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https://math.stackexchange.com/questions/1834472/approximate-greatest-common-divisor
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# approximate greatest common divisor
I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers.
For example, I have the following numbers:
399, 710, 105, 891, 402, 102, 397, ...
As you can see, the average gcd is approximately 100, but how to compute it ?
more details:
I try to find the carrier signal length of a HF signal. This signal is an alternation of high levels and low levels.
eg. ----__------____------__-- ...
I have the duration of each level, but this time is not accurate.
My aim is to find as quick as possible the base time of the signal.
My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate.
With a perfect signal I would have gcd(400, 700, 100, 900, 400, 100, 400) = 100
• What is the average gcd? 397 is a prime, so its gcd with any of the other numbers in the series is 1. Jun 21, 2016 at 13:24
• Does computing all gcds then taking the average not work..? Jun 21, 2016 at 13:25
• @MattSamuel, all gcds give very low values, eg. (399, 710) => 1, etc..., then the average will be far of the expected value Jun 21, 2016 at 13:28
• @Soubok: No matter which word you use, it looks like you will have to describe what it is you want, with greater detail and specificity than just choosing a word to use for it. Jun 21, 2016 at 13:46
• Try to reformulate then, it might be an interesting problem. You'll need some kind of norm or measure, like the maximum of $\gcd(a_1+e_1,\dots,a_n+e_n)$ where $e_1+\cdots e_n<N$...
– Lehs
Jun 21, 2016 at 14:25
I made a similar question here, where I propose a partial solution.
How to find the approximate basic frequency or GCD of a list of numbers?
In summary, I came with this
• being $v$ the list $\{v_1, v_2, \ldots, v_n\}$,
• $\operatorname{mean}_{\sin}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\sin(2\pi v_i/x)$
• $\operatorname{mean}_{\cos}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\cos(2\pi v_i/x)$
• $\operatorname{gcd}_{appeal}(v, x)$ = $1 - \frac{1}{2}\sqrt{\operatorname{mean}_{\sin}(v, x)^2 + (\operatorname{mean}_{\cos}(v, x) - 1)^2}$
And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$. Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123:
testSeq = [399, 710, 105, 891, 402, 102, 397]
gcd = calculateGCDAppeal(x, testSeq)
find_local_maximum(gcd,90,110)
plot(gcd,(x, 10, 200), scale = "semilogx")
One approach: take the minimum of all your numbers, here $102$ as the first trial. Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval. Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers?
If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well.
This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it.
Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd.
Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$|r_i|$$ is very small, say, smaller than $$50$$.
Then, you can take two of those values, say, $$x_1$$ and $$x_2$$, and compute $$d_{a, b} = \gcd(x_1 - a, x_2 - b)$$ for all $$a, b \in [-50, 50]\cap\mathbb{Z}$$. Notice that this step costs only $$50^2$$ gcds computations, which is very cheap for any computer.
It is guaranteed that one $$d_{a,b}$$ is equal to $$p$$. Thus, to check which one is the correct one, just compute the "centered modular reduction" $$r_i' := x_i \bmod d_{a,b} \in [-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$ for $$i \ge 3$$.
If $$d_{a,b} = p$$, then each $$r_i'$$ equals $$r_i$$, therefore, all the values $$r_i'$$ that you get are small (e.g., smaller than 50). Otherwise, the values $$r_i'$$ will look randomly distributed in $$[-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$.
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# approximate greatest common divisor. I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers.. For example, I have the following numbers:. 399, 710, 105, 891, 402, 102, 397, .... As you can see, the average gcd is approximately 100, but how to compute it ?. more details:. I try to find the carrier signal length of a HF signal. This signal is an alternation of high levels and low levels.. eg. ----__------____------__-- .... I have the duration of each level, but this time is not accurate.. My aim is to find as quick as possible the base time of the signal.. My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate.. With a perfect signal I would have gcd(400, 700, 100, 900, 400, 100, 400) = 100. • What is the average gcd? 397 is a prime, so its gcd with any of the other numbers in the series is 1. Jun 21, 2016 at 13:24. • Does computing all gcds then taking the average not work..? Jun 21, 2016 at 13:25. • @MattSamuel, all gcds give very low values, eg. (399, 710) => 1, etc..., then the average will be far of the expected value Jun 21, 2016 at 13:28. • @Soubok: No matter which word you use, it looks like you will have to describe what it is you want, with greater detail and specificity than just choosing a word to use for it. Jun 21, 2016 at 13:46. • Try to reformulate then, it might be an interesting problem. You'll need some kind of norm or measure, like the maximum of $\gcd(a_1+e_1,\dots,a_n+e_n)$ where $e_1+\cdots e_n<N$.... – Lehs. Jun 21, 2016 at 14:25. I made a similar question here, where I propose a partial solution.. How to find the approximate basic frequency or GCD of a list of numbers?. In summary, I came with this. • being $v$ the list $\{v_1, v_2, \ldots, v_n\}$,. • $\operatorname{mean}_{\sin}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\sin(2\pi v_i/x)$. • $\operatorname{mean}_{\cos}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\cos(2\pi v_i/x)$. • $\operatorname{gcd}_{appeal}(v, x)$ = $1 - \frac{1}{2}\sqrt{\operatorname{mean}_{\sin}(v, x)^2 + (\operatorname{mean}_{\cos}(v, x) - 1)^2}$. And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$. Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123:. testSeq = [399, 710, 105, 891, 402, 102, 397]. gcd = calculateGCDAppeal(x, testSeq). find_local_maximum(gcd,90,110). plot(gcd,(x, 10, 200), scale = "semilogx"). One approach: take the minimum of all your numbers, here $102$ as the first trial. Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval. Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers?. If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well.. This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it.. Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd.. Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$|r_i|$$ is very small, say, smaller than $$50$$.. Then, you can take two of those values, say, $$x_1$$ and $$x_2$$, and compute $$d_{a, b} = \gcd(x_1 - a, x_2 - b)$$ for all $$a, b \in [-50, 50]\cap\mathbb{Z}$$. Notice that this step costs only $$50^2$$ gcds computations, which is very cheap for any computer.
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It is guaranteed that one $$d_{a,b}$$ is equal to $$p$$. Thus, to check which one is the correct one, just compute the "centered modular reduction" $$r_i' := x_i \bmod d_{a,b} \in [-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$ for $$i \ge 3$$.. If $$d_{a,b} = p$$, then each $$r_i'$$ equals $$r_i$$, therefore, all the values $$r_i'$$ that you get are small (e.g., smaller than 50). Otherwise, the values $$r_i'$$ will look randomly distributed in $$[-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$.
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View Question
Question
Subject: Java Programming - Quadratic Equations Category: Computers > Programming Asked by: java_design-ga List Price: \$15.00 Posted: 29 Nov 2005 05:34 PST Expires: 30 Nov 2005 09:47 PST Question ID: 598931
```Design and develop a Java program that continuously computes and displays value(s) for x, given quadratic equations (i.e. a second-order polynomials) of the form: ax2 + bx + c = 0 where the values for the coefficients a, b and c are supplied by the user, and are assumed to be integers within the range of -100 to 100. To control the loop use a menu interface. The menu should include two options: "Calculate quadratic" and "End". Note that to solve a quadratic equation we must calculate the roots. This can be done using the quadratic formula: root 1 = (-b + sqrt(b2-4ac)) / 2a root2 = (-b - sqrt(b2-4ac)) / 2a Example: x2 + 2x - 8 = 0 a= 1, b = 2, c = -8 roots = (-2 +or- sqrt(22-4x1x-8)) / 2x1 = (-2 +or- sqrt(4+32)) / 2 root1 = (-2 + 6)/2 = 4/2 = 2.0 root2 = (-2 - 6)/2 = -8/2 = -4.0 x = 2.0 or -4.0 However, there are certain special consideration to be taken into account: If a and b are both zero there is no solution (this is referred to as the degenerate case): -8 = 0? a= 0, b = 0, c = -8 (degenerate case) If a is zero and b is non zero the equation becomes a linear equation. 2x - 8 = 0 a= 0, b = 2, c = -8 (Linear equation) root = -c/b = 8/2 = 4.0 x = 4.0 If the value for the term b2 - 4ac (the discriminant) is negative there is no solution (conventionally we cannot find the square root of a negative number!): x2 + 2x + 8 = 0 a= 1, b = 2, c = 8 roots = (-2 +or- sqrt(22-4x1x8)) / 2x1 = (-2 +or- sqrt(4-32)) / 2 = (-2 +or- sqrt(-28)) Negative discriminant therefore no solution. If the discriminant is 0 then there are two identical solutions, i.e. only one solution (root) need be calculated: x2 + 4x + 4 = 0 a= 1, b = 4, c = 4 roots = (-4 +or- sqrt(42-4x1x4)) / 2x1 = (-4 +or- sqrt(16-16)) / 2 (Discriminant = 0, there fore only one solution) root = -4/2 = -2 x = -2.0 Output, where appropriate, should be accurate to at least several decimal places. Please try to include explanations where appropriate.``` Clarification of Question by java_design-ga on 29 Nov 2005 05:39 PST `The program must be written in Java 1.5 !!!`
`I'd do it for \$200. I wouldn't do it for \$15.`
```import java.io.*; public class quadratic { public int a = 0,b = 0,c = 0; public int flag=0; public double r1=0,r2=0; public quadratic() { do { System.out.println("\n\n\n\n\nType 1 to Calculate quadratic equation"); System.out.println("Type 3 to END"); int choice = getChoice(); switch(choice) { case 1: a = inputABC("a"); b = inputABC("b"); c = inputABC("c"); if((Math.pow(b,2)-4*a*c)<0) {System.out.println("\nNegative discriminant therefore no solution!");} else if((Math.pow(b,2)-4*a*c)==0) { r1 = getRoot1(a,b,c); System.out.print("x= "+r1); } else if(a==0&&b==0) {System.out.println("\nDegenerate Case");} else if(a==0) {double u = -c/b; System.out.println("The root is "+u);} else { r1 = getRoot1(a,b,c); r2 = getRoot2(a,b,c); System.out.print("x= "+r1+" or "+r2); } break; case 3: flag = 1; break; default: System.out.println("\nThat is not an option! Try Again."); break; } }while(flag==0); } public int inputABC(String s2) { InputStreamReader stdin = new InputStreamReader(System.in); BufferedReader console = new BufferedReader(stdin); int i1=0; String s1; try { System.out.print("Enter interger for "+s2+": "); s1 = console.readLine(); i1 = Integer.parseInt(s1); } catch(IOException ioex) { System.out.println("\nInput error"); System.exit(1); } catch(NumberFormatException nfex) { System.out.println("\"" + nfex.getMessage() + "\" is not numeric"); System.exit(1); } return(i1); } public int getChoice() { InputStreamReader stdin = new InputStreamReader(System.in); BufferedReader console = new BufferedReader(stdin); int i2=0; String s2; try { System.out.print("User's Choice: "); s2 = console.readLine(); i2 = Integer.parseInt(s2); } catch(IOException ioex) { System.out.println("Input error"); System.exit(1); } catch(NumberFormatException nfex) { System.out.println("\"" + nfex.getMessage() + "\" is not numeric"); System.exit(1); } return(i2); } public double getRoot1(int x, int y, int z) { double root1 = (-y + Math.sqrt((Math.pow(b,2))-(4*x*z)))/(2*x); return(root1); } public double getRoot2(int x, int y, int z) { double root2 = (-y - Math.sqrt((Math.pow(b,2))-(4*x*z)))/(2*x); return(root2); } public static void main(String[] args) { quadratic qd = new quadratic(); } }```
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View Question. Question. Subject: Java Programming - Quadratic Equations Category: Computers > Programming Asked by: java_design-ga List Price: \$15.00 Posted: 29 Nov 2005 05:34 PST Expires: 30 Nov 2005 09:47 PST Question ID: 598931. ```Design and develop a Java program that continuously computes and displays value(s) for x, given quadratic equations (i.e. a second-order polynomials) of the form: ax2 + bx + c = 0 where the values for the coefficients a, b and c are supplied by the user, and are assumed to be integers within the range of -100 to 100. To control the loop use a menu interface. The menu should include two options: "Calculate quadratic" and "End". Note that to solve a quadratic equation we must calculate the roots. This can be done using the quadratic formula: root 1 = (-b + sqrt(b2-4ac)) / 2a root2 = (-b - sqrt(b2-4ac)) / 2a Example: x2 + 2x - 8 = 0 a= 1, b = 2, c = -8 roots = (-2 +or- sqrt(22-4x1x-8)) / 2x1 = (-2 +or- sqrt(4+32)) / 2 root1 = (-2 + 6)/2 = 4/2 = 2.0 root2 = (-2 - 6)/2 = -8/2 = -4.0 x = 2.0 or -4.0 However, there are certain special consideration to be taken into account: If a and b are both zero there is no solution (this is referred to as the degenerate case): -8 = 0? a= 0, b = 0, c = -8 (degenerate case) If a is zero and b is non zero the equation becomes a linear equation. 2x - 8 = 0 a= 0, b = 2, c = -8 (Linear equation) root = -c/b = 8/2 = 4.0 x = 4.0 If the value for the term b2 - 4ac (the discriminant) is negative there is no solution (conventionally we cannot find the square root of a negative number!): x2 + 2x + 8 = 0 a= 1, b = 2, c = 8 roots = (-2 +or- sqrt(22-4x1x8)) / 2x1 = (-2 +or- sqrt(4-32)) / 2 = (-2 +or- sqrt(-28)) Negative discriminant therefore no solution. If the discriminant is 0 then there are two identical solutions, i.e. only one solution (root) need be calculated: x2 + 4x + 4 = 0 a= 1, b = 4, c = 4 roots = (-4 +or- sqrt(42-4x1x4)) / 2x1 = (-4 +or- sqrt(16-16)) / 2 (Discriminant = 0, there fore only one solution) root = -4/2 = -2 x = -2.0 Output, where appropriate, should be accurate to at least several decimal places. Please try to include explanations where appropriate.``` Clarification of Question by java_design-ga on 29 Nov 2005 05:39 PST `The program must be written in Java 1.5 !!!`. `I'd do it for \$200.
|
I wouldn't do it for \$15.`. ```import java.io.*; public class quadratic { public int a = 0,b = 0,c = 0; public int flag=0; public double r1=0,r2=0; public quadratic() { do { System.out.println("\n\n\n\n\nType 1 to Calculate quadratic equation"); System.out.println("Type 3 to END"); int choice = getChoice(); switch(choice) { case 1: a = inputABC("a"); b = inputABC("b"); c = inputABC("c"); if((Math.pow(b,2)-4*a*c)<0) {System.out.println("\nNegative discriminant therefore no solution!");} else if((Math.pow(b,2)-4*a*c)==0) { r1 = getRoot1(a,b,c); System.out.print("x= "+r1); } else if(a==0&&b==0) {System.out.println("\nDegenerate Case");} else if(a==0) {double u = -c/b; System.out.println("The root is "+u);} else { r1 = getRoot1(a,b,c); r2 = getRoot2(a,b,c); System.out.print("x= "+r1+" or "+r2); } break; case 3: flag = 1; break; default: System.out.println("\nThat is not an option! Try Again."); break; } }while(flag==0); } public int inputABC(String s2) { InputStreamReader stdin = new InputStreamReader(System.in); BufferedReader console = new BufferedReader(stdin); int i1=0; String s1; try { System.out.print("Enter interger for "+s2+": "); s1 = console.readLine(); i1 = Integer.parseInt(s1); } catch(IOException ioex) { System.out.println("\nInput error"); System.exit(1); } catch(NumberFormatException nfex) { System.out.println("\"" + nfex.getMessage() + "\" is not numeric"); System.exit(1); } return(i1); } public int getChoice() { InputStreamReader stdin = new InputStreamReader(System.in); BufferedReader console = new BufferedReader(stdin); int i2=0; String s2; try { System.out.print("User's Choice: "); s2 = console.readLine(); i2 = Integer.parseInt(s2); } catch(IOException ioex) { System.out.println("Input error"); System.exit(1); } catch(NumberFormatException nfex) { System.out.println("\"" + nfex.getMessage() + "\" is not numeric"); System.exit(1); } return(i2); } public double getRoot1(int x, int y, int z) { double root1 = (-y + Math.sqrt((Math.pow(b,2))-(4*x*z)))/(2*x); return(root1); } public double getRoot2(int x, int y, int z) { double root2 = (-y - Math.sqrt((Math.pow(b,2))-(4*x*z)))/(2*x); return(root2); } public static void main(String[] args) { quadratic qd = new quadratic(); } }```.
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1. ## havent a clue!
havent a clue!
6x2+x-15=0
2. Originally Posted by jenko
havent a clue!
6x2+x-15=0
Let's try this using the method I posted in your other thread.
Mulitply 6 and -15: -90
Now list all pairs of factors of -90:
1, -90
2, -45
3, -30
6, -15
9, -10
10, -9
15, -6
30, -3
45, -2
90, -1
Now which of these pairs add to 1? The 10 and -9, of course. So 1 = 10 - 9:
$6x^2 + x - 15 = 0$
$6x^2 + (10 - 9)x - 15 = 0$
$6x^2 + 10x - 9x - 15 = 0$
$(6x^2 + 10x) + (-9x - 15) = 0$
$2x(3x + 5) - 3(3x + 5) = 0$
$(2x - 3)(3x + 5) = 0$
So set each factor equal to 0:
$2x - 3 = 0 \implies x = \frac{3}{2}$
or
$3x + 5 = 0 \implies x = -\frac{5}{3}$
-Dan
3. BTW, after a lot of practice of factorials, you will be able to look at them for a couple seconds and know what the constants will be.
