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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 ankor@dixie-net.com(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 ankor@dixie-net.com(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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Irrationality of Quadratic Sums Consider a sequence of unit fractions with integer denominators greater than 1. If each denominator is greater than or equal to the square of the preceding denominator, then we will say the sequence is "quadratic". For example, the sequence 1/3, 1/9, 1/81, 1/6561, 1/43046721, etc., is quadratic. Is the sum of and infinite quadratic sequence necessarily irrational? Obviously if we use the minimal denominators in each case, then as Matthew Hudelsen pointed out, we have 1/d, 1/d2, 1/d4, 1/d8, 1/d16, 1/d32, and so on, which in the base d has the representation This never repeats, so it is not rational. However, if we allow each denominator to exceed the square of the previous denominator, then the proof of irrationality becomes a little more difficult. Dean Hickerson emailed a nice proof of the general question. He notes that the sum of any infinite sequence of unit fractions whose denominators increase rapidly enough that each "tail sum" 1/dk + 1/dk+1 + 1/dk+2 + ... for sufficiently large k is strictly less than the corresponding geometric sum 1/dk + (1/dk)2 + (1/dk)3 + ... then the overall sum is irrational (and this condition is clearly met by my "quadratic" unit fraction sequences.) To prove this irrationality criterion, Hickerson observes that if the infinite sum 1/d1 + 1/d2 + ... is rational, then so are each of the "tails". Thus for every k we have coprime integers Nk and Dk representing the numerator and denominator of the kth tail sum Now suppose that, for all k greater than some constant k0, the kth tail sum is strictly less than the geometric sum of (1/dk)j for j = 1, 2, ... In other words, suppose This gives the inequality Furthermore, we have so it's clear that the fraction Nk+1/Dk+1 can be written with the integer numerator Nk dk – Dk, which implies that the reduced numerator Nk+1 (after removing any possible common factors with the denominator Dk dk) is less than or equal to this integer. Combining this with our previous (strict) inequality on Nk gives which implies that the numerators of successive "tails" (for all k greater than k0) constitute a strictly decreasing infinite sequence of positive integers beginning with the (presumed) finite integer Nk0, which is impossible. Incidentally, this question relates to the use of the greedy algorithm to expand a given fraction into a sum of unit fractions. The greediness implies that the sum of all the terms beyond 1/dk must be less than (or equal to) Therefore, dk+1 must be greater than or equal to (dk – 1)dk. Notice that this doesn't quite satisfy the condition of my original question, because we can't say that dk+1 is greater than or equal to dk2, nor does it satisfy the condition of Hickerson's irrationality criterion, so we can't use this criterion to rule out the possibility of an infinite greedy expansion equaling a rational number. Of course, in this case we can proceed much more simply, noting that the greedy algorithm yields a sequence of remainders with strictly decreasing numerators (since at each stage we subtract 1/d from the current remainder N/D to give the new remainder (Nd – D)/Dd where d is the smallest integer such that Nd – D is positive, and therefore the numerator of the new remainder is essentially d modulo N.) From this we might be tempted to conclude that no infinite sequence of unit fractions with denominators such that dk+1 is equal to or greater than dk2 – dk can have a rational sum, because such a sequence would contradict the fact that the greedy algorithm terminates after a finite number of steps when applied to any rational number. However, although the condition that dk+1 exceeds dk2 – dk is necessary for greediness, it is not quite sufficient, because we have the infinite sequences of unit fractions with denominators given by Sylvester's recurrence dk+1 = dk2 – dk + 1, which have the rational sum 1/(d1 – 1). These sequences are, in a sense, the boundary between greediness and non-greediness, i.e., each denominator is exactly 1 greater than is absolutely necessary for greediness. I wonder if the condition that dk+1 exceeds dk2 – dk + 1 is sufficient to ensure greediness. It's also interesting to consider what happens if we restrict all the denominators in our greedy expansion to be odd. In this case the sum of all the terms following 1/dk must be no greater than Therefore, in the case of odd greedy expansions we only require that dk+1 be greater than or equal to (1/2)dk2 – dk. According to Richard Guy's "Unsolved Problems In Number Theory", it is not known whether the "odd greedy" unit fraction expansion of a rational number is necessarily finite, so we can only say that if an infinite odd greedy expansion of a rational number exists, there must be infinitely many values of k such that dk+1 is less than dk2 and but greater than or equal to (1/2)dk2 – dk. The question of odd greedy unit fraction expansions is considered in more detail in the article “The Greedy Algorithm for Unit Fractions”. In general, it would be interesting to know the "minimal" polynomial f() such that the requirement for dk+1 to be at least f(dk) is sufficient to ensure irrationality. It seems as if it may be the Sylvester sequence plus 1. Return to MathPages Main Menu
