed (as more general), distance from ##B## to ##C## is ##BC \approx 340 \times 1 \enspace m = 340 \enspace m## would be sufficient. Last edited: Infrared Gold Member @Fred Wright Looks good now, but in order to interchange the sum and integral at the end, you should establish uniform convergence (or else use some other criterion). Fortunately this is easy in this case (e.g. use Weierstrass M-test) Consider the function ##f(x) = x^{s/r}## where ##s>r## are non-zero real numbers. We analyze the convexity. Taking the second derivative we find: $$f''(x) = \frac{s}{r}\left(\frac{s}{r} - 1\right)x^{\frac{s}{r} - 2} = \frac{s(s-r)}{r^2}x^{\frac{s}{r}-2}$$ Since the power function is strictly positive for ##x>0##, we find that ##f(x)## is convex for ##s>0## and concave for ##s<0##. We will then use the Jensen's inequality for this function, and weights as they're defined in the exercise. Case 1: ##s>0## $$\left(\sum_{k=1}^n \omega_k x^r_k\right)^{\frac{s}{r}} \leq \sum_{k=1}^n \omega_k (x^r_k)^{\frac{s}{r}} =\sum_{k=1}^n \omega_k x^s_k$$ Taking the ##\frac{1}{s}## - th power of both sides, we obtain the inequality as required. Case 2: ##s<0## If ##s>r## then ##r<0##, as well. Using case 1, we find(##-s<-r##): $$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{-r}} \geq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{-s}}$$ The inequality sign will reverse when we raise to the power of -1, so we get: $$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{r}} \leq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{s}}$$ But now without loss of generality we can apply the same inequality to the set of numbers ##y_k = \frac{1}{x_k}##, which gives us the desired inequality. Since sets ##x_k## and ##y_k## both satisfy the same constraints(that they are positive), this substitution gives equivalent inequality to the one above. This completes the proof. Mentor Consider the function ##f(x) = x^{s/r}## where ##s>r## are non-zero real numbers. We analyze the convexity. Taking the second derivative we find: $$f''(x) = \frac{s}{r}\left(\frac{s}{r} - 1\right)x^{\frac{s}{r} - 2} = \frac{s(s-r)}{r^2}x^{\frac{s}{r}-2}$$ Since the power function is strictly positive for ##x>0##, we find that ##f(x)## is convex for ##s>0## and concave for ##s<0##. We will then use the Jensen's inequality for this function, and weights as they're defined in the exercise. Case 1: ##s>0## $$\left(\sum_{k=1}^n \omega_k x^r_k\right)^{\frac{s}{r}} \leq \sum_{k=1}^n \omega_k (x^r_k)^{\frac{s}{r}} =\sum_{k=1}^n \omega_k x^s_k$$ Taking the ##\frac{1}{s}## - th power of both sides, we obtain the inequality as required. Case 2: ##s<0## If ##s>r## then ##r<0##, as well. Using case 1, we find(##-s<-r##): $$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{-r}} \geq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{-s}}$$ The inequality sign will reverse when we raise to the power of -1, so we get: $$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{r}} \leq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{s}}$$ But now without loss of generality we can apply the same inequality to the set of numbers ##y_k = \frac{1}{x_k}##, which gives us the desired inequality. Since sets ##x_k## and ##y_k## both satisfy the same constraints(that they are positive), this substitution gives equivalent inequality to the one above. This completes the proof. What about the cases ##r<0<s## and ##r\cdot s = 0##? What about the cases ##r<0<s## and ##r\cdot s = 0##? Case ##r<0<s## is covered in case 1, since the negative value of ##r## doesn't change the procedure based on Jensen's inequality, in this case(convexity analysis is the same, and nowhere in the inequality derivation is negative power taken that would reverse the sign of inequality). As for ##r \cdot s = 0##, thank you for reminding me, I'll add it here. Case 3: ##r = 0##, ##s>0##. We look at the definition of the mean for ##r=0##, we find: $$\prod_{k=1}^n x_k^{\omega_k} = \exp\left({\ln{\prod_{k=1}^n x_k^{\omega_k}}}\right)= \exp\left(\sum_{k=1}^n \omega_k \ln{x_k}\right)$$ By Jensen inequality for logarithm function which is concave on the the whole domain, we know that: $$\sum_{k=1}^n \omega_k \ln{x_k} \leq \ln{\left(\sum_{k=1}^n \omega_k x_k \right)}$$ Combining the last two equations we find(exponential function is strictly raising so it doesn't change the inequality): $$\prod_{k=1}^n x_k^{\omega_k} \leq \sum_{k=1}^n \omega_k x_k$$ Now we use the same trick as in case 2. We make a substitution that gives us equivalent inequality: ##x_k \rightarrow y_k = x_k^s##. Taking ##\tfrac{1}{s}## - th power of both sides of this inequality(which doesn't reverse inequality sign since ##s>0## so this power function is strictly raising on positive reals), we find the desired inequality: $$\prod_{k=1}^n x_k^{\omega_k} \leq \left(\sum_{k=1}^n \omega_k x_k^s\right)^{\frac{1}{s}}$$ Case 4: ##s=0##, ##r<0## Looking at case 3 for ##-r>s=0## we find the inequality: $$\prod_{k=1}^n x_k^{\omega_k} \leq \left(\sum_{k=1}^n \omega_k x_k^{-r}\right)^{\frac{1}{-r}}$$ Now raising to the power of -1 reverses the sign of inequality, and then the equivalent substitution ##x_k \rightarrow x_k^{-1}## obtains the desired inequality(the product gets simply raised to the power of 1 in total by these two transformations, while the inequality reverses sign, and the right side changes to the desired form). fresh_42 14. Mr. Smith on a full up flight with 50 passengers on a CRJ 100 had lost his boarding pass. The flight attendant tells him to sit anywhere. All other passengers sit on their booked seats, unless it is already occupied, in which case they randomly choose another seat just like Mr. Smith did. What are the chances that the last passenger gets the seat printed on his boarding pass? Should we assume that Mr. Smith is equally likely to be the 1st, 2nd, 3rd, .., 50th passenger to board the flight, so the probability of his being the ##i^{th}## passenger to board the flight is 1/50? If so, here is an attempt to solve the problem. There could be a much more intuitive and simpler way to solve the problem but this is the best approach I could get to. Let ##P_i## be the probability of last boarding passenger getting the seat printed on his/her boarding pass when Smith is the ##i^{th}## passenger to board the flight. ##P_{50} = 1## because if Smith is the last passenger to board, just one seat would be left and it should be the one originally allotted to him since all other passengers have already occupied their original seats. ##P_{49} = 1/2## because 2 seats are free when Smith boards and the probability of his choosing his allotted seat, and so not displacing the last and 50th passenger, is 1/2 Notice that if Smith is the ##i^{th}## passenger to board and i < 50, the last passenger gets his/her original seat either if Smith picks his (Smith's) own original seat (the "no displaced passenger" case) or if Smith picks the seat of the ##j^{th}## passenger ##j \in {i, (i+1)..., 49}## and the chain of displacements caused does not involve someone between ##i^{th}## and ##50^{th}## passengers occupying the last passenger's seat. If Smith is the 40th passenger to board and randomly chooses the seat of say the 46th passenger, then passengers 41 to 45 (as per order of boarding) occupy their original seats, 46th passenger has to pick a seat randomly and so the chain of displacements starts at 46 and in this case the (conditional) probability of the last passenger getting his/her original seat is the same as it would have been if instead Smith was the 46th passenger to board. ##P_{48} =## Probability of Smith picking his original seat + (Probability of Smith picking the seat of 49th passenger x Probability that 49th passenger does not choose 50th passenger's seat) = ##1/3 + 1/3 \times 1/2 = 1/2## ##P_{47} = ## Probability of Smith picking his original seat + (Probability of Smith picking 48th passenger's seat x ##P_{48}##) + (Probability of Smith picking 49th passenger's seat x ##P_{49}##) = ##1/4 + 1/4 \times 1/2 + 1/4 \times 1/2 = 1/2## It is now easy to show that ##P_{46}, P_{45}, .., P_{2}, P_{1}## are all equal to 1/2. Now, the final unconditional probability of last passenger getting his/her originally allotted seat is ##1/50 \times P_{1} + 1/50 \times P_{2} + .. + 1/50 \times P_{49} + 1/50 \times P_{50} = 49/50 \times 1/2 + 1/50 \times 1 = 51/100 = 0.51## Mentor Should we assume that Mr. Smith is equally likely to be the 1st, 2nd, 3rd, .., 50th passenger to board the flight, so the probability of his being the ##i^{th}## passenger to board the flight is 1/50? If so, here is an attempt to solve the problem. There could be a much more intuitive and simpler way to solve the problem but this is the best approach I could get to. Let ##P_i## be the probability of last boarding passenger getting the seat printed on his/her boarding pass when Smith is the ##i^{th}## passenger to board the flight. ##P_{50} = 1## because if Smith is the last passenger to board, just one seat would be left and it should be the one originally allotted to him since all other passengers have already occupied their original seats. ##P_{49} = 1/2## because 2 seats are free when Smith boards and the probability of his choosing his allotted seat, and so not displacing the last and 50th passenger, is 1/2 Notice that if Smith is the ##i^{th}## passenger to board and i < 50, the last passenger gets his/her original seat either if Smith picks his (Smith's) own original seat (the "no displaced passenger" case) or if Smith picks the seat of the ##j^{th}## passenger ##j \in {i, (i+1)..., 49}## and the chain of displacements caused does not involve someone between ##i^{th}## and ##50^{th}## passengers occupying the last passenger's seat. If Smith is the 40th passenger to board and randomly chooses the seat of say the 46th passenger, then passengers 41 to 45 (as per order of boarding) occupy their original seats, 46th passenger has to pick a seat randomly and so the chain of displacements starts at 46 and in this case the (conditional) probability of the last passenger getting his/her original seat is the same as it would have been if instead Smith was the 46th passenger to board. ##P_{48} =## Probability of Smith picking his original seat + (Probability of Smith picking the seat of 49th passenger x Probability that 49th passenger does not choose 50th passenger's seat) = ##1/3 + 1/3 \times 1/2 = 1/2## ##P_{47} = ## Probability of Smith picking his original seat + (Probability of Smith picking 48th passenger's seat x ##P_{48}##) + (Probability of Smith picking 49th passenger's seat x ##P_{49}##) = ##1/4 + 1/4 \times 1/2 + 1/4 \times 1/2 = 1/2## It is now easy to show that ##P_{46}, P_{45}, .., P_{2}, P_{1}## are all equal to 1/2. Now, the final unconditional probability of last passenger getting his/her originally allotted seat is ##1/50 \times P_{1} + 1/50 \times P_{2} + .. + 1/50 \times P_{49} + 1/50 \times P_{50} = 49/50 \times 1/2 + 1/50 \times 1 = 51/100 = 0.51## Close, but no. You used Bayes' theorem but didn't formalize it, so I have difficulties to pin down the error. My guess is, that e.g. if Mr. Smith boards first, then there aren't 50 choices left, only 49. It can be solved without Bayes, or better with far fewer cases. Close, but no. You used Bayes' theorem but didn't formalize it, so I have difficulties to pin down the error. My guess is, that e.g. if Mr. Smith boards first, then there aren't 50 choices left, only 49. It can be solved without Bayes, or better with far fewer cases. Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem. Since you mentioned that my previous answer was close, I make one more attempt at solving the problem and along the same lines as before. Smith will definitely have the choice of 50 seats if he is the 1st passenger to board, so I guess it should be assumed that Smith is NOT the last passenger to board, either because it is a trivial case and there is no choice to be made, or because the wording of the question says "printed on his boarding pass", so the last passenger has a pass and therefore is not Mr. Smith. If that assumption is made, only the final step in my earlier solution, and the probability of Smith being the ##i^{th}## passenger change. The probability of Smith being the ##i^{th}## passenger would be ##1/49## where ##i \in {1, 2, .., 49}##. The final step therefore becomes: Final unconditional probability of last passenger getting his/her originally allotted seat is ##\sum_{i=1}^{49} 1/49 \times P_{i} = 49 \times 1/49 \times 1/2 = 1/2 = 0.5## Note that ##P_{50} = 1## is still used in deriving ##P_{49}, .., P_{1}## and is a valid value because it represents the situation where the NONE of the passengers before the 50th passenger have occupied a seat that is different from what was originally allotted to them. When I get some free time, I will continue pondering over this problem in the hope of finding a simpler, more elegant way to solve the problem. It requires more creative thinking and that doesn't come naturally to all fresh_42 Mentor Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem. Well, probabilities isn't my favorite subject either. I tend to count wrongly. The solution I have also uses more words than formulas, so yours is not bad, and 0.5 is the correct answer. Math_QED Homework Helper 2019 Award Well, probabilities isn't my favorite subject either. I tend to count wrongly. The solution I have also uses more words than formulas, so yours is not bad, and 0.5 is the correct answer. Yes, this is also what I hate about probability. The analytical side of the subject is nice though. Last edited: Based on the equivalence relation given in the exercise, we find that this ##n##-dimensional projective space is quotient space of ##\mathbb{R}^{n+1}## where points on every ray that is given by a ##(n+1)##-dimensional vector from the origin are equivalent. Therefore, every equivalence class is one ray in ##\mathbb{R}^{n+1}##, and quotient map with respect to which we define the topology, maps every point in ##\mathbb{R}^{n+1}## onto it's equivalence class. It is trivial to see that this map is surjective. We will use this map to induce quotient topology on the projective space. 'Geometrically', this space will then be given by a certain set of points on ##S^n##, the ##n##-dimensional hypersphere of unit radius with the center at the origin. Each point on the hypersphere will be mapped with inverse quotient map into a ray that is defined by the vector in ##\mathbb{R}^{n+1}##. We notice that the points which are connected by the diameter of the hypersphere are also identified, that is, rays of opposite directions are also identified. So, bearing this in mind, we will define geometrical representation of ##\mathbb{R}P^n## in ##\mathbb{R}^{n+1}## with the following convention(it will make the treatment more palpable, at least to me): 1. We take ##(0, 0, \dots , 0, 1)## to be the north pole of ##S^n##(All points/vectors we are mentioning are ##\mathbb{R}^{n+1}## coordinates notation). 2. Then ##\mathbb{R}P^n## is represented by a set of points on ##S^n## with positive coordinate ##0 < x_{n+1} \leq 1##(northern hemisphere), and points ##x_{n+1} = 0## on ##S^n##(equator), for which ##x_1>0##, and points ##x_{n+1}=0 , x_1 = 0## for which ##x_2>0##, etc. inductively, up until the point ##(0, 0, \dots , 0, 1, 0)##. Three-dimensional analog of this would be north hemisphere of a 3-dimensional sphere with equator circle given by points defined by ##\phi \in [0,\pi)##, where ##\phi## is the azimuthal angle on the equator with ##\phi=0## defined to be measured counterclockwise from the point ##(x =1 , y = 0, z=0)##. We will call this geometrical representation of ##\mathbb{R}P^n##, ##S##, and it is homeomorphic to ##\mathbb{R}P^n##. The quotient topology is then induced in a standard way. Open sets in ##R\mathbb{P}^n## are sets who's inverse images under quotient map are open in ##\mathbb{R}^{n+1}##. Let's take ##P## to be the set of all ##n##-dimensional hyperplanes in ##\mathbb{R}^{n+1}## that intersect ##S^n##. Their intersections will be ##(n-1)##-dimensional spheres on the surface of this hypersphere. These ##(n-1)##-dimensional spheres are contained in ##S##, or they're intersecting ##S## or they're disjunct from ##S##. We will consider the first two cases, since the third one isn't interesting to us. In the first case, if we name interiors of those ##(n-1)##-dimensional spheres, ##U_i##, we find that inverse image of ##U_i## is the space bounded by a circular conical surface in ##\mathbb{R}^{n+1}##, whose axis direction is given by the direction of the vector perpendicular to the hyperplane that generated the boundary of ##U_i## by intersecting with ##S##. This set is obviously open in ##\mathbb{R}^{n+1}##, so it should be open in the quotient topology. In the second case, the ##(n-1)##-dimensional hypersphere is intersecting ##S## and the intersection is an ##(n-1)## dimensional 'arc', with ends at the equator(If it's end is on the open-ended part of the equator, the arc is open-ended on that side, while if it's on the part of the equator that is included in ##S##, then that end of the arc is closed(not in the sense of topology but in the sense that it includes the point from the equator)). The equator of ##S^n##(or of ##S##) is mapped by inverse quotient map into a hyperplane passing through the origin, which is open in ##\mathbb{R}^{n+1}##, hence the equator of ##S## is open in quotient topology. However, the interior bounded by the arc and the equator, which we may call ##W_i## is mapped with inverse quotient map into space that is bounded on one side by part of a conical surface that is the image of the arc the same way as in case 1, and the hyperplane that is the image of the equator. Also, it can be that the arc is actually half of the 'great circle' of ##S##, in which case it is mapped into a hyperplane in ##\mathbb{R}^{n+1}## so in that case the inverse image of ##W_i## is the space between two intersecting hyperplanes passing through the origin(the direction of 'between' is given by any vector whose end is mapped into ##W_i##). So this set in the second case is also open in ##\mathbb{R}^{n+1}##, and therefore should be open in the quotient topology. So the union of ##W_i## and the part of the equator that is the boundary of ##W_i##, which we can call ##V_i## is open in ##\mathbb{R}^{n+1}##. We assert that the sets ##U_i## and ##V_i## consist the basis of quotient topology. We need to check the two conditions for the basis: 1. For every ##x \in S## we have a basis element that contains it. If we take an arbitrary ##x \in S##, then the vector given by coordinates of ##x## in ##\mathbb{R}^{n+1}## defines the hyperplane that is perpendicular to it. We choose this hyperplane to intersect ##S^n## at the half the length of this vector, and obtain one of the two intersections mentioned above. Then such an intersection with ##S## defines a basis element ##U_i## or ##V_i## that contains ##x##. The notion of length of this vector is not needed for this claim, since the point of intersection of the chosen hyperplane with the ray is defined with all the coordinates of ##x## halved. 2. If ##x## belongs to the intersection of two basis elements ##B_1## and ##B_2##, then there exists a basis element ##B_3 \subset B_1 \cap B_2## containing ##x##. Take two basis elements containing a chosen point ##x## in ##S##, such that they are not subsets of one another(in which case the condition follows trivially). Every basis element can be defined by the hyperplane that generated it, to which there corresponds a unique point in ##S## whose inverse image is a ray perpendicular to this hyperplane. Choose a hyperplane corresponding to ##x##, such that it is tangent to the boundary of ##B_1##. Name the basis element generated by it ##B'_1##. Choose another hyperplane corresponding to ##x## such that it is tangent to the boundary of ##B_2##. Name the basis element generated by it ##B'_2##. Then both ##B'_1## and ##B'_2## contain ##x##, and either ##B'_1 \subseteq B'2## or ##B'_2 \subseteq B'_1##, since they are generated by two parallel hyperplanes. Without loss of generality, choose the first option(proving the condition in the second option is the same). Since the boundary of ##B'_2## is tangent to the boundary of ##B_2##, it may intersect the boundary ##B_1##. But then since ##B'_1 \subseteq B'_2##, it cannot have a non-empty intersection with ##B_2##. Since it's boundary tangents the boundary of ##B_1##, it's interior is then contained in ##B_1 \cap B_2##. Therefore ##B'_1## is the sought for basis element, by construction. In the second case it would be ##B'_2##. This has proven that our collection of sets ##U_i## and ##V_i## is a basis, and therefore, the quotient topology is generated by it. To prove that ##\mathbb{R}P^n## is Hausdorff, we need to show that two points ##x_1## and ##x_2## of ##S## may have disjunct neighbourhoods. Choose two points ##x_1## and ##x_2## and hyperplanes corresponding to them in such a way, that those hyperplanes intersect outside ##S##, and generate basis elements belonging to the collection ##U_i##. Then those two basis elements ##U_1## and ##U_2## are disjunct trivially. To prove that ##\mathbb{R}P^n## is second-countable, we may parametrize every basis element with the radius of the ##(n-1)##-dimensional sphere that is it's boundary. Then we can choose for the basis of the quotient topology those basis elements that have rational radii. Such a basis is countable, so ##\mathbb{R}P^n## is second-countable. I might have overcomplicated the solution, but I hope it is correct. I've only recently started improving in topology, so this was a nice exercise. Last edited: Math_QED Homework Helper 2019 Award Based on the equivalence relation given in the exercise, we find that this ##n##-dimensional projective space is quotient space of ##\mathbb{R}^{n+1}## where points on every ray that is given by a ##(n+1)##-dimensional vector from the origin are equivalent. Therefore, every equivalence class is one ray in ##\mathbb{R}^{n+1}##, and quotient map with respect to which we define the topology, maps every point in ##\mathbb{R}^{n+1}## onto it's equivalence class. It is trivial to see that this map is surjective. We will use this map to induce quotient topology on the projective space. 