I understand your plight. When I was a kid learning about factoring, my teacher didn't even show me about the additives and multiples that make up the constants right away. Needless to say, I taught it to myself.
4. i know the basics of mathematics e.g multiples etc it jus some of the terminologies that are used which i have never heared of or are unsure of ite been a while since i have done this last time i done then wen i was in high school two years ago and ever since that day i forgot it all as i thought i would never use it on it later life which i realy regret otherwise i wouldnt be here now or better still would be here with even more harder questions for you guys to help me with!
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1. ## havent a clue!. havent a clue!. 6x2+x-15=0. 2. Originally Posted by jenko. havent a clue!. 6x2+x-15=0. Let's try this using the method I posted in your other thread.. Mulitply 6 and -15: -90. Now list all pairs of factors of -90:. 1, -90. 2, -45. 3, -30. 6, -15. 9, -10. 10, -9. 15, -6. 30, -3. 45, -2. 90, -1. Now which of these pairs add to 1? The 10 and -9, of course. So 1 = 10 - 9:. $6x^2 + x - 15 = 0$. $6x^2 + (10 - 9)x - 15 = 0$. $6x^2 + 10x - 9x - 15 = 0$. $(6x^2 + 10x) + (-9x - 15) = 0$. $2x(3x + 5) - 3(3x + 5) = 0$. $(2x - 3)(3x + 5) = 0$. So set each factor equal to 0:. $2x - 3 = 0 \implies x = \frac{3}{2}$. or. $3x + 5 = 0 \implies x = -\frac{5}{3}$. -Dan. 3. BTW, after a lot of practice of factorials, you will be able to look at them for a couple seconds and know what the constants will be.. I understand your plight. When I was a kid learning about factoring, my teacher didn't even show me about the additives and multiples that make up the constants right away. Needless to say, I taught it to myself.
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4. i know the basics of mathematics e.g multiples etc it jus some of the terminologies that are used which i have never heared of or are unsure of ite been a while since i have done this last time i done then wen i was in high school two years ago and ever since that day i forgot it all as i thought i would never use it on it later life which i realy regret otherwise i wouldnt be here now or better still would be here with even more harder questions for you guys to help me with!.
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## Got It
Got It is an adding game for two players. You can play against the computer or with a friend. It is a version of a well known game called Nim.
Start with the Got It target $23$.
The first player chooses a whole number from $1$ to $4$ .
Players take turns to add a whole number from $1$ to $4$ to the running total.
The player who hits the target of $23$ wins the game.
Play the game several times.
Can you find a winning strategy?
Can you always win?
Does your strategy depend on whether or not you go first?
Full screen version
This text is usually replaced by the Flash movie.
To change the game, choose a new Got It target or a new range of numbers to add on.
Test out the strategy you found earlier. Does it need adapting?
Can you work out a winning strategy for any target?
Can you work out a winning strategy for any range of numbers?
Is it best to start the game? Always?
Away from the computer, challenge your friends:
One of you names the target and range and lets the other player start.
Extensions:
Can you play without writing anything down?
Consider playing the game where a player CANNOT add the same number as that used previously by the opponent.
### Why play this game?
Got It is a motivating context in which learners can apply simple addition and subtraction. However, the real challenge here is to find a winning strategy that always works, and this involves working systematically, conjecturing, refining ideas, generalising, and using knowledge of factors and multiples.
### Possible approach
All the notes that follow assume that the game's default setting is a target of $23$ using the numbers $1$ to $4$.
Introduce the game to the class by inviting a volunteer to play against the computer. Do this a couple of times, giving them the option of going first or second each time (you can use the "Change settings" button to do this).
Ask the students to play the game in pairs, either at computers or on paper. Challenge them to find a strategy for beating the computer. As they play, circulate around the classroom and ask them what they think is important so far. Some might suggest that in order to win, they must be on $18$. Others may have thought further back and have ideas about how they can make sure they get to $18$, and therefore $23$.
After a suitable length of time bring the whole class together and invite one pair to demonstrate their strategy, explaining their decisions as they go along. Use other ideas to refine the strategy.
Demonstrate how you can vary the game by choosing different targets and different ranges of numbers. Ask the students to play the game in pairs, either at computers or on paper, using settings of their own choice. Challenge them to find a winning strategy that will ensure they will always win, whatever the setting.
### Key questions
How can I work out the 'stepping stones' that I must 'hit' on my way to the target?
Is there an efficient way of finding the first 'stepping stone'?
When is it better to go first and when is it better to let the computer go first?
If the computer says $1$, I say...?
If the computer says $2$, I say...?
If the computer says $3$, I say...?
...
### Possible extension
Two more demanding games, requiring similar strategic thinking, are
Got a Strategy for Last Biscuit?
Nim-interactive
### Possible support
You could demonstrate the game a few more times at the start. Alter the settings on the game to have a lower target and a shorter range of numbers (for example a target of $10$ using the numbers $1$ and $2$). As you play, note down the running totals to refer back to later.
Here you can find a photocopiable version of both the problem and the teacher's notes.
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## Got It. Got It is an adding game for two players. You can play against the computer or with a friend. It is a version of a well known game called Nim.. Start with the Got It target $23$.. The first player chooses a whole number from $1$ to $4$ .. Players take turns to add a whole number from $1$ to $4$ to the running total.. The player who hits the target of $23$ wins the game.. Play the game several times.. Can you find a winning strategy?. Can you always win?. Does your strategy depend on whether or not you go first?. Full screen version. This text is usually replaced by the Flash movie.. To change the game, choose a new Got It target or a new range of numbers to add on.. Test out the strategy you found earlier. Does it need adapting?. Can you work out a winning strategy for any target?. Can you work out a winning strategy for any range of numbers?. Is it best to start the game? Always?. Away from the computer, challenge your friends:. One of you names the target and range and lets the other player start.. Extensions:. Can you play without writing anything down?. Consider playing the game where a player CANNOT add the same number as that used previously by the opponent.. ### Why play this game?. Got It is a motivating context in which learners can apply simple addition and subtraction. However, the real challenge here is to find a winning strategy that always works, and this involves working systematically, conjecturing, refining ideas, generalising, and using knowledge of factors and multiples.. ### Possible approach. All the notes that follow assume that the game's default setting is a target of $23$ using the numbers $1$ to $4$.. Introduce the game to the class by inviting a volunteer to play against the computer. Do this a couple of times, giving them the option of going first or second each time (you can use the "Change settings" button to do this).. Ask the students to play the game in pairs, either at computers or on paper. Challenge them to find a strategy for beating the computer. As they play, circulate around the classroom and ask them what they think is important so far. Some might suggest that in order to win, they must be on $18$. Others may have thought further back and have ideas about how they can make sure they get to $18$, and therefore $23$.. After a suitable length of time bring the whole class together and invite one pair to demonstrate their strategy, explaining their decisions as they go along. Use other ideas to refine the strategy.. Demonstrate how you can vary the game by choosing different targets and different ranges of numbers. Ask the students to play the game in pairs, either at computers or on paper, using settings of their own choice. Challenge them to find a winning strategy that will ensure they will always win, whatever the setting.. ### Key questions. How can I work out the 'stepping stones' that I must 'hit' on my way to the target?. Is there an efficient way of finding the first 'stepping stone'?. When is it better to go first and when is it better to let the computer go first?. If the computer says $1$, I say...?. If the computer says $2$, I say...?. If the computer says $3$, I say...?. .... ### Possible extension. Two more demanding games, requiring similar strategic thinking, are. Got a Strategy for Last Biscuit?. Nim-interactive. ### Possible support. You could demonstrate the game a few more times at the start.
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Alter the settings on the game to have a lower target and a shorter range of numbers (for example a target of $10$ using the numbers $1$ and $2$). As you play, note down the running totals to refer back to later.. Here you can find a photocopiable version of both the problem and the teacher's notes.
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Three common formats for numbers room fractions, decimals, and percents.
You are watching: How do you write 0.6 as a fraction
Percents are regularly used to connect a family member amount. You have probably seen them supplied for discounts, where the percent the discount can apply to different prices. Percents are likewise used when stating taxes and also interest rates on savings and also loans.
A percent is a proportion of a number come 100. Every cent way “per 100,” or “how numerous out that 100.” You usage the price % ~ a number to show percent.
Notice the 12 that the 100 squares in the grid below have to be shaded green. This to represent 12 percent (12 per 100).
12% = 12 percent = 12 parts out the 100 =
How countless of the squares in the grid over are unshaded? due to the fact that 12 space shaded and there room a full of 100 squares, 88 space unshaded. The unshaded section of the totality grid is 88 components out of 100, or 88% of the grid. Notice that the shaded and also unshaded portions together make 100% of the network (100 out of 100 squares).
Example Problem What percent of the network is shaded? The net is separated into 100 smaller sized squares, v 10 squares in each row. 23 squares out of 100 squares room shaded. Answer 23% of the network is shaded.
Example Problem What percent of the big square is shaded? The net is divided into 10 rectangles. For percents, you need to look at 100 equal-sized parts of the whole. You can divide each of the 10 rectangles right into 10 pieces, providing 100 parts. 30 tiny squares out of 100 space shaded. Answer 30% of the huge square is shaded.
What percent that this network is shaded?
A) 3%
B) 11%
C) 38%
D) 62%
A) 3%
Incorrect. Three complete columns that 10 squares space shaded, plus an additional 8 squares indigenous the next column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the large square. The exactly answer is 38%.
B) 11%
Incorrect. Three complete columns of 10 squares room shaded, plus another 8 squares native the following column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the big square. The correct answer is 38%.
C) 38%
Correct. Three complete columns that 10 squares room shaded, plus an additional 8 squares from the next column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the large square. This means 38% that the huge square is shaded.
D) 62%
Incorrect. There room 62 tiny unshaded squares out of the 100 in the large square, therefore the percent of the large square the is unshaded is 62%. However, the inquiry asked what percent is shaded. There room 38 shaded squares that the 100 squares in the huge square, therefore the exactly answer is 38%.
Rewriting Percents, Decimals, and also Fractions
It is often helpful to readjust the layout of a number. Because that example, girlfriend may find it less complicated to add decimals 보다 to include fractions. If you have the right to write the fractions together decimals, girlfriend can add them as decimals. Then you can rewrite your decimal amount as a fraction, if necessary.
Percents can be composed as fractions and also decimals in very few steps.
Example Problem Write 25% as a simplified portion and as a decimal. Write together a fraction. 25% = Since % method “out the 100,” 25% way 25 the end of 100. You create this together a fraction, utilizing 100 together the denominator. Simplify the portion by separating the numerator and denominator through the usual factor 25. Write together a decimal. 25% = = 0.25 You can also just move the decimal suggest in the whole number 25 two locations to the left to acquire 0.25. Answer 25% = = 0.25
Notice in the diagram listed below that 25% that a network is additionally of the grid, as you found in the example.
Notice that in the vault example, rewriting a percent together a decimal takes just a shift of the decimal point. You have the right to use fractions to recognize why this is the case. Any type of percentage x can be represented as the portion , and any fraction can be created as a decimal by moving the decimal suggest in x two places to the left. For example, 81% can be composed as
, and dividing 81 by 100 results in 0.81. People often skip end the intermediary portion step and just convert a percent come a decimal by relocating the decimal suggest two places to the left.
In the same way, rewriting a decimal together a percent (or as a fraction) requires few steps.
Example Problem Write 0.6 together a percent and also as a streamlined fraction. Write as a percent. 0.6 = 0.60 = 60% Write 0.6 as 0.60, i beg your pardon is 60 hundredths. 60 hundredths is 60 percent. You can additionally move the decimal point two places to the appropriate to discover the percent equivalent. Write as a fraction. 0.6 = To compose 0.6 together a fraction, you read the decimal, 6 tenths, and write 6 tenths in fraction form. Simplify the portion by splitting the numerator and also denominator by 2, a usual factor. Answer 0.6 = 60% =
In this example, the percent is not a entirety number. You can handle this in the exact same way, yet it’s usually less complicated to convert the percent come a decimal and also then transform the decimal to a fraction.
Example Problem Write 5.6% as a decimal and also as a streamlined fraction. Write as a decimal. 5.6% = 0.056 Move the decimal point two locations to the left. In this case, insert a 0 in front of the 5 (05.6) in bespeak to be able to move the decimal come the left two places. Write as a fraction. 0.056 = Write the portion as friend would check out the decimal. The last digit is in the thousandths place, therefore the denominator is 1,000. Simplify the portion by separating the numerator and also denominator through 8, a usual factor. Answer 5.6% = = 0.056
Write 0.645 together a percent and as a simplified fraction. A) 64.5% and B) 0.645% and also C) 645% and also D) 64.5% and also Show/Hide Answer A) 64.5% and Correct. 0.645 = 64.5% = . B) 0.645% and also Incorrect. 0.645 = 64.5%, not 0.645%. Psychic that when you convert a decimal to a percent you have to move the decimal suggest two locations to the right. The correct answer is 64.5% and . C) 645% and Incorrect. 0.645 = 64.5%, not 645%. Remember that when you convert a decimal to a percent you need to move the decimal point two areas to the right. The correct answer is 64.5% and . D) 64.5% and also Incorrect. To create 0.645 as a percent, move the decimal ar two locations to the right: 64.5%. To create 0.645 together a fraction, usage 645 as the numerator. The place value that the critical digit (the 5) is thousandths, for this reason the denominator is 1,000. The fraction is . The greatest common factor the 645 and also 1,000 is 5, therefore you deserve to divide the numerator and denominator by 5 to gain . The exactly answer is 64.5% and also .
In stimulate to create a portion as a decimal or a percent, you have the right to write the fraction as an equivalent fraction with a denominator that 10 (or any kind of other strength of 10 such as 100 or 1,000), which deserve to be then converted to a decimal and then a percent.
Example Problem Write as a decimal and as a percent. Write together a decimal. Find one equivalent portion with 10, 100, 1,000, or various other power that 10 in the denominator. Due to the fact that 100 is a lot of of 4, you deserve to multiply 4 through 25 to obtain 100. Multiply both the numerator and also the denominator by 25. = 0.75 Write the fraction as a decimal through the 5 in the percentage percent place. Write as a percent. 0.75 = 75% To create the decimal together a percent, move the decimal suggest two locations to the right. Answer = 0.75 = 75%
If that is complicated to find an equivalent portion with a denominator of 10, 100, 1,000, and so on, friend can constantly divide the molecule by the denominator to discover the decimal equivalent.
Example Problem Write as a decimal and also as a percent. Write as a decimal. Divide the molecule by the denominator. 3 ÷ 8 = 0.375. Write as a percent. 0.375 = 37.5% To create the decimal together a percent, relocate the decimal allude two areas to the right. Answer = 0.375 = 37.5%
Write as a decimal and also as a percent.
A) 80.0 and 0.8%
B) 0.4 and 4%
C) 0.8 and also 80%
D) 0.8 and also 8%
A) 80.0 and also 0.8%
Incorrect. An alert that 10 is a multiple of 5, so you have the right to rewrite using 10 together the denominator. Main point the numerator and also denominator by 2 to gain . The indistinguishable decimal is 0.8. You deserve to write this together a percent by moving the decimal suggest two places to the right. Since 0.8 has only one location to the right, encompass 0 in the percentage percent place: 0.8 = 0.80 = 80%. The correct answer is 0.8 and also 80%.
B) 0.4 and also 4%
Incorrect. To uncover a decimal indistinguishable for , an initial convert the portion to tenths. Multiply the numerator and denominator by 2 to acquire . The tantamount decimal is 0.8. So, and 0.4 space not indistinguishable quantities. The correct answer is 0.8 and 80%.
C) 0.8 and also 80%
Correct. The price is = 0.8 = 80%.
D) 0.8 and also 8%
Incorrect. It is true the = 0.8, however this does not equal 8%. To create 0.8 as a percent, relocate the decimal allude two locations to the right: 0.8 = 0.80 = 80%. The correct answer is 0.8 and also 80%.
Mixed Numbers
All the previous examples involve fractions and also decimals less than 1, so all of the percents you have actually seen so far have been much less than 100%.
Percents greater than 100% are feasible as well. Percents more than 100% are used to describe instances where there is more than one entirety (fractions and decimals higher than 1 are offered for the same reason).
In the diagram below, 115% is shaded. Every grid is taken into consideration a whole, and also you require two grids for 115%.
Expressed as a decimal, the percent 115% is 1.15; together a fraction, that is
, or
. Notice that you deserve to still convert amongst percents, fractions, and also decimals once the quantity is higher than one whole.
Numbers better than one that incorporate a fractional component can be composed as the sum of a totality number and also the fractional part. For instance, the mixed number is the amount of the entirety number 3 and the portion . = 3 + .