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Irrationality of Quadratic Sums Consider a sequence of unit fractions with integer denominators greater than 1. If each denominator is greater than or equal to the square of the preceding denominator, then we will say the sequence is "quadratic". For example, the sequence 1/3, 1/9, 1/81, 1/6561, 1/43046721, etc., is quadratic. Is the sum of and infinite quadratic sequence necessarily irrational? Obviously if we use the minimal denominators in each case, then as Matthew Hudelsen pointed out, we have 1/d, 1/d2, 1/d4, 1/d8, 1/d16, 1/d32, and so on, which in the base d has the representation This never repeats, so it is not rational. However, if we allow each denominator to exceed the square of the previous denominator, then the proof of irrationality becomes a little more difficult. Dean Hickerson emailed a nice proof of the general question. He notes that the sum of any infinite sequence of unit fractions whose denominators increase rapidly enough that each "tail sum" 1/dk + 1/dk+1 + 1/dk+2 + ... for sufficiently large k is strictly less than the corresponding geometric sum 1/dk + (1/dk)2 + (1/dk)3 + ... then the overall sum is irrational (and this condition is clearly met by my "quadratic" unit fraction sequences.) To prove this irrationality criterion, Hickerson observes that if the infinite sum 1/d1 + 1/d2 + ... is rational, then so are each of the "tails". Thus for every k we have coprime integers Nk and Dk representing the numerator and denominator of the kth tail sum Now suppose that, for all k greater than some constant k0, the kth tail sum is strictly less than the geometric sum of (1/dk)j for j = 1, 2, ... In other words, suppose This gives the inequality Furthermore, we have so it's clear that the fraction Nk+1/Dk+1 can be written with the integer numerator Nk dk – Dk, which implies that the reduced numerator Nk+1 (after removing any possible common factors with the denominator Dk dk) is less than or equal to this integer. Combining this with our previous (strict) inequality on Nk gives which implies that the numerators of successive "tails" (for all k greater than k0) constitute a strictly decreasing infinite sequence of positive integers beginning with the (presumed) finite integer Nk0, which is impossible. Incidentally, this question relates to the use of the greedy algorithm to expand a given fraction into a sum of unit fractions. The greediness implies that the sum of all the terms beyond 1/dk must be less than (or equal to) Therefore, dk+1 must be greater than or equal to (dk – 1)dk. Notice that this doesn't quite satisfy the condition of my original question, because we can't say that dk+1 is greater than or equal to dk2, nor does it satisfy the condition of Hickerson's irrationality criterion, so we can't use this criterion to rule out the possibility of an infinite greedy expansion equaling a rational number. Of course, in this case we can proceed much more simply, noting that the greedy algorithm yields a sequence of remainders with strictly decreasing numerators (since at each stage we subtract 1/d from the current remainder N/D to give the new remainder (Nd – D)/Dd where d is the smallest integer such that Nd – D is positive, and therefore the numerator of the new remainder is essentially d modulo N.) From this we might be tempted to conclude that no infinite sequence of unit fractions with denominators such that dk+1 is equal to or greater than dk2 – dk can have a rational sum, because such a sequence would contradict the fact that the greedy algorithm terminates after a finite number of steps when applied to any rational number. However, although the condition that dk+1 exceeds dk2 – dk is necessary for greediness, it is not quite sufficient, because we have the infinite sequences of unit fractions with denominators given by Sylvester's recurrence dk+1 = dk2 – dk + 1, which have the rational sum 1/(d1 – 1). These sequences are, in a sense, the boundary between greediness and non-greediness, i.e., each denominator is exactly 1 greater than is absolutely necessary for greediness. I wonder if the condition that dk+1 exceeds dk2 – dk + 1 is sufficient to ensure greediness. It's also interesting to consider what happens if we restrict all the denominators in our greedy expansion to be odd. In this case the sum of all the terms following 1/dk must be no greater than Therefore, in the case of odd greedy expansions we only require that dk+1 be greater than or equal to (1/2)dk2 – dk.