'Geometrically', this space will then be given by a certain set of points on ##S^n##, the ##n##-dimensional hypersphere of unit radius with the center at the origin. Each point on the hypersphere will be mapped with inverse quotient map into a ray that is defined by the vector in ##\mathbb{R}^{n+1}##. We notice that the points which are connected by the diameter of the hypersphere are also identified, that is, rays of opposite directions are also identified. So, bearing this in mind, we will define geometrical representation of ##\mathbb{R}P^n## in ##\mathbb{R}^{n+1}## with the following convention(it will make the treatment more palpable, at least to me): 1. We take ##(0, 0, \dots , 0, 1)## to be the north pole of ##S^n##(All points/vectors we are mentioning are ##\mathbb{R}^{n+1}## coordinates notation). 2. Then ##\mathbb{R}P^n## is represented by a set of points on ##S^n## with positive coordinate ##0 < x_{n+1} \leq 1##(northern hemisphere), and points ##x_{n+1} = 0## on ##S^n##(equator), for which ##x_1>0##, and points ##x_{n+1}=0 , x_1 = 0## for which ##x_2>0##, etc. inductively, up until the point ##(0, 0, \dots , 0, 1, 0)##. Three-dimensional analog of this would be north hemisphere of a 3-dimensional sphere with equator circle given by points defined by ##\phi \in [0,\pi)##, where ##\phi## is the azimuthal angle on the equator with ##\phi=0## defined to be measured counterclockwise from the point ##(x =1 , y = 0, z=0)##. We will call this geometrical representation of ##\mathbb{R}P^n##, ##S##, and it is homeomorphic to ##\mathbb{R}P^n##. The quotient topology is then induced in a standard way. Open sets in ##R\mathbb{P}^n## are sets who's inverse images under quotient map are open in ##\mathbb{R}^{n+1}##. Let's take ##P## to be the set of all ##n##-dimensional hyperplanes in ##\mathbb{R}^{n+1}## that intersect ##S^n##. Their intersections will be ##(n-1)##-dimensional spheres on the surface of this hypersphere. These ##(n-1)##-dimensional spheres are contained in ##S##, or they're intersecting ##S## or they're disjunct from ##S##. We will consider the first two cases, since the third one isn't interesting to us. In the first case, if we name interiors of those ##(n-1)##-dimensional spheres, ##U_i##, we find that inverse image of ##U_i## is the space bounded by a circular conical surface in ##\mathbb{R}^{n+1}##, whose axis direction is given by the direction of the vector perpendicular to the hyperplane that generated the boundary of ##U_i## by intersecting with ##S##. This set is obviously open in ##\mathbb{R}^{n+1}##, so it should be open in the quotient topology. In the second case, the ##(n-1)##-dimensional hypersphere is intersecting ##S## and the intersection is an ##(n-1)## dimensional 'arc', with ends at the equator(If it's end is on the open-ended part of the equator, the arc is open-ended on that side, while if it's on the part of the equator that is included in ##S##, then that end of the arc is closed(not in the sense of topology but in the sense that it includes the point from the equator)). The equator of ##S^n##(or of ##S##) is mapped by inverse quotient map into a hyperplane passing through the origin, which is open in ##\mathbb{R}^{n+1}##, hence the equator of ##S## is open in quotient topology. However, the interior bounded by the arc and the equator, which we may call ##W_i## is mapped with inverse quotient map into space that is bounded on one side by part of a conical surface that is the image of the arc the same way as in case 1, and the hyperplane that is the image of the equator. Also, it can be that the arc is actually half of the 'great circle' of ##S##, in which case it is mapped into a hyperplane in ##\mathbb{R}^{n+1}## so in that case the inverse image of ##W_i## is the space between two intersecting hyperplanes passing through the origin(the direction of 'between' is given by any vector whose end is mapped into ##W_i##). So this set in the second case is also open in ##\mathbb{R}^{n+1}##, and therefore should be open in the quotient topology. So the union of ##W_i## and the part of the equator that is the boundary of ##W_i##, which we can call ##V_i## is open in ##\mathbb{R}^{n+1}##. We assert that the sets ##U_i## and ##V_i## consist the basis of quotient topology. We need to check the two conditions for the basis: 1. For every ##x \in S## we have a basis element that contains it. If we take an arbitrary ##x \in S##, then the vector given by coordinates of ##x## in ##\mathbb{R}^{n+1}## defines the hyperplane that is perpendicular to it. We choose this hyperplane to intersect ##S^n## at the half the length of this vector, and obtain one of the two intersections mentioned above. Then such an intersection with ##S## defines a basis element ##U_i## or ##V_i## that contains ##x##. The notion of length of this vector is not needed for this claim, since the point of intersection of the chosen hyperplane with the ray is defined with all the coordinates of ##x## halved. 2. If ##x## belongs to the intersection of two basis elements ##B_1## and ##B_2##, then there exists a basis element ##B_3 \subset B_1 \cap B_2## containing ##x##. Take two basis elements containing a chosen point ##x## in ##S##, such that they are not subsets of one another(in which case the condition follows trivially). Every basis element can be defined by the hyperplane that generated it, to which there corresponds a unique point in ##S## whose inverse image is a ray perpendicular to this hyperplane. Choose a hyperplane corresponding to ##x##, such that it is tangent to the boundary of ##B_1##. Name the basis element generated by it ##B'_1##. Choose another hyperplane corresponding to ##x## such that it is tangent to the boundary of ##B_2##. Name the basis element generated by it ##B'_2##. Then both ##B'_1## and ##B'_2## contain ##x##, and either ##B'_1 \subseteq B'2## or ##B'_2 \subseteq B'_1##, since they are generated by two parallel hyperplanes. Without loss of generality, choose the first option(proving the condition in the second option is the same). Since the boundary of ##B'_2## is tangent to the boundary of ##B_2##, it may intersect the boundary ##B_1##. But then since ##B'_1 \subseteq B'_2##, it cannot have a non-empty intersection with ##B_2##. Since it's boundary tangents the boundary of ##B_1##, it's interior is then contained in ##B_1 \cap B_2##. Therefore ##B'_1## is the sought for basis element, by construction. In the second case it would be ##B'_2##. This has proven that our collection of sets ##U_i## and ##V_i## is a basis, and therefore, the quotient topology is generated by it. To prove that ##\mathbb{R}P^n## is Hausdorff, we need to show that two points ##x_1## and ##x_2## of ##S## may have disjunct neighbourhoods. Choose two points ##x_1## and ##x_2## and hyperplanes corresponding to them in such a way, that those hyperplanes intersect outside ##S##, and generate basis elements belonging to the collection ##U_i##. Then those two basis elements ##U_1## and ##U_2## are disjunct trivially. To prove that ##\mathbb{R}P^n## is second-countable, we may parametrize every basis element with the radius of the ##(n-1)##-dimensional sphere that is it's boundary. Then we can choose for the basis of the quotient topology those basis elements that have rational radii. Such a basis is countable, so ##\mathbb{R}P^n## is second-countable. I might have overcomplicated the solution, but I hope it is correct. I've only recently started improving in topology, so this was a nice exercise. Thanks for the answer! Looking at the projective space as a sphere with antipodal points identified is good intuition. My answer does not use this though but I looked up other answers and this is definitely a way to approach the problem. For starters, it needs a proof that this construct ##S## is homeomorphic to my definition of the projective space. To be quite explicit, describe the sphere with antipodal points identified somewhat more carefully (I guess you will need an equivalence relation for this), the topology you place on it and exhibit an explicit homeomorphism between your construct ##S## and the projective space. Last edited: PeroK Homework Helper Gold Member Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem. Since you mentioned that my previous answer was close, I make one more attempt at solving the problem and along the same lines as before. Smith will definitely have the choice of 50 seats if he is the 1st passenger to board, so I guess it should be assumed that Smith is NOT the last passenger to board, either because it is a trivial case and there is no choice to be made, or because the wording of the question says "printed on his boarding pass", so the last passenger has a pass and therefore is not Mr. Smith. If that assumption is made, only the final step in my earlier solution, and the probability of Smith being the ##i^{th}## passenger change. The probability of Smith being the ##i^{th}## passenger would be ##1/49## where ##i \in {1, 2, .., 49}##. The final step therefore becomes: Final unconditional probability of last passenger getting his/her originally allotted seat is ##\sum_{i=1}^{49} 1/49 \times P_{i} = 49 \times 1/49 \times 1/2 = 1/2 = 0.5## Note that ##P_{50} = 1## is still used in deriving ##P_{49}, .., P_{1}## and is a valid value because it represents the situation where the NONE of the passengers before the 50th passenger have occupied a seat that is different from what was originally allotted to them. When I get some free time, I will continue pondering over this problem in the hope of finding a simpler, more elegant way to solve the problem. It requires more creative thinking and that doesn't come naturally to all If it's not too late: For a plane with any number of seats. Let's call Mr Smith's seat ##S## and the last to board's seat ##X##. When Mr Smith enters the plane, some seats are taken. But not ##S## or ##X##. If Mr Smith sits elsewhere, then one and only one of seats ##S## or ##X## is left for the last passenger. To see this, imagine Mr Smith sits in seat ##Y##. There are no problems until passenger ##Y## boards. If he/she sits in another seat ##Z##, then there are no problems until passenger ##Z## boards. Sooner or later one of these displaced passengers sits in seat ##S## or ##X## and then there are no problems until ##X## boards. As ##S## and ##X## are equally likely to be the one taken, the probability is equal that ##X## has his seat or Mr Smith's left. If Mr Smith sits in his seat, ##S##, then ##X## gets his correct seat. But, if Mr Smith sits in seat ##X##, then ##X## gets Mr Smith's seat. As these happen with equal likelihood (whenever Mr Smith boards), again ##X## gets his seat or Mr Smith's with equal probability in these cases. This assumes, as above, that Mr Smith is not the last passenger. Not anonymous and fresh_42	11,846	40,639	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.609375	4	CC-MAIN-2020-45	longest	en	0.497307	# Math Challenge - November 2019. • Challenge. • Featured. Unfortunately, the sum under your integral sign doesn't converge (for any ##\theta##). I think you lost your factorial term.. The sum should have been$$I(\theta)=\int_0^{2\pi} \sum_{m=0}^{\infty}\frac{cos(m\theta)}{m!}d\theta$$I was to hasty with typing latex. I think that sum converges.. StoneTemplePython. Gold Member. 2019 Award. The proof as of now is incomplete because it relies on Bonferonni's Inequalities, which is a refinement of inclusion-exclusion. I plan on dropping in a reasonably nice proof of Bonferonni in the near future to complete this.. It is too late, I know, but it belatedly occurs to me that I should have just ditched this Bonferonni stuff and instead mimicked the proof for the Bernouli Inequality.. for ##a \in (-1,0)##. claim:. ##\big(1-\vert a \vert\big)^n \leq 1 - n\vert a \vert + \binom{n}{2}a^2##. for natural number ##n \geq 2##. Base Case:. ##n=2##. ##\big(1-\vert a \vert\big)^2 = 1-2\vert a \vert + \vert a \vert^2 \leq 1 - 2\vert a \vert + \binom{2}{2}a^2##. (i.e. met with equality). Inductive Case:. suppose holds for ##n-1##, need to show this implies it holds for n. ##\big(1 + a \big)^n ##. ##\big(1-\vert a \vert\big)^n ##. ##=\big(1-\vert a \vert\big) \cdot \big(1-\vert a \vert\big)^{n-1} ##. ##\leq \big(1-\vert a \vert\big)\big(1 - (n-1)\vert a \vert + \binom{n-1}{2}\vert a\vert ^2\big)## (inductive hypothesis, note ##\big(1-\vert a \vert\big)\gt 0##). ##= \big(1 - (n-1)\vert a \vert + \binom{n-1}{2}a^2\big) + \big(-\vert a \vert + (n-1) \vert a \vert^2 -\binom{n-1}{2}\vert a\vert^3\big)##. ##= 1 - n\vert a \vert + \big\{\binom{n-1}{2}a^2 + (n-1) a^2\big\} -\binom{n-1}{2}\vert a\vert^3##. ##= 1 - n\vert a \vert + \big\{\frac{(n-1)(n-2)}{2}a^2 + \frac{2n-2}{2} a^2\big\} -\binom{n-1}{2}\vert a\vert^3##. ##= 1 - n\vert a \vert + \big\{\frac{n^2-3n + 2}{2}a^2 + \frac{2n-2}{2} a^2\big\} -\binom{n-1}{2}\vert a\vert^3##. ##= 1 - n\vert a \vert + \big\{\frac{n^2-n}{2}a^2 \big\} -\binom{n-1}{2}\vert a\vert^3##. ##= 1 - n\vert a \vert + \binom{n}{2}a^2 -\binom{n-1}{2}\vert a\vert^3##. ##\leq 1 - n\vert a \vert + \binom{n}{2}a^2##. ##= 1 + n\cdot a + \binom{n}{2}a^2##. now for any. ##x \in (-\infty, 0)##. we can select. ##a:= \frac{x}{n}##. for all large enough ##n## (i.e. ##n \gt \vert x \vert## ). giving. ## \big(1 + \frac{x}{n} \big)^n ##. ##=\big(1 + a \big)^n ##. ##\leq 1 + n\cdot a + \binom{n}{2}a^2 ##. ##= 1 + n\cdot \frac{x}{n} + \binom{n}{2}\frac{x^2}{n^2} ##. ##\leq 1 + x + \frac{n^2}{2}\frac{x^2}{n^2} ##. ## = 1 + x + \frac{x^2}{2} ##. now passing limits gives the result, i.e.. ##e^{x} = \lim_{n\to \infty} \big(1 + \frac{x}{n} \big)^n \leq 1 + x + \frac{x^2}{2} ##. for ##x \in (-\infty, 0)##. QuantumQuest. Gold Member. I assume that the kid throws every ball with the same speed. Without that assumption, there would be innumerable valid solution - the kid could alternate between 2 speeds for the throw for odd and even numbered throws and achieve a rate of 2 balls per seconds. But with the same speed assumption, every ball must be reaching its maximum height 0.5 seconds after its throw.. Gravitational acceleration ##g = 9.8 \ m / s^2##. When ball reaches maximum height, its speed becomes 0. If initial speed of ball is ##s## and speed becomes 0 after 0.5 seconds due to deceleration ##g##, ##0 = s - 0.5 \times g \Rightarrow s = 4.9 \ m /s##. Maximum height reached by ball ##D## = Distance traveled by ball in 0.5 seconds from the time of throw. Using the equation for distance as a function of velocity ##v## and time ##t##, and knowing that velocity of the ball ##t## seconds after its throw is ##v = (s - g t)## we get. $$D = \int_0^{0.5} v \, dt = \int_0^{0.5} (s - gt) \, dt =\left. st - \frac {gt^2} {2} \right\|_0^{0.5} = 0.5 s - \frac {9.8 \times {0.5}^2} {2} = 1.225 \ m$$. Well done. Just one comment: letter ##s## is usually for distance, so I think that it's best to denote initial velocity as ##\upsilon_0## just to avoid potential looking back and forth from readers.. QuantumQuest. Gold Member. 13. A radio station at point A transmits a signal which is received by the receivers B and C. A listener located at B is listening to the signal via his receiver and after one second hears the signal via receiver C which has a strong loudspeaker. What is the distance between B and C?. Answer: The distance between B and C would be approximately equal to the distance traveled by sound in air in 1 second, i.e. around 340 m.. To prove this, let us denote by ##d_1##, ##d_2## and ##d_3## the distances between A & B, A & C, and B & C respectively. Let ##T_0## be the time instant at which radio signal was emitted from A, ##T_1## be the time instant at which it was received at A and ##T_2## be the instant at which the retransmitted (as sound) signal from C is heard at B. Let ##v_l## and ##v_s## be the speed of radio waves (which is same as speed of light) and speed of sound respectively. I assume that the signal is emitted at sound from C instantaneously on receiving the signal from A, i.e. there is no time lag between signal reception and sound emission at C. It is given that ##T_2 - T_1 = 1##. $$T_1 = T_0 + \frac {d_1} {v_l} \\ T_2 = T_0 + \frac {d_2} {v_l} + \frac {d_3} {v_s} \\ T_2 - T_1 = 1 \Rightarrow \frac {d_2} {v_l} + \frac {d_3} {v_s} - \frac {d_1} {v_l} = 1 \\ \Rightarrow d_3 = v_s + v_s \frac {d_1 - d_2} {v_l}$$. By triangle inequality, ##d_2 \le d_1 + d_3 \Rightarrow 0 \le d_2 \le d_1 + d_3## (lower bound is 0 as d_2 is a length). If ##d_2 = 0##, substituting in the earlier equation gives ##d_3 = v_s + v_s \frac {d_1} {v_l}##, but this is also case where points A and C coincide, so distance between B & C = distance between B & A ##\Rightarrow d_3 = v_s + v_s \frac {d_3} {v_l} \Rightarrow d_3 = \frac {v_s} {1 - \frac {v_s} {v_l}}##. If on the other hand, ##d_2 = d_1 + d_3##, the earlier equation becomes. $$d_3 = v_s + v_s \frac {- d_3} {v_l} \Rightarrow d_3 = \frac {v_s} {1 + \frac {v_s} {v_l}}$$. Therefore ##d_3 \in (\frac {v_s} {1 + \frac {v_s} {v_l}}, \frac {v_s} {1 - \frac {v_s} {v_l}}) ##. Since speed of sound is very small compared to speed of radio waves ##\frac {v_s} {v_l} \approx 0 \Rightarrow## denominators of both lower and upper bounds are almost equal to 1, so the value ##d_3 \approx v_s \approx 340 \ m##. Well done. I want to point out here that what is asked from the potential solver of the question is, essentially, to recognize that he / she has to use the speed of sound in order to find the distance asked. So, an answer along these lines:. Because radio waves have speed ##\approx 300,000 \frac{km}{sec}##, the signal is received almost simultaneously by ##B## and ##C## although they are at different distances from ##A##. But sound travels from ##C## to ##B## at ##\approx 330 \frac{m}{sec}## (##0^o C##) or ##\approx 340 \frac{m}{sec}## (##20^o C##). So, taking the second speed (as more general), distance from ##B## to ##C## is ##BC \approx 340 \times 1 \enspace m = 340 \enspace m##. would be sufficient.. Last edited:. Infrared. Gold Member. @Fred Wright Looks good now, but in order to interchange the sum and integral at the end, you should establish uniform convergence (or else use some other criterion). Fortunately this is easy in this case (e.g. use Weierstrass M-test). Consider the function ##f(x) = x^{s/r}## where ##s>r## are non-zero real numbers. We analyze the convexity. Taking the second derivative we find:. $$f''(x) = \frac{s}{r}\left(\frac{s}{r} - 1\right)x^{\frac{s}{r} - 2} = \frac{s(s-r)}{r^2}x^{\frac{s}{r}-2}$$. Since the power function is strictly positive for ##x>0##, we find that ##f(x)## is convex for ##s>0## and concave for ##s<0##. We will then use the Jensen's inequality for this function, and weights as they're defined in the exercise.. Case 1: ##s>0##. $$\left(\sum_{k=1}^n \omega_k x^r_k\right)^{\frac{s}{r}} \leq \sum_{k=1}^n \omega_k (x^r_k)^{\frac{s}{r}} =\sum_{k=1}^n \omega_k x^s_k$$. Taking the ##\frac{1}{s}## - th power of both sides, we obtain the inequality as required.. Case 2: ##s<0##. If ##s>r## then ##r<0##, as well. Using case 1, we find(##-s<-r##):. $$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{-r}} \geq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{-s}}$$. The inequality sign will reverse when we raise to the power of -1, so we get:. $$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{r}} \leq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{s}}$$. But now without loss of generality we can apply the same inequality to the set of numbers ##y_k = \frac{1}{x_k}##, which gives us the desired inequality. Since sets ##x_k## and ##y_k## both satisfy the same constraints(that they are positive), this substitution gives equivalent inequality to the one above.. This completes the proof.. Mentor. Consider the function ##f(x) = x^{s/r}## where ##s>r## are non-zero real numbers. We analyze the convexity. Taking the second derivative we find:. $$f''(x) = \frac{s}{r}\left(\frac{s}{r} - 1\right)x^{\frac{s}{r} - 2} = \frac{s(s-r)}{r^2}x^{\frac{s}{r}-2}$$. Since the power function is strictly positive for ##x>0##, we find that ##f(x)## is convex for ##s>0## and concave for ##s<0##. We will then use the Jensen's inequality for this function, and weights as they're defined in the exercise.. Case 1: ##s>0##. $$\left(\sum_{k=1}^n \omega_k x^r_k\right)^{\frac{s}{r}} \leq \sum_{k=1}^n \omega_k (x^r_k)^{\frac{s}{r}} =\sum_{k=1}^n \omega_k x^s_k$$. Taking the ##\frac{1}{s}## - th power of both sides, we obtain the inequality as required.. Case 2: ##s<0##. If ##s>r## then ##r<0##, as well. Using case 1, we find(##-s<-r##):. $$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{-r}} \geq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{-s}}$$. The inequality sign will reverse when we raise to the power of -1, so we get:. $$\left(\sum_{k=1}^n \omega_k x^{-r}_k\right)^{\frac{1}{r}} \leq \left(\sum_{k=1}^n \omega_k x^{-s}_k\right)^{\frac{1}{s}}$$. But now without loss of generality we can apply the same inequality to the set of numbers ##y_k = \frac{1}{x_k}##, which gives us the desired inequality. Since sets ##x_k## and ##y_k## both satisfy the same constraints(that they are positive), this substitution gives equivalent inequality to the one above.. This completes the proof.. What about the cases ##r<0<s## and ##r\cdot s = 0##?. What about the cases ##r<0<s## and ##r\cdot s = 0##?. Case ##r<0<s## is covered in case 1, since the negative value of ##r## doesn't change the procedure based on Jensen's inequality, in this case(convexity analysis is the same, and nowhere in the inequality derivation is negative power taken that would reverse the sign of inequality).. As for ##r \cdot s = 0##, thank you for reminding me, I'll add it here.. Case 3: ##r = 0##, ##s>0##.. We look at the definition of the mean for ##r=0##, we find:. $$\prod_{k=1}^n x_k^{\omega_k} = \exp\left({\ln{\prod_{k=1}^n x_k^{\omega_k}}}\right)= \exp\left(\sum_{k=1}^n \omega_k \ln{x_k}\right)$$. By Jensen inequality for logarithm function which is concave on the the whole domain, we know that:. $$\sum_{k=1}^n \omega_k \ln{x_k} \leq \ln{\left(\sum_{k=1}^n \omega_k x_k \right)}$$. Combining the last two equations we find(exponential function is strictly raising so it doesn't change the inequality):. $$\prod_{k=1}^n x_k^{\omega_k} \leq \sum_{k=1}^n \omega_k x_k$$. Now we use the same trick as in case 2. We make a substitution that gives us equivalent inequality: ##x_k \rightarrow y_k = x_k^s##. Taking ##\tfrac{1}{s}## - th power of both sides of this inequality(which doesn't reverse inequality sign since ##s>0## so this power function is strictly raising on positive reals), we find the desired inequality:. $$\prod_{k=1}^n x_k^{\omega_k} \leq \left(\sum_{k=1}^n \omega_k x_k^s\right)^{\frac{1}{s}}$$. Case 4: ##s=0##, ##r<0##. Looking at case 3 for ##-r>s=0## we find the inequality:. $$\prod_{k=1}^n x_k^{\omega_k} \leq \left(\sum_{k=1}^n \omega_k x_k^{-r}\right)^{\frac{1}{-r}}$$. Now raising to the power of -1 reverses the sign of inequality, and then the equivalent substitution ##x_k \rightarrow x_k^{-1}## obtains the desired inequality(the product gets simply raised to the power of 1 in total by these two transformations, while the inequality reverses sign, and the right side changes to the desired form).. fresh_42. 