Example Problem Write as a decimal and as a percent. Write the mixed fraction as 2 wholes to add the spring part. Write together a decimal. Write the fractional component as a decimal by splitting the numerator by the denominator. 7 ÷ 8 = 0.875. Add 2 come the decimal. Write together a percent. 2.875 = 287.5% Now you can move the decimal allude two areas to the right to create the decimal as a percent. Answer = 2.875 = 287.5%
Note that a totality number can be created as a percent. 100% means one whole; so 2 wholes would certainly be 200%.
Example Problem Write 375% as a decimal and also as a streamlined fraction. Write together a decimal. 375% = 3.75 Move the decimal allude two areas to the left. Keep in mind that over there is a entirety number together with the decimal together the percent is more than 100%. Write as a fraction. 3.75 = 3 + 0.75 Write the decimal as a amount of the totality number and also the fractional part. 0.75 = Write the decimal part as a fraction. Simplify the portion by splitting the numerator and also denominator by a usual factor that 25. 3 + = Add the totality number component to the fraction. Answer 375% = 3.75=
Write 4.12 as a percent and also as a simplified fraction. A) 0.0412% and B) 412% and also C) 412% and also D) 4.12% and also Show/Hide Answer A) 0.0412% and Incorrect. To convert 4.12 to a percent, move the decimal suggest two locations to the right, not the left. The exactly answer is 412% and also . B) 412% and also Correct. 4.12 amounts to 412%, and also the simplified kind of is . C) 412% and Incorrect. 4.12 does equal 412%, yet it is also equivalent to , no . The correct answer is 412% and also . D) 4.12% and also Incorrect. To transform 4.12 come a percent, relocate the decimal allude two places to the right. The exactly answer is 412% and .See more: How To Build An Indoor Pitching Mound Plans: Step By Step Instructions Summary Percents space a common means to stand for fractional amounts, simply as decimals and fractions are. Any number that deserve to be composed as a decimal, fraction, or percent can also be written utilizing the various other two representations. .tags a { color: #fff; background: #909295; padding: 3px 10px; border-radius: 10px; font-size: 13px; line-height: 30px; white-space: nowrap; } .tags a:hover { background: #818182; } Home Contact - Advertising Copyright © 2022 dearteassociazione.org #footer {font-size: 14px;background: #ffffff;padding: 10px;text-align: center;} #footer a {color: #2c2b2b;margin-right: 10px;}
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Three common formats for numbers room fractions, decimals, and percents.. You are watching: How do you write 0.6 as a fraction. Percents are regularly used to connect a family member amount. You have probably seen them supplied for discounts, where the percent the discount can apply to different prices. Percents are likewise used when stating taxes and also interest rates on savings and also loans.. A percent is a proportion of a number come 100. Every cent way “per 100,” or “how numerous out that 100.” You usage the price % ~ a number to show percent.. Notice the 12 that the 100 squares in the grid below have to be shaded green. This to represent 12 percent (12 per 100).. 12% = 12 percent = 12 parts out the 100 =. How countless of the squares in the grid over are unshaded? due to the fact that 12 space shaded and there room a full of 100 squares, 88 space unshaded. The unshaded section of the totality grid is 88 components out of 100, or 88% of the grid. Notice that the shaded and also unshaded portions together make 100% of the network (100 out of 100 squares).. Example Problem What percent of the network is shaded? The net is separated into 100 smaller sized squares, v 10 squares in each row. 23 squares out of 100 squares room shaded. Answer 23% of the network is shaded.. Example Problem What percent of the big square is shaded? The net is divided into 10 rectangles. For percents, you need to look at 100 equal-sized parts of the whole. You can divide each of the 10 rectangles right into 10 pieces, providing 100 parts. 30 tiny squares out of 100 space shaded. Answer 30% of the huge square is shaded.. What percent that this network is shaded?. A) 3%. B) 11%. C) 38%. D) 62%. A) 3%. Incorrect. Three complete columns that 10 squares space shaded, plus an additional 8 squares indigenous the next column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the large square. The exactly answer is 38%.. B) 11%. Incorrect. Three complete columns of 10 squares room shaded, plus another 8 squares native the following column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the big square. The correct answer is 38%.. C) 38%. Correct. Three complete columns that 10 squares room shaded, plus an additional 8 squares from the next column. So, there space 30 + 8, or 38, squares shaded the end of the 100 squares in the large square. This means 38% that the huge square is shaded.. D) 62%. Incorrect. There room 62 tiny unshaded squares out of the 100 in the large square, therefore the percent of the large square the is unshaded is 62%. However, the inquiry asked what percent is shaded. There room 38 shaded squares that the 100 squares in the huge square, therefore the exactly answer is 38%.. Rewriting Percents, Decimals, and also Fractions. It is often helpful to readjust the layout of a number. Because that example, girlfriend may find it less complicated to add decimals 보다 to include fractions. If you have the right to write the fractions together decimals, girlfriend can add them as decimals. Then you can rewrite your decimal amount as a fraction, if necessary.. Percents can be composed as fractions and also decimals in very few steps.. Example Problem Write 25% as a simplified portion and as a decimal. Write together a fraction. 25% = Since % method “out the 100,” 25% way 25 the end of 100. You create this together a fraction, utilizing 100 together the denominator. Simplify the portion by separating the numerator and denominator through the usual factor 25. Write together a decimal. 25% = = 0.25 You can also just move the decimal suggest in the whole number 25 two locations to the left to acquire 0.25. Answer 25% = = 0.25. Notice in the diagram listed below that 25% that a network is additionally of the grid, as you found in the example.. Notice that in the vault example, rewriting a percent together a decimal takes just a shift of the decimal point. You have the right to use fractions to recognize why this is the case. Any type of percentage x can be represented as the portion , and any fraction can be created as a decimal by moving the decimal suggest in x two places to the left. For example, 81% can be composed as. , and dividing 81 by 100 results in 0.81. People often skip end the intermediary portion step and just convert a percent come a decimal by relocating the decimal suggest two places to the left.. In the same way, rewriting a decimal together a percent (or as a fraction) requires few steps.. Example Problem Write 0.6 together a percent and also as a streamlined fraction. Write as a percent. 0.6 = 0.60 = 60% Write 0.6 as 0.60, i beg your pardon is 60 hundredths. 60 hundredths is 60 percent. You can additionally move the decimal point two places to the appropriate to discover the percent equivalent. Write as a fraction. 0.6 = To compose 0.6 together a fraction, you read the decimal, 6 tenths, and write 6 tenths in fraction form. Simplify the portion by splitting the numerator and also denominator by 2, a usual factor. Answer 0.6 = 60% =. In this example, the percent is not a entirety number. You can handle this in the exact same way, yet it’s usually less complicated to convert the percent come a decimal and also then transform the decimal to a fraction.. Example Problem Write 5.6% as a decimal and also as a streamlined fraction. Write as a decimal. 5.6% = 0.056 Move the decimal point two locations to the left. In this case, insert a 0 in front of the 5 (05.6) in bespeak to be able to move the decimal come the left two places. Write as a fraction. 0.056 = Write the portion as friend would check out the decimal. The last digit is in the thousandths place, therefore the denominator is 1,000. Simplify the portion by separating the numerator and also denominator through 8, a usual factor. Answer 5.6% = = 0.056. Write 0.645 together a percent and as a simplified fraction. A) 64.5% and B) 0.645% and also C) 645% and also D) 64.5% and also Show/Hide Answer A) 64.5% and Correct. 0.645 = 64.5% = . B) 0.645% and also Incorrect. 0.645 = 64.5%, not 0.645%. Psychic that when you convert a decimal to a percent you have to move the decimal suggest two locations to the right. The correct answer is 64.5% and . C) 645% and Incorrect. 0.645 = 64.5%, not 645%. Remember that when you convert a decimal to a percent you need to move the decimal point two areas to the right. The correct answer is 64.5% and . D) 64.5% and also Incorrect. To create 0.645 as a percent, move the decimal ar two locations to the right: 64.5%. To create 0.645 together a fraction, usage 645 as the numerator. The place value that the critical digit (the 5) is thousandths, for this reason the denominator is 1,000. The fraction is . The greatest common factor the 645 and also 1,000 is 5, therefore you deserve to divide the numerator and denominator by 5 to gain . The exactly answer is 64.5% and also .. In stimulate to create a portion as a decimal or a percent, you have the right to write the fraction as an equivalent fraction with a denominator that 10 (or any kind of other strength of 10 such as 100 or 1,000), which deserve to be then converted to a decimal and then a percent.. Example Problem Write as a decimal and as a percent. Write together a decimal. Find one equivalent portion with 10, 100, 1,000, or various other power that 10 in the denominator. Due to the fact that 100 is a lot of of 4, you deserve to multiply 4 through 25 to obtain 100. Multiply both the numerator and also the denominator by 25. = 0.75 Write the fraction as a decimal through the 5 in the percentage percent place. Write as a percent. 0.75 = 75% To create the decimal together a percent, move the decimal suggest two locations to the right. Answer = 0.75 = 75%. If that is complicated to find an equivalent portion with a denominator of 10, 100, 1,000, and so on, friend can constantly divide the molecule by the denominator to discover the decimal equivalent.. Example Problem Write as a decimal and also as a percent. Write as a decimal. Divide the molecule by the denominator. 3 ÷ 8 = 0.375. Write as a percent. 0.375 = 37.5% To create the decimal together a percent, relocate the decimal allude two areas to the right. Answer = 0.375 = 37.5%. Write as a decimal and also as a percent.. A) 80.0 and 0.8%. B) 0.4 and 4%. C) 0.8 and also 80%. D) 0.8 and also 8%. A) 80.0 and also 0.8%. Incorrect. An alert that 10 is a multiple of 5, so you have the right to rewrite using 10 together the denominator. Main point the numerator and also denominator by 2 to gain . The indistinguishable decimal is 0.8. You deserve to write this together a percent by moving the decimal suggest two places to the right. Since 0.8 has only one location to the right, encompass 0 in the percentage percent place: 0.8 = 0.80 = 80%. The correct answer is 0.8 and also 80%.. B) 0.4 and also 4%. Incorrect. To uncover a decimal indistinguishable for , an initial convert the portion to tenths. Multiply the numerator and denominator by 2 to acquire . The tantamount decimal is 0.8. So, and 0.4 space not indistinguishable quantities. The correct answer is 0.8 and 80%.. C) 0.8 and also 80%. Correct. The price is = 0.8 = 80%.. D) 0.8 and also 8%. Incorrect. It is true the = 0.8, however this does not equal 8%. To create 0.8 as a percent, relocate the decimal allude two locations to the right: 0.8 = 0.80 = 80%. The correct answer is 0.8 and also 80%.. Mixed Numbers. All the previous examples involve fractions and also decimals less than 1, so all of the percents you have actually seen so far have been much less than 100%.. Percents greater than 100% are feasible as well. Percents more than 100% are used to describe instances where there is more than one entirety (fractions and decimals higher than 1 are offered for the same reason).. In the diagram below, 115% is shaded. Every grid is taken into consideration a whole, and also you require two grids for 115%.. Expressed as a decimal, the percent 115% is 1.15; together a fraction, that is. , or. . Notice that you deserve to still convert amongst percents, fractions, and also decimals once the quantity is higher than one whole.. Numbers better than one that incorporate a fractional component can be composed as the sum of a totality number and also the fractional part. For instance, the mixed number is the amount of the entirety number 3 and the portion . = 3 + .. Example Problem Write as a decimal and as a percent. Write the mixed fraction as 2 wholes to add the spring part. Write together a decimal. Write the fractional component as a decimal by splitting the numerator by the denominator. 7 ÷ 8 = 0.875. Add 2 come the decimal. Write together a percent. 2.875 = 287.5% Now you can move the decimal allude two areas to the right to create the decimal as a percent. Answer = 2.875 = 287.5%. Note that a totality number can be created as a percent. 100% means one whole; so 2 wholes would certainly be 200%.. Example Problem Write 375% as a decimal and also as a streamlined fraction. Write together a decimal. 375% = 3.75 Move the decimal allude two areas to the left. Keep in mind that over there is a entirety number together with the decimal together the percent is more than 100%. Write as a fraction. 3.75 = 3 + 0.75 Write the decimal as a amount of the totality number and also the fractional part. 0.75 = Write the decimal part as a fraction. Simplify the portion by splitting the numerator and also denominator by a usual factor that 25. 3 + = Add the totality number component to the fraction. Answer 375% = 3.75=. Write 4.12 as a percent and also as a simplified fraction. A) 0.0412% and B) 412% and also C) 412% and also D) 4.12% and also Show/Hide Answer A) 0.0412% and Incorrect. To convert 4.12 to a percent, move the decimal suggest two locations to the right, not the left. The exactly answer is 412% and also . B) 412% and also Correct. 4.12 amounts to 412%, and also the simplified kind of is . C) 412% and Incorrect. 4.12 does equal 412%, yet it is also equivalent to , no . The correct answer is 412% and also . D) 4.12% and also Incorrect. To transform 4.12 come a percent, relocate the decimal allude two places to the right. The exactly answer is 412% and .See more: How To Build An Indoor Pitching Mound Plans: Step By Step Instructions Summary Percents space a common means to stand for fractional amounts, simply as decimals and fractions are.
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Any number that deserve to be composed as a decimal, fraction, or percent can also be written utilizing the various other two representations. .tags a { color: #fff; background: #909295; padding: 3px 10px; border-radius: 10px; font-size: 13px; line-height: 30px; white-space: nowrap; } .tags a:hover { background: #818182; } Home Contact - Advertising Copyright © 2022 dearteassociazione.org #footer {font-size: 14px;background: #ffffff;padding: 10px;text-align: center;} #footer a {color: #2c2b2b;margin-right: 10px;}.
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http://tasks.illustrativemathematics.org/content-standards/5/NF/B/tasks/882
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# Painting a Wall
Alignments to Content Standards: 5.NF.B
Nicolas is helping to paint a wall at a park near his house as part of a community service project. He had painted half of the wall yellow when the park director walked by and said,
This wall is supposed to be painted red.
Nicolas immediately started painting over the yellow portion of the wall. By the end of the day, he had repainted $\frac56$ of the yellow portion red.
What fraction of the entire wall is painted red at the end of the day?
## IM Commentary
The purpose of this task is for students to find the answer to a question in context that can be represented by fraction multiplication. This task is appropriate for either instruction or assessment depending on how it is used and where students are in their understanding of fraction multiplication. If used in instruction, it can provide a lead-in to the meaning of fraction multiplication. If used for assessment, it can help teachers see whether students readily see that this is can be solved by multiplying $\frac56\times \frac12$ or not, which can help diagnose their comfort level with the meaning of fraction multiplication.
The teacher might need to emphasize that the task is asking for what portion of the total wall is red, it is not asking what portion of the yellow has been repainted.
## Solutions
Solution: Solution 1
In order to see what fraction of the wall is red we need to find out what $\frac56$ of $\frac12$ is. To do this we can multiply the fractions together like so:
$\frac56 \times \frac12 = \frac{5 \times 1}{6 \times 2} = \frac{5}{12}$
So we can see that $\frac{5}{12}$ of the wall is red.
Solution: Solution 2
The solution can also be represented with pictures. Here we see the wall right before the park director walks by:
And now we can break up the yellow portion into 6 equally sized parts:
Now we can show what the wall looked like at the end of the day by shading 5 out of those 6 parts red.
And finally, we can see that if we had broken up the wall into 12 equally sized pieces from the beginning, that finding the fraction of the wall that is red would be just a matter of counting the number of red pieces and comparing them to the total.
And so, since 5 pieces of the total 12 are red, we can see that $\frac{5}{12}$ of the wall is red at the end of the day.
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# Painting a Wall. Alignments to Content Standards: 5.NF.B. Nicolas is helping to paint a wall at a park near his house as part of a community service project. He had painted half of the wall yellow when the park director walked by and said,. This wall is supposed to be painted red.. Nicolas immediately started painting over the yellow portion of the wall. By the end of the day, he had repainted $\frac56$ of the yellow portion red.. What fraction of the entire wall is painted red at the end of the day?. ## IM Commentary. The purpose of this task is for students to find the answer to a question in context that can be represented by fraction multiplication. This task is appropriate for either instruction or assessment depending on how it is used and where students are in their understanding of fraction multiplication. If used in instruction, it can provide a lead-in to the meaning of fraction multiplication. If used for assessment, it can help teachers see whether students readily see that this is can be solved by multiplying $\frac56\times \frac12$ or not, which can help diagnose their comfort level with the meaning of fraction multiplication.. The teacher might need to emphasize that the task is asking for what portion of the total wall is red, it is not asking what portion of the yellow has been repainted.. ## Solutions. Solution: Solution 1. In order to see what fraction of the wall is red we need to find out what $\frac56$ of $\frac12$ is. To do this we can multiply the fractions together like so:. $\frac56 \times \frac12 = \frac{5 \times 1}{6 \times 2} = \frac{5}{12}$. So we can see that $\frac{5}{12}$ of the wall is red.. Solution: Solution 2. The solution can also be represented with pictures. Here we see the wall right before the park director walks by:. And now we can break up the yellow portion into 6 equally sized parts:. Now we can show what the wall looked like at the end of the day by shading 5 out of those 6 parts red.
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And finally, we can see that if we had broken up the wall into 12 equally sized pieces from the beginning, that finding the fraction of the wall that is red would be just a matter of counting the number of red pieces and comparing them to the total.. And so, since 5 pieces of the total 12 are red, we can see that $\frac{5}{12}$ of the wall is red at the end of the day.
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http://mathoverflow.net/questions/104212/adjoining-a-new-isolated-point-without-changing-the-space/104214
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# Adjoining a new isolated point without changing the space
Suppose $X$ is a $T_1$ space with an infinite set of isolated points. Show that if $X^\sharp = X \cup \lbrace \infty \rbrace$ is obtained by adding a single new isolated point, then $X$ and $X^\sharp$ are homeomorphic.
I am almost embarrased to raise this, which seems obvious. The proof must be simple, but it eludes me for now. Maybe it is an exercise in some textbook. You can clearly establish a 1-1 equivalence between the isolated points of $X$ and those of $X^\sharp$. But it is not clear how this equivalence would extend to the closure of the isolated points.
The theorem is easy when $X$ is compact $T_2$ and $cl(D) = \beta(D)$, where $D$ is the set of isolated points.
-
In my answer to http://mathoverflow.net/questions/26414 I descibed a somewhat simpler-looking example than Nik's, but proving that it works may be harder. Take two copies of $\beta\mathbb N$ and glue each non-isolated point of one copy to the corresponding point of the other copy. Any way of "absorbing" a new isolated point into the two copies of $\mathbb N$ forces a relative shift of those two copies, which forces corresponding shifts of the non-isolated points, which in turn conflicts with the gluing. The perhaps surprising thing about this example is that, if you add two isolated points, the result is (easily) homeomorphic to the original.
Well, I think this is false. Start with a family of $2^{2^{\aleph_0}}$ mutually non-homeomorphic connected spaces, and attach them to the non-isolated points of $\beta {\bf N}$. (I.e., start with the disjoint union of $\beta {\bf N}$ and the other spaces, and factor out an equivalence relation which identifies each point of $\beta {\bf N} - {\bf N}$ with a point of one of the other spaces.) Any homeomorphism between $X$ and $X^\sharp$ has to take isolated points to isolated points; taking closures, it takes $\beta{\bf N}$ onto itself; and by connectedness it takes each of the extra spaces onto itself. So it has to fix each point of $\beta {\bf N} - {\bf N}$. Now the question is whether a bijection between ${\bf N}$ and ${\bf N}$ minus a point can fix $\beta {\bf N} - {\bf N}$ pointwise. The answer is no because iterating the map, starting on the missing point, yields a sequence within ${\bf N}$ that gets shifted by the map, and it is easy to see that this shift does not fix the ultrafilters supported on that sequence.
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# Adjoining a new isolated point without changing the space. Suppose $X$ is a $T_1$ space with an infinite set of isolated points. Show that if $X^\sharp = X \cup \lbrace \infty \rbrace$ is obtained by adding a single new isolated point, then $X$ and $X^\sharp$ are homeomorphic.. I am almost embarrased to raise this, which seems obvious. The proof must be simple, but it eludes me for now. Maybe it is an exercise in some textbook. You can clearly establish a 1-1 equivalence between the isolated points of $X$ and those of $X^\sharp$. But it is not clear how this equivalence would extend to the closure of the isolated points.. The theorem is easy when $X$ is compact $T_2$ and $cl(D) = \beta(D)$, where $D$ is the set of isolated points.. -. In my answer to http://mathoverflow.net/questions/26414 I descibed a somewhat simpler-looking example than Nik's, but proving that it works may be harder. Take two copies of $\beta\mathbb N$ and glue each non-isolated point of one copy to the corresponding point of the other copy. Any way of "absorbing" a new isolated point into the two copies of $\mathbb N$ forces a relative shift of those two copies, which forces corresponding shifts of the non-isolated points, which in turn conflicts with the gluing. The perhaps surprising thing about this example is that, if you add two isolated points, the result is (easily) homeomorphic to the original.. Well, I think this is false. Start with a family of $2^{2^{\aleph_0}}$ mutually non-homeomorphic connected spaces, and attach them to the non-isolated points of $\beta {\bf N}$. (I.e., start with the disjoint union of $\beta {\bf N}$ and the other spaces, and factor out an equivalence relation which identifies each point of $\beta {\bf N} - {\bf N}$ with a point of one of the other spaces.) Any homeomorphism between $X$ and $X^\sharp$ has to take isolated points to isolated points; taking closures, it takes $\beta{\bf N}$ onto itself; and by connectedness it takes each of the extra spaces onto itself.
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So it has to fix each point of $\beta {\bf N} - {\bf N}$. Now the question is whether a bijection between ${\bf N}$ and ${\bf N}$ minus a point can fix $\beta {\bf N} - {\bf N}$ pointwise. The answer is no because iterating the map, starting on the missing point, yields a sequence within ${\bf N}$ that gets shifted by the map, and it is easy to see that this shift does not fix the ultrafilters supported on that sequence.
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# The Human Number Line
Today I’m sharing a fun activity that I like to do with my students at the beginning of the year in Algebra 2. At the beginning of the year, I teach about subsets of real numbers.
First, get a stack of blank index cards. I choose as many cards as I have students, so that everyone can participate. Then, I write numbers in all forms on the cards. Notice that some numbers that are the same, just written in a different form (ex. 1/4 and 0.25). Also include radicals that aren’t reduced (with perfect squares as the radicand) and irrational numbers. Really, you can do anything you want. This should take you about 5 minutes.
Human Number Line
During class, give each student a card. Then, direct them to put themselves into a number line. I usually tell my students where zero is and which direction is positive and which direction is negative. This helps the students get organized faster.
Once the students think they are in order, I start asking them questions. I make sure I ask questions about repeating decimals. I also ask about which number being bigger, pi or 3.14.
You could also only make half the amount of cards and have the students without cards direct and “coach” the students with the cards.
Sets of Real Numbers Sorting
Then, I have the students sort themselves into groups by their card. First, I have them move into rational and irrational numbers. Then, I’ll have them move into other sets (integers and non-integers, etc.).
If I have a class that seems very bright, I may try to stump them. I’ll give them categories like real numbers and integers. Students that are paying attention won’t know which category they fall into.
This is just a fun idea that helps liven up class at the beginning of the year. It takes very little prep work, but gets the students talking.
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# The Human Number Line. Today I’m sharing a fun activity that I like to do with my students at the beginning of the year in Algebra 2. At the beginning of the year, I teach about subsets of real numbers.. First, get a stack of blank index cards. I choose as many cards as I have students, so that everyone can participate. Then, I write numbers in all forms on the cards. Notice that some numbers that are the same, just written in a different form (ex. 1/4 and 0.25). Also include radicals that aren’t reduced (with perfect squares as the radicand) and irrational numbers. Really, you can do anything you want. This should take you about 5 minutes.. Human Number Line. During class, give each student a card. Then, direct them to put themselves into a number line. I usually tell my students where zero is and which direction is positive and which direction is negative. This helps the students get organized faster.. Once the students think they are in order, I start asking them questions. I make sure I ask questions about repeating decimals. I also ask about which number being bigger, pi or 3.14.. You could also only make half the amount of cards and have the students without cards direct and “coach” the students with the cards.. Sets of Real Numbers Sorting. Then, I have the students sort themselves into groups by their card. First, I have them move into rational and irrational numbers. Then, I’ll have them move into other sets (integers and non-integers, etc.).
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If I have a class that seems very bright, I may try to stump them. I’ll give them categories like real numbers and integers. Students that are paying attention won’t know which category they fall into.. This is just a fun idea that helps liven up class at the beginning of the year. It takes very little prep work, but gets the students talking.
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https://gmatclub.com/forum/which-pair-of-points-could-both-appear-on-the-same-line-165122.html
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# Which pair of points could both appear on the same line
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Which pair of points could both appear on the same line [#permalink]
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27 Dec 2013, 15:51
Which pair of points could both appear on the same line, if that line passes through the origin? Please make one selection in each column, with column A containing the point with the lesser x-value and column B containing the point with the greater x-value.
Col (A) Col (B)
( ) ( ) (-7, -4)
( ) ( ) (-6, 5)
( ) ( ) (0, -3)
( ) ( ) (4, 0)
( ) ( ) (8, -3)
[Reveal] Spoiler:
MY Question: I wonder if the two points chosen should have the same slope. What do you guys think?
OE:
For a line to pass through the origin, there are really four options for the quadrants it can pass through. It can pass through quadrants I and III; it can pass through II and IV; it can pass directly through the x-axis; and it can pass directly through the y-axis. Point (-7, -4) is in quadrant III, and there isn't a matching quadrant I to go with it. Point (0, -3) is on the y-axis, and there isn't another point on the y-axis to match it. Similar for point (4, 0) which is on the x-axis with no pairing. Point (-6, 5) is in quadrant II and point (8, -3) is in quadrant III, so these two points could be on the same line that passes through the origin.
Source: VeritasPrep
Last edited by nechets on 29 Dec 2013, 06:01, edited 1 time in total.
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Re: Which pair of points could both appear on the same line [#permalink]
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27 Dec 2013, 19:49
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nechets wrote:
Which pair of points could both appear on the same line, if that line passes through the origin? Please make one selection in each column, with column A containing the point with the lesser x-value and column B containing the point with the greater x-value.
Col (A) Col (B)
( ) ( ) (-7, -4)
( ) ( ) (-6, 5)
( ) ( ) (0, -3)
( ) ( ) (4, 0)
( ) ( ) (8, -3)
[Reveal] Spoiler:
MY Question: I wonder if the two points choosed should have the same slope. What do you guys think?
OE:
For a line to pass through the origin, there are really four options for the quadrants it can pass through. It can pass through quadrants I and III; it can pass through II and IV; it can pass directly through the x-axis; and it can pass directly through the y-axis. Point (-7, -4) is in quadrant III, and there isn't a matching quadrant I to go with it. Point (0, -3) is on the y-axis, and there isn't another point on the y-axis to match it. Similar for point (4, 0) which is on the x-axis with no pairing. Point (-6, 5) is in quadrant II and point (8, -3) is in quadrant III, so these two points could be on the same line that passes through the origin.
Source: VeritasPrep
This looks liked a bad question to me. I agree with you that the both points chosen should have the same slope to the origin. Interestingly, their answer of (-6, 5) and (8, -3) have different slopes to the origin... meaning if you draw a line between these two points, it will not pass through the origin.
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Re: Which pair of points could both appear on the same line [#permalink]
### Show Tags
27 Feb 2014, 12:33
What exactly is the answer? are there 3 answers instead of 2? (8,-3) (-6,5) (0,-3) ?
I thought the question was asking which 2 of the points cross the origin if both of the points were on the same line. Someone please explain.
Thanks!
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Re: Which pair of points could both appear on the same line [#permalink] 27 Feb 2014, 12:33
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# Which pair of points could both appear on the same line
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It is currently 22 Oct 2017, 23:48. ### 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. Your Progress. 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. # Which pair of points could both appear on the same line. new topic post reply Question banks Downloads My Bookmarks Reviews Important topics. Author Message. Manager. Joined: 04 Oct 2013. Posts: 87. Kudos [?]: 86 [0], given: 45. Location: Brazil. GMAT 1: 660 Q45 V35. GMAT 2: 710 Q49 V38. Which pair of points could both appear on the same line [#permalink]. ### Show Tags. 27 Dec 2013, 15:51. Which pair of points could both appear on the same line, if that line passes through the origin? Please make one selection in each column, with column A containing the point with the lesser x-value and column B containing the point with the greater x-value.. Col (A) Col (B). ( ) ( ) (-7, -4). ( ) ( ) (-6, 5). ( ) ( ) (0, -3). ( ) ( ) (4, 0). ( ) ( ) (8, -3). [Reveal] Spoiler:. MY Question: I wonder if the two points chosen should have the same slope. What do you guys think?. OE:. For a line to pass through the origin, there are really four options for the quadrants it can pass through. It can pass through quadrants I and III; it can pass through II and IV; it can pass directly through the x-axis; and it can pass directly through the y-axis. Point (-7, -4) is in quadrant III, and there isn't a matching quadrant I to go with it. Point (0, -3) is on the y-axis, and there isn't another point on the y-axis to match it. Similar for point (4, 0) which is on the x-axis with no pairing. Point (-6, 5) is in quadrant II and point (8, -3) is in quadrant III, so these two points could be on the same line that passes through the origin.. Source: VeritasPrep. Last edited by nechets on 29 Dec 2013, 06:01, edited 1 time in total.. Kudos [?]: 86 [0], given: 45. Senior Manager. Status: busyness school student. Joined: 09 Sep 2013. Posts: 419. Kudos [?]: 306 [1], given: 145. Location: United States. Schools: Tepper '16 (M). GMAT 1: 730 Q52 V37. Re: Which pair of points could both appear on the same line [#permalink]. ### Show Tags. 27 Dec 2013, 19:49. 1. This post received. KUDOS. nechets wrote:. Which pair of points could both appear on the same line, if that line passes through the origin? Please make one selection in each column, with column A containing the point with the lesser x-value and column B containing the point with the greater x-value.. Col (A) Col (B). ( ) ( ) (-7, -4). ( ) ( ) (-6, 5). ( ) ( ) (0, -3). ( ) ( ) (4, 0). ( ) ( ) (8, -3). [Reveal] Spoiler:. MY Question: I wonder if the two points choosed should have the same slope. What do you guys think?. OE:. For a line to pass through the origin, there are really four options for the quadrants it can pass through. It can pass through quadrants I and III; it can pass through II and IV; it can pass directly through the x-axis; and it can pass directly through the y-axis. Point (-7, -4) is in quadrant III, and there isn't a matching quadrant I to go with it. Point (0, -3) is on the y-axis, and there isn't another point on the y-axis to match it. Similar for point (4, 0) which is on the x-axis with no pairing. Point (-6, 5) is in quadrant II and point (8, -3) is in quadrant III, so these two points could be on the same line that passes through the origin.. Source: VeritasPrep. This looks liked a bad question to me. I agree with you that the both points chosen should have the same slope to the origin. Interestingly, their answer of (-6, 5) and (8, -3) have different slopes to the origin... meaning if you draw a line between these two points, it will not pass through the origin.. _________________. My review of some of the CAT practice exams. You can usually find me on gmatclub's chat.. Kudos [?]: 306 [1], given: 145. Intern. Joined: 08 Dec 2013. Posts: 10. Kudos [?]: 8 [0], given: 3. Re: Which pair of points could both appear on the same line [#permalink]. ### Show Tags. 27 Feb 2014, 12:33. What exactly is the answer? are there 3 answers instead of 2? (8,-3) (-6,5) (0,-3) ?. I thought the question was asking which 2 of the points cross the origin if both of the points were on the same line. Someone please explain.. Thanks!. Kudos [?]: 8 [0], given: 3. Re: Which pair of points could both appear on the same line [#permalink] 27 Feb 2014, 12:33. Display posts from previous: Sort by. # Which pair of points could both appear on the same line.
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Content
# Mathematics of Sudoku
The general problem of solving Sudoku puzzles on n2 x n2 boards of n x n blocks is known to be NP-complete. This gives some indication of why Sudoku is difficult to solve, although on boards of finite size the problem is finite and can be solved by a deterministic finite automaton that knows the entire game tree.
Solving Sudoku puzzles (as well as any other NP-hard problem) can be expressed as a graph colouring problem. The aim of the puzzle in its standard form is to construct a proper 9-colouring of a particular graph, given a partial 9-colouring. The graph in question has 81 vertices, one vertex for each cell of the grid. The vertices can be labelled with the ordered pairs , where x and y are integers between 1 and 9.
The puzzle is then completed by assigning an integer between 1 and 9 to each vertex, in such a way that vertices that are joined by an edge do not have the same integer assigned to them.
A valid Sudoku solution grid is also a Latin square. There are significantly fewer valid Sudoku solution grids than Latin squares because Sudoku imposes the additional regional constraint. Nonetheless, the number of valid Sudoku solution grids for the standard 9×9 grid was calculated by Bertram Felgenhauer in 2005 to be 6,670,903,752,021,072,936,960 (sequence A107739 in OEIS). This number is equal to 9! × 722 × 27 × 27,704,267,971, the last factor of which is prime. The result was derived through logic and brute force computation. The derivation of this result was considerably simplified by analysis provided by Frazer Jarvis and the figure has been confirmed independently by Ed Russell. Russell and Jarvis also showed that when symmetries were taken into account, there were 5,472,730,538 solutions (sequence A109741 in OEIS). The number of valid Sudoku solution grids for the 16×16 derivation is not known.
The maximum number of givens that can be provided while still not rendering the solution unique is four short of a full grid; if two instances of two numbers each are missing and the cells they are to occupy form the corners of an orthogonal rectangle, and exactly two of these cells are within one region, there are two ways the numbers can be assigned. Since this applies to Latin squares in general, most variants of Sudoku have the same maximum. The inverse problem—the fewest givens that render a solution unique—is unsolved, although the lowest number yet found for the standard variation without a symmetry constraint is 17, a number of which have been found by Japanese puzzle enthusiasts, and 18 with the givens in rotationally symmetric cells.
It is possible to set starting grids with more than one solution and to set grids with no solution, but such are not considered proper Sudoku puzzles; as in most other pure-logic puzzles, a unique solution is expected.
Building a Sudoku puzzle by hand can be performed efficiently by pre-determining the locations of the givens and assigning them values only as needed to make deductive progress. Such an undefined given can be assumed to not hold any particular value as long as it is given a different value before construction is completed; the solver will be able to make the same deductions stemming from such assumptions, as at that point the given is very much defined as something else. This technique gives the constructor greater control over the flow of puzzle solving, leading the solver along the same path the compiler used in building the puzzle. (This technique is adaptable to composing puzzles other than Sudoku as well.) Great caution is required, however, as failing to recognize where a number can be logically deduced at any point in construction-regardless of how tortuous that logic may be-can result in an unsolvable puzzle when defining a future given contradicts what has already been built. Building a Sudoku with symmetrical givens is a simple matter of placing the undefined givens in a symmetrical pattern to begin with.
It is commonly believed that Dell Number Place puzzles are computer-generated; they typically have over 30 givens placed in an apparently random scatter, some of which can possibly be deduced from other givens. They also have no authoring credits - that is, the name of the constructor is not printed with any puzzle. Wei-Hwa Huang claims that he was commissioned by Dell to write a Number Place puzzle generator in the winter of 2000; prior to that, he was told, the puzzles were hand-made. The puzzle generator was written with Visual C++, and although it had options to generate a more Japanese-style puzzle, with symmetry constraints and fewer numbers, Dell opted not to use those features, at least not until their recent publication of Sudoku-only magazines.
Nikoli Sudoku are hand-constructed, with the author being credited; the givens are always found in a symmetrical pattern. Dell Number Place Challenger puzzles also list authors . The Sudoku puzzles printed in most UK newspapers are apparently computer-generated but employ symmetrical givens; The Guardian licenses and publishes Nikoli-constructed Sudoku puzzles, though it does not include credits. The Guardian famously claimed that because they were hand-constructed, their puzzles would contain "imperceptible witticisms" that would be very unlikely in computer-generated Sudoku. The challenge to Sudoku programmers is teaching a program how to build clever puzzles, such that they may be indistinguishable from those constructed by humans; Wayne Gould required six years of tweaking his popular program before he believed he achieved that level.
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Content. # Mathematics of Sudoku. The general problem of solving Sudoku puzzles on n2 x n2 boards of n x n blocks is known to be NP-complete. This gives some indication of why Sudoku is difficult to solve, although on boards of finite size the problem is finite and can be solved by a deterministic finite automaton that knows the entire game tree.. Solving Sudoku puzzles (as well as any other NP-hard problem) can be expressed as a graph colouring problem. The aim of the puzzle in its standard form is to construct a proper 9-colouring of a particular graph, given a partial 9-colouring. The graph in question has 81 vertices, one vertex for each cell of the grid. The vertices can be labelled with the ordered pairs , where x and y are integers between 1 and 9.. The puzzle is then completed by assigning an integer between 1 and 9 to each vertex, in such a way that vertices that are joined by an edge do not have the same integer assigned to them.. A valid Sudoku solution grid is also a Latin square. There are significantly fewer valid Sudoku solution grids than Latin squares because Sudoku imposes the additional regional constraint. Nonetheless, the number of valid Sudoku solution grids for the standard 9×9 grid was calculated by Bertram Felgenhauer in 2005 to be 6,670,903,752,021,072,936,960 (sequence A107739 in OEIS). This number is equal to 9! × 722 × 27 × 27,704,267,971, the last factor of which is prime. The result was derived through logic and brute force computation. The derivation of this result was considerably simplified by analysis provided by Frazer Jarvis and the figure has been confirmed independently by Ed Russell. Russell and Jarvis also showed that when symmetries were taken into account, there were 5,472,730,538 solutions (sequence A109741 in OEIS). The number of valid Sudoku solution grids for the 16×16 derivation is not known.. The maximum number of givens that can be provided while still not rendering the solution unique is four short of a full grid; if two instances of two numbers each are missing and the cells they are to occupy form the corners of an orthogonal rectangle, and exactly two of these cells are within one region, there are two ways the numbers can be assigned. Since this applies to Latin squares in general, most variants of Sudoku have the same maximum. The inverse problem—the fewest givens that render a solution unique—is unsolved, although the lowest number yet found for the standard variation without a symmetry constraint is 17, a number of which have been found by Japanese puzzle enthusiasts, and 18 with the givens in rotationally symmetric cells.. It is possible to set starting grids with more than one solution and to set grids with no solution, but such are not considered proper Sudoku puzzles; as in most other pure-logic puzzles, a unique solution is expected.. Building a Sudoku puzzle by hand can be performed efficiently by pre-determining the locations of the givens and assigning them values only as needed to make deductive progress. Such an undefined given can be assumed to not hold any particular value as long as it is given a different value before construction is completed; the solver will be able to make the same deductions stemming from such assumptions, as at that point the given is very much defined as something else. This technique gives the constructor greater control over the flow of puzzle solving, leading the solver along the same path the compiler used in building the puzzle. (This technique is adaptable to composing puzzles other than Sudoku as well.) Great caution is required, however, as failing to recognize where a number can be logically deduced at any point in construction-regardless of how tortuous that logic may be-can result in an unsolvable puzzle when defining a future given contradicts what has already been built. Building a Sudoku with symmetrical givens is a simple matter of placing the undefined givens in a symmetrical pattern to begin with.. It is commonly believed that Dell Number Place puzzles are computer-generated; they typically have over 30 givens placed in an apparently random scatter, some of which can possibly be deduced from other givens. They also have no authoring credits - that is, the name of the constructor is not printed with any puzzle. Wei-Hwa Huang claims that he was commissioned by Dell to write a Number Place puzzle generator in the winter of 2000; prior to that, he was told, the puzzles were hand-made. The puzzle generator was written with Visual C++, and although it had options to generate a more Japanese-style puzzle, with symmetry constraints and fewer numbers, Dell opted not to use those features, at least not until their recent publication of Sudoku-only magazines.. Nikoli Sudoku are hand-constructed, with the author being credited; the givens are always found in a symmetrical pattern. Dell Number Place Challenger puzzles also list authors . The Sudoku puzzles printed in most UK newspapers are apparently computer-generated but employ symmetrical givens; The Guardian licenses and publishes Nikoli-constructed Sudoku puzzles, though it does not include credits.
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The Guardian famously claimed that because they were hand-constructed, their puzzles would contain "imperceptible witticisms" that would be very unlikely in computer-generated Sudoku. The challenge to Sudoku programmers is teaching a program how to build clever puzzles, such that they may be indistinguishable from those constructed by humans; Wayne Gould required six years of tweaking his popular program before he believed he achieved that level.
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# Álgebra Lineal y Algunas de sus Aplicaciones by L.I. GOLOVINA
By L.I. GOLOVINA
HARDCOVER,page edges yellowed excellent
Similar linear books
Quaternions and rotation sequences: a primer with applications to orbits, aerospace, and virtual reality
Ever because the Irish mathematician William Rowan Hamilton brought quaternions within the 19th century--a feat he celebrated by way of carving the founding equations right into a stone bridge--mathematicians and engineers were thinking about those mathematical gadgets. this present day, they're utilized in purposes as numerous as describing the geometry of spacetime, guiding the gap trip, and constructing desktop purposes in digital fact.
Instructor's Solution Manual for "Applied Linear Algebra" (with Errata)
Resolution guide for the booklet utilized Linear Algebra via Peter J. Olver and Chehrzad Shakiban
Extra info for Álgebra Lineal y Algunas de sus Aplicaciones
Sample text
The decrease in population is gradual, however, and over the 100-year time span shown here it decreases only from 3000 to 1148. 9830 times that of the previous year. 9830) ≈ 411 years. One could try to find the matrix element a32 that results in the largest magnitude eigenvalue of A being exactly 1, and while this is an interesting mathematical problem, from the point of view of population control, it would likely be an exercise in futility. Clearly, this model is highly simplified, and even if the assumptions about survival rates and fecundity rates held initially, they might well change over time.
4 ECOLOGICAL MODELS Computational biology is a growing field of application for numerical methods. In this section, we explore a simplified example from ecology. Suppose we wish to study the population of a certain species of bird. These birds are born in the spring and live at most 3 years. We will keep track of the population just before breeding, when there will be three classes of birds, based on age: Age 0 (born the previous spring), Age 1, and Age 2. Let , and represent the number of females in each age class in Year n.
Both deterministic methods and Monte Carlo methods are sometimes used in this case. The picture to the left is a finite element discretization used in a deterministic model for radiation transport problems. (Image reproduced with the kind permission of the Applied Modelling and Computational Group at Imperial College London and EDF Energy. 3 MODELING IN SPORTS In FIFA World Cup soccer matches, soccer balls curve and swerve through the air, in the players’ attempts to confuse goalkeepers and send the ball sailing to the back of the net.
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# Álgebra Lineal y Algunas de sus Aplicaciones by L.I. GOLOVINA. By L.I. GOLOVINA. HARDCOVER,page edges yellowed excellent. Similar linear books. Quaternions and rotation sequences: a primer with applications to orbits, aerospace, and virtual reality. Ever because the Irish mathematician William Rowan Hamilton brought quaternions within the 19th century--a feat he celebrated by way of carving the founding equations right into a stone bridge--mathematicians and engineers were thinking about those mathematical gadgets. this present day, they're utilized in purposes as numerous as describing the geometry of spacetime, guiding the gap trip, and constructing desktop purposes in digital fact.. Instructor's Solution Manual for "Applied Linear Algebra" (with Errata). Resolution guide for the booklet utilized Linear Algebra via Peter J. Olver and Chehrzad Shakiban. Extra info for Álgebra Lineal y Algunas de sus Aplicaciones. Sample text. The decrease in population is gradual, however, and over the 100-year time span shown here it decreases only from 3000 to 1148. 9830 times that of the previous year. 9830) ≈ 411 years. One could try to find the matrix element a32 that results in the largest magnitude eigenvalue of A being exactly 1, and while this is an interesting mathematical problem, from the point of view of population control, it would likely be an exercise in futility. Clearly, this model is highly simplified, and even if the assumptions about survival rates and fecundity rates held initially, they might well change over time.. 4 ECOLOGICAL MODELS Computational biology is a growing field of application for numerical methods. In this section, we explore a simplified example from ecology. Suppose we wish to study the population of a certain species of bird. These birds are born in the spring and live at most 3 years. We will keep track of the population just before breeding, when there will be three classes of birds, based on age: Age 0 (born the previous spring), Age 1, and Age 2. Let , and represent the number of females in each age class in Year n.. Both deterministic methods and Monte Carlo methods are sometimes used in this case. The picture to the left is a finite element discretization used in a deterministic model for radiation transport problems.
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(Image reproduced with the kind permission of the Applied Modelling and Computational Group at Imperial College London and EDF Energy. 3 MODELING IN SPORTS In FIFA World Cup soccer matches, soccer balls curve and swerve through the air, in the players’ attempts to confuse goalkeepers and send the ball sailing to the back of the net.
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https://math.stackexchange.com/questions/851072/theorem-on-giuga-number/851114
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# Theorem on Giuga number
Giuga number : $n$ is a Giuga number $\iff$ For every prime factor $p$ of $n$ , $p | (\frac{n}{p}-1)$
How to prove the following theorem on Giuga numbers
$n$ is a giuga number $\iff$ $\sum_{i=1}^{n-1} i^{\phi(n)} \equiv -1 \mod {n}$
## 1 Answer
The $\Rightarrow$ part. For first, a giuga number must be squarefree, since, by assuming $p^2\mid n$, we have that $p$ divides two consecutive numbers, $\frac{n}{p}$ and $\frac{n}{p}-1$, that is clearly impossible. So we have: $$n = \prod_{i=1}^{k} p_i$$ that implies: $$\phi(n) = \prod_{i=1}^{k} (p_i-1).$$ By considering the sum $$\sum_{i=0}^{n-1}i^{\phi(n)}$$ $\pmod{p_i}$ we have that all the terms contribute with a $1$, except the multiples of $p_i$ that contribute with a zero. This gives: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv n-\frac{n}{p_i}\equiv (n-1)\pmod{p_i}\tag{1}$$ that holds for any $i\in[1,k]$. The chinese theorem now give: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv n-1\pmod{\prod_{i=1}^{k}p_i}$$ that is just: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv -1\pmod{n}$$ as claimed. For the $\Leftarrow$ part, we have that the congruence $\!\!\!\pmod{n}$ implies the congruence $\!\!\!\pmod{p_i}$, hence $(1)$ must hold, so we must have: $$\frac{n}{p_i}\equiv 1\pmod{p_i}$$ that is equivalent to $p_i\mid\left(\frac{n}{p_i}-1\right).$
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# Theorem on Giuga number. Giuga number : $n$ is a Giuga number $\iff$ For every prime factor $p$ of $n$ , $p | (\frac{n}{p}-1)$. How to prove the following theorem on Giuga numbers. $n$ is a giuga number $\iff$ $\sum_{i=1}^{n-1} i^{\phi(n)} \equiv -1 \mod {n}$. ## 1 Answer. The $\Rightarrow$ part. For first, a giuga number must be squarefree, since, by assuming $p^2\mid n$, we have that $p$ divides two consecutive numbers, $\frac{n}{p}$ and $\frac{n}{p}-1$, that is clearly impossible. So we have: $$n = \prod_{i=1}^{k} p_i$$ that implies: $$\phi(n) = \prod_{i=1}^{k} (p_i-1).$$ By considering the sum $$\sum_{i=0}^{n-1}i^{\phi(n)}$$ $\pmod{p_i}$ we have that all the terms contribute with a $1$, except the multiples of $p_i$ that contribute with a zero.
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This gives: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv n-\frac{n}{p_i}\equiv (n-1)\pmod{p_i}\tag{1}$$ that holds for any $i\in[1,k]$. The chinese theorem now give: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv n-1\pmod{\prod_{i=1}^{k}p_i}$$ that is just: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv -1\pmod{n}$$ as claimed. For the $\Leftarrow$ part, we have that the congruence $\!\!\!\pmod{n}$ implies the congruence $\!\!\!\pmod{p_i}$, hence $(1)$ must hold, so we must have: $$\frac{n}{p_i}\equiv 1\pmod{p_i}$$ that is equivalent to $p_i\mid\left(\frac{n}{p_i}-1\right).$.
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Algebra -> Algebra -> Quadratic Equations and Parabolas -> Quadratic Equations Lessons -> Questions on Algebra: Quadratic Equation answered by real tutors! Log On
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Question 372797: Jack wanted to throw an apple to Lauren, who was on a balcony 40 feet above him, so he tossed it upward with an initial speed of 56 ft/s. Lauren missed it on the way up, but then caught it on the way down. How long was the apple in the air? Would I use the equation h(t) = -16t^2 + 56t +40? (where h = height and t = time) I'm not very good at solving quadratic word problems :( Click here to see answer by nerdybill(6959)
Question 372841: I have pulled my hair out trying to learn this stuff!! If anyone could help me I'd so appreciate it!! thank you!! Solve using the elimination method. Show your work. If the system has no solution or an infinite number of solutions, state this. 3x + 11y = 34.5 9x – 7y = -16.5 Click here to see answer by josmiceli(9681)
Question 372800: The number of miles, M, that a car can travel on a gallon of gas during a trip depends on the average speed, v, of the car on the trip according to the model M = -.011v^2 + 1.3v. Find the average speed that will minimize the costs for a trip. A person wants to take a 1900 mile trip in the car and wants to spend less than \$55 for gas. If gas costs \$1.10, how fast can the person drive? Click here to see answer by [email protected](15652)
Question 372988: To buy both a new car and a new house, Tina sought two loans totalling \$191,610. The simple interest rate on the first loan was 6.3%, while the simple interest rate on the second loan was 8.8%. At the end of the first year, Tina paid a combined interest payment of \$16,681.66. What were the amounts of the two loans? Click here to see answer by mananth(12270)
Question 373133: I have to do a project and find two irrational solutions of: 4x^2+bx+25 The project also tells us to solve for: two imaginary solutions, two rational solutions, and one rational solution too, and the only one I solved was the one rational solution. can you please help! I am completely stumped! Thank you for your help :) Click here to see answer by Fombitz(13828)
Question 373177: Write a quadratic equation for which the following are solution for x: 7 and -3. Click here to see answer by Fombitz(13828)
Question 373397: How do you find the vertex, axis of symmetry, minimum or maximum value of a quadratic function? Click here to see answer by kac123kac(8)
Question 373414: how do you find the (a) quantity from a graph that has the vertex at 4,0 and the y intercept at 200 Click here to see answer by scott8148(6628)
Question 373661: I need to solve these quadratic equations using the method that came from India. I am confused. The sample problem is: xsquared + 3x -10 = 0 The solution reads as follows, but I do not understand it. x squared + 3x = 10 4x squared + 12x = 40 4x squared + 12x +9 = 40 + 9 4x squared + 12x +9 = 49 2x + 3 = +-7 2x + 3 = 7 2x = 4 x = 2 and 2x + 3 = -7 2x = -10 x = -5 Is there anyway you can explain this method to me? I do not understand where the x squared turned into the 4x squared and where the 12x went later in the problem. Thank you in advance for your help. Click here to see answer by solver91311(16897)
Question 373705: How to solve 2x^2 + 7 = 61 and 5z^2-200 = 0 Thank you very much Click here to see answer by ewatrrr(10682)
Question 373715: using the discriminant, how many times does the graph of this equation cross the x-axis? 2x^2+6x-3=0 im not looking for the answer just explain the steps please? Click here to see answer by Fombitz(13828)
Question 373895: For all values of n and r, where r ≤ n, does nCr always equal nCn-r? Why or why not? NO? I am so confused Click here to see answer by robertb(4012)
Question 373865: Find a value of the variable that shows that the two expressions are not equivalent. Answers may vary. a + 7 / 7 ;a Click here to see answer by Fombitz(13828)
Question 373884: A positive integer is 11 more than 5 times another. The product is 988. Find the two integers using the quadratic formula. Click here to see answer by mananth(12270)
Question 373971: help me solve this, please. 7x+x(x-5)=0. I'm not sure what to put in place of the x Click here to see answer by nerdybill(6959)
Question 374113: Can you help me solve this problem. Our topic is about quadratic equations by completing the square. The product of two numbers is 10 less than 16 times the smaller.If twice the smaller is 5 more than the larger number, find the two numbers. Click here to see answer by scott8148(6628)
Question 374191: When the square of a certain number is diminished by 9 times the number the result is 36. Find the number. Click here to see answer by Bebita1992(4)
Question 374107: the area of a carpet is 308 sq m .If its length is 2 meters less and its width 6 meters more,the carpet will be square.Find the dimensions of the carpet. P.S Solving Quadratic Equations by Completing the Square is the topic. Click here to see answer by amoresroy(333)
Question 374235: Can you help me solve this problem please! topic:quadratic equations by completing the square marivic is five years older than michelle. in three years , the prdocut of her age and michelle's age five years ago will be 90 years.Find their present ages. Click here to see answer by mananth(12270)
Question 374365: Determine the nature of the solutions of the equation X2-3x+9=0 A: one real solution B: two non-real solutions C: two real solutions Click here to see answer by jsmallt9(3296)
Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560, 16561..16605, 16606..16650, 16651..16695, 16696..16740, 16741..16785, 16786..16830, 16831..16875, 16876..16920, 16921..16965, 16966..17010, 17011..17055, 17056..17100, 17101..17145, 17146..17190, 17191..17235, 17236..17280, 17281..17325, 17326..17370, 17371..17415, 17416..17460, 17461..17505, 17506..17550, 17551..17595, 17596..17640, 17641..17685, 17686..17730, 17731..17775, 17776..17820, 17821..17865, 17866..17910, 17911..17955, 17956..18000, 18001..18045, 18046..18090, 18091..18135, 18136..18180, 18181..18225, 18226..18270, 18271..18315, 18316..18360, 18361..18405, 18406..18450, 18451..18495, 18496..18540, 18541..18585, 18586..18630, 18631..18675, 18676..18720, 18721..18765, 18766..18810, 18811..18855, 18856..18900, 18901..18945, 18946..18990, 18991..19035, 19036..19080, 19081..19125, 19126..19170, 19171..19215, 19216..19260, 19261..19305, 19306..19350, 19351..19395, 19396..19440, 19441..19485, 19486..19530, 19531..19575, 19576..19620, 19621..19665, 19666..19710, 19711..19755, 19756..19800, 19801..19845, 19846..19890, 19891..19935, 19936..19980, 19981..20025, 20026..20070, 20071..20115, 20116..20160, 20161..20205, 20206..20250
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Algebra -> Algebra -> Quadratic Equations and Parabolas -> Quadratic Equations Lessons -> Questions on Algebra: Quadratic Equation answered by real tutors! Log On. Ad: You enter your algebra equation or inequality - Algebrator solves it step-by-step while providing clear explanations. Free on-line demo . 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!. Question 372487: I need to use quadratic equations to find two real numbers that satisfy the sum of 15 and the product 36 Click here to see answer by amoresroy(333). Question 372559: How to solve the quadratic equation for x by factoring? For Instance, 6x^2 + 3x = 0 Click here to see answer by CharlesG2(828). Question 372585: Compute the value of the discriminant and give the number of real solutions to the quadratic equation. -2x^2+3x+1=0 discriminant= number of real numbers = Click here to see answer by Fombitz(13828). Question 372514: Use the quadratic formula to solve x2 + 8x + 9 = 0. Estimate irrational solutions to the nearest tenth. Click here to see answer by Fombitz(13828). Question 372622: Determine the coordinates of the x-intercepts for the quadratic equation 4x2 – 3x - 1 = 0. Click here to see answer by ewatrrr(10682). Question 372621: Determine the coordinates of the x-intercepts for the quadratic equation 5x2 – 21x + 4 = 0. Click here to see answer by ewatrrr(10682). Question 372615: Determine the coordinates of the x-intercepts for the quadratic equation 6x2 - 7x + 2 = 0. Click here to see answer by ewatrrr(10682). Question 372626: The product of twenty less than eight times a number and thirty less than two times a number equals zero. Find the number(s). Click here to see answer by Fombitz(13828). Question 372664: The Length of a rectangular flower bed is 3 times the width. The area of the bed is 108 square meters. What are the dimensions of the bed? Please Help me. Thank You Click here to see answer by checkley79(3050). Question 372627: A basketball was thrown from one end of a court to the other. It had an initial upward velocity of 48 feet per second. How long did it take for the ball to hit the ground? Use the formula h = vt – 16t2. Click here to see answer by Fombitz(13828). Question 372769: 15. Which two lines are parallel? I. 5y= -3x-5 II. 5y= -1 - 3x III. 3y- 2x= -1 (1 point) Click here to see answer by Alan3354(30993). Question 372797: Jack wanted to throw an apple to Lauren, who was on a balcony 40 feet above him, so he tossed it upward with an initial speed of 56 ft/s. Lauren missed it on the way up, but then caught it on the way down. How long was the apple in the air? Would I use the equation h(t) = -16t^2 + 56t +40? (where h = height and t = time) I'm not very good at solving quadratic word problems :( Click here to see answer by nerdybill(6959). Question 372841: I have pulled my hair out trying to learn this stuff!! If anyone could help me I'd so appreciate it!! thank you!! Solve using the elimination method. Show your work. If the system has no solution or an infinite number of solutions, state this. 3x + 11y = 34.5 9x – 7y = -16.5 Click here to see answer by josmiceli(9681). Question 372800: The number of miles, M, that a car can travel on a gallon of gas during a trip depends on the average speed, v, of the car on the trip according to the model M = -.011v^2 + 1.3v. Find the average speed that will minimize the costs for a trip. A person wants to take a 1900 mile trip in the car and wants to spend less than \$55 for gas. If gas costs \$1.10, how fast can the person drive? Click here to see answer by [email protected](15652). Question 372988: To buy both a new car and a new house, Tina sought two loans totalling \$191,610. The simple interest rate on the first loan was 6.3%, while the simple interest rate on the second loan was 8.8%. At the end of the first year, Tina paid a combined interest payment of \$16,681.66. What were the amounts of the two loans? Click here to see answer by mananth(12270). Question 373133: I have to do a project and find two irrational solutions of: 4x^2+bx+25 The project also tells us to solve for: two imaginary solutions, two rational solutions, and one rational solution too, and the only one I solved was the one rational solution. can you please help! I am completely stumped! Thank you for your help :) Click here to see answer by Fombitz(13828). Question 373177: Write a quadratic equation for which the following are solution for x: 7 and -3. Click here to see answer by Fombitz(13828). Question 373397: How do you find the vertex, axis of symmetry, minimum or maximum value of a quadratic function? Click here to see answer by kac123kac(8). Question 373414: how do you find the (a) quantity from a graph that has the vertex at 4,0 and the y intercept at 200 Click here to see answer by scott8148(6628). Question 373661: I need to solve these quadratic equations using the method that came from India. I am confused. The sample problem is: xsquared + 3x -10 = 0 The solution reads as follows, but I do not understand it. x squared + 3x = 10 4x squared + 12x = 40 4x squared + 12x +9 = 40 + 9 4x squared + 12x +9 = 49 2x + 3 = +-7 2x + 3 = 7 2x = 4 x = 2 and 2x + 3 = -7 2x = -10 x = -5 Is there anyway you can explain this method to me? I do not understand where the x squared turned into the 4x squared and where the 12x went later in the problem. Thank you in advance for your help. Click here to see answer by solver91311(16897). Question 373705: How to solve 2x^2 + 7 = 61 and 5z^2-200 = 0 Thank you very much Click here to see answer by ewatrrr(10682). Question 373715: using the discriminant, how many times does the graph of this equation cross the x-axis? 2x^2+6x-3=0 im not looking for the answer just explain the steps please? Click here to see answer by Fombitz(13828). Question 373895: For all values of n and r, where r ≤ n, does nCr always equal nCn-r? Why or why not? NO? I am so confused Click here to see answer by robertb(4012). Question 373865: Find a value of the variable that shows that the two expressions are not equivalent. Answers may vary. a + 7 / 7 ;a Click here to see answer by Fombitz(13828). Question 373884: A positive integer is 11 more than 5 times another. The product is 988. Find the two integers using the quadratic formula. Click here to see answer by mananth(12270). Question 373971: help me solve this, please. 7x+x(x-5)=0. I'm not sure what to put in place of the x Click here to see answer by nerdybill(6959). Question 374113: Can you help me solve this problem. Our topic is about quadratic equations by completing the square. The product of two numbers is 10 less than 16 times the smaller.If twice the smaller is 5 more than the larger number, find the two numbers. Click here to see answer by scott8148(6628). Question 374191: When the square of a certain number is diminished by 9 times the number the result is 36. Find the number. Click here to see answer by Bebita1992(4). Question 374107: the area of a carpet is 308 sq m .If its length is 2 meters less and its width 6 meters more,the carpet will be square.Find the dimensions of the carpet. P.S Solving Quadratic Equations by Completing the Square is the topic. Click here to see answer by amoresroy(333). Question 374235: Can you help me solve this problem please! topic:quadratic equations by completing the square marivic is five years older than michelle. in three years , the prdocut of her age and michelle's age five years ago will be 90 years.Find their present ages. Click here to see answer by mananth(12270).
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Question 374365: Determine the nature of the solutions of the equation X2-3x+9=0 A: one real solution B: two non-real solutions C: two real solutions Click here to see answer by jsmallt9(3296). Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560, 16561..16605, 16606..16650, 16651..16695, 16696..16740, 16741..16785, 16786..16830, 16831..16875, 16876..16920, 16921..16965, 16966..17010, 17011..17055, 17056..17100, 17101..17145, 17146..17190, 17191..17235, 17236..17280, 17281..17325, 17326..17370, 17371..17415, 17416..17460, 17461..17505, 17506..17550, 17551..17595, 17596..17640, 17641..17685, 17686..17730, 17731..17775, 17776..17820, 17821..17865, 17866..17910, 17911..17955, 17956..18000, 18001..18045, 18046..18090, 18091..18135, 18136..18180, 18181..18225, 18226..18270, 18271..18315, 18316..18360, 18361..18405, 18406..18450, 18451..18495, 18496..18540, 18541..18585, 18586..18630, 18631..18675, 18676..18720, 18721..18765, 18766..18810, 18811..18855, 18856..18900, 18901..18945, 18946..18990, 18991..19035, 19036..19080, 19081..19125, 19126..19170, 19171..19215, 19216..19260, 19261..19305, 19306..19350, 19351..19395, 19396..19440, 19441..19485, 19486..19530, 19531..19575, 19576..19620, 19621..19665, 19666..19710, 19711..19755, 19756..19800, 19801..19845, 19846..19890, 19891..19935, 19936..19980, 19981..20025, 20026..20070, 20071..20115, 20116..20160, 20161..20205, 20206..20250.
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# The moon's mass is 7.34x10-kg and it is 3.8x10m away from earth. Calculate the gravitational force of attraction between earth and moon
2
by poolruthpepe
and it says is 3.8x10m away from earth
it's 380m?
I'll just google the distances and weights and give you an answer
show me the work
okay
2014-11-28T15:14:21-05:00
Again I think you did not give the right constants. So I would use the correct constants for mass of moon and distance from earth to moon.
The formula for force of attraction between any two bodies in the universe
F = GMm / r^2. (Newton's Universal law of Gravitation).
G = Universal gravitational constant, G = 6.67 * 10 ^ -11 Nm^2 / kg^2.
M = Mass of Earth. = 5.97 x 10^24 kg.
m = mass of moon = 7.34 x 10^22 kg.
r = distance apart, between centers = in this case it is the distance from Earth to the Moon
= 3.8 x 10^8 m.
(Sorry I could not assume with the values you gave, they are wrong, and if we use them we would be insulting Physics).
So F = ((6.67 * 10 ^ -11)*(5.97 x 10^24)*(7.34 * 10^22)) / (3.8 x 10^8)^2.
Punch it all up in your calculator.
I used a Casio 991 calculator, it should be one of the best in the world.Really lovely calculator, that has helped me a lot in computations like this. I am thankful for the Calculator.
F = 2.0240 * 10^ 20 N.
So that's our answer.
Hurray!!
I decided to use the right values. I don't think your teacher would want you to use an assumed value. Please confirm from him or her.
2014-11-28T15:16:51-05:00
Using the gravity equation: force of gravity = (G*mass of object 1*mass of object 2)/distance between the objects^2, we can plug in the masses of the objects (5.97x10^24kg for earth, and 7.34x10^22kg for the moon) and the distance between the objects (3.8x10^8metres) and (6.67x10^-11 gravitational constant) for G, we get (6.67x10^-11*5.97x10^24*7.34^22)/(3.8x10^8)^2 which equals =2.02x10^20 which is the force of gravity
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# The moon's mass is 7.34x10-kg and it is 3.8x10m away from earth. Calculate the gravitational force of attraction between earth and moon. 2. by poolruthpepe. and it says is 3.8x10m away from earth. it's 380m?. I'll just google the distances and weights and give you an answer. show me the work. okay. 2014-11-28T15:14:21-05:00. Again I think you did not give the right constants. So I would use the correct constants for mass of moon and distance from earth to moon.. The formula for force of attraction between any two bodies in the universe. F = GMm / r^2. (Newton's Universal law of Gravitation).. G = Universal gravitational constant, G = 6.67 * 10 ^ -11 Nm^2 / kg^2.. M = Mass of Earth. = 5.97 x 10^24 kg.. m = mass of moon = 7.34 x 10^22 kg.. r = distance apart, between centers = in this case it is the distance from Earth to the Moon. = 3.8 x 10^8 m.. (Sorry I could not assume with the values you gave, they are wrong, and if we use them we would be insulting Physics).. So F = ((6.67 * 10 ^ -11)*(5.97 x 10^24)*(7.34 * 10^22)) / (3.8 x 10^8)^2.. Punch it all up in your calculator.. I used a Casio 991 calculator, it should be one of the best in the world.Really lovely calculator, that has helped me a lot in computations like this. I am thankful for the Calculator.. F = 2.0240 * 10^ 20 N.. So that's our answer.. Hurray!!. I decided to use the right values. I don't think your teacher would want you to use an assumed value. Please confirm from him or her.
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2014-11-28T15:16:51-05:00. Using the gravity equation: force of gravity = (G*mass of object 1*mass of object 2)/distance between the objects^2, we can plug in the masses of the objects (5.97x10^24kg for earth, and 7.34x10^22kg for the moon) and the distance between the objects (3.8x10^8metres) and (6.67x10^-11 gravitational constant) for G, we get (6.67x10^-11*5.97x10^24*7.34^22)/(3.8x10^8)^2 which equals =2.02x10^20 which is the force of gravity.
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# Master integration by parts pdf
This unit derives and illustrates this rule with a number of examples. Archimedes is the founder of surface areas and volumes of solids such as the sphere and the cone. Spare parts are components that are kept in your inventory as spare. Math 105 921 solutions to integration exercises ubc math. To evaluate that integral, you can apply integration by parts again. Live demonstration of a parts integration with nlp the conflict between making choices and staying with what you have in career is usually a common and tough one. The integration enables users to reduce the amount of time they spend trying to find and share design information, as well as eliminate unnecessary change orders by ensuring that everyone is working from the latest design information. Apart from that, but more importantly, if you want to master taking derivatives of functions, and integration, youll need to devote yourself to practice, and lots of it. This is illustrated using as an example the twoloop sunset diagram with onshell. An intuitive and geometric explanation sahand rabbani the formula for integration by parts is given below. Now, integration by parts produces first use of integration by parts this first use of integration by parts has succeeded in simplifying the original integral, but the integral on the right still doesnt fit a basic integration rule. That is, we want to compute z px qx dx where p, q are polynomials. Integration by parts is like the reverse of the product formula. Methods of integration william gunther june 15, 2011 in this we will go over some of the techniques of integration, and when to apply them.
Material is procured from external or internal sources on the basis of the requirements determined by material requirements planning. Sometimes integration by parts must be repeated to obtain an answer. A firm grounding in differential calculus is a must. For the following problems, indicate whether you would use integration by parts with your choices of u and dv, substitution with your choice of u, or neither. Integration by parts mctyparts20091 a special rule, integrationbyparts, is available for integrating products of two functions. That is integration, and it is the goal of integral calculus. I can sit for hours and do a 1,000, 2,000 or 5,000piece jigsaw puzzle. And just observe each and every solved problem in textbook. If nothing else works, convert everything to sines and cosines. When choosing u and dv, u should get \simpler with di erentiation and you should be able to integrate dv. During all transactions, inventory management accesses both master data such as material master data and transaction data such as purchasing documents shared by all logistics components.
Createdisplay a location is a virtual record of the location where an equipment is installed. We show that the new relation between master integrals recently obtained in ref. In order to master the techniques explained here it is vital that you undertake. Example 1 z f xg0xdx f xgx z gxf 0xdx or z udv uv z vdu. The integration by parts formula we need to make use of the integration by parts formula which states. Integration of parts follows a common negotiating technique. Integration by parts examples, tricks and a secret howto. Jul, 2016 refurbished spare parts are parts that can be repaired either internally or sent out to be repaired by an external vendor.
Make sure you identify the parts clearly, and understand the nature of the conflict. Calculus ii integration strategy pauls online math notes. Introduction to integration by parts unlike the previous method, we already know everything we need to to under stand integration by parts. If youre behind a web filter, please make sure that the domains. If youre seeing this message, it means were having trouble loading external resources on our website.
Integration techniques integral calculus 2017 edition. This sdk supports connecting to and commanding aris explorer and aris voyager sonars. Several methods of calculation of master integrals also. Integrationbyparts procedure with effective mass article pdf available in physics letters b 7123 february 2012 with 33 reads how we measure reads.
When you have the product of two xterms in which one term is not the derivative of the other, this is the most common situation and special integrals like. This is unfortunate because tabular integration by parts is not only a valuable tool for finding integrals but can also be applied to more advanced topics including the derivations of some important. This uses one of the problems from the collection and gives you an idea of what the grading rubric looks like. The function being integrated, fx, is called the integrand. Integration by parts easy method i liate i integral uv i. Integral calculus 2017 edition integration techniques. Jan 21, 2017 integration by parts shortcut method in hindi i liate i integral uv i class 12 ncert. The following are solutions to the integration by parts practice problems posted november 9.
For instance, a substitution may lead to using integration by parts or partial fractions integral. Tabular integration by parts when integration by parts is needed more than once you are actually doing integration by parts recursively. Notice that we needed to use integration by parts twice to solve this problem. With, and, the rule is also written more compactly as 2 equation 1 comes from the product rule. Find, read and cite all the research you need on researchgate. This is illustrated using as an example the twoloop sunset diagram with onshell kinematics. Using integration by parts might not always be the correct or best solution. Thats the quick waybut do bear in mind that, typically, an online editor isnt as fully featured as its desktop counterpart, plus the file is exposed to the internet which might be of. Here are the tricks and secrets you need to know to master this technique of integration. Some of these are online pdf editors that work right in your web browser, so all you have to do is upload your pdf file to the website, make the changes you want, and then save it back to your computer. To develop competence and mastery, you need to do math, and not just read about it. This process is necessary for parts that are no longer commercially available and for parts that can be overhauled and reused at a fraction of the price of a.
Inventory management is part of the materials management module and is fully integrated in the entire logistics system. Then there are trig and hyperbolic functions, they can be utilised by considering their derivatives backward, etc. When one party needs the deal less than the other party, they have the power and ability to walk away. Then z exsinxdx exsinx z excosxdx now we need to use integration by parts on the second integral. Lot of people just seem to ignore the solved problems, they jump to solving exercises, this is foolishness, when there ar. Navigation refresher learning objectives in this lesson, you will. You will see plenty of examples soon, but first let us see the rule. Integration by parts is a technique for evaluating integrals whose integrand is the product of two functions. Integration by parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways. We want to choose u u and d v d v so that when we compute d u d u and v v and plugging everything into the integration by parts formula the new integral we get is one that we can do. Using the catia integration, design teams can quickly search for catia parts, assemblies and drawings. Refurbished spare parts are parts that can be repaired either internally or sent out to be repaired by an external vendor. Mar 09, 2016 best choice will be to start with the state board textbooks.
The parts are then placed back into inventory for future use. Using repeated applications of integration by parts. You have two parties that come together for a negotiation. Sumdi erence r fx gx dx r fxdx r gx dx scalar multiplication r cfx. An lloop diagram has l loop integration momenta k1. Finney,calculus and analytic geometry,addisonwesley, reading, ma 1988. Z vdu 1 while most texts derive this equation from the product rule of di. With that in mind it looks like the following choices for u u and d v d v should work for us. Pdf integration by parts is used to reduce scalar feynman integrals to master integrals. Bonus evaluate r 1 0 x 5e x using integration by parts. C is an arbitrary constant called the constant of integration.
Well learn that integration and di erentiation are inverse operations of each other. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. One strategy is the might makes right strategy, which says whoever can exert the most force will win the negotiation. Nabeel khan 61 daud mirza 57 danish mirza 58 fawad usman 66 amir mughal 72 m. The integration by parts formula for indefinite integrals is given by. Integration by parts a special rule, integration by parts, is available for integrating products of two functions. Integrationbyparts procedure with effective mass article pdf available in physics letters b 7123 february 2012 with 33. Best choice will be to start with the state board textbooks. Have the part, which represents the unwanted state or behaviour come out on the hand first. At first it appears that integration by parts does not apply, but let. For example, if integrating the function fx with respect to x.
541 1442 856 9 444 1323 1017 687 602 551 1096 1494 101 1318 770 1320 272 138 979 141 382 172 1150 518 1038 1283 509 585 262 1439 510 1263 1057 987 1299 737 719 493 1161 301 1257 916 42 868
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# Master integration by parts pdf. This unit derives and illustrates this rule with a number of examples. Archimedes is the founder of surface areas and volumes of solids such as the sphere and the cone. Spare parts are components that are kept in your inventory as spare. Math 105 921 solutions to integration exercises ubc math. To evaluate that integral, you can apply integration by parts again. Live demonstration of a parts integration with nlp the conflict between making choices and staying with what you have in career is usually a common and tough one. The integration enables users to reduce the amount of time they spend trying to find and share design information, as well as eliminate unnecessary change orders by ensuring that everyone is working from the latest design information. Apart from that, but more importantly, if you want to master taking derivatives of functions, and integration, youll need to devote yourself to practice, and lots of it. This is illustrated using as an example the twoloop sunset diagram with onshell. An intuitive and geometric explanation sahand rabbani the formula for integration by parts is given below. Now, integration by parts produces first use of integration by parts this first use of integration by parts has succeeded in simplifying the original integral, but the integral on the right still doesnt fit a basic integration rule. That is, we want to compute z px qx dx where p, q are polynomials. Integration by parts is like the reverse of the product formula. Methods of integration william gunther june 15, 2011 in this we will go over some of the techniques of integration, and when to apply them.. Material is procured from external or internal sources on the basis of the requirements determined by material requirements planning. Sometimes integration by parts must be repeated to obtain an answer. A firm grounding in differential calculus is a must. For the following problems, indicate whether you would use integration by parts with your choices of u and dv, substitution with your choice of u, or neither. Integration by parts mctyparts20091 a special rule, integrationbyparts, is available for integrating products of two functions. That is integration, and it is the goal of integral calculus. I can sit for hours and do a 1,000, 2,000 or 5,000piece jigsaw puzzle. And just observe each and every solved problem in textbook. If nothing else works, convert everything to sines and cosines. When choosing u and dv, u should get \simpler with di erentiation and you should be able to integrate dv. During all transactions, inventory management accesses both master data such as material master data and transaction data such as purchasing documents shared by all logistics components.. Createdisplay a location is a virtual record of the location where an equipment is installed. We show that the new relation between master integrals recently obtained in ref. In order to master the techniques explained here it is vital that you undertake. Example 1 z f xg0xdx f xgx z gxf 0xdx or z udv uv z vdu. The integration by parts formula we need to make use of the integration by parts formula which states. Integration of parts follows a common negotiating technique. Integration by parts examples, tricks and a secret howto. Jul, 2016 refurbished spare parts are parts that can be repaired either internally or sent out to be repaired by an external vendor.. Make sure you identify the parts clearly, and understand the nature of the conflict. Calculus ii integration strategy pauls online math notes. Introduction to integration by parts unlike the previous method, we already know everything we need to to under stand integration by parts. If youre behind a web filter, please make sure that the domains. If youre seeing this message, it means were having trouble loading external resources on our website.. Integration techniques integral calculus 2017 edition. This sdk supports connecting to and commanding aris explorer and aris voyager sonars. Several methods of calculation of master integrals also. Integrationbyparts procedure with effective mass article pdf available in physics letters b 7123 february 2012 with 33 reads how we measure reads.. When you have the product of two xterms in which one term is not the derivative of the other, this is the most common situation and special integrals like. This is unfortunate because tabular integration by parts is not only a valuable tool for finding integrals but can also be applied to more advanced topics including the derivations of some important. This uses one of the problems from the collection and gives you an idea of what the grading rubric looks like. The function being integrated, fx, is called the integrand. Integration by parts easy method i liate i integral uv i. Integral calculus 2017 edition integration techniques. Jan 21, 2017 integration by parts shortcut method in hindi i liate i integral uv i class 12 ncert. The following are solutions to the integration by parts practice problems posted november 9.. For instance, a substitution may lead to using integration by parts or partial fractions integral. Tabular integration by parts when integration by parts is needed more than once you are actually doing integration by parts recursively. Notice that we needed to use integration by parts twice to solve this problem. With, and, the rule is also written more compactly as 2 equation 1 comes from the product rule. Find, read and cite all the research you need on researchgate. This is illustrated using as an example the twoloop sunset diagram with onshell kinematics. Using integration by parts might not always be the correct or best solution. Thats the quick waybut do bear in mind that, typically, an online editor isnt as fully featured as its desktop counterpart, plus the file is exposed to the internet which might be of. Here are the tricks and secrets you need to know to master this technique of integration. Some of these are online pdf editors that work right in your web browser, so all you have to do is upload your pdf file to the website, make the changes you want, and then save it back to your computer. To develop competence and mastery, you need to do math, and not just read about it. This process is necessary for parts that are no longer commercially available and for parts that can be overhauled and reused at a fraction of the price of a.. Inventory management is part of the materials management module and is fully integrated in the entire logistics system. Then there are trig and hyperbolic functions, they can be utilised by considering their derivatives backward, etc. When one party needs the deal less than the other party, they have the power and ability to walk away. Then z exsinxdx exsinx z excosxdx now we need to use integration by parts on the second integral. Lot of people just seem to ignore the solved problems, they jump to solving exercises, this is foolishness, when there ar. Navigation refresher learning objectives in this lesson, you will. You will see plenty of examples soon, but first let us see the rule. Integration by parts is a technique for evaluating integrals whose integrand is the product of two functions. Integration by parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways. We want to choose u u and d v d v so that when we compute d u d u and v v and plugging everything into the integration by parts formula the new integral we get is one that we can do. Using the catia integration, design teams can quickly search for catia parts, assemblies and drawings. Refurbished spare parts are parts that can be repaired either internally or sent out to be repaired by an external vendor. Mar 09, 2016 best choice will be to start with the state board textbooks.. The parts are then placed back into inventory for future use. Using repeated applications of integration by parts. You have two parties that come together for a negotiation. Sumdi erence r fx gx dx r fxdx r gx dx scalar multiplication r cfx. An lloop diagram has l loop integration momenta k1. Finney,calculus and analytic geometry,addisonwesley, reading, ma 1988. Z vdu 1 while most texts derive this equation from the product rule of di. With that in mind it looks like the following choices for u u and d v d v should work for us. Pdf integration by parts is used to reduce scalar feynman integrals to master integrals. Bonus evaluate r 1 0 x 5e x using integration by parts. C is an arbitrary constant called the constant of integration.. Well learn that integration and di erentiation are inverse operations of each other. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. One strategy is the might makes right strategy, which says whoever can exert the most force will win the negotiation. Nabeel khan 61 daud mirza 57 danish mirza 58 fawad usman 66 amir mughal 72 m. The integration by parts formula for indefinite integrals is given by. Integration by parts a special rule, integration by parts, is available for integrating products of two functions. Integrationbyparts procedure with effective mass article pdf available in physics letters b 7123 february 2012 with 33. Best choice will be to start with the state board textbooks. Have the part, which represents the unwanted state or behaviour come out on the hand first. At first it appears that integration by parts does not apply, but let.
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For example, if integrating the function fx with respect to x.. 541 1442 856 9 444 1323 1017 687 602 551 1096 1494 101 1318 770 1320 272 138 979 141 382 172 1150 518 1038 1283 509 585 262 1439 510 1263 1057 987 1299 737 719 493 1161 301 1257 916 42 868.
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https://tarungehlots.blogspot.com/2014/08/chapter-6-worked-out-examples.html
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tgt
## Saturday, 2 August 2014
### CHAPTER 6- Worked Out Examples
Example: 1
If ${x^2} + {y^2} = t + \dfrac{1}{t}\,\,$ and $\,{x^4} + {y^4} = {t^2} + \dfrac{1}{{{t^2}}}$, then prove that $\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^3}y}}$
Solution: 1
We first try to use the two given relations to get rid of the parameter $t$, so that we obtain a (implicit) relation between $x$ and $y$.
${x^2} + {y^2} = t + \dfrac{1}{t}$
Squaring, we get
${x^4} + {y^4} + 2{x^2}{y^2} = {t^2} + \dfrac{1}{{{t^2}}} + 2$ $\ldots(i)$
Using the second relation in ($i$), we get
$2{x^2}{y^2} = 2$ $\Rightarrow \,\, {y^2} = \dfrac{1}{{{x^2}}}$
Differentiating both sides w.r.t $x$, we get
$2y\dfrac{{dy}}{{dx}} = \dfrac{{ - 2}}{{{x^3}}}$ $\Rightarrow \,\, \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^3}y}}$
Example: 2
If ${y^2} = {a^2}{\cos ^2}x + {b^2}{\sin ^2}x$, then prove that
$\dfrac{{{d^2}y}}{{d{x^2}}} + y = \dfrac{{{a^2}{b^2}}}{{{y^3}}}$
Solution: 2
The final relation that we need to obtain is independent of $\sin x$ and $\cos x$; this gives us a hint that using the given relation, we must first get rid of $\sin x$and $\cos x$:
${y^2} = {a^2}{\cos ^2}x + {b^2}{\sin ^2}x$ $= \dfrac{1}{2}\left\{ {{a^2}\left( {2{{\cos }^2}x} \right) + {b^2}\left( {2{{\sin }^2}x} \right)} \right\}$ $= \dfrac{1}{2}\left\{ {{a^2}\left( {1 + \cos 2x} \right) + {b^2}\left( {1 - \cos 2x} \right)} \right\}$ $= \dfrac{1}{2}\left\{ {\left( {{a^2} + {b^2}} \right) + \left( {{a^2} - {b^2}} \right)\cos 2x} \right\}$ $\Rightarrow \, 2{y^2} - \left( {{a^2} + {b^2}} \right) = \left( {{a^2} - {b^2}} \right)\cos 2x$ $\ldots(i)$
Differentiating both sides of ($i$) w.r.t $x$, we get
$4y\dfrac{{dy}}{{dx}} = - 2\left( {{a^2} - {b^2}} \right)\sin 2x$ $\Rightarrow \,\, - 2y\dfrac{{dy}}{{dx}} = \left( {{a^2} - {b^2}} \right)\sin 2x$ $\ldots(ii)$
We see now that squaring ($i$) and ($ii$) and adding them will lead to an expression independent of the trig. terms:
${\left( {2{y^2} - \left( {{a^2} + {b^2}} \right)} \right)^2} + 4{y^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {\left( {{a^2} - {b^2}} \right)^2}$
A slight rearrangement gives:
${\left( {\dfrac{{dy}}{{dx}}} \right)^2} + {y^2} - \left( {{a^2} + {b^2}} \right) = - \dfrac{{{a^2}{b^2}}}{{{y^2}}}$ $\ldots(iii)$
Differentiating both sides of ($iii$) w.r.t $x$:
$2\left( {\dfrac{{dy}}{{dx}}} \right)\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) + 2y\dfrac{{dy}}{{dx}} = \dfrac{{2{a^2}{b^2}}}{{{y^3}}}\dfrac{{dy}}{{dx}}$ $\Rightarrow \,\, \dfrac{{{d^2}y}}{{d{x^2}}} + y = \dfrac{{{a^2}{b^2}}}{{{y^3}}}$
Example:3
If the derivatives of $f(x)$ and $g(x)$ are known, find the derivative of $y = {\left\{ {f\left( x \right)} \right\}^{g\left( x \right)}}$
Solution: 3
We cannot directly differentiate the given relation since no rule tells us how to differentiate a term ${p^q}$ where both $p$ and $q$ are variables.
What we can instead do is take the logarithm of both sides of the given relation:
$y = {\left\{ {f\left( x \right)} \right\}^{g\left( x \right)}}$ $\Rightarrow \,\, \ln y = g\left( x \right)\ln \left( {f\left( x \right)} \right)$
Now we differentiate both sides w.r.t $x$:
$\Rightarrow \,\, \dfrac{1}{y}\dfrac{{dy}}{{dx}} = g\left( x \right) \cdot \dfrac{1}{{f\left( x \right)}} \cdot f'\left( x \right) + \ln \left( {f\left( x \right)} \right) \cdot g'\left( x \right)$ $\Rightarrow \,\, \dfrac{{dy}}{{dx}} = y\left\{ {\dfrac{{f'\left( x \right)g\left( x \right)}}{{f\left( x \right)}} + g'\left( x \right)\ln \left( {f\left( x \right)} \right)} \right.$ $= {\left( {f\left( x \right)} \right)^{g\left( x \right)}}\left\{ {\dfrac{{f'\left( x \right)g\left( x \right)}}{{f\left( x \right)}} + g'\left( x \right)\ln \left( {f\left( x \right)} \right)} \right\}$
As a simple example, suppose we have to differentiate $y = {x^x}$:
$\ln y = x\ln x$ $\Rightarrow \,\, \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \cdot \dfrac{1}{x} + \ln x \cdot 1$ $= 1 + \ln x$ $\Rightarrow \,\, \dfrac{{dy}}{{dx}} = y\left( {1 + \ln x} \right)$ $= {x^x}\left( {1 + \ln x} \right)$
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tgt. ## Saturday, 2 August 2014. ### CHAPTER 6- Worked Out Examples. Example: 1. If ${x^2} + {y^2} = t + \dfrac{1}{t}\,\,$ and $\,{x^4} + {y^4} = {t^2} + \dfrac{1}{{{t^2}}}$, then prove that $\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^3}y}}$. Solution: 1. We first try to use the two given relations to get rid of the parameter $t$, so that we obtain a (implicit) relation between $x$ and $y$.. ${x^2} + {y^2} = t + \dfrac{1}{t}$. Squaring, we get. ${x^4} + {y^4} + 2{x^2}{y^2} = {t^2} + \dfrac{1}{{{t^2}}} + 2$ $\ldots(i)$. Using the second relation in ($i$), we get. $2{x^2}{y^2} = 2$ $\Rightarrow \,\, {y^2} = \dfrac{1}{{{x^2}}}$. Differentiating both sides w.r.t $x$, we get. $2y\dfrac{{dy}}{{dx}} = \dfrac{{ - 2}}{{{x^3}}}$ $\Rightarrow \,\, \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^3}y}}$. Example: 2. If ${y^2} = {a^2}{\cos ^2}x + {b^2}{\sin ^2}x$, then prove that. $\dfrac{{{d^2}y}}{{d{x^2}}} + y = \dfrac{{{a^2}{b^2}}}{{{y^3}}}$. Solution: 2. The final relation that we need to obtain is independent of $\sin x$ and $\cos x$; this gives us a hint that using the given relation, we must first get rid of $\sin x$and $\cos x$:. ${y^2} = {a^2}{\cos ^2}x + {b^2}{\sin ^2}x$ $= \dfrac{1}{2}\left\{ {{a^2}\left( {2{{\cos }^2}x} \right) + {b^2}\left( {2{{\sin }^2}x} \right)} \right\}$ $= \dfrac{1}{2}\left\{ {{a^2}\left( {1 + \cos 2x} \right) + {b^2}\left( {1 - \cos 2x} \right)} \right\}$ $= \dfrac{1}{2}\left\{ {\left( {{a^2} + {b^2}} \right) + \left( {{a^2} - {b^2}} \right)\cos 2x} \right\}$ $\Rightarrow \, 2{y^2} - \left( {{a^2} + {b^2}} \right) = \left( {{a^2} - {b^2}} \right)\cos 2x$ $\ldots(i)$. Differentiating both sides of ($i$) w.r.t $x$, we get. $4y\dfrac{{dy}}{{dx}} = - 2\left( {{a^2} - {b^2}} \right)\sin 2x$ $\Rightarrow \,\, - 2y\dfrac{{dy}}{{dx}} = \left( {{a^2} - {b^2}} \right)\sin 2x$ $\ldots(ii)$. We see now that squaring ($i$) and ($ii$) and adding them will lead to an expression independent of the trig. terms:. ${\left( {2{y^2} - \left( {{a^2} + {b^2}} \right)} \right)^2} + 4{y^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {\left( {{a^2} - {b^2}} \right)^2}$. A slight rearrangement gives:. ${\left( {\dfrac{{dy}}{{dx}}} \right)^2} + {y^2} - \left( {{a^2} + {b^2}} \right) = - \dfrac{{{a^2}{b^2}}}{{{y^2}}}$ $\ldots(iii)$. Differentiating both sides of ($iii$) w.r.t $x$:. $2\left( {\dfrac{{dy}}{{dx}}} \right)\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) + 2y\dfrac{{dy}}{{dx}} = \dfrac{{2{a^2}{b^2}}}{{{y^3}}}\dfrac{{dy}}{{dx}}$ $\Rightarrow \,\, \dfrac{{{d^2}y}}{{d{x^2}}} + y = \dfrac{{{a^2}{b^2}}}{{{y^3}}}$. Example:3. If the derivatives of $f(x)$ and $g(x)$ are known, find the derivative of $y = {\left\{ {f\left( x \right)} \right\}^{g\left( x \right)}}$. Solution: 3. We cannot directly differentiate the given relation since no rule tells us how to differentiate a term ${p^q}$ where both $p$ and $q$ are variables.. What we can instead do is take the logarithm of both sides of the given relation:. $y = {\left\{ {f\left( x \right)} \right\}^{g\left( x \right)}}$ $\Rightarrow \,\, \ln y = g\left( x \right)\ln \left( {f\left( x \right)} \right)$. Now we differentiate both sides w.r.t $x$:. $\Rightarrow \,\, \dfrac{1}{y}\dfrac{{dy}}{{dx}} = g\left( x \right) \cdot \dfrac{1}{{f\left( x \right)}} \cdot f'\left( x \right) + \ln \left( {f\left( x \right)} \right) \cdot g'\left( x \right)$ $\Rightarrow \,\, \dfrac{{dy}}{{dx}} = y\left\{ {\dfrac{{f'\left( x \right)g\left( x \right)}}{{f\left( x \right)}} + g'\left( x \right)\ln \left( {f\left( x \right)} \right)} \right.$ $= {\left( {f\left( x \right)} \right)^{g\left( x \right)}}\left\{ {\dfrac{{f'\left( x \right)g\left( x \right)}}{{f\left( x \right)}} + g'\left( x \right)\ln \left( {f\left( x \right)} \right)} \right\}$.
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As a simple example, suppose we have to differentiate $y = {x^x}$:. $\ln y = x\ln x$ $\Rightarrow \,\, \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \cdot \dfrac{1}{x} + \ln x \cdot 1$ $= 1 + \ln x$ $\Rightarrow \,\, \dfrac{{dy}}{{dx}} = y\left( {1 + \ln x} \right)$ $= {x^x}\left( {1 + \ln x} \right)$.
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https://www.teachoo.com/3171/685/Example-11---Two-farmers-Ramkishan-and-Gurcharan-Singh-cultivates/category/Examples/
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Examples
Chapter 3 Class 12 Matrices
Serial order wise
Get live Maths 1-on-1 Classs - Class 6 to 12
Transcript
Example 11 (i) Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B. September Sales (in Rupees) Basmati Permal Naura A = [■8(10,000&20,000&30,[email protected],000&30,000&10,000)] October Sales (in Rupees) Basmati Permal Naura B = [■8(5000&10,000&[email protected],000&10,000&10,000)] (i) Find the combined sales in September and October for each farmer in each variety. Guracharan Singh Guracharan Singh Combined sales in September and October will be A + B A + B = [■8(10,000&20,000&30,[email protected],000&30,000&10,000)] + [■8(5000&10,000&[email protected],000&10,000&10,000)] = [■8(10000+5000&20000+10000&[email protected]+20000&30000+ 10,000&10000+10000)] = [■8(15,000&30,000&36,[email protected],000&40,000&20,000)] Hence, Total Sales = [■8(15,000&30,000&36,[email protected]70,000&40,000&20,000)] Guracharan singh Example 11 (ii) Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B. September Sales (in Rupees) Basmati Permal Naura A = [■8(10,000&20,000&30,[email protected],000&30,000&10,000)] October Sales (in Rupees) Basmati Permal Naura B = [■8(5000&10,000&[email protected],000&10,000&10,000)] (ii) Find the decrease in sales from September to October. Decrease in sales from September to October will be A – B A – B = [■8(10,000&20,000&30,[email protected],000&30,000&10,000)] – [■8(5000&10,000&[email protected],000&10,000&10,000)] = [■8(10000−5000&20000−10000&30000−[email protected]−20000&30000−10,000&10000−10000)] = [■8(5000&10,000&24,[email protected],000&20,000&0)] Hence, Decrease in sales from September to October = [■8(5000&10,000&24,[email protected],000&20,000&0)] Example 11 (iii) Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B. September Sales (in Rupees) Basmati Permal Naura A = [■8(10,000&20,000&30,[email protected],000&30,000&10,000)] October Sales (in Rupees) Basmati Permal Naura B = [■8(5000&10,000&[email protected],000&10,000&10,000)] (iii) If both farmers receive 2% profit on gross sales, compute the profit for each farmer and for each variety sold in October. Profit = 2% × (Sales on October) = 2/100 × B = 0.02 × B = 0.02 × [■8(5000&10,000&[email protected],000&10,000&10,000)] = [■8(0.02×5000&0.02×10,000&0.02×[email protected]×20,000&0.02×10,000&0.02×10,000)] = [■8(100&200&[email protected]&200&200)] Hence, Profit = [■8(100& 200& [email protected]& 200& 200)] Hence, Ramkishan receives Rs 100 profit on sale of Basmati rice , Rs 200 profit on sale of Permal & Rs 120 profit in the sale of Naura in the month of October Hence, Gurcharan Singh receives Rs 400 profit on sale of Basmati rice , Rs 200 profit on sale of Permal & Rs 200 profit in the sale of Naura in the month of October
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Examples. Chapter 3 Class 12 Matrices. Serial order wise. Get live Maths 1-on-1 Classs - Class 6 to 12. Transcript. Example 11 (i) Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B. September Sales (in Rupees) Basmati Permal Naura A = [■8(10,000&20,000&30,[email protected],000&30,000&10,000)] October Sales (in Rupees) Basmati Permal Naura B = [■8(5000&10,000&[email protected],000&10,000&10,000)] (i) Find the combined sales in September and October for each farmer in each variety. Guracharan Singh Guracharan Singh Combined sales in September and October will be A + B A + B = [■8(10,000&20,000&30,[email protected],000&30,000&10,000)] + [■8(5000&10,000&[email protected],000&10,000&10,000)] = [■8(10000+5000&20000+10000&[email protected]+20000&30000+ 10,000&10000+10000)] = [■8(15,000&30,000&36,[email protected],000&40,000&20,000)] Hence, Total Sales = [■8(15,000&30,000&36,[email protected]70,000&40,000&20,000)] Guracharan singh Example 11 (ii) Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B. September Sales (in Rupees) Basmati Permal Naura A = [■8(10,000&20,000&30,[email protected],000&30,000&10,000)] October Sales (in Rupees) Basmati Permal Naura B = [■8(5000&10,000&[email protected],000&10,000&10,000)] (ii) Find the decrease in sales from September to October. Decrease in sales from September to October will be A – B A – B = [■8(10,000&20,000&30,[email protected],000&30,000&10,000)] – [■8(5000&10,000&[email protected],000&10,000&10,000)] = [■8(10000−5000&20000−10000&30000−[email protected]−20000&30000−10,000&10000−10000)] = [■8(5000&10,000&24,[email protected],000&20,000&0)] Hence, Decrease in sales from September to October = [■8(5000&10,000&24,[email protected],000&20,000&0)] Example 11 (iii) Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B.
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September Sales (in Rupees) Basmati Permal Naura A = [■8(10,000&20,000&30,[email protected],000&30,000&10,000)] October Sales (in Rupees) Basmati Permal Naura B = [■8(5000&10,000&[email protected],000&10,000&10,000)] (iii) If both farmers receive 2% profit on gross sales, compute the profit for each farmer and for each variety sold in October. Profit = 2% × (Sales on October) = 2/100 × B = 0.02 × B = 0.02 × [■8(5000&10,000&[email protected],000&10,000&10,000)] = [■8(0.02×5000&0.02×10,000&0.02×[email protected]×20,000&0.02×10,000&0.02×10,000)] = [■8(100&200&[email protected]&200&200)] Hence, Profit = [■8(100& 200& [email protected]& 200& 200)] Hence, Ramkishan receives Rs 100 profit on sale of Basmati rice , Rs 200 profit on sale of Permal & Rs 120 profit in the sale of Naura in the month of October Hence, Gurcharan Singh receives Rs 400 profit on sale of Basmati rice , Rs 200 profit on sale of Permal & Rs 200 profit in the sale of Naura in the month of October.
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https://www.prokidsedu.com/given-right-triangle-jkm-which-correctly-describes-the-locations-of-the-sides-in-relation-to-j
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# Given Right Triangle Jkm, Which Correctly Describes The Locations Of The Sides In Relation To ∠J?
Given Right Triangle Jkm, Which Correctly Describes The Locations Of The Sides In Relation To ∠J?. Given the right triangle, jkm describes the location of sides in relation to angle j. 4) a is adjacent, b is opposite, c is the hypotenuse.
Triangle j k m is shown. Given right triangle xyz, which correctly describes the locations of the sides in relation to ∠y? Sin (c) =root of 3/2.
### Given Right Triangle Mno, Which Represents The Value Of Cos(M)?
Given the right triangle, jkm describes the location of sides in relation to angle j. Given the right triangle jkm, which correctly describes the locations of the sides in relation to ∠j? (1) a is the hypotenuse, b is adjacent, c is opposite (2) a is the hypotenuse, b is opposite, c is adjacent (3) a is adjacent, b is opposite, c is the hypotenuse (4) a is opposite, b is the hypotenuse, c is adjacent
### Given Right Triangle Jkm, Which Correctly Describes The Locations Of The Sides In Relation To ∠J?
The length of hypotenuse k j is a, the length of j m is b, and the length of k m is c. A is the hypotenuse, b is adjacent, c is opposite. A is adjacent, b is opposite, c is the hypotenuse.
See Also : How Far Out From The End Of The Pipe Is The Point Where The Stream Of Water Meets The Creek?
### 4) A Is Adjacent, B Is Opposite, C Is The Hypotenuse.
2) a is the hypotenuse, b is opposite, c is adjacent. Given right triangle pqr, which represents the value of sin(p)? 1) a is opposite, b is adjacent, c is the hypotenuse.
### Given Right Triangle Jkm, Which Correctly Describes The Locations Of The Sides In Relation To ∠J?
The length of hypotenuse k j is a, the length of j m is b, and the length of k m is c. Angle j m k is a right angle. Given right triangle jkm, which correctly describes the locations of the sides in relation to ∠j?
### Given The Right Triangle, Jkm Describes The Location Of Sides In Relation To Angle J.
Given right triangle xyz, which correctly describes the locations of the sides in relation to ∠y? Given right triangle jkm, which correctly describes the locations of the sides in relation to ∠j? Which trigonometric ratios are correct for triangle abc?
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# Given Right Triangle Jkm, Which Correctly Describes The Locations Of The Sides In Relation To ∠J?. Given Right Triangle Jkm, Which Correctly Describes The Locations Of The Sides In Relation To ∠J?. Given the right triangle, jkm describes the location of sides in relation to angle j. 4) a is adjacent, b is opposite, c is the hypotenuse.. Triangle j k m is shown. Given right triangle xyz, which correctly describes the locations of the sides in relation to ∠y? Sin (c) =root of 3/2.. ### Given Right Triangle Mno, Which Represents The Value Of Cos(M)?. Given the right triangle, jkm describes the location of sides in relation to angle j. Given the right triangle jkm, which correctly describes the locations of the sides in relation to ∠j? (1) a is the hypotenuse, b is adjacent, c is opposite (2) a is the hypotenuse, b is opposite, c is adjacent (3) a is adjacent, b is opposite, c is the hypotenuse (4) a is opposite, b is the hypotenuse, c is adjacent. ### Given Right Triangle Jkm, Which Correctly Describes The Locations Of The Sides In Relation To ∠J?. The length of hypotenuse k j is a, the length of j m is b, and the length of k m is c. A is the hypotenuse, b is adjacent, c is opposite. A is adjacent, b is opposite, c is the hypotenuse.. See Also : How Far Out From The End Of The Pipe Is The Point Where The Stream Of Water Meets The Creek?. ### 4) A Is Adjacent, B Is Opposite, C Is The Hypotenuse.. 2) a is the hypotenuse, b is opposite, c is adjacent. Given right triangle pqr, which represents the value of sin(p)? 1) a is opposite, b is adjacent, c is the hypotenuse.. ### Given Right Triangle Jkm, Which Correctly Describes The Locations Of The Sides In Relation To ∠J?. The length of hypotenuse k j is a, the length of j m is b, and the length of k m is c. Angle j m k is a right angle. Given right triangle jkm, which correctly describes the locations of the sides in relation to ∠j?.
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### Given The Right Triangle, Jkm Describes The Location Of Sides In Relation To Angle J.. Given right triangle xyz, which correctly describes the locations of the sides in relation to ∠y? Given right triangle jkm, which correctly describes the locations of the sides in relation to ∠j? Which trigonometric ratios are correct for triangle abc?.
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