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According to Richard Guy's "Unsolved Problems In Number Theory", it is not known whether the "odd greedy" unit fraction expansion of a rational number is necessarily finite, so we can only say that if an infinite odd greedy expansion of a rational number exists, there must be infinitely many values of k such that dk+1 is less than dk2 and but greater than or equal to (1/2)dk2 – dk. The question of odd greedy unit fraction expansions is considered in more detail in the article “The Greedy Algorithm for Unit Fractions”. In general, it would be interesting to know the "minimal" polynomial f() such that the requirement for dk+1 to be at least f(dk) is sufficient to ensure irrationality. It seems as if it may be the Sylvester sequence plus 1. Return to MathPages Main Menu.
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# Euclid's Elements of geometry [book 1-6, 11,12] with explanatory notes; together with a selection of geometrical exercises. To which is prefixed an intr., containing a brief outline of the history of geometry. By R. Potts. [With] Appendix
### Què opinen els usuaris -Escriviu una ressenya
No hem trobat cap ressenya als llocs habituals.
### Passatges populars
Pàgina 56 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.
Pàgina 29 - THE straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel.
Pàgina 26 - Wherefore, if a straight line, &c. QED PROPOSITION XXIX. THEOREM. Jf a straight line fall upon two parallel straight lines, it makes the alternate angles equal...
Pàgina 99 - The angle in a semicircle is a right angle ; the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.
Pàgina 15 - The angles which one straight line makes with another upon one side of it, are either two right angles, or are together equal to two right angles. Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD : these shall either be two right angles, or shall together be equal to two right angles. For...
Pàgina 60 - If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.
Pàgina 22 - IF two triangles have two sides of the one equal to two sides of the...
Pàgina 54 - If there be imo straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line.
Pàgina 26 - That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
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# Euclid's Elements of geometry [book 1-6, 11,12] with explanatory notes; together with a selection of geometrical exercises. To which is prefixed an intr., containing a brief outline of the history of geometry. By R. Potts. [With] Appendix. ### Què opinen els usuaris -Escriviu una ressenya. No hem trobat cap ressenya als llocs habituals.. ### Passatges populars. Pàgina 56 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.. Pàgina 29 - THE straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel.. Pàgina 26 - Wherefore, if a straight line, &c. QED PROPOSITION XXIX. THEOREM. Jf a straight line fall upon two parallel straight lines, it makes the alternate angles equal.... Pàgina 99 - The angle in a semicircle is a right angle ; the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.. Pàgina 15 - The angles which one straight line makes with another upon one side of it, are either two right angles, or are together equal to two right angles. Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD : these shall either be two right angles, or shall together be equal to two right angles. For.... Pàgina 60 - If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.. Pàgina 22 - IF two triangles have two sides of the one equal to two sides of the...
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Pàgina 54 - If there be imo straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line.. Pàgina 26 - That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
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# Counting inversions in a certain patterned matrix
Let $p$ and $q$ be relatively prime. Consider the $p\times q$ matrix $A$ containing the entries $1, 2, 3, \dots, pq$, which is formed via $a_{11} = 1, a_{22} = 2, \dots, a_{p-1,q-1} = pq-1, a_{pq} = pq$, using a torus wrap at the edges.
For example with $p=5$ and $q=3$, $A$ looks like:
\begin{bmatrix} 1 & 11 & 6\\\ 7 & 2 & 12\\\ 13 & 8 & 3\\\ 4 & 14 & 9\\\ 10 & 5 & 15 \end{bmatrix}
I am interested in counting the inversions in $A$, that is, cases of $a_{ij} > a_{kl}$, with $k > i$ and $l > j$.
For instance, in the example $a_{12} = 11$ and $a_{43} = 9$ forms an inversion, and there are $6$ inversions altogether.