14. Mr. Smith on a full up flight with 50 passengers on a CRJ 100 had lost his boarding pass. The flight attendant tells him to sit anywhere. All other passengers sit on their booked seats, unless it is already occupied, in which case they randomly choose another seat just like Mr. Smith did. What are the chances that the last passenger gets the seat printed on his boarding pass?. Should we assume that Mr. Smith is equally likely to be the 1st, 2nd, 3rd, .., 50th passenger to board the flight, so the probability of his being the ##i^{th}## passenger to board the flight is 1/50? If so, here is an attempt to solve the problem. There could be a much more intuitive and simpler way to solve the problem but this is the best approach I could get to.. Let ##P_i## be the probability of last boarding passenger getting the seat printed on his/her boarding pass when Smith is the ##i^{th}## passenger to board the flight.. ##P_{50} = 1## because if Smith is the last passenger to board, just one seat would be left and it should be the one originally allotted to him since all other passengers have already occupied their original seats.. ##P_{49} = 1/2## because 2 seats are free when Smith boards and the probability of his choosing his allotted seat, and so not displacing the last and 50th passenger, is 1/2. Notice that if Smith is the ##i^{th}## passenger to board and i < 50, the last passenger gets his/her original seat either if Smith picks his (Smith's) own original seat (the "no displaced passenger" case) or if Smith picks the seat of the ##j^{th}## passenger ##j \in {i, (i+1)..., 49}## and the chain of displacements caused does not involve someone between ##i^{th}## and ##50^{th}## passengers occupying the last passenger's seat. If Smith is the 40th passenger to board and randomly chooses the seat of say the 46th passenger, then passengers 41 to 45 (as per order of boarding) occupy their original seats, 46th passenger has to pick a seat randomly and so the chain of displacements starts at 46 and in this case the (conditional) probability of the last passenger getting his/her original seat is the same as it would have been if instead Smith was the 46th passenger to board.. ##P_{48} =## Probability of Smith picking his original seat + (Probability of Smith picking the seat of 49th passenger x Probability that 49th passenger does not choose 50th passenger's seat) = ##1/3 + 1/3 \times 1/2 = 1/2##. ##P_{47} = ## Probability of Smith picking his original seat + (Probability of Smith picking 48th passenger's seat x ##P_{48}##) + (Probability of Smith picking 49th passenger's seat x ##P_{49}##) = ##1/4 + 1/4 \times 1/2 + 1/4 \times 1/2 = 1/2##. It is now easy to show that ##P_{46}, P_{45}, .., P_{2}, P_{1}## are all equal to 1/2.. Now, the final unconditional probability of last passenger getting his/her originally allotted seat is ##1/50 \times P_{1} + 1/50 \times P_{2} + .. + 1/50 \times P_{49} + 1/50 \times P_{50} = 49/50 \times 1/2 + 1/50 \times 1 = 51/100 = 0.51##. Mentor. Should we assume that Mr.	Smith is equally likely to be the 1st, 2nd, 3rd, .., 50th passenger to board the flight, so the probability of his being the ##i^{th}## passenger to board the flight is 1/50? If so, here is an attempt to solve the problem. There could be a much more intuitive and simpler way to solve the problem but this is the best approach I could get to.. Let ##P_i## be the probability of last boarding passenger getting the seat printed on his/her boarding pass when Smith is the ##i^{th}## passenger to board the flight.. ##P_{50} = 1## because if Smith is the last passenger to board, just one seat would be left and it should be the one originally allotted to him since all other passengers have already occupied their original seats.. ##P_{49} = 1/2## because 2 seats are free when Smith boards and the probability of his choosing his allotted seat, and so not displacing the last and 50th passenger, is 1/2. Notice that if Smith is the ##i^{th}## passenger to board and i < 50, the last passenger gets his/her original seat either if Smith picks his (Smith's) own original seat (the "no displaced passenger" case) or if Smith picks the seat of the ##j^{th}## passenger ##j \in {i, (i+1)..., 49}## and the chain of displacements caused does not involve someone between ##i^{th}## and ##50^{th}## passengers occupying the last passenger's seat. If Smith is the 40th passenger to board and randomly chooses the seat of say the 46th passenger, then passengers 41 to 45 (as per order of boarding) occupy their original seats, 46th passenger has to pick a seat randomly and so the chain of displacements starts at 46 and in this case the (conditional) probability of the last passenger getting his/her original seat is the same as it would have been if instead Smith was the 46th passenger to board.. ##P_{48} =## Probability of Smith picking his original seat + (Probability of Smith picking the seat of 49th passenger x Probability that 49th passenger does not choose 50th passenger's seat) = ##1/3 + 1/3 \times 1/2 = 1/2##. ##P_{47} = ## Probability of Smith picking his original seat + (Probability of Smith picking 48th passenger's seat x ##P_{48}##) + (Probability of Smith picking 49th passenger's seat x ##P_{49}##) = ##1/4 + 1/4 \times 1/2 + 1/4 \times 1/2 = 1/2##. It is now easy to show that ##P_{46}, P_{45}, .., P_{2}, P_{1}## are all equal to 1/2.. Now, the final unconditional probability of last passenger getting his/her originally allotted seat is ##1/50 \times P_{1} + 1/50 \times P_{2} + .. + 1/50 \times P_{49} + 1/50 \times P_{50} = 49/50 \times 1/2 + 1/50 \times 1 = 51/100 = 0.51##. Close, but no. You used Bayes' theorem but didn't formalize it, so I have difficulties to pin down the error. My guess is, that e.g. if Mr. Smith boards first, then there aren't 50 choices left, only 49. It can be solved without Bayes, or better with far fewer cases.. Close, but no. You used Bayes' theorem but didn't formalize it, so I have difficulties to pin down the error. My guess is, that e.g. if Mr. Smith boards first, then there aren't 50 choices left, only 49. It can be solved without Bayes, or better with far fewer cases.. Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem. Since you mentioned that my previous answer was close, I make one more attempt at solving the problem and along the same lines as before. Smith will definitely have the choice of 50 seats if he is the 1st passenger to board, so I guess it should be assumed that Smith is NOT the last passenger to board, either because it is a trivial case and there is no choice to be made, or because the wording of the question says "printed on his boarding pass", so the last passenger has a pass and therefore is not Mr. Smith. If that assumption is made, only the final step in my earlier solution, and the probability of Smith being the ##i^{th}## passenger change. The probability of Smith being the ##i^{th}## passenger would be ##1/49## where ##i \in {1, 2, .., 49}##. The final step therefore becomes:. Final unconditional probability of last passenger getting his/her originally allotted seat is ##\sum_{i=1}^{49} 1/49 \times P_{i} = 49 \times 1/49 \times 1/2 = 1/2 = 0.5##. Note that ##P_{50} = 1## is still used in deriving ##P_{49}, .., P_{1}## and is a valid value because it represents the situation where the NONE of the passengers before the 50th passenger have occupied a seat that is different from what was originally allotted to them.. When I get some free time, I will continue pondering over this problem in the hope of finding a simpler, more elegant way to solve the problem. It requires more creative thinking and that doesn't come naturally to all. fresh_42. Mentor. Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem.. Well, probabilities isn't my favorite subject either. I tend to count wrongly. The solution I have also uses more words than formulas, so yours is not bad, and 0.5 is the correct answer.. Math_QED. Homework Helper. 2019 Award. Well, probabilities isn't my favorite subject either. I tend to count wrongly. The solution I have also uses more words than formulas, so yours is not bad, and 0.5 is the correct answer.. Yes, this is also what I hate about probability. The analytical side of the subject is nice though.. Last edited:. Based on the equivalence relation given in the exercise, we find that this ##n##-dimensional projective space is quotient space of ##\mathbb{R}^{n+1}## where points on every ray that is given by a ##(n+1)##-dimensional vector from the origin are equivalent. Therefore, every equivalence class is one ray in ##\mathbb{R}^{n+1}##, and quotient map with respect to which we define the topology, maps every point in ##\mathbb{R}^{n+1}## onto it's equivalence class. It is trivial to see that this map is surjective. We will use this map to induce quotient topology on the projective space. 'Geometrically', this space will then be given by a certain set of points on ##S^n##, the ##n##-dimensional hypersphere of unit radius with the center at the origin. Each point on the hypersphere will be mapped with inverse quotient map into a ray that is defined by the vector in ##\mathbb{R}^{n+1}##. We notice that the points which are connected by the diameter of the hypersphere are also identified, that is, rays of opposite directions are also identified. So, bearing this in mind, we will define geometrical representation of ##\mathbb{R}P^n## in ##\mathbb{R}^{n+1}## with the following convention(it will make the treatment more palpable, at least to me):. 1. We take ##(0, 0, \dots , 0, 1)## to be the north pole of ##S^n##(All points/vectors we are mentioning are ##\mathbb{R}^{n+1}## coordinates notation).. 2. Then ##\mathbb{R}P^n## is represented by a set of points on ##S^n## with positive coordinate ##0 < x_{n+1} \leq 1##(northern hemisphere), and points ##x_{n+1} = 0## on ##S^n##(equator), for which ##x_1>0##, and points ##x_{n+1}=0 , x_1 = 0## for which ##x_2>0##, etc. inductively, up until the point ##(0, 0, \dots , 0, 1, 0)##.. Three-dimensional analog of this would be north hemisphere of a 3-dimensional sphere with equator circle given by points defined by ##\phi \in [0,\pi)##, where ##\phi## is the azimuthal angle on the equator with ##\phi=0## defined to be measured counterclockwise from the point ##(x =1 , y = 0, z=0)##.. We will call this geometrical representation of ##\mathbb{R}P^n##, ##S##, and it is homeomorphic to ##\mathbb{R}P^n##.. The quotient topology is then induced in a standard way. Open sets in ##R\mathbb{P}^n## are sets who's inverse images under quotient map are open in ##\mathbb{R}^{n+1}##. Let's take ##P## to be the set of all ##n##-dimensional hyperplanes in ##\mathbb{R}^{n+1}## that intersect ##S^n##. Their intersections will be ##(n-1)##-dimensional spheres on the surface of this hypersphere. These ##(n-1)##-dimensional spheres are contained in ##S##, or they're intersecting ##S## or they're disjunct from ##S##. We will consider the first two cases, since the third one isn't interesting to us.. In the first case, if we name interiors of those ##(n-1)##-dimensional spheres, ##U_i##, we find that inverse image of ##U_i## is the space bounded by a circular conical surface in ##\mathbb{R}^{n+1}##, whose axis direction is given by the direction of the vector perpendicular to the hyperplane that generated the boundary of ##U_i## by intersecting with ##S##. This set is obviously open in ##\mathbb{R}^{n+1}##, so it should be open in the quotient topology.. In the second case, the ##(n-1)##-dimensional hypersphere is intersecting ##S## and the intersection is an ##(n-1)## dimensional 'arc', with ends at the equator(If it's end is on the open-ended part of the equator, the arc is open-ended on that side, while if it's on the part of the equator that is included in ##S##, then that end of the arc is closed(not in the sense of topology but in the sense that it includes the point from the equator)). The equator of ##S^n##(or of ##S##) is mapped by inverse quotient map into a hyperplane passing through the origin, which is open in ##\mathbb{R}^{n+1}##, hence the equator of ##S## is open in quotient topology. However, the interior bounded by the arc and the equator, which we may call ##W_i## is mapped with inverse quotient map into space that is bounded on one side by part of a conical surface that is the image of the arc the same way as in case 1, and the hyperplane that is the image of the equator.. Also, it can be that the arc is actually half of the 'great circle' of ##S##, in which case it is mapped into a hyperplane in ##\mathbb{R}^{n+1}## so in that case the inverse image of ##W_i## is the space between two intersecting hyperplanes passing through the origin(the direction of 'between' is given by any vector whose end is mapped into ##W_i##). So this set in the second case is also open in ##\mathbb{R}^{n+1}##, and therefore should be open in the quotient topology. So the union of ##W_i## and the part of the equator that is the boundary of ##W_i##, which we can call ##V_i## is open in ##\mathbb{R}^{n+1}##. We assert that the sets ##U_i## and ##V_i## consist the basis of quotient topology.. We need to check the two conditions for the basis:. 1. For every ##x \in S## we have a basis element that contains it.. If we take an arbitrary ##x \in S##, then the vector given by coordinates of ##x## in ##\mathbb{R}^{n+1}## defines the hyperplane that is perpendicular to it. We choose this hyperplane to intersect ##S^n## at the half the length of this vector, and obtain one of the two intersections mentioned above. Then such an intersection with ##S## defines a basis element ##U_i## or ##V_i## that contains ##x##. The notion of length of this vector is not needed for this claim, since the point of intersection of the chosen hyperplane with the ray is defined with all the coordinates of ##x## halved.. 2. If ##x## belongs to the intersection of two basis elements ##B_1## and ##B_2##, then there exists a basis element ##B_3 \subset B_1 \cap B_2## containing ##x##.. Take two basis elements containing a chosen point ##x## in ##S##, such that they are not subsets of one another(in which case the condition follows trivially). Every basis element can be defined by the hyperplane that generated it, to which there corresponds a unique point in ##S## whose inverse image is a ray perpendicular to this hyperplane. Choose a hyperplane corresponding to ##x##, such that it is tangent to the boundary of ##B_1##. Name the basis element generated by it ##B'_1##. Choose another hyperplane corresponding to ##x## such that it is tangent to the boundary of ##B_2##. Name the basis element generated by it ##B'_2##. Then both ##B'_1## and ##B'_2## contain ##x##, and either ##B'_1 \subseteq B'2## or ##B'_2 \subseteq B'_1##, since they are generated by two parallel hyperplanes. Without loss of generality, choose the first option(proving the condition in the second option is the same). Since the boundary of ##B'_2## is tangent to the boundary of ##B_2##, it may intersect the boundary ##B_1##. But then since ##B'_1 \subseteq B'_2##, it cannot have a non-empty intersection with ##B_2##. Since it's boundary tangents the boundary of ##B_1##, it's interior is then contained in ##B_1 \cap B_2##. Therefore ##B'_1## is the sought for basis element, by construction. In the second case it would be ##B'_2##.. This has proven that our collection of sets ##U_i## and ##V_i## is a basis, and therefore, the quotient topology is generated by it.. To prove that ##\mathbb{R}P^n## is Hausdorff, we need to show that two points ##x_1## and ##x_2## of ##S## may have disjunct neighbourhoods.. Choose two points ##x_1## and ##x_2## and hyperplanes corresponding to them in such a way, that those hyperplanes intersect outside ##S##, and generate basis elements belonging to the collection ##U_i##. Then those two basis elements ##U_1## and ##U_2## are disjunct trivially.. To prove that ##\mathbb{R}P^n## is second-countable, we may parametrize every basis element with the radius of the ##(n-1)##-dimensional sphere that is it's boundary. Then we can choose for the basis of the quotient topology those basis elements that have rational radii. Such a basis is countable, so ##\mathbb{R}P^n## is second-countable.. I might have overcomplicated the solution, but I hope it is correct. I've only recently started improving in topology, so this was a nice exercise.. Last edited:. Math_QED. Homework Helper. 2019 Award. Based on the equivalence relation given in the exercise, we find that this ##n##-dimensional projective space is quotient space of ##\mathbb{R}^{n+1}## where points on every ray that is given by a ##(n+1)##-dimensional vector from the origin are equivalent. Therefore, every equivalence class is one ray in ##\mathbb{R}^{n+1}##, and quotient map with respect to which we define the topology, maps every point in ##\mathbb{R}^{n+1}## onto it's equivalence class. It is trivial to see that this map is surjective. We will use this map to induce quotient topology on the projective space. 'Geometrically', this space will then be given by a certain set of points on ##S^n##, the ##n##-dimensional hypersphere of unit radius with the center at the origin. Each point on the hypersphere will be mapped with inverse quotient map into a ray that is defined by the vector in ##\mathbb{R}^{n+1}##. We notice that the points which are connected by the diameter of the hypersphere are also identified, that is, rays of opposite directions are also identified. So, bearing this in mind, we will define geometrical representation of ##\mathbb{R}P^n## in ##\mathbb{R}^{n+1}## with the following convention(it will make the treatment more palpable, at least to me):. 1. We take ##(0, 0, \dots , 0, 1)## to be the north pole of ##S^n##(All points/vectors we are mentioning are ##\mathbb{R}^{n+1}## coordinates notation).. 2. Then ##\mathbb{R}P^n## is represented by a set of points on ##S^n## with positive coordinate ##0 < x_{n+1} \leq 1##(northern hemisphere), and points ##x_{n+1} = 0## on ##S^n##(equator), for which ##x_1>0##, and points ##x_{n+1}=0 , x_1 = 0## for which ##x_2>0##, etc. inductively, up until the point ##(0, 0, \dots , 0, 1, 0)##.. Three-dimensional analog of this would be north hemisphere of a 3-dimensional sphere with equator circle given by points defined by ##\phi \in [0,\pi)##, where ##\phi## is the azimuthal angle on the equator with ##\phi=0## defined to be measured counterclockwise from the point ##(x =1 , y = 0, z=0)##.. We will call this geometrical representation of ##\mathbb{R}P^n##, ##S##, and it is homeomorphic to ##\mathbb{R}P^n##.. The quotient topology is then induced in a standard way. Open sets in ##R\mathbb{P}^n## are sets who's inverse images under quotient map are open in ##\mathbb{R}^{n+1}##. Let's take ##P## to be the set of all ##n##-dimensional hyperplanes in ##\mathbb{R}^{n+1}## that intersect ##S^n##. Their intersections will be ##(n-1)##-dimensional spheres on the surface of this hypersphere. These ##(n-1)##-dimensional spheres are contained in ##S##, or they're intersecting ##S## or they're disjunct from ##S##. We will consider the first two cases, since the third one isn't interesting to us.. In the first case, if we name interiors of those ##(n-1)##-dimensional spheres, ##U_i##, we find that inverse image of ##U_i## is the space bounded by a circular conical surface in ##\mathbb{R}^{n+1}##, whose axis direction is given by the direction of the vector perpendicular to the hyperplane that generated the boundary of ##U_i## by intersecting with ##S##. This set is obviously open in ##\mathbb{R}^{n+1}##, so it should be open in the quotient topology.. In the second case, the ##(n-1)##-dimensional hypersphere is intersecting ##S## and the intersection is an ##(n-1)## dimensional 'arc', with ends at the equator(If it's end is on the open-ended part of the equator, the arc is open-ended on that side, while if it's on the part of the equator that is included in ##S##, then that end of the arc is closed(not in the sense of topology but in the sense that it includes the point from the equator)). The equator of ##S^n##(or of ##S##) is mapped by inverse quotient map into a hyperplane passing through the origin, which is open in ##\mathbb{R}^{n+1}##, hence the equator of ##S## is open in quotient topology. However, the interior bounded by the arc and the equator, which we may call ##W_i## is mapped with inverse quotient map into space that is bounded on one side by part of a conical surface that is the image of the arc the same way as in case 1, and the hyperplane that is the image of the equator.. Also, it can be that the arc is actually half of the 'great circle' of ##S##, in which case it is mapped into a hyperplane in ##\mathbb{R}^{n+1}## so in that case the inverse image of ##W_i## is the space between two intersecting hyperplanes passing through the origin(the direction of 'between' is given by any vector whose end is mapped into ##W_i##). So this set in the second case is also open in ##\mathbb{R}^{n+1}##, and therefore should be open in the quotient topology. So the union of ##W_i## and the part of the equator that is the boundary of ##W_i##, which we can call ##V_i## is open in ##\mathbb{R}^{n+1}##. We assert that the sets ##U_i## and ##V_i## consist the basis of quotient topology.. We need to check the two conditions for the basis:. 1. For every ##x \in S## we have a basis element that contains it.. If we take an arbitrary ##x \in S##, then the vector given by coordinates of ##x## in ##\mathbb{R}^{n+1}## defines the hyperplane that is perpendicular to it. We choose this hyperplane to intersect ##S^n## at the half the length of this vector, and obtain one of the two intersections mentioned above. Then such an intersection with ##S## defines a basis element ##U_i## or ##V_i## that contains ##x##. The notion of length of this vector is not needed for this claim, since the point of intersection of the chosen hyperplane with the ray is defined with all the coordinates of ##x## halved.. 2. If ##x## belongs to the intersection of two basis elements ##B_1## and ##B_2##, then there exists a basis element ##B_3 \subset B_1 \cap B_2## containing ##x##.. Take two basis elements containing a chosen point ##x## in ##S##, such that they are not subsets of one another(in which case the condition follows trivially). Every basis element can be defined by the hyperplane that generated it, to which there corresponds a unique point in ##S## whose inverse image is a ray perpendicular to this hyperplane. Choose a hyperplane corresponding to ##x##, such that it is tangent to the boundary of ##B_1##. Name the basis element generated by it ##B'_1##. Choose another hyperplane corresponding to ##x## such that it is tangent to the boundary of ##B_2##. Name the basis element generated by it ##B'_2##. Then both ##B'_1## and ##B'_2## contain ##x##, and either ##B'_1 \subseteq B'2## or ##B'_2 \subseteq B'_1##, since they are generated by two parallel hyperplanes. Without loss of generality, choose the first option(proving the condition in the second option is the same). Since the boundary of ##B'_2## is tangent to the boundary of ##B_2##, it may intersect the boundary ##B_1##. But then since ##B'_1 \subseteq B'_2##, it cannot have a non-empty intersection with ##B_2##. Since it's boundary tangents the boundary of ##B_1##, it's interior is then contained in ##B_1 \cap B_2##. Therefore ##B'_1## is the sought for basis element, by construction. In the second case it would be ##B'_2##.. This has proven that our collection of sets ##U_i## and ##V_i## is a basis, and therefore, the quotient topology is generated by it.. To prove that ##\mathbb{R}P^n## is Hausdorff, we need to show that two points ##x_1## and ##x_2## of ##S## may have disjunct neighbourhoods.. Choose two points ##x_1## and ##x_2## and hyperplanes corresponding to them in such a way, that those hyperplanes intersect outside ##S##, and generate basis elements belonging to the collection ##U_i##. Then those two basis elements ##U_1## and ##U_2## are disjunct trivially.. To prove that ##\mathbb{R}P^n## is second-countable, we may parametrize every basis element with the radius of the ##(n-1)##-dimensional sphere that is it's boundary. Then we can choose for the basis of the quotient topology those basis elements that have rational radii. Such a basis is countable, so ##\mathbb{R}P^n## is second-countable.. I might have overcomplicated the solution, but I hope it is correct. I've only recently started improving in topology, so this was a nice exercise.. Thanks for the answer! Looking at the projective space as a sphere with antipodal points identified is good intuition. My answer does not use this though but I looked up other answers and this is definitely a way to approach the problem.. For starters, it needs a proof that this construct ##S## is homeomorphic to my definition of the projective space.. To be quite explicit, describe the sphere with antipodal points identified somewhat more carefully (I guess you will need an equivalence relation for this), the topology you place on it and exhibit an explicit homeomorphism between your construct ##S## and the projective space.. Last edited:. PeroK. Homework Helper. Gold Member. Thanks fresh_42, for reviewing my answer! I am not as strong on intuitive thinking as other people on this forum and will need to think a lot more to find a simpler approach that doesn't use Bayes' theorem. Since you mentioned that my previous answer was close, I make one more attempt at solving the problem and along the same lines as before. Smith will definitely have the choice of 50 seats if he is the 1st passenger to board, so I guess it should be assumed that Smith is NOT the last passenger to board, either because it is a trivial case and there is no choice to be made, or because the wording of the question says "printed on his boarding pass", so the last passenger has a pass and therefore is not Mr. Smith. If that assumption is made, only the final step in my earlier solution, and the probability of Smith being the ##i^{th}## passenger change. The probability of Smith being the ##i^{th}## passenger would be ##1/49## where ##i \in {1, 2, .., 49}##. The final step therefore becomes:. Final unconditional probability of last passenger getting his/her originally allotted seat is ##\sum_{i=1}^{49} 1/49 \times P_{i} = 49 \times 1/49 \times 1/2 = 1/2 = 0.5##. Note that ##P_{50} = 1## is still used in deriving ##P_{49}, .., P_{1}## and is a valid value because it represents the situation where the NONE of the passengers before the 50th passenger have occupied a seat that is different from what was originally allotted to them.. When I get some free time, I will continue pondering over this problem in the hope of finding a simpler, more elegant way to solve the problem. It requires more creative thinking and that doesn't come naturally to all. If it's not too late:. For a plane with any number of seats. Let's call Mr Smith's seat ##S## and the last to board's seat ##X##.. When Mr Smith enters the plane, some seats are taken. But not ##S## or ##X##. If Mr Smith sits elsewhere, then one and only one of seats ##S## or ##X## is left for the last passenger. To see this, imagine Mr Smith sits in seat ##Y##. There are no problems until passenger ##Y## boards. If he/she sits in another seat ##Z##, then there are no problems until passenger ##Z## boards. Sooner or later one of these displaced passengers sits in seat ##S## or ##X## and then there are no problems until ##X## boards.. As ##S## and ##X## are equally likely to be the one taken, the probability is equal that ##X## has his seat or Mr Smith's left.. If Mr Smith sits in his seat, ##S##, then ##X## gets his correct seat. But, if Mr Smith sits in seat ##X##, then ##X## gets Mr Smith's seat. As these happen with equal likelihood (whenever Mr Smith boards), again ##X## gets his seat or Mr Smith's with equal probability in these cases.. This assumes, as above, that Mr Smith is not the last passenger.. Not anonymous and fresh_42.
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Donate. # CS201 Assignment#03 Solution & Discussion Due Date: 13-06-2011 Views: 255 Attachments: ### Replies to This Discussion CS201 Assignment 3 Spring 2011 #include <iostream> #include <cmath> #include <iomanip> using namespace std; int main() { float fAreaCircle; float fCircumferenceCircle; cout "\nEnter the radius of circle: " ; fCircumferenceCircle = (2.0 * M_PI) * radius; cout "The area of the circle is: " fAreaCircle endl; cout "The circumference of the circle is: " fCircumferenceCircle endl; system("Pause"); return 0; } In instruction radius 3.5 is a sample output. Sorry, Bhai Adnan Ayub ke solution mein execute kartay hoye thori si problem hai. Correct solution is #include <iostream> #include <cmath> #include <iomanip> using namespace std; #define PI 3.14 class Circle { public: Circle(); Circle(double); double calculateArea(double); double calculateCircum(double); void display(); }; Circle::Circle() { } Circle::Circle(double r) { } double Circle::calculateArea(double r) { return (PIrr); } double Circle::calculateCircum(double r) { return (2PIr); } void Circle::display() { cout "Area of the Circle is " calculateArea(Circle::radius) endl; cout "Circumference of the Circle is " calculateCircum(Circle::radius) endl endl; } main() { Circle circle1; Circle circle2(3.5); circle1.display(); circle2.display(); system("pause"); } Correct Solution is attached Attachments: which one is correct............ any one explain? Correct Solution is attached Attachments: which one is correct? :( CS201 Assingment no 3 Solution Spring 2011 Attachments: CS201 Assignment No. 3 solution #include<iostream.h> #define PI 3.14 class Circle { public: Circle(); Circle(double); double calculateArea(double); double calculateCircum(double); void display(); }; Circle::Circle() { } Circle::Circle(double r) { } double Circle::calculateArea(double r) { return (PIrr); } double Circle::calculateCircum(double r) { return (2PIr); } void Circle::display() { cout "Area of the Circle is " calculateArea(Circle::radius) endl; cout "Circumference of the Circle is " calculateCircum(Circle::radius) endl endl; } main() { Circle circle1; Circle circle2(3.5); circle1.display(); circle2.display(); system("pause"); } CS201 Assignment No. 3 solution #include<iostream.h> #define PI 3.14 class Circle { public: Circle(); Circle(double); double calculateArea(double); double calculateCircum(double); void display(); }; Circle::Circle() { } Circle::Circle(double r) { } double Circle::calculateArea(double r) { return (PIrr); } double Circle::calculateCircum(double r) { return (2PIr); } void Circle::display() { cout "Area of the Circle is " calculateArea(Circle::radius) endl; cout "Circumference of the Circle is " calculateCircum(Circle::radius) endl endl; } main() { Circle circle1; Circle circle2(3.5); circle1.display(); circle2.display(); system("pause"); } #include<iostream.h> #include <cmath> #include <iomanip> using namespace std; #define PI 3.14 //definig the macro for PI class circle //declaration of the class { private: circle(); circle(float); float circlearea(float); float circlecircumference(float); void display(); }; circle::circle() { } circle::circle(float i) { } float circle::circlearea (float i) { return (PIii); } float circle::circlecircumference (float i) { return (2PIi); } void circle::display() { cout"Area of the circle is=" circlearea(circle::radius)endl; cout"circumference of the circle is "circlecircumference(circle::radius)endl; } main() { circle circle1; circle circle2(3.5); circle1.display(); circle2.display(); } 1 2 3 4 5 ## VIP Member Badge & Others ------------------------------------ ## Latest Activity Fatima replied to Fatima's discussion eco403 assignment solution 2 in the group ECO403 Macroeconomics 6 minutes ago vu student updated their profile 7 minutes ago ### Jannah ? Hold you deen tight. 39 minutes ago 1 hour ago 3 hours ago Muhammad Bilal replied to BBA(6th smes)'s discussion assignment 1 spring 2021 in the group MGT402 Cost & Management Accounting 3 hours ago ★---★ ⱮԱϚҠȺហ ȺచȺហ ★---★ updated their profile 4 hours ago Mani Siddiqui posted a discussion 5 hours ago	1,140	4,374	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2021-31	latest	en	0.611646	www.vustudents.ning.com. We non-commercial site working hard since 2009 to facilitate learning Read More. We can't keep up without your support. Donate.. # CS201 Assignment#03 Solution & Discussion Due Date: 13-06-2011. Views: 255. Attachments:. ### Replies to This Discussion. CS201 Assignment 3 Spring 2011. #include <iostream>. #include <cmath>. #include <iomanip>. using namespace std;. int main(). {. float fAreaCircle;. float fCircumferenceCircle;. cout "\nEnter the radius of circle: " ;. fCircumferenceCircle = (2.0 * M_PI) * radius;. cout "The area of the circle is: " fAreaCircle endl;. cout "The circumference of the circle is: " fCircumferenceCircle endl;. system("Pause");. return 0;. }. In instruction radius 3.5 is a sample output.. Sorry, Bhai Adnan Ayub ke solution mein execute kartay hoye thori si problem hai.. Correct solution is. #include <iostream>. #include <cmath>. #include <iomanip>. using namespace std;. #define PI 3.14. class Circle. {. public:. Circle();. Circle(double);. double calculateArea(double);. double calculateCircum(double);. void display();. };. Circle::Circle(). {. }. Circle::Circle(double r). {. }. double Circle::calculateArea(double r). {. return (PIrr);. }. double Circle::calculateCircum(double r). {. return (2PIr);. }. void Circle::display(). {. cout "Area of the Circle is " calculateArea(Circle::radius) endl;. cout "Circumference of the Circle is " calculateCircum(Circle::radius) endl endl;. }. main(). {. Circle circle1;. Circle circle2(3.5);. circle1.display();. circle2.display();. system("pause");. }. Correct Solution is attached. Attachments:. which one is correct............ any one explain?. Correct Solution is attached. Attachments:. which one is correct? :(. CS201 Assingment no 3 Solution Spring 2011. Attachments:. CS201 Assignment No. 3 solution. #include<iostream.h>. #define PI 3.14. class Circle. {. public:. Circle();. Circle(double);. double calculateArea(double);. double calculateCircum(double);. void display();. };. Circle::Circle(). {. }. Circle::Circle(double r). {. }. double Circle::calculateArea(double r). {. return (PIrr);. }. double Circle::calculateCircum(double r). {. return (2PIr);. }. void Circle::display(). {. cout "Area of the Circle is " calculateArea(Circle::radius) endl;. cout "Circumference of the Circle is " calculateCircum(Circle::radius) endl. endl;. }. main(). {.	Circle circle1;. Circle circle2(3.5);. circle1.display();. circle2.display();. system("pause");. }. CS201 Assignment No. 3 solution. #include<iostream.h>. #define PI 3.14. class Circle. {. public:. Circle();. Circle(double);. double calculateArea(double);. double calculateCircum(double);. void display();. };. Circle::Circle(). {. }. Circle::Circle(double r). {. }. double Circle::calculateArea(double r). {. return (PIrr);. }. double Circle::calculateCircum(double r). {. return (2PIr);. }. void Circle::display(). {. cout "Area of the Circle is " calculateArea(Circle::radius) endl;. cout "Circumference of the Circle is " calculateCircum(Circle::radius) endl. endl;. }. main(). {. Circle circle1;. Circle circle2(3.5);. circle1.display();. circle2.display();. system("pause");. }. #include<iostream.h>. #include <cmath>. #include <iomanip>. using namespace std;. #define PI 3.14 //definig the macro for PI. class circle //declaration of the class. {. private:. circle();. circle(float);. float circlearea(float);. float circlecircumference(float);. void display();. };. circle::circle(). {. }. circle::circle(float i). {. }. float circle::circlearea (float i). {. return (PIii);. }. float circle::circlecircumference (float i). {. return (2PIi);. }. void circle::display(). {. cout"Area of the circle is=" circlearea(circle::radius)endl;. cout"circumference of the circle is "circlecircumference(circle::radius)endl;. }. main(). {. circle circle1;. circle circle2(3.5);. circle1.display();. circle2.display();. }. 1. 2. 3. 4. 5. ## VIP Member Badge & Others. ------------------------------------. ## Latest Activity. Fatima replied to Fatima's discussion eco403 assignment solution 2 in the group ECO403 Macroeconomics. 6 minutes ago. vu student updated their profile. 7 minutes ago. ### Jannah ? Hold you deen tight.. 39 minutes ago. 1 hour ago. 3 hours ago. Muhammad Bilal replied to BBA(6th smes)'s discussion assignment 1 spring 2021 in the group MGT402 Cost & Management Accounting. 3 hours ago. ★---★ ⱮԱϚҠȺហ ȺచȺហ ★---★ updated their profile. 4 hours ago. Mani Siddiqui posted a discussion. 5 hours ago.
https://brainly.com/question/319107	1,485,236,511,000,000,000	text/html	crawl-data/CC-MAIN-2017-04/segments/1484560284270.95/warc/CC-MAIN-20170116095124-00395-ip-10-171-10-70.ec2.internal.warc.gz	787,113,636	9,017	2015-02-23T16:16:54-05:00 Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest. OK. We could have gotten along nicely without the comment. -5v - 6 = 3v + 10 I think what you're trying to find is the number that ' v ' must be in order to make the equation a true statement. That number is called the 'solution' of the equation. To find it, you have to massage and manipulate the equation around until you have ' v ' and nothing else on one side. You can do anything you want to a whole side of the equation, but whatever you do to one side, you must immediately do to the other side. Here's one way you -5v - 6 = 3v + 10 Add 5v to each side: -6 = 8v + 10 Subtract 10 from each side: -16 = 8v Divide each side by 8 : -2 = v Now, was that so painful ? no im sorry i just dont like math but i get it thabk-you :-)	304	1,083	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.859375	4	CC-MAIN-2017-04	latest	en	0.915636	2015-02-23T16:16:54-05:00. Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.. OK. We could have gotten along nicely without the comment.. -5v - 6 = 3v + 10. I think what you're trying to find is the number that ' v ' must be. in order to make the equation a true statement. That number. is called the 'solution' of the equation.. To find it, you have to massage and manipulate the equation around. until you have ' v ' and nothing else on one side. You can do anything.	you want to a whole side of the equation, but whatever you do to one. side, you must immediately do to the other side. Here's one way you. -5v - 6 = 3v + 10. Add 5v to each side: -6 = 8v + 10. Subtract 10 from each side: -16 = 8v. Divide each side by 8 : -2 = v. Now, was that so painful ?. no im sorry i just dont like math but i get it thabk-you. :-).
https://www.gbhatnagar.com/2002/08/experience-mathematics-9-prime-numbers.html	1,685,679,980,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224648322.84/warc/CC-MAIN-20230602040003-20230602070003-00099.warc.gz	846,153,868	11,730	## Friday, August 16, 2002 ### Experience Mathematics #9 -- Prime numbers The prime numbers are numbers that only have $1$ and themselves as factors. The first few prime numbers are $2, 3, 5, 7, 11, 13, 17, 19, \dots$. One of the most important ideas in the theory of numbers is that a number can be written in a unique way as a product of prime powers. For example, $4=2^2$, $15$ is $3$ times $5$, and $20$ is $2\times 5$. The uniqueness of the prime factorization is used to show that the square root of $2$ is not a rational number. A rational number is a number in the form $p/q$, where both $p$ and $q$ are integers, and $q$ is not equal to zero. By doing the following questions, you can prove that the square root of two is an irrational number. The proof is by contradiction. We assume that the square root of $2$ can be written as a fraction, and then show our assumption implies a false statement. Q1. Suppose the square root of $2$ equals $p/q$ for some integers $p$ and $q$. Show that $2q^2=p^2$. Q2. Show that the power of $2$ in the prime factorization of the number $2q^2$ is an odd number. Q3. Show that the power of $2$ in the prime factorization of the number $p^2$ is an even number. The above two statements are contradictory. Thus our assumption, that the square root of $2$ is a rational number, must be false. Can you generalize this proof to prove that the square root of $3$ and $5$ are also irrational numbers?	395	1,444	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.34375	4	CC-MAIN-2023-23	latest	en	0.943679	## Friday, August 16, 2002. ### Experience Mathematics #9 -- Prime numbers. The prime numbers are numbers that only have $1$ and themselves as factors. The first few prime numbers are $2, 3, 5, 7, 11, 13, 17, 19, \dots$. One of the most important ideas in the theory of numbers is that a number can be written in a unique way as a product of prime powers. For example,. $4=2^2$, $15$ is $3$ times $5$, and $20$ is $2\times 5$.. The uniqueness of the prime factorization is used to show that the square root of $2$ is not a rational number. A rational number is a number in the form $p/q$, where both $p$ and $q$ are integers, and $q$ is not equal to zero. By doing the following questions, you can prove that the square root of two is an irrational number. The proof is by contradiction. We assume that the square root of $2$ can be written as a fraction, and then show our assumption implies a false statement.	Q1. Suppose the square root of $2$ equals $p/q$ for some integers $p$ and $q$. Show that $2q^2=p^2$.. Q2. Show that the power of $2$ in the prime factorization of the number $2q^2$ is an odd number.. Q3. Show that the power of $2$ in the prime factorization of the number $p^2$ is an even number.. The above two statements are contradictory. Thus our assumption, that the square root of $2$ is a rational number, must be false.. Can you generalize this proof to prove that the square root of $3$ and $5$ are also irrational numbers?.
https://kipdf.com/chapter-5-elasticity-and-its-application_5b0009de8ead0e469e8b4594.html	1,723,266,775,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00895.warc.gz	263,916,157	9,121	## Chapter 5: Elasticity and Its Application Economics 2010-100 Soojae Moon University of Colorado Fall 2011 Chapter 5: Elasticity and Its Application     What is elasticity? What kids of ... Economics 2010-100 Soojae Moon University of Colorado Fall 2011 Chapter 5: Elasticity and Its Application     What is elasticity? What kids of issues can elasticity help us understand? What is the price elasticity of demand? How is it related to the demand curve? How is it related to revenue & expenditure? What is the price elasticity of supply? How is it related to the supply curve? What are the income and cross-price elasticities of demand? ___________________________: a measure of the responsiveness of quantity demanded or quantity supplied to one of its determinants ELASTICITY OF DEMAND Law of Demand states a fall in price increases the quantity demanded. By how much? ____________________________: a measure of how much the quantity demanded of a good responds to a change in the price of that good Price elasticity of demand = Percent change in Qd/% change in Price o Examples: Price of Butterfinger Bites falls 20% & causes an increase in Qd of 25%, Price of insulin increases 15% and the Qd of insulin decreases 2.5% Computing the Price Elasticity of Demand The Midpoint Method of Calculating the Elasticity of Demand o Elasticity is ratio of percent changes, therefore going from point A to B will have different elasticity than going from B to A. o Get around this problem with the midpoint method  Note: Standard method of computing Percent change is (end value - start value)/start value100  Midpoint method: (end value – start value)/mid point 100 (Q2  Q1 ) (Q2  Q1 ) %Qd 2 d   ( P2  P1 ) %P ( P2  P1 ) 2 Example: Price of hotel rooms increases from \$70 to \$90. The Qd decreases from 5000 to 3000. Economics 2010-100 Soojae Moon University of Colorado Fall 2011 Note: Because of the law of demand quantity demanded will always move in the opposite direction to that of price change so the price elasticity of demand will always be negative. It is common to ignore the negative sign and report only the absolute number. What determines Price Elasticity of Demand? - Availability of Close Substitutes (Breakfast Cereal vs. Sunscreen) o Goods with close substitutes have more elastic demand (because consumers can easily switch goods) - Necessities vs. Luxuries (Insulin vs. Caribbean Cruises) o Necessities have inelastic demand o Luxuries have elastic demand - Definition of the Market (blue Jeans vs. Clothing) o narrowly defined markets have elastic demand o broadly defined markets have inelastic demand - Time Horizon (Gasoline in the SR vs. Gasoline in the LR) o longer periods of time have elastic demand, with time more substitutes become available What do these numbers mean?      Graphs PERFECTLY INELASTIC if ε=0 INELASTIC if ε1 PERFECTLY ELASTIC if ε=infinity Economics 2010-100 Soojae Moon University of Colorado Fall 2011 Elasticity of a Linear Demand Curve Even though the slope (change in y/change in x) is constant, the elasticity (ratio of % changes) is not - Low price & high Qd inelastic demand - High price & low Qd elastic. P Elasticity >1 (_________) Unit Elastic Elasticity 1→%ΔQd>%ΔP  Price ↑→Qd↓→TR↓  ↓P→Qd↑→↑TR Demand inelastic→E%ΔQd  Price ↑→Qd↓→TR↑  ↓P→Qd↑→↓TR  Example: A. Pharmacies raise the price of insulin by 10%. Does total expenditure on insulin rise or fall? B. As a result of a fare war, the price of a luxury cruise falls 20%. Does luxury cruise companies’ total revenue rise or fall? Economics 2010-100 Soojae Moon University of Colorado Fall 2011 PRICE ELASTICITY OF SUPPLY price elasticity of supply: a measure of how much the quantity supplied of a good responds to a change in the price of that good. %Qs s  %P Same as Demand: Perfectly inelastic: E=0 Inelastic: E1 Perfectly elastic: E=infinity The determinants of supply curve  More easily change the Q sellers produce, greater price elasticity of supply  Greater in the LR Elasticity changes along the supply curve. Economics 2010-100 Soojae Moon University of Colorado Fall 2011 Other Elasticities: _______________________________: a measure of how much the quantity demanded of a good responds to a change in consumers’ income Income elasticity of demand = Percent change in Qd/% change in Income d    %Qd %Y Normal goods have positive income Elasticities Inferior goods have negative income elasticities Example: Jessica’s income increases from \$1500/mo to \$2000/mo. Her consumption of Jamba Juices increases from 20/mo to 27/mo. Is Jamba a normal or inferior good for Jessica? What is her income elasticity? _________________________________: a measure of how much the quantity demanded of a good responds to a change in the price of another good Cross-Price elasticity of demand = Percent change in Qd good 1/ % change in P good 2 d    %Qd1 %P2 Substitutes have positive cross-price Elasticities Complements have negative cross-price elasticities Example: The price of a burrito falls from \$23 to \$17. As a result, Mansour’s quantity demanded of beer falls from 3 to 2. Are these goods complements or substitutes? What is the cross-price elasticity? Suggested Problems: Problems and Applications – 1, 2, 5	1,381	5,313	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2024-33	latest	en	0.860852	## Chapter 5: Elasticity and Its Application. Economics 2010-100 Soojae Moon University of Colorado Fall 2011 Chapter 5: Elasticity and Its Application     What is elasticity? What kids of .... Economics 2010-100 Soojae Moon. University of Colorado Fall 2011. Chapter 5: Elasticity and Its Application    . What is elasticity? What kids of issues can elasticity help us understand? What is the price elasticity of demand? How is it related to the demand curve? How is it related to revenue & expenditure? What is the price elasticity of supply? How is it related to the supply curve? What are the income and cross-price elasticities of demand?. ___________________________: a measure of the responsiveness of quantity demanded or quantity supplied to one of its determinants ELASTICITY OF DEMAND Law of Demand states a fall in price increases the quantity demanded. By how much? ____________________________: a measure of how much the quantity demanded of a good responds to a change in the price of that good Price elasticity of demand = Percent change in Qd/% change in Price o Examples: Price of Butterfinger Bites falls 20% & causes an increase in Qd of 25%, Price of insulin increases 15% and the Qd of insulin decreases 2.5% Computing the Price Elasticity of Demand The Midpoint Method of Calculating the Elasticity of Demand o Elasticity is ratio of percent changes, therefore going from point A to B will have different elasticity than going from B to A. o Get around this problem with the midpoint method  Note: Standard method of computing Percent change is (end value - start value)/start value100  Midpoint method: (end value – start value)/mid point 100 (Q2  Q1 ) (Q2  Q1 ) %Qd 2 d   ( P2  P1 ) %P ( P2  P1 ) 2. Example: Price of hotel rooms increases from \$70 to \$90. The Qd decreases from 5000 to 3000.. Economics 2010-100 Soojae Moon. University of Colorado Fall 2011. Note: Because of the law of demand quantity demanded will always move in the opposite direction to that of price change so the price elasticity of demand will always be negative. It is common to ignore the negative sign and report only the absolute number. What determines Price Elasticity of Demand? -. Availability of Close Substitutes (Breakfast Cereal vs. Sunscreen) o Goods with close substitutes have more elastic demand (because consumers can easily switch goods). -. Necessities vs. Luxuries (Insulin vs. Caribbean Cruises) o Necessities have inelastic demand o Luxuries have elastic demand. -. Definition of the Market (blue Jeans vs. Clothing) o narrowly defined markets have elastic demand o broadly defined markets have inelastic demand. -. Time Horizon (Gasoline in the SR vs. Gasoline in the LR) o longer periods of time have elastic demand, with time more substitutes become available. What do these numbers mean?      Graphs. PERFECTLY INELASTIC if ε=0 INELASTIC if ε1 PERFECTLY ELASTIC if ε=infinity.	Economics 2010-100 Soojae Moon. University of Colorado Fall 2011. Elasticity of a Linear Demand Curve Even though the slope (change in y/change in x) is constant, the elasticity (ratio of % changes) is not - Low price & high Qd inelastic demand - High price & low Qd elastic. P. Elasticity >1 (_________) Unit Elastic Elasticity 1→%ΔQd>%ΔP  Price ↑→Qd↓→TR↓  ↓P→Qd↑→↑TR Demand inelastic→E%ΔQd  Price ↑→Qd↓→TR↑  ↓P→Qd↑→↓TR.  Example: A. Pharmacies raise the price of insulin by 10%. Does total expenditure on insulin rise or fall? B. As a result of a fare war, the price of a luxury cruise falls 20%. Does luxury cruise companies’ total revenue rise or fall?. Economics 2010-100 Soojae Moon. University of Colorado Fall 2011. PRICE ELASTICITY OF SUPPLY price elasticity of supply: a measure of how much the quantity supplied of a good responds to a change in the price of that good. %Qs s  %P Same as Demand: Perfectly inelastic: E=0 Inelastic: E1 Perfectly elastic: E=infinity The determinants of supply curve  More easily change the Q sellers produce, greater price elasticity of supply  Greater in the LR Elasticity changes along the supply curve.. Economics 2010-100 Soojae Moon. University of Colorado Fall 2011. Other Elasticities: _______________________________: a measure of how much the quantity demanded of a good responds to a change in consumers’ income Income elasticity of demand = Percent change in Qd/% change in Income. d   . %Qd %Y. Normal goods have positive income Elasticities Inferior goods have negative income elasticities. Example: Jessica’s income increases from \$1500/mo to \$2000/mo. Her consumption of Jamba Juices increases from 20/mo to 27/mo. Is Jamba a normal or inferior good for Jessica? What is her income elasticity?. _________________________________: a measure of how much the quantity demanded of a good responds to a change in the price of another good Cross-Price elasticity of demand = Percent change in Qd good 1/ % change in P good 2. d   . %Qd1 %P2. Substitutes have positive cross-price Elasticities Complements have negative cross-price elasticities. Example: The price of a burrito falls from \$23 to \$17. As a result, Mansour’s quantity demanded of beer falls from 3 to 2. Are these goods complements or substitutes? What is the cross-price elasticity?. Suggested Problems: Problems and Applications – 1, 2, 5.
https://math.stackexchange.com/questions/2377120/solution-to-a-complex-first-order-cubic-nonlinear-ode	1,563,249,248,000,000,000	text/html	crawl-data/CC-MAIN-2019-30/segments/1563195524502.23/warc/CC-MAIN-20190716035206-20190716061206-00133.warc.gz	463,757,584	34,752	# Solution to a complex first order cubic nonlinear ODE I'm attempting to solve the differential equation $$\left(i\omega+\sigma\right)\dot{y}+\mu\left\|y\right\|^2y=0$$ with the initial condition $y(0)=0$ I've tried to use separation of variables which means I have the following integral to solve $$t-t_0=-\frac{i\omega+\sigma}{\mu}\int_{y_0}^{y}\frac{dz}{\left\|z\right\|^2z}.$$ I guess the general solution to this is $$\left\|y\right\|^2=y_0^2\frac{i\omega+\sigma}{y_0^2\mu (t-t_0)+i\omega+\sigma},$$ but I know somewhere in there something isn't quite right (maybe because uniqueness isn't guaranteed). Does someone know of a different way to find the solution(s) to this equation for y instead of $\|y\|^2$? For simplicity, we denote $a+ib=\mu/(\sigma+i\omega)$. Assume $y$ is not the trivial solution $y=0$. Letting $y(t)=\rho(t) e^{i\theta(t)}$ be its polar representation, your equation $$\dot y+(a+ib)\|y\|^2y=0$$ becomes $$e^{i\theta}(i\rho\dot \theta+\dot\rho+(a+ib)\rho^3)=0.$$ Equating real and imaginary parts to zero gives a system for $\rho(t)$ and $\theta(t)$: \begin{align} \dot\rho+a\rho^3&=0,\\ \dot\theta+b\rho^2&=0. \end{align} The first equation is decoupled, so the second equation is linear in $\theta$ once $\rho$ is found.	403	1,243	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0}	3.96875	4	CC-MAIN-2019-30	latest	en	0.80333	# Solution to a complex first order cubic nonlinear ODE. I'm attempting to solve the differential equation $$\left(i\omega+\sigma\right)\dot{y}+\mu\left\|y\right\|^2y=0$$ with the initial condition $y(0)=0$ I've tried to use separation of variables which means I have the following integral to solve $$t-t_0=-\frac{i\omega+\sigma}{\mu}\int_{y_0}^{y}\frac{dz}{\left\|z\right\|^2z}.$$ I guess the general solution to this is $$\left\|y\right\|^2=y_0^2\frac{i\omega+\sigma}{y_0^2\mu (t-t_0)+i\omega+\sigma},$$ but I know somewhere in there something isn't quite right (maybe because uniqueness isn't guaranteed). Does someone know of a different way to find the solution(s) to this equation for y instead of $\|y\|^2$?. For simplicity, we denote $a+ib=\mu/(\sigma+i\omega)$. Assume $y$ is not the trivial solution $y=0$. Letting $y(t)=\rho(t) e^{i\theta(t)}$ be its polar representation, your equation $$\dot y+(a+ib)\|y\|^2y=0$$ becomes $$e^{i\theta}(i\rho\dot \theta+\dot\rho+(a+ib)\rho^3)=0.$$ Equating real and imaginary parts to zero gives a system for $\rho(t)$ and $\theta(t)$: \begin{align} \dot\rho+a\rho^3&=0,\\ \dot\theta+b\rho^2&=0.	\end{align} The first equation is decoupled, so the second equation is linear in $\theta$ once $\rho$ is found.
https://socratic.org/questions/how-do-you-find-the-equation-of-the-tangent-line-and-normal-line-to-the-curve-at	1,611,774,808,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610704832583.88/warc/CC-MAIN-20210127183317-20210127213317-00358.warc.gz	555,976,528	5,900	# How do you find the equation of the tangent line and normal line to the curve at the given point y = (1+2x)^2, (1,9)? Sep 16, 2015 equation of tangent is $y = 12 x - 3$ equation of normal line is $12 y = 109 - x$ #### Explanation: the slope of tangent at (1,9) is $2 \left(1 + 2 x\right) \cdot 2$ by normal differentiation at x=1 the slope value is 12 let the equation of tangent is y=mx+c substituiting points 9=12+c c=-3 equation of tangent is y=12x-3 for normal line slope =-1/12 from the line formula y=mx+c substituiting points 9=-1/12+c c=109/12 equation is 12y = 109-x	209	582	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.4375	4	CC-MAIN-2021-04	latest	en	0.841514	# How do you find the equation of the tangent line and normal line to the curve at the given point y = (1+2x)^2, (1,9)?. Sep 16, 2015. equation of tangent is $y = 12 x - 3$. equation of normal line is $12 y = 109 - x$. #### Explanation:. the slope of tangent at (1,9) is $2 \left(1 + 2 x\right) \cdot 2$ by normal differentiation. at x=1 the slope value is 12. let the equation of tangent is y=mx+c. substituiting points. 9=12+c. c=-3.	equation of tangent is y=12x-3. for normal line. slope =-1/12. from the line formula y=mx+c. substituiting points. 9=-1/12+c. c=109/12. equation is. 12y = 109-x.
https://www.jiskha.com/display.cgi?id=1354292234	1,503,476,236,000,000,000	text/html	crawl-data/CC-MAIN-2017-34/segments/1502886117911.49/warc/CC-MAIN-20170823074634-20170823094634-00464.warc.gz	913,468,898	3,973	# AP CALCULUS posted by . Using the mean value theorem; F'(x) = f(b)-f(a) / b-a f(x)=x^2-8x+3; interval [-1,6] • AP CALCULUS - I assume you want to find c such that f'(c) = (f(6)-f(-1))/7 nothing simpler: f'(x) = 2x-8 f(6) = -9 f(-1) = 12 so, we want f'(c) = -21/7 = -3 2x-8 = -3 x = 5/2 ## Similar Questions 1. ### calculus Let f(x) = (x+1)/(x-1). Show that there are no vlue of c such that f(2)-f(0) =f'(c)(2-0). Why does this not contradict the Mean Value Theorem? 2. ### calculus Verify that the hypotheses of the Mean-Value Theorem are satisfied on the given interval, and find all values of c in that interval that satisfy the conclusion of the theorem. f(x)=x^2-3x; [-2,6] 3. ### calculus verify that the function satisfies the hypothesis of the mean value theorem on the given interval. then find all numbers c that satisfy the conclusion of the mean value theorem. f(x) = x/(x+2) , [1,4] 4. ### Calculus Verify that the hypotheses of the Mean-Value Theorem are satisfied for f(x) = √(16-x^2 ) on the interval [-4,1] and find all values of C in this interval that satisfy the conclusion of the theorem. 5. ### Math - Calculus Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2]. Perhaps Rolle's Theorem, Mean Value Theorem, or Intermediate Value Theorem hold clues? 6. ### Math - Calculus Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2]. Perhaps Rolle's Theorem, Mean Value Theorem, or Intermediate Value Theorem hold clues? 7. ### Calculus Verify the hypothesis of the mean value theorem for each function below defined on the indicated interval. Then find the value “C” referred to by the theorem. Q1a) h(x)=√(x+1 ) [3,8] Q1b) K(x)=(x-1)/(x=1) [0,4] Q1c) Explain … 8. ### calculus determine whether the mean value theorem can be applied to f on the closed interval [a,b]. If the Mean Value Theorem can be applied, find all values of c in the open interval (a,b) such that f(c) =f(b) - f(a) / b - a 9. ### Calculus Given f(x) = -1/x, find all c in the interval [-3, -½] that satisfies the Mean Value Theorem. A. c= -sqrt(3/2) B. c= +or- sqrt(3/2) C. The Mean Value Theorem doesn’t apply because f is not continuous at x=0 D. The Mean Value Theorem … 10. ### Calculus Let f(x)=αx^2+βx+γ be a quadratic function, so α≠0, and let I=[a,b]. a) Check f satisfies the hypothesis of the Mean Value Theorem. b)Show that the number c ∈ (a,b) in the Mean Value Theorem is the midpoint of the interval … More Similar Questions	787	2,502	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.953125	4	CC-MAIN-2017-34	latest	en	0.748398	# AP CALCULUS. posted by .. Using the mean value theorem;. F'(x) = f(b)-f(a) / b-a. f(x)=x^2-8x+3; interval [-1,6]. • AP CALCULUS -. I assume you want to find c such that f'(c) = (f(6)-f(-1))/7. nothing simpler:. f'(x) = 2x-8. f(6) = -9. f(-1) = 12. so, we want f'(c) = -21/7 = -3. 2x-8 = -3. x = 5/2. ## Similar Questions. 1. ### calculus. Let f(x) = (x+1)/(x-1). Show that there are no vlue of c such that f(2)-f(0) =f'(c)(2-0). Why does this not contradict the Mean Value Theorem?. 2. ### calculus. Verify that the hypotheses of the Mean-Value Theorem are satisfied on the given interval, and find all values of c in that interval that satisfy the conclusion of the theorem. f(x)=x^2-3x; [-2,6]. 3. ### calculus. verify that the function satisfies the hypothesis of the mean value theorem on the given interval. then find all numbers c that satisfy the conclusion of the mean value theorem. f(x) = x/(x+2) , [1,4]. 4. ### Calculus.	Verify that the hypotheses of the Mean-Value Theorem are satisfied for f(x) = √(16-x^2 ) on the interval [-4,1] and find all values of C in this interval that satisfy the conclusion of the theorem.. 5. ### Math - Calculus. Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2]. Perhaps Rolle's Theorem, Mean Value Theorem, or Intermediate Value Theorem hold clues?. 6. ### Math - Calculus. Show that the equation x^3-15x+c=0 has at most one root in the interval [-2,2]. Perhaps Rolle's Theorem, Mean Value Theorem, or Intermediate Value Theorem hold clues?. 7. ### Calculus. Verify the hypothesis of the mean value theorem for each function below defined on the indicated interval. Then find the value “C” referred to by the theorem. Q1a) h(x)=√(x+1 ) [3,8] Q1b) K(x)=(x-1)/(x=1) [0,4] Q1c) Explain …. 8. ### calculus. determine whether the mean value theorem can be applied to f on the closed interval [a,b]. If the Mean Value Theorem can be applied, find all values of c in the open interval (a,b) such that f(c) =f(b) - f(a) / b - a. 9. ### Calculus. Given f(x) = -1/x, find all c in the interval [-3, -½] that satisfies the Mean Value Theorem. A. c= -sqrt(3/2) B. c= +or- sqrt(3/2) C. The Mean Value Theorem doesn’t apply because f is not continuous at x=0 D. The Mean Value Theorem …. 10. ### Calculus. Let f(x)=αx^2+βx+γ be a quadratic function, so α≠0, and let I=[a,b]. a) Check f satisfies the hypothesis of the Mean Value Theorem. b)Show that the number c ∈ (a,b) in the Mean Value Theorem is the midpoint of the interval …. More Similar Questions.
https://www.meritnation.com/ask-answer/question/a-chord-of-circle-of-radius-10-cm-subtends-a-right-angle-at/circles/1818819	1,618,532,244,000,000,000	text/html	crawl-data/CC-MAIN-2021-17/segments/1618038088264.43/warc/CC-MAIN-20210415222106-20210416012106-00129.warc.gz	920,034,429	12,075	# A chord of circle of radius 10 cm subtends a right angle at the centre.Find the area of the corresponding minor segment and hence find the area of the major sector? Let r be the radius of the circle and θ be the angle subtended by the chord at the centre. Given, r = 10 cm and θ = 90° Area of the minor segment = Area of sector OAPB – Area of ∆AOB Area of the major sector OAQB = Area of circle – Area of sector OAPB • 74 What are you looking for?	124	456	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.84375	4	CC-MAIN-2021-17	latest	en	0.85211	# A chord of circle of radius 10 cm subtends a right angle at the centre.Find the area of the corresponding minor segment and hence find the area of the major sector?. Let r be the radius of the circle and θ be the angle subtended by the chord at the centre.. Given, r = 10 cm and θ = 90°. Area of the minor segment. = Area of sector OAPB – Area of ∆AOB.	Area of the major sector OAQB. = Area of circle – Area of sector OAPB. • 74. What are you looking for?.
https://www.studysmarter.co.uk/explanations/math/probability-and-statistics/quantile-function/	1,721,292,381,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00120.warc.gz	865,684,065	68,535	Quantile function The quantile function, essential in statistical analysis, inversely calculates the value at a given probability in a distribution's cumulative distribution function (CDF). Serving as a cornerstone in statistical modelling, it plays a pivotal role in identifying data thresholds and assessing statistical outcomes, catering to diverse fields such as finance, meteorology, and health sciences. Memorising the quantitative function's applications enhances your understanding of probability distributions and their practical implications in real-world scenarios. Create learning materials about Quantile function with our free learning app! • Flashcards, notes, mock-exams and more • Everything you need to ace your exams What is a Quantile Function? Quantile function plays a crucial role in statistics, serving as a tool to understand and interpret the distribution of data. Through this function, one can determine the value below which a given percentage of observations in a group of observations falls. Breaking Down the Quantile Function Definition Quantile Function: A mathematical function that, for a given probability, returns the value of a random variable below which the given percentage of observations fall. Quantile functions are the inverse of cumulative distribution functions (CDFs). While CDFs answer the question, "What is the probability that the random variable X is less than or equal to a certain value?", quantile functions ask, "For a given probability, what value does the random variable X not exceed?" This transition from probabilities to values is what makes quantile functions especially valuable in data analysis and probability theory.It is important to differentiate between quantiles and the quantile function. Quantiles are specific points in a distribution that divide the data into equal parts, while the quantile function enables one to find these points for continuous distributions. If you are using the quantile function to find the median of a set of data, you are looking for the 50th percentile. This means that you are searching for a value such that 50% of the data points are below it. In mathematical terms, this is represented as: egin{equation} Q(p) = F^{-1}(0.5), egin{equation} where $$Q(p)$$ is the quantile function and $$F^{-1}(0.5)$$ represents the inverse of the cumulative distribution function at 0.5 (or 50%). The term 'percentile' is often used interchangeably with 'quantile', though percentile specifically refers to the quantile that divides the data into 100 equally sized intervals. The Mathematics Behind Quantile Functions To deeply understand the mathematics behind quantile functions, it's essential to familiarise oneself with the concept of the Cumulative Distribution Function (CDF) and its inverse. The CDF of a random variable X is a function that maps from a value to its percentile rank. In mathematical terms, it is defined as: egin{equation} F(x) = P(X egin{equation} where $$F(x)$$ represents the CDF of X, and $$P(X < x)$$ is the probability that the random variable X takes on a value less than x. Finding the quantile function of a distribution requires calculating the inverse of its CDF. For many distributions, especially continuous ones, this process involves solving equations that may not have a straightforward solution. As a result, numerical methods or approximations are often employed. This exploration into computing quantile functions not only aids in understanding the distribution's characteristics but also showcases the intersection of statistical theory and computational mathematics.A fascinating case is the normal distribution, widely used across various fields. Its quantile function doesn’t have an explicit formula in elementary functions, leading to the development of tables and algorithms for its approximation. This highlights the practical necessity of quantile functions in statistical applications and computational statistics. Exploring Quantile Function Examples The quantile function, a fundamental concept in statistics, finds its application across various scenarios, allowing for the determination of specific data points within a distribution. This section will delve into two specific types of quantile functions, each instrumental in interpreting data sets more efficiently. Empirical Quantile Function Explained An empirical quantile function estimates the quantile function from a set of observed data points. Unlike its theoretical counterpart, which assumes a known distribution, the empirical quantile function is used when the distribution of data is unknown or when dealing with real-world data. Here, the calculation of quantiles relies on the actual data set's distribution. Consider a set of test scores from a class of 30 students. If you want to determine the score corresponding to the 90th percentile, you sort the scores in ascending order and calculate the position using the formula: egin{equation}P = rac{(n + 1) imes q}{100} egin{equation} where $$P$$ is the position, $$n$$ the number of data points (30 in this case), and $$q$$ the desired quantile (90th percentile). This method interpolates between observed data points to estimate the value. In practice, various methods for estimating the quantile from empirical data exist, each handling ties and gaps between observations differently. Conditional Quantile Function: A Detailed Look The conditional quantile function extends the concept of quantiles into the realm of conditional distributions. It describes the quantile of a random variable conditional on another variable. In other words, it allows us to explore how the quantiles of one variable change as another variable varies. Conditional Quantile Function: For random variables X and Y, the conditional quantile function of Y given X is denoted as Q(Y\|X) and is defined as: egin{equation}Q_{Y\|X}(p) = infigrace{y: F_{Y\|X}(y) igrace}. egin{equation}This essentially specifies the value of Y such that the probability of Y being less than or equal to this value, given X, is p. Imagine analyzing the relationship between students' studying hours (X) and their exam scores (Y). The conditional quantile function can help understand how the expected exam scores (at various quantiles) change with an increase in studying hours. If you're interested in the median score given studying hours, you would be looking at the 50th percentile of the conditional distribution of Y given X. The conditional quantile function is especially useful in econometrics and finance, where variables often exhibit non-linear relationships. For example, it can illuminate how the quantiles of wage distributions change with education level or how the risk of financial loss varies across different quantiles of investment portfolios. This concept is also the backbone of quantile regression, a statistical technique that estimates the conditional median or other quantiles instead of the mean. Quantile regression provides a more comprehensive view of the potential outcomes, especially in the presence of heteroscedasticity or outliers, thus offering a nuanced understanding of the data beyond the average or mean relationship. Interpreting Quantile Functions Interpreting quantile functions is essential for understanding the distribution of data within a dataset. This process involves using the quantile function to find values corresponding to specific probabilities or percentages of the data. The interpretation provides insights into the probability distribution of a dataset and helps in making informed decisions based on statistical analysis.The ability to interpret quantile functions is especially valuable in fields that rely on uncertainty and risk assessment, such as finance, meteorology, and health sciences. By understanding the distribution of data through quantile functions, you can predict outcomes and make decisions that are informed by the likelihood of specific events. Quantile Function Interpretation Steps Interpreting a quantile function typically involves several key steps: • Identify the probability or quantile of interest. • Apply the quantile function to find the corresponding data value. • Analyse the result in the context of the dataset and the question at hand. For example, to find the 90th percentile of a dataset using its quantile function, you would input 0.9 into the function. The output is the value below which 90% of the data lies. This value is crucial for understanding the upper range of the dataset and identifying outliers or extreme values. Let's imagine you're analysing the annual rainfall in a particular region over the past decade and you wish to understand the distribution of rainfall amounts. If the quantile function for the dataset is given by $Q(p) = 100 + 200p^2$ where $$p$$ is the probability, to find the rainfall amount that was not exceeded 75% of the time, you would substitute $$p = 0.75$$ into the equation to get: $Q(0.75) = 100 + 200(0.75)^2 = 250$ millimetres. This means that 75% of the time, the annual rainfall did not exceed 250 millimetres. Real-Life Applications of Quantile Function Interpretation Quantile function interpretation finds its application in various real-life scenarios. Its utility spans across disciplines, offering valuable insights into data distribution and aiding in decision-making: • In finance, quantile functions are used to assess risk and return profiles, informing investment strategies by evaluating the potential downside or upside of an asset. • In healthcare, interpreting quantile functions of patient data helps in understanding disease prevalence and outcomes, guiding treatment protocols. • Environmental scientists utilise quantile functions to predict extreme weather conditions, contributing to disaster preparedness and climate research. By leveraging the quantile function, professionals in these fields can quantify the likelihood of specified events, making it a powerful tool for analysis and planning. When interpreting quantile functions, remember that the outcome is highly dependent on the accuracy and distribution of the underlying data. Anomalies or biases in the dataset can lead to misinterpretations. A fascinating application of quantile function interpretation is in the field of educational research, where it's used to analyse test scores and achievement levels across different student populations. By examining the quantiles of test score distributions, educators and policymakers can identify achievement gaps and tailor interventions to support underperforming groups.This approach allows for a nuanced understanding of educational outcomes beyond mean or median scores, highlighting variability and providing insights into the tails of the distribution. As a result, strategies can be more precisely targeted to uplift students who may be struggling and ensure resources are allocated where they are most needed, ultimately contributing to more equitable educational outcomes. Practising with Quantile Function Exercises Quantile function exercises are an excellent way to deepen your understanding of statistical distributions and their applications. These exercises range from basic problems suitable for beginners to more advanced scenarios that challenge even experienced statisticians. Basic Quantile Function Exercises for Beginners If you're new to quantile functions, starting with basic exercises can help you grasp their fundamental concepts and applications. These exercises often involve identifying values at specified quantiles and understanding the relationship between quantile functions and probability distributions.Basic exercises are crafted to reinforce the understanding of quantile functions as inverses of cumulative distribution functions (CDFs). By practising with these problems, you'll become more comfortable in handling various types of data distributions and making predictions based on statistical analysis. Calculate the 25th percentile ( extit{first quartile}) for a dataset using the quantile function $Q(p) = 15 + 5p^3$.To solve this, substitute $$p = 0.25$$ into the quantile function: $Q(0.25) = 15 + 5(0.25)^3 = 15 + 5(0.015625) = 15.078125$.This result indicates that 25% of the data falls below 15.078125. Remember that the percentile is simply the quantile multiplied by 100. So, the 25th percentile corresponds to the 0.25 quantile. For those already familiar with the basics of quantile functions, advanced exercises present more complex scenarios that require a deeper understanding of statistical distributions and their analysis. These exercises may involve conditional quantile functions, non-linear distributions, and the application of quantile regression techniques.Advanced exercises aim to challenge your analytical skills and encourage you to apply quantile functions in more sophisticated contexts. By tackling these problems, you'll refine your ability to make precise statistical inferences and predictions. Assume that the quantile function for a dataset is given by $Q(p) = rac{1}{ig(2p + 0.5ig)^2}$.Calculate the value at the 90th percentile (p = 0.9).Substituting $$p = 0.9$$ into the quantile function gives: $Q(0.9) = rac{1}{ig(2(0.9) + 0.5ig)^2} = rac{1}{(1.8 + 0.5)^2} = rac{1}{(2.3)^2} = rac{1}{5.29}$.This result provides the value below which 90% of the data lies, according to the given quantile function. Advanced exercises often lead to the exploration of conditional quantile functions, which offer insights into the relationship between two variables. For instance, in modelling the impact of education on income, a conditional quantile function can reveal how the distribution of income changes at different levels of education.Understanding these relationships through quantile functions can yield actionable insights in fields such as economics, meteorology, and health sciences. This goes beyond basic descriptive statistics to allow for a nuanced view of how outcomes change across different segments of a population or under varying conditions. Quantile function - Key takeaways • Quantile Function Definition: A function that identifies the value below which a given percentage of observations in a data set lies. • Quantile vs. Quantile Function: Quantiles divide data into equal parts, whereas the quantile function finds these points for continuous distributions. • Empirical Quantile Function: Estimates quantiles using the actual data set's distribution when the distribution is unknown. • Conditional Quantile Function: Determines quantiles of a random variable conditional on another variable, enabling analysis of how one variable affects the distribution of another. • Interpreting Quantile Functions: Essential for understanding a dataset's distribution, it involves identifying values tied to specific probabilities or percentages within the data. Learn with 0 Quantile function flashcards in the free StudySmarter app We have 14,000 flashcards about Dynamic Landscapes. What is the definition of a quantile function in statistics? A quantile function, in statistics, is the inverse of the cumulative distribution function (CDF). It provides the value below which a given percentage of observations in a group of observations falls. How is the quantile function related to the cumulative distribution function? The quantile function is the inverse of the cumulative distribution function (CDF) for continuous distributions. It maps a probability $$p$$ to the value $$x$$ such that the probability of the random variable being less than or equal to $$x$$ is $$p$$. What are the applications of a quantile function in data analysis? In data analysis, quantile functions are crucial for estimating the distribution of a data set, setting thresholds for outlier detection, performing probabilistic forecasting, and simulating data samples from a theoretical distribution to test hypotheses or models. They facilitate comparison between different datasets by enabling the calculation of quantiles. How do you calculate the value of a quantile function for a given probability? To calculate the value of a quantile function for a given probability, locate the probability value in the cumulative distribution function (CDF) and identify the corresponding value in the domain of the distribution; this value is the quantile for the specified probability. What are the differences between a quantile function and a percentile function? A quantile function determines the value below which a certain proportion of data falls, applicable for any fraction. A percentile function is a specific type of quantile function for divisions into 100 equal parts, focusing on the percent below which a certain proportion of data is found. StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. 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StudySmarter Editorial Team Team Math Teachers • Checked by StudySmarter Editorial Team	3,407	17,625	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.46875	4	CC-MAIN-2024-30	latest	en	0.869091	Quantile function. The quantile function, essential in statistical analysis, inversely calculates the value at a given probability in a distribution's cumulative distribution function (CDF). Serving as a cornerstone in statistical modelling, it plays a pivotal role in identifying data thresholds and assessing statistical outcomes, catering to diverse fields such as finance, meteorology, and health sciences. Memorising the quantitative function's applications enhances your understanding of probability distributions and their practical implications in real-world scenarios.. Create learning materials about Quantile function with our free learning app!. • Flashcards, notes, mock-exams and more. • Everything you need to ace your exams. What is a Quantile Function?. Quantile function plays a crucial role in statistics, serving as a tool to understand and interpret the distribution of data. Through this function, one can determine the value below which a given percentage of observations in a group of observations falls.. Breaking Down the Quantile Function Definition. Quantile Function: A mathematical function that, for a given probability, returns the value of a random variable below which the given percentage of observations fall.. Quantile functions are the inverse of cumulative distribution functions (CDFs). While CDFs answer the question, "What is the probability that the random variable X is less than or equal to a certain value?", quantile functions ask, "For a given probability, what value does the random variable X not exceed?" This transition from probabilities to values is what makes quantile functions especially valuable in data analysis and probability theory.It is important to differentiate between quantiles and the quantile function. Quantiles are specific points in a distribution that divide the data into equal parts, while the quantile function enables one to find these points for continuous distributions.. If you are using the quantile function to find the median of a set of data, you are looking for the 50th percentile. This means that you are searching for a value such that 50% of the data points are below it. In mathematical terms, this is represented as: egin{equation} Q(p) = F^{-1}(0.5), egin{equation} where $$Q(p)$$ is the quantile function and $$F^{-1}(0.5)$$ represents the inverse of the cumulative distribution function at 0.5 (or 50%).. The term 'percentile' is often used interchangeably with 'quantile', though percentile specifically refers to the quantile that divides the data into 100 equally sized intervals.. The Mathematics Behind Quantile Functions. To deeply understand the mathematics behind quantile functions, it's essential to familiarise oneself with the concept of the Cumulative Distribution Function (CDF) and its inverse. The CDF of a random variable X is a function that maps from a value to its percentile rank. In mathematical terms, it is defined as: egin{equation} F(x) = P(X egin{equation} where $$F(x)$$ represents the CDF of X, and $$P(X < x)$$ is the probability that the random variable X takes on a value less than x.. Finding the quantile function of a distribution requires calculating the inverse of its CDF. For many distributions, especially continuous ones, this process involves solving equations that may not have a straightforward solution. As a result, numerical methods or approximations are often employed. This exploration into computing quantile functions not only aids in understanding the distribution's characteristics but also showcases the intersection of statistical theory and computational mathematics.A fascinating case is the normal distribution, widely used across various fields. Its quantile function doesn’t have an explicit formula in elementary functions, leading to the development of tables and algorithms for its approximation. This highlights the practical necessity of quantile functions in statistical applications and computational statistics.. Exploring Quantile Function Examples. The quantile function, a fundamental concept in statistics, finds its application across various scenarios, allowing for the determination of specific data points within a distribution. This section will delve into two specific types of quantile functions, each instrumental in interpreting data sets more efficiently.. Empirical Quantile Function Explained. An empirical quantile function estimates the quantile function from a set of observed data points. Unlike its theoretical counterpart, which assumes a known distribution, the empirical quantile function is used when the distribution of data is unknown or when dealing with real-world data. Here, the calculation of quantiles relies on the actual data set's distribution.. Consider a set of test scores from a class of 30 students. If you want to determine the score corresponding to the 90th percentile, you sort the scores in ascending order and calculate the position using the formula: egin{equation}P = rac{(n + 1) imes q}{100} egin{equation} where $$P$$ is the position, $$n$$ the number of data points (30 in this case), and $$q$$ the desired quantile (90th percentile). This method interpolates between observed data points to estimate the value.. In practice, various methods for estimating the quantile from empirical data exist, each handling ties and gaps between observations differently.. Conditional Quantile Function: A Detailed Look. The conditional quantile function extends the concept of quantiles into the realm of conditional distributions. It describes the quantile of a random variable conditional on another variable. In other words, it allows us to explore how the quantiles of one variable change as another variable varies.. Conditional Quantile Function: For random variables X and Y, the conditional quantile function of Y given X is denoted as Q(Y\|X) and is defined as: egin{equation}Q_{Y\|X}(p) = infigrace{y: F_{Y\|X}(y) igrace}. egin{equation}This essentially specifies the value of Y such that the probability of Y being less than or equal to this value, given X, is p.. Imagine analyzing the relationship between students' studying hours (X) and their exam scores (Y). The conditional quantile function can help understand how the expected exam scores (at various quantiles) change with an increase in studying hours. If you're interested in the median score given studying hours, you would be looking at the 50th percentile of the conditional distribution of Y given X.. The conditional quantile function is especially useful in econometrics and finance, where variables often exhibit non-linear relationships. For example, it can illuminate how the quantiles of wage distributions change with education level or how the risk of financial loss varies across different quantiles of investment portfolios. This concept is also the backbone of quantile regression, a statistical technique that estimates the conditional median or other quantiles instead of the mean. Quantile regression provides a more comprehensive view of the potential outcomes, especially in the presence of heteroscedasticity or outliers, thus offering a nuanced understanding of the data beyond the average or mean relationship.. Interpreting Quantile Functions. Interpreting quantile functions is essential for understanding the distribution of data within a dataset. This process involves using the quantile function to find values corresponding to specific probabilities or percentages of the data. The interpretation provides insights into the probability distribution of a dataset and helps in making informed decisions based on statistical analysis.The ability to interpret quantile functions is especially valuable in fields that rely on uncertainty and risk assessment, such as finance, meteorology, and health sciences. By understanding the distribution of data through quantile functions, you can predict outcomes and make decisions that are informed by the likelihood of specific events.. Quantile Function Interpretation Steps. Interpreting a quantile function typically involves several key steps:. • Identify the probability or quantile of interest.. • Apply the quantile function to find the corresponding data value.. • Analyse the result in the context of the dataset and the question at hand.. For example, to find the 90th percentile of a dataset using its quantile function, you would input 0.9 into the function. The output is the value below which 90% of the data lies. This value is crucial for understanding the upper range of the dataset and identifying outliers or extreme values.. Let's imagine you're analysing the annual rainfall in a particular region over the past decade and you wish to understand the distribution of rainfall amounts.	If the quantile function for the dataset is given by $Q(p) = 100 + 200p^2$ where $$p$$ is the probability, to find the rainfall amount that was not exceeded 75% of the time, you would substitute $$p = 0.75$$ into the equation to get: $Q(0.75) = 100 + 200(0.75)^2 = 250$ millimetres. This means that 75% of the time, the annual rainfall did not exceed 250 millimetres.. Real-Life Applications of Quantile Function Interpretation. Quantile function interpretation finds its application in various real-life scenarios. Its utility spans across disciplines, offering valuable insights into data distribution and aiding in decision-making:. • In finance, quantile functions are used to assess risk and return profiles, informing investment strategies by evaluating the potential downside or upside of an asset.. • In healthcare, interpreting quantile functions of patient data helps in understanding disease prevalence and outcomes, guiding treatment protocols.. • Environmental scientists utilise quantile functions to predict extreme weather conditions, contributing to disaster preparedness and climate research.. By leveraging the quantile function, professionals in these fields can quantify the likelihood of specified events, making it a powerful tool for analysis and planning.. When interpreting quantile functions, remember that the outcome is highly dependent on the accuracy and distribution of the underlying data. Anomalies or biases in the dataset can lead to misinterpretations.. A fascinating application of quantile function interpretation is in the field of educational research, where it's used to analyse test scores and achievement levels across different student populations. By examining the quantiles of test score distributions, educators and policymakers can identify achievement gaps and tailor interventions to support underperforming groups.This approach allows for a nuanced understanding of educational outcomes beyond mean or median scores, highlighting variability and providing insights into the tails of the distribution. As a result, strategies can be more precisely targeted to uplift students who may be struggling and ensure resources are allocated where they are most needed, ultimately contributing to more equitable educational outcomes.. Practising with Quantile Function Exercises. Quantile function exercises are an excellent way to deepen your understanding of statistical distributions and their applications. These exercises range from basic problems suitable for beginners to more advanced scenarios that challenge even experienced statisticians.. Basic Quantile Function Exercises for Beginners. If you're new to quantile functions, starting with basic exercises can help you grasp their fundamental concepts and applications. These exercises often involve identifying values at specified quantiles and understanding the relationship between quantile functions and probability distributions.Basic exercises are crafted to reinforce the understanding of quantile functions as inverses of cumulative distribution functions (CDFs). By practising with these problems, you'll become more comfortable in handling various types of data distributions and making predictions based on statistical analysis.. Calculate the 25th percentile ( extit{first quartile}) for a dataset using the quantile function $Q(p) = 15 + 5p^3$.To solve this, substitute $$p = 0.25$$ into the quantile function: $Q(0.25) = 15 + 5(0.25)^3 = 15 + 5(0.015625) = 15.078125$.This result indicates that 25% of the data falls below 15.078125.. Remember that the percentile is simply the quantile multiplied by 100. So, the 25th percentile corresponds to the 0.25 quantile.. For those already familiar with the basics of quantile functions, advanced exercises present more complex scenarios that require a deeper understanding of statistical distributions and their analysis. These exercises may involve conditional quantile functions, non-linear distributions, and the application of quantile regression techniques.Advanced exercises aim to challenge your analytical skills and encourage you to apply quantile functions in more sophisticated contexts. By tackling these problems, you'll refine your ability to make precise statistical inferences and predictions.. Assume that the quantile function for a dataset is given by $Q(p) = rac{1}{ig(2p + 0.5ig)^2}$.Calculate the value at the 90th percentile (p = 0.9).Substituting $$p = 0.9$$ into the quantile function gives: $Q(0.9) = rac{1}{ig(2(0.9) + 0.5ig)^2} = rac{1}{(1.8 + 0.5)^2} = rac{1}{(2.3)^2} = rac{1}{5.29}$.This result provides the value below which 90% of the data lies, according to the given quantile function.. Advanced exercises often lead to the exploration of conditional quantile functions, which offer insights into the relationship between two variables. For instance, in modelling the impact of education on income, a conditional quantile function can reveal how the distribution of income changes at different levels of education.Understanding these relationships through quantile functions can yield actionable insights in fields such as economics, meteorology, and health sciences. This goes beyond basic descriptive statistics to allow for a nuanced view of how outcomes change across different segments of a population or under varying conditions.. Quantile function - Key takeaways. • Quantile Function Definition: A function that identifies the value below which a given percentage of observations in a data set lies.. • Quantile vs. Quantile Function: Quantiles divide data into equal parts, whereas the quantile function finds these points for continuous distributions.. • Empirical Quantile Function: Estimates quantiles using the actual data set's distribution when the distribution is unknown.. • Conditional Quantile Function: Determines quantiles of a random variable conditional on another variable, enabling analysis of how one variable affects the distribution of another.. • Interpreting Quantile Functions: Essential for understanding a dataset's distribution, it involves identifying values tied to specific probabilities or percentages within the data.. Learn with 0 Quantile function flashcards in the free StudySmarter app. We have 14,000 flashcards about Dynamic Landscapes.. What is the definition of a quantile function in statistics?. A quantile function, in statistics, is the inverse of the cumulative distribution function (CDF). It provides the value below which a given percentage of observations in a group of observations falls.. How is the quantile function related to the cumulative distribution function?. The quantile function is the inverse of the cumulative distribution function (CDF) for continuous distributions. It maps a probability $$p$$ to the value $$x$$ such that the probability of the random variable being less than or equal to $$x$$ is $$p$$.. What are the applications of a quantile function in data analysis?. In data analysis, quantile functions are crucial for estimating the distribution of a data set, setting thresholds for outlier detection, performing probabilistic forecasting, and simulating data samples from a theoretical distribution to test hypotheses or models. They facilitate comparison between different datasets by enabling the calculation of quantiles.. How do you calculate the value of a quantile function for a given probability?. To calculate the value of a quantile function for a given probability, locate the probability value in the cumulative distribution function (CDF) and identify the corresponding value in the domain of the distribution; this value is the quantile for the specified probability.. What are the differences between a quantile function and a percentile function?. A quantile function determines the value below which a certain proportion of data falls, applicable for any fraction. A percentile function is a specific type of quantile function for divisions into 100 equal parts, focusing on the percent below which a certain proportion of data is found.. StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.. StudySmarter Editorial Team. Team Math Teachers. • Checked by StudySmarter Editorial Team.
http://www.dummies.com/how-to/content/how-to-add-mixed-numbers.html	1,469,487,554,000,000,000	text/html	crawl-data/CC-MAIN-2016-30/segments/1469257824395.52/warc/CC-MAIN-20160723071024-00026-ip-10-185-27-174.ec2.internal.warc.gz	419,157,051	15,001	Whether mixed numbers have the same denominators or different denominators, adding them is a lot like adding whole numbers: you stack them up one on top of the other, draw a line, and add. For this reason, some students feel more comfortable adding mixed numbers than adding fractions. Here’s how to add two mixed numbers: 1. Add the fractional parts and if necessary, change this sum to a mixed number and reduce it. 2. If the answer you found in Step 1 is an improper fraction, change it to a mixed number, write down the fractional part, and carry the whole number part to the whole number column. 3. Add the whole number parts (including any number carried). Your answer may also need to be reduced to lowest terms. The examples that follow show you everything you need to know. ## Add mixed numbers with the same denominators As with any problem involving fractions, adding is always easier when the denominators are the same. For example, suppose you want to add 3 1/3 + 5 1/3. Doing mixed number problems is often easier if you place one number above the other: As you can see, this arrangement is similar to how you add whole numbers, but it includes an extra column for fractions. Here’s how you add these two mixed numbers step by step: Because 2/3 is a proper fraction, you don’t have to change it. 3. Add the whole number parts. 3 + 5 = 8 Here’s how your problem looks in column form: This problem is about as simple as they get. In this case, all three steps are pretty easy. But sometimes, Step 2 requires more attention. For example, suppose you want to add 8 3/5 + 6 4/5. Here’s how you do it: 2. Switch improper fractions to mixed numbers, write down the fractional part, and carry over the whole number. Because the sum is an improper fraction, convert it to the mixed number 1 2/5. Write down 2/5 and carry the 1 over to the whole-number column. 3. Add the whole number parts, including any whole numbers you carried over when you switched to a mixed number. 1 + 8 + 6 = 15 Here’s how the solved problem looks in column form. (Be sure to line up the whole numbers in one column and the fractions in another.) As with any other problems involving fractions, sometimes you need to reduce at the end of Step 1. ## Add mixed numbers with different denominators The most difficult type of mixed number addition is when the denominators of the fractions are different. This difference doesn’t change Steps 2 or 3, but it does make Step 1 tougher. For example, suppose you want to add 16 3/5 and 7 7/9. 2. Switch improper fractions to mixed numbers, write down the fractional part, and carry over the whole number. This fraction is improper, so change it to the mixed number 1 17/45. Fortunately, the fractional part of this mixed number isn’t reducible. Write down the 17/45 and carry over the 1 to the whole-number column. 1 + 16 + 7 = 24 Here’s how the completed problem looks:	685	2,919	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.96875	5	CC-MAIN-2016-30	latest	en	0.894207	Whether mixed numbers have the same denominators or different denominators, adding them is a lot like adding whole numbers: you stack them up one on top of the other, draw a line, and add. For this reason, some students feel more comfortable adding mixed numbers than adding fractions.. Here’s how to add two mixed numbers:. 1. Add the fractional parts and if necessary, change this sum to a mixed number and reduce it.. 2. If the answer you found in Step 1 is an improper fraction, change it to a mixed number, write down the fractional part, and carry the whole number part to the whole number column.. 3. Add the whole number parts (including any number carried).. Your answer may also need to be reduced to lowest terms. The examples that follow show you everything you need to know.. ## Add mixed numbers with the same denominators. As with any problem involving fractions, adding is always easier when the denominators are the same. For example, suppose you want to add 3 1/3 + 5 1/3. Doing mixed number problems is often easier if you place one number above the other:. As you can see, this arrangement is similar to how you add whole numbers, but it includes an extra column for fractions. Here’s how you add these two mixed numbers step by step:. Because 2/3 is a proper fraction, you don’t have to change it.. 3. Add the whole number parts.. 3 + 5 = 8. Here’s how your problem looks in column form:. This problem is about as simple as they get. In this case, all three steps are pretty easy. But sometimes, Step 2 requires more attention.	For example, suppose you want to add 8 3/5 + 6 4/5. Here’s how you do it:. 2. Switch improper fractions to mixed numbers, write down the fractional part, and carry over the whole number.. Because the sum is an improper fraction, convert it to the mixed number 1 2/5. Write down 2/5 and carry the 1 over to the whole-number column.. 3. Add the whole number parts, including any whole numbers you carried over when you switched to a mixed number.. 1 + 8 + 6 = 15. Here’s how the solved problem looks in column form. (Be sure to line up the whole numbers in one column and the fractions in another.). As with any other problems involving fractions, sometimes you need to reduce at the end of Step 1.. ## Add mixed numbers with different denominators. The most difficult type of mixed number addition is when the denominators of the fractions are different. This difference doesn’t change Steps 2 or 3, but it does make Step 1 tougher.. For example, suppose you want to add 16 3/5 and 7 7/9.. 2. Switch improper fractions to mixed numbers, write down the fractional part, and carry over the whole number.. This fraction is improper, so change it to the mixed number 1 17/45. Fortunately, the fractional part of this mixed number isn’t reducible.. Write down the 17/45 and carry over the 1 to the whole-number column.. 1 + 16 + 7 = 24. Here’s how the completed problem looks:.
https://byjus.com/cbse-notes/cbse-class-8-science-notes-chapter-12-friction/	1,695,717,997,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233510179.22/warc/CC-MAIN-20230926075508-20230926105508-00577.warc.gz	177,661,061	118,603	# Friction Class 8 Notes - Chapter 12 According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9. The Class 8 Science chapter 12 Friction discusses friction and various factor affecting it. The force which opposes the relative motion between two surfaces in contact is known as friction. It acts on both surfaces. Listed below are factors that affect friction. ## What Friction? ### Friction force • The external force that opposes relative motion between 2 surfaces in contact. • Friction acts on the surface of contact between both bodies. To know more about Friction, visit here. ### Relative motion When one object moves relative to another, it is called relative motion. ## Why Friction? ### Cause of friction • Friction occurs due to surface irregularities of the two objects in contact. • Adhesive forces between surfaces in contact. • Plowing effect. ### Surface irregularities • All surfaces, when zoomed into a microscopic level, contain hills and valleys that interlock when they move or rub on top of each other. • This unevenness of the surface is called surface irregularities or roughness. • Rough surfaces have larger irregularities, while smoother surfaces have lesser irregularities. • When two surfaces are in contact, they start to form bonds and begin to stick to each other. This phenomenon is called Adhesion. • When we try to move objects that are on top of one another, we are basically breaking the bonds or overcoming the adhesive forces. ### Plowing effect • When surfaces are soft or can change their shape easily, they get deformed when they come in contact with another object. Ex: carpets, when a heavy object is placed on them, it looks like a valley that is caused by the deformation of the shape. • This effect of the surfaces sinking into each other is known as the Plowing effect. ## Factors Affecting Friction ### Factors affecting friction Depends on the nature of the surfaces in contact. (Friction exists between two surfaces), For example, glass and rubber ### Nature of surface in contact • Friction depends on how hard the two surfaces are pressed together; as more surfaces are in contact and more bonds are formed→ more bonds to break → means more friction. • Only the normal reaction force (exactly perpendicular ) to the two surfaces increases friction. ### Calculating frictional force using a spring balance • Using a spring balance, we can find the frictional forces opposed by different materials. • Sandpaper gives a higher reading as compared to stainless steel. ### Polishing surfaces in contact to change friction • Polishing surface reduces irregularities and therefore makes the surface smooth. • Reduces friction. ### Normal reaction force • Force applied that is exactly perpendicular to the surfaces in contact is called normal reaction force. • It increases the frictional force. ### Static Friction Friction due to a body at rest with the surface in contact is called Static friction. ### Kinetic Friction • The friction that comes into play when objects are in motion is called as kinetic friction. • Kinetic friction: * Sliding * Rolling friction To know more about Types of Friction, visit here. ## Friction a Frenemy? ### How does friction produce heat? As friction involves breaking bonds, they make the particles vibrate → increase kinetic energy and therefore increase heat. ### Applications of friction Writing, walking, running, tyres on a car, a nail stays in the wall due to friction, usage of a matchstick. ## Reinvent the Wheel ### Rolling and using treads to change friction • Using ball bearings reduces friction as rolling friction is< other types of friction. • Treads on tyres help expunge water and give better grip by increasing friction. ### Rolling friction • Rolling provides less friction as compared to sliding. • Rolling friction < Sliding friction. • Machines use ball bearings to reduce the friction of moving parts. ## Skydiving Cat ### Drag force • The frictional force exerted by fluids is called drag. • The drag force on an object depends on speed as well as the shape of the body and the nature of the fluid. Also Access NCERT Solutions for Class 8 Science Chapter 12 NCERT Exemplar for Class 8 Science Chapter 12 Learn more about the different types of friction and other related topics, including class 8 Science notes, at BYJU’S. Friction – A necessary evil Types of Friction ## Frequently Asked Questions on CBSE Class 8 Science Notes Chapter 12 Friction Q1 ### What is friction? Friction is referred to as a force acting between two moving or sliding surfaces. Q2 ### What are the benefits of friction? The benefits of friction are that objects can be piled up without sliding, enabling us to walk with a grip, and helping in the transfer of energy from one form to another. Q3 ### What are the examples of adhesive forces? Examples of adhesive forces are painting, icing on cakes, make-up, butter/jam on bread	1,048	5,000	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2023-40	latest	en	0.938536	# Friction Class 8 Notes - Chapter 12. According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9.. The Class 8 Science chapter 12 Friction discusses friction and various factor affecting it. The force which opposes the relative motion between two surfaces in contact is known as friction. It acts on both surfaces. Listed below are factors that affect friction.. ## What Friction?. ### Friction force. • The external force that opposes relative motion between 2 surfaces in contact.. • Friction acts on the surface of contact between both bodies.. To know more about Friction, visit here.. ### Relative motion. When one object moves relative to another, it is called relative motion.. ## Why Friction?. ### Cause of friction. • Friction occurs due to surface irregularities of the two objects in contact.. • Adhesive forces between surfaces in contact.. • Plowing effect.. ### Surface irregularities. • All surfaces, when zoomed into a microscopic level, contain hills and valleys that interlock when they move or rub on top of each other.. • This unevenness of the surface is called surface irregularities or roughness.. • Rough surfaces have larger irregularities, while smoother surfaces have lesser irregularities.. • When two surfaces are in contact, they start to form bonds and begin to stick to each other. This phenomenon is called Adhesion.. • When we try to move objects that are on top of one another, we are basically breaking the bonds or overcoming the adhesive forces.. ### Plowing effect. • When surfaces are soft or can change their shape easily, they get deformed when they come in contact with another object. Ex: carpets, when a heavy object is placed on them, it looks like a valley that is caused by the deformation of the shape.. • This effect of the surfaces sinking into each other is known as the Plowing effect.. ## Factors Affecting Friction. ### Factors affecting friction. Depends on the nature of the surfaces in contact. (Friction exists between two surfaces), For example, glass and rubber. ### Nature of surface in contact. • Friction depends on how hard the two surfaces are pressed together; as more surfaces are in contact and more bonds are formed→ more bonds to break → means more friction.. • Only the normal reaction force (exactly perpendicular ) to the two surfaces increases friction.. ### Calculating frictional force using a spring balance. • Using a spring balance, we can find the frictional forces opposed by different materials.. • Sandpaper gives a higher reading as compared to stainless steel.. ### Polishing surfaces in contact to change friction. • Polishing surface reduces irregularities and therefore makes the surface smooth.. • Reduces friction.	### Normal reaction force. • Force applied that is exactly perpendicular to the surfaces in contact is called normal reaction force.. • It increases the frictional force.. ### Static Friction. Friction due to a body at rest with the surface in contact is called Static friction.. ### Kinetic Friction. • The friction that comes into play when objects are in motion is called as kinetic friction.. • Kinetic friction:. * Sliding. * Rolling friction. To know more about Types of Friction, visit here.. ## Friction a Frenemy?. ### How does friction produce heat?. As friction involves breaking bonds, they make the particles vibrate → increase kinetic energy and therefore increase heat.. ### Applications of friction. Writing, walking, running, tyres on a car, a nail stays in the wall due to friction, usage of a matchstick.. ## Reinvent the Wheel. ### Rolling and using treads to change friction. • Using ball bearings reduces friction as rolling friction is< other types of friction.. • Treads on tyres help expunge water and give better grip by increasing friction.. ### Rolling friction. • Rolling provides less friction as compared to sliding.. • Rolling friction < Sliding friction.. • Machines use ball bearings to reduce the friction of moving parts.. ## Skydiving Cat. ### Drag force. • The frictional force exerted by fluids is called drag.. • The drag force on an object depends on speed as well as the shape of the body and the nature of the fluid.. Also Access NCERT Solutions for Class 8 Science Chapter 12 NCERT Exemplar for Class 8 Science Chapter 12. Learn more about the different types of friction and other related topics, including class 8 Science notes, at BYJU’S.. Friction – A necessary evil Types of Friction. ## Frequently Asked Questions on CBSE Class 8 Science Notes Chapter 12 Friction. Q1. ### What is friction?. Friction is referred to as a force acting between two moving or sliding surfaces.. Q2. ### What are the benefits of friction?. The benefits of friction are that objects can be piled up without sliding, enabling us to walk with a grip, and helping in the transfer of energy from one form to another.. Q3. ### What are the examples of adhesive forces?. Examples of adhesive forces are painting, icing on cakes, make-up, butter/jam on bread.
https://testbook.com/question-answer/two-bikes-cost-rs-40000-each-one-of-them-is-sold--5ea941ddf60d5d33d95543b2	1,628,063,188,000,000,000	text/html	crawl-data/CC-MAIN-2021-31/segments/1627046154796.71/warc/CC-MAIN-20210804045226-20210804075226-00497.warc.gz	518,679,745	20,911	# Two bikes cost Rs. 40000 each. One of them is sold for Rs. 48000. At what price should the other be sold to get an overall profit of 25%? Free Practice With Testbook Mock Tests ## Options: 1. Rs. 48,000 2. Rs. 52,000 3. Rs. 50,000 4. Rs. 60,000 ### Correct Answer: Option 2 (Solution Below) This question was previously asked in SSC MTS Previous Paper 27 (Held On: 16 August 2019 Shift 3) ## Solution: Cost of bike = 40,000 Total cost price of both bikes = 40,000 × 2 = 80,000 Total profit = 80,000 × 25/100 = 20,000 Profit on first bike = 48,000 – 40,000 = 8,000 Profit on second bike should be = 20,000 – 8,000 = 12,000 Selling price of the second bike = 40,000 + 12,000 = 52,000	244	699	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.96875	4	CC-MAIN-2021-31	latest	en	0.930919	# Two bikes cost Rs. 40000 each. One of them is sold for Rs. 48000. At what price should the other be sold to get an overall profit of 25%?. Free Practice With Testbook Mock Tests. ## Options:. 1. Rs. 48,000. 2. Rs. 52,000. 3. Rs.	50,000. 4. Rs. 60,000. ### Correct Answer: Option 2 (Solution Below). This question was previously asked in. SSC MTS Previous Paper 27 (Held On: 16 August 2019 Shift 3). ## Solution:. Cost of bike = 40,000. Total cost price of both bikes = 40,000 × 2 = 80,000. Total profit = 80,000 × 25/100 = 20,000. Profit on first bike = 48,000 – 40,000 = 8,000. Profit on second bike should be = 20,000 – 8,000 = 12,000. Selling price of the second bike = 40,000 + 12,000 = 52,000.
http://www.numbersaplenty.com/121233011	1,582,577,131,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875145981.35/warc/CC-MAIN-20200224193815-20200224223815-00379.warc.gz	219,988,086	3,450	Search a number 121233011 = 1091111121 BaseRepresentation bin1110011100111… …01111001110011 322110010021102112 413032131321303 5222013424021 620010235535 73001315004 oct716357163 9273107375 10121233011 1162484179 1234725bab 131c169213 141215b1ab hex739de73 121233011 has 4 divisors (see below), whose sum is σ = 121345224. Its totient is φ = 121120800. The previous prime is 121232987. The next prime is 121233019. The reversal of 121233011 is 110332121. Adding to 121233011 its reverse (110332121), we get a palindrome (231565132). It is a semiprime because it is the product of two primes. It is a cyclic number. It is not a de Polignac number, because 121233011 - 210 = 121231987 is a prime. It is a super-2 number, since 2×1212330112 = 29394885912252242, which contains 22 as substring. It is a Duffinian number. It is not an unprimeable number, because it can be changed into a prime (121233019) by changing a digit. It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 54470 + ... + 56651. It is an arithmetic number, because the mean of its divisors is an integer number (30336306). Almost surely, 2121233011 is an apocalyptic number. 121233011 is a deficient number, since it is larger than the sum of its proper divisors (112213). 121233011 is a wasteful number, since it uses less digits than its factorization. 121233011 is an evil number, because the sum of its binary digits is even. The sum of its prime factors is 112212. The product of its (nonzero) digits is 36, while the sum is 14. The square root of 121233011 is about 11010.5863149970. The cubic root of 121233011 is about 494.9260318682. The spelling of 121233011 in words is "one hundred twenty-one million, two hundred thirty-three thousand, eleven". Divisors: 1 1091 111121 121233011	546	1,829	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2020-10	latest	en	0.842788	Search a number. 121233011 = 1091111121. BaseRepresentation. bin1110011100111…. …01111001110011. 322110010021102112. 413032131321303. 5222013424021. 620010235535. 73001315004. oct716357163. 9273107375. 10121233011. 1162484179. 1234725bab. 131c169213. 141215b1ab. hex739de73. 121233011 has 4 divisors (see below), whose sum is σ = 121345224. Its totient is φ = 121120800.. The previous prime is 121232987. The next prime is 121233019.	The reversal of 121233011 is 110332121.. Adding to 121233011 its reverse (110332121), we get a palindrome (231565132).. It is a semiprime because it is the product of two primes.. It is a cyclic number.. It is not a de Polignac number, because 121233011 - 210 = 121231987 is a prime.. It is a super-2 number, since 2×1212330112 = 29394885912252242, which contains 22 as substring.. It is a Duffinian number.. It is not an unprimeable number, because it can be changed into a prime (121233019) by changing a digit.. It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 54470 + ... + 56651.. It is an arithmetic number, because the mean of its divisors is an integer number (30336306).. Almost surely, 2121233011 is an apocalyptic number.. 121233011 is a deficient number, since it is larger than the sum of its proper divisors (112213).. 121233011 is a wasteful number, since it uses less digits than its factorization.. 121233011 is an evil number, because the sum of its binary digits is even.. The sum of its prime factors is 112212.. The product of its (nonzero) digits is 36, while the sum is 14.. The square root of 121233011 is about 11010.5863149970. The cubic root of 121233011 is about 494.9260318682.. The spelling of 121233011 in words is "one hundred twenty-one million, two hundred thirty-three thousand, eleven".. Divisors: 1 1091 111121 121233011.
https://sa.wikisource.org/wiki/%E0%A4%AA%E0%A5%83%E0%A4%B7%E0%A5%8D%E0%A4%A0%E0%A4%AE%E0%A5%8D:%E0%A4%97%E0%A4%A3%E0%A4%BF%E0%A4%A4%E0%A4%B8%E0%A4%BE%E0%A4%B0%E0%A4%B8%E0%A4%99%E0%A5%8D%E0%A4%97%E0%A5%8D%E0%A4%B0%E0%A4%B9%E0%A4%83%E0%A5%92%E0%A4%B0%E0%A4%99%E0%A5%8D%E0%A4%97%E0%A4%BE%E0%A4%9A%E0%A4%BE%E0%A4%B0%E0%A5%8D%E0%A4%AF%E0%A5%87%E0%A4%A3%E0%A4%BE%E0%A4%A8%E0%A5%82%E0%A4%A6%E0%A4%BF%E0%A4%A4%E0%A4%83%E0%A5%92%E0%A5%A7%E0%A5%AF%E0%A5%A7%E0%A5%A8.djvu/%E0%A5%A9%E0%A5%A8%E0%A5%AE	1,597,037,617,000,000,000	text/html	crawl-data/CC-MAIN-2020-34/segments/1596439738609.73/warc/CC-MAIN-20200810042140-20200810072140-00072.warc.gz	470,025,572	11,932	# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३२८ एतत् पृष्ठम् परिष्कृतम् अस्ति 132 GAŅITASĀRASAṄGRAHA. An example illustration thereof. 144${\displaystyle {\tfrac {1}{2}}}$ and 145${\displaystyle {\tfrac {1}{2}}}$. There are here fragrant citrons, plantains, wood-apples and pomegranates mixed up (in three heaps). The number of fruits in the first (heap) is 21, in the second 22, and in the third 23. The combined price of each of these (leaps) is 73. What is the number of tho (various) fruits (in each of the heaps), and what the price (of the different varieties of fruits) ? The rule for arriving at the numerical value of the prices of dearer and cheaper things (respectively) from the given mixed value (of their total price):- 146${\displaystyle {\tfrac {1}{2}}}$.[*] Divide (the rate-quantities of the given things) by their rate-prices. Diminish (these resulting quantities separately) by the least among them. Then multiply by the least (of the above mentioned quotient-quantities) the given mixed price of all the things and subtract (this product) from the given (total number of the various) things. Then split up (this remainder optionally) into as many (bits as there are remainders of the above quotient quantities left after subtraction); and then divide (these bits by those remainders of the quotient-quantities. Thus the prices of the various cheaper things are arrived at). These, separated from the total price, give rise to the price of the dearest article of purchase. Examples in illustration thereof. 147${\displaystyle {\tfrac {1}{2}}}$ to 149. "In accordance with the rates of 3 peacocks for 2 paņas 4 pigeons for 3 paņas,5 swans for 4 paņas, and 6 sārasa 146${\displaystyle {\tfrac {1}{2}}}$.^ The rule will be clear from the following working of the problem given in 147${\displaystyle {\tfrac {1}{2}}}$-149:- Divide the ratio-quantities 3, 4, 5, 6 by the respective rate-prices 2, 3, 4, 5; thus we have ${\displaystyle {\tfrac {3}{2}},{\tfrac {4}{3}},{\tfrac {5}{4}},{\tfrac {6}{5}}}$. Subtract the least of these ${\displaystyle {\tfrac {6}{5}}}$ from each of the other three. we get ${\displaystyle {\tfrac {3}{10}},{\tfrac {2}{15}},{\tfrac {1}{20}}}$. By multiplying the given mixed price, 56 by the above mentioned least quantity ${\displaystyle {\tfrac {6}{5}}}$, we have ${\displaystyle 56\times {\tfrac {6}{5}}}$. Subtract this from the total number of birds, 72. Split up the remainder ${\displaystyle {\tfrac {24}{5}}}$ into any three parts, say ${\displaystyle {\tfrac {7}{5}},{\tfrac {6}{5}},{\tfrac {9}{5}}}$. Dividing these respectively by ${\displaystyle {\tfrac {3}{10}},{\tfrac {2}{15}},{\tfrac {1}{20}}}$ we get the prices of the first three kinds of birds,${\displaystyle {\tfrac {14}{3}},12,36}$. The price of the fourth variety of birds can be found out by subtracting all these three prices from the total 56.	878	2,883	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.859375	4	CC-MAIN-2020-34	latest	en	0.868412	# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३२८. एतत् पृष्ठम् परिष्कृतम् अस्ति. 132. GAŅITASĀRASAṄGRAHA.. An example illustration thereof.. 144${\displaystyle {\tfrac {1}{2}}}$ and 145${\displaystyle {\tfrac {1}{2}}}$. There are here fragrant citrons, plantains, wood-apples and pomegranates mixed up (in three heaps). The number of fruits in the first (heap) is 21, in the second 22, and in the third 23. The combined price of each of these (leaps) is 73. What is the number of tho (various) fruits (in each of the heaps), and what the price (of the different varieties of fruits) ?. The rule for arriving at the numerical value of the prices of dearer and cheaper things (respectively) from the given mixed value (of their total price):-. 146${\displaystyle {\tfrac {1}{2}}}$.[*] Divide (the rate-quantities of the given things) by their rate-prices. Diminish (these resulting quantities separately) by the least among them.	Then multiply by the least (of the above mentioned quotient-quantities) the given mixed price of all the things and subtract (this product) from the given (total number of the various) things. Then split up (this remainder optionally) into as many (bits as there are remainders of the above quotient quantities left after subtraction); and then divide (these bits by those remainders of the quotient-quantities. Thus the prices of the various cheaper things are arrived at). These, separated from the total price, give rise to the price of the dearest article of purchase.. Examples in illustration thereof.. 147${\displaystyle {\tfrac {1}{2}}}$ to 149. "In accordance with the rates of 3 peacocks for 2 paņas 4 pigeons for 3 paņas,5 swans for 4 paņas, and 6 sārasa. 146${\displaystyle {\tfrac {1}{2}}}$.^ The rule will be clear from the following working of the problem given in 147${\displaystyle {\tfrac {1}{2}}}$-149:-. Divide the ratio-quantities 3, 4, 5, 6 by the respective rate-prices 2, 3, 4, 5; thus we have ${\displaystyle {\tfrac {3}{2}},{\tfrac {4}{3}},{\tfrac {5}{4}},{\tfrac {6}{5}}}$. Subtract the least of these ${\displaystyle {\tfrac {6}{5}}}$ from each of the other three. we get ${\displaystyle {\tfrac {3}{10}},{\tfrac {2}{15}},{\tfrac {1}{20}}}$. By multiplying the given mixed price, 56 by the above mentioned least quantity ${\displaystyle {\tfrac {6}{5}}}$, we have ${\displaystyle 56\times {\tfrac {6}{5}}}$. Subtract this from the total number of birds, 72. Split up the remainder ${\displaystyle {\tfrac {24}{5}}}$ into any three parts, say ${\displaystyle {\tfrac {7}{5}},{\tfrac {6}{5}},{\tfrac {9}{5}}}$. Dividing these respectively by ${\displaystyle {\tfrac {3}{10}},{\tfrac {2}{15}},{\tfrac {1}{20}}}$ we get the prices of the first three kinds of birds,${\displaystyle {\tfrac {14}{3}},12,36}$. The price of the fourth variety of birds can be found out by subtracting all these three prices from the total 56.
https://gmatclub.com/forum/a-ds-question-from-gmatprep-53868.html?fl=similar	1,508,622,211,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187824899.43/warc/CC-MAIN-20171021205648-20171021225648-00682.warc.gz	703,380,496	42,838	It is currently 21 Oct 2017, 14:43 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # A DS question from GMATPrep Author Message Manager Joined: 13 Mar 2007 Posts: 98 Kudos [?]: 36 [0], given: 0 A DS question from GMATPrep [#permalink] ### Show Tags 13 Oct 2007, 16:25 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 0 sessions ### HideShow timer Statistics This topic is locked. If you want to discuss this question please re-post it in the respective forum. If you know how to solve this problem and you can explain your answer, please let me know. If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10? 1) n = 2 2) m = 1 Here what I did: According to a formula that found online, whenever we have x divided by y we can write it as: x = y + remainder. Now we have: 3^(4n+2) + m = 10 + remainder, thus we need to k now n and m in order to find remainder. However, according to the solution, we only need to know m. So, I am getting confuse about this question. Please help me if you know how to solve this problem. Thanks! Kudos [?]: 36 [0], given: 0 VP Joined: 09 Jul 2007 Posts: 1098 Kudos [?]: 141 [0], given: 0 Location: London Re: A DS question from GMATPrep [#permalink] ### Show Tags 13 Oct 2007, 16:40 GMAT_700 wrote: If you know how to solve this problem and you can explain your answer, please let me know. If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10? 1) n = 2 2) m = 1 Here what I did: According to a formula that found online, whenever we have x divided by y we can write it as: x = y + remainder. Now we have: 3^(4n+2) + m = 10 + remainder, thus we need to k now n and m in order to find remainder. However, according to the solution, we only need to know m. So, I am getting confuse about this question. Please help me if you know how to solve this problem. Thanks! Gmat does not throw questions that can not be solved by following easy steps. 3^(4n+2)+m 1. n=2, 3^10+m we cannot say anything without knowing the value, it is smth. (...7+m)/10 m can be anything and can yield any remainder. 2. m=1, 3^(4n+2)+1 as we know n is a positive integer. thus put any number from 1 to million notice 3^1=3 3^2=9 3^3=27 3^4=81 3^5=243 3^6=...9 when n=1 3^7=..7 3^8=..1 3^9=..3 3^10=..9 when n=2 3^14=..9 when n=3 and so on so there is no remainder when 3^(4n+2)+1 as it is smth ...0/10 B Kudos [?]: 141 [0], given: 0 CEO Joined: 29 Mar 2007 Posts: 2554 Kudos [?]: 516 [0], given: 0 Re: A DS question from GMATPrep [#permalink] ### Show Tags 13 Oct 2007, 23:15 Ravshonbek wrote: GMAT_700 wrote: If you know how to solve this problem and you can explain your answer, please let me know. If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10? 1) n = 2 2) m = 1 Here what I did: According to a formula that found online, whenever we have x divided by y we can write it as: x = y + remainder. Now we have: 3^(4n+2) + m = 10 + remainder, thus we need to k now n and m in order to find remainder. However, according to the solution, we only need to know m. So, I am getting confuse about this question. Please help me if you know how to solve this problem. Thanks! Gmat does not throw questions that can not be solved by following easy steps. 3^(4n+2)+m 1. n=2, 3^10+m we cannot say anything without knowing the value, it is smth. (...7+m)/10 m can be anything and can yield any remainder. 2. m=1, 3^(4n+2)+1 as we know n is a positive integer. thus put any number from 1 to million notice 3^1=3 3^2=9 3^3=27 3^4=81 3^5=243 3^6=...9 when n=1 3^7=..7 3^8=..1 3^9=..3 3^10=..9 when n=2 3^14=..9 when n=3 and so on so there is no remainder when 3^(4n+2)+1 as it is smth ...0/10 B Well said. There is a thread on this problem already b/c I remember writing on it. Kudos [?]: 516 [0], given: 0 Re: A DS question from GMATPrep [#permalink] 13 Oct 2007, 23:15 Display posts from previous: Sort by	1,448	4,527	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.03125	4	CC-MAIN-2017-43	latest	en	0.916819	It is currently 21 Oct 2017, 14:43. ### GMAT Club Daily Prep. #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. # Events & Promotions. ###### Events & Promotions in June. Open Detailed Calendar. # A DS question from GMATPrep. Author Message. Manager. Joined: 13 Mar 2007. Posts: 98. Kudos [?]: 36 [0], given: 0. A DS question from GMATPrep [#permalink]. ### Show Tags. 13 Oct 2007, 16:25. 00:00. Difficulty:. (N/A). Question Stats:. 0% (00:00) correct 0% (00:00) wrong based on 0 sessions. ### HideShow timer Statistics. This topic is locked. If you want to discuss this question please re-post it in the respective forum.. If you know how to solve this problem and you can explain your answer, please let me know.. If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10?. 1) n = 2. 2) m = 1. Here what I did:. According to a formula that found online, whenever we have x divided by y we can write it as: x = y + remainder.. Now we have: 3^(4n+2) + m = 10 + remainder, thus we need to k now n and m in order to find remainder. However, according to the solution, we only need to know m. So, I am getting confuse about this question. Please help me if you know how to solve this problem. Thanks!. Kudos [?]: 36 [0], given: 0. VP. Joined: 09 Jul 2007. Posts: 1098. Kudos [?]: 141 [0], given: 0. Location: London. Re: A DS question from GMATPrep [#permalink]. ### Show Tags. 13 Oct 2007, 16:40. GMAT_700 wrote:. If you know how to solve this problem and you can explain your answer, please let me know.. If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10?. 1) n = 2. 2) m = 1. Here what I did:. According to a formula that found online, whenever we have x divided by y we can write it as: x = y + remainder.. Now we have: 3^(4n+2) + m = 10 + remainder, thus we need to k now n and m in order to find remainder. However, according to the solution, we only need to know m. So, I am getting confuse about this question. Please help me if you know how to solve this problem. Thanks!. Gmat does not throw questions that can not be solved by following easy steps.. 3^(4n+2)+m. 1. n=2, 3^10+m we cannot say anything without knowing the value, it is smth. (...7+m)/10 m can be anything and can yield any remainder.. 2. m=1, 3^(4n+2)+1.	as we know n is a positive integer. thus put any number from 1 to million. notice. 3^1=3. 3^2=9. 3^3=27. 3^4=81. 3^5=243. 3^6=...9 when n=1. 3^7=..7. 3^8=..1. 3^9=..3. 3^10=..9 when n=2. 3^14=..9. when n=3. and so on. so there is no remainder when 3^(4n+2)+1 as it is smth ...0/10. B. Kudos [?]: 141 [0], given: 0. CEO. Joined: 29 Mar 2007. Posts: 2554. Kudos [?]: 516 [0], given: 0. Re: A DS question from GMATPrep [#permalink]. ### Show Tags. 13 Oct 2007, 23:15. Ravshonbek wrote:. GMAT_700 wrote:. If you know how to solve this problem and you can explain your answer, please let me know.. If n and m are positive integers, what is the remainder when 3^(4n+2) + m is divided by 10?. 1) n = 2. 2) m = 1. Here what I did:. According to a formula that found online, whenever we have x divided by y we can write it as: x = y + remainder.. Now we have: 3^(4n+2) + m = 10 + remainder, thus we need to k now n and m in order to find remainder. However, according to the solution, we only need to know m. So, I am getting confuse about this question. Please help me if you know how to solve this problem. Thanks!. Gmat does not throw questions that can not be solved by following easy steps.. 3^(4n+2)+m. 1. n=2, 3^10+m we cannot say anything without knowing the value, it is smth. (...7+m)/10 m can be anything and can yield any remainder.. 2. m=1, 3^(4n+2)+1. as we know n is a positive integer. thus put any number from 1 to million. notice. 3^1=3. 3^2=9. 3^3=27. 3^4=81. 3^5=243. 3^6=...9 when n=1. 3^7=..7. 3^8=..1. 3^9=..3. 3^10=..9 when n=2. 3^14=..9. when n=3. and so on. so there is no remainder when 3^(4n+2)+1 as it is smth ...0/10. B. Well said.. There is a thread on this problem already b/c I remember writing on it.. Kudos [?]: 516 [0], given: 0. Re: A DS question from GMATPrep [#permalink] 13 Oct 2007, 23:15. Display posts from previous: Sort by.

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