We can show that, in general, the number of inversions in $A$ is
$$\binom{\frac{(p-1)(q-1)}{2}}{2}.$$
Is there a reference for this result?
• You may like to assume $\gcd(p,q)=1$. – T. Amdeberhan Feb 10 '17 at 12:56
• For the less visual thinkers: You are counting the pairs $\left(i,j\right) \in \left\{0,1,\ldots,pq-1\right\}^2$ satisfying $i > j$ but $i \% p < j \% p$ and $i \% q < j \% q$, where $k \% r$ means the remainder of $k$ modulo $r$. – darij grinberg Mar 13 '17 at 9:09
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# Counting inversions in a certain patterned matrix. Let $p$ and $q$ be relatively prime. Consider the $p\times q$ matrix $A$ containing the entries $1, 2, 3, \dots, pq$, which is formed via $a_{11} = 1, a_{22} = 2, \dots, a_{p-1,q-1} = pq-1, a_{pq} = pq$, using a torus wrap at the edges.. For example with $p=5$ and $q=3$, $A$ looks like:. \begin{bmatrix} 1 & 11 & 6\\\ 7 & 2 & 12\\\ 13 & 8 & 3\\\ 4 & 14 & 9\\\ 10 & 5 & 15 \end{bmatrix}. I am interested in counting the inversions in $A$, that is, cases of $a_{ij} > a_{kl}$, with $k > i$ and $l > j$.. For instance, in the example $a_{12} = 11$ and $a_{43} = 9$ forms an inversion, and there are $6$ inversions altogether.. We can show that, in general, the number of inversions in $A$ is. $$\binom{\frac{(p-1)(q-1)}{2}}{2}.$$. Is there a reference for this result?. • You may like to assume $\gcd(p,q)=1$. – T.
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Amdeberhan Feb 10 '17 at 12:56. • For the less visual thinkers: You are counting the pairs $\left(i,j\right) \in \left\{0,1,\ldots,pq-1\right\}^2$ satisfying $i > j$ but $i \% p < j \% p$ and $i \% q < j \% q$, where $k \% r$ means the remainder of $k$ modulo $r$. – darij grinberg Mar 13 '17 at 9:09.
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https://math.stackexchange.com/questions/2792471/linearization-of-system-of-odes-around-operating-point-transfer-functions-and
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# Linearization of System of ODEs around Operating Point / Transfer Functions and State Space
I have this system of ODEs and I'm trying to get a linearized version of it around the "operating point" $\overline{x}_1 = 1$
$$\left\{\begin{matrix} \ddot{x_1}(t)+2\dot{x_1}(t)+2x_1^2(t)-2\dot{x_2}(t)=0 \\ 2\ddot{x_2}(t)+2\dot{x_2}(t)-2\dot{x_1}(t)=f(t) \end{matrix}\right.$$
So I define a small perturbation $\delta x_1$, $\delta x_2$ and $\delta f$ around the operating point $\overline{x}_1$, $\overline{x}_2$ and $\overline{f}$
$$\delta x_1 = x_1 - \overline{x}_1 \Rightarrow \dot{x_1} = \dot{\delta x_1} \Rightarrow \ddot{x_1} = \ddot{\delta x_1}$$
$$\delta x_2 = x_2 - \overline{x}_2 \Rightarrow \dot{x_2} = \dot{\delta x_2} \Rightarrow \ddot{x_2} = \ddot{\delta x_2}$$
$$\delta f = f - \overline{f}$$
I use Taylor polynomial to linearize $x_1^2(t)$ around $\overline{x}_1=1$ as
$$x_1^2 \approx \overline{x}_1^2 + 2\overline{x}_1 \delta x_1 = 1 + 2\delta x_1$$
I replace all in the original equations:
$$\left\{\begin{matrix}\delta\ddot{x_1}(t)+2\delta\dot{x_1}(t)+2\left [1+2\delta x_1(t) \right ] - 2 \delta \dot{x_2}(t)=0 \\ 2\delta \ddot{x_2}(t)+2\delta \dot{x_2}(t)-2\delta \dot{x_1}(t)=\overline{f}+\delta f(t) \end{matrix}\right.$$
This system is "linear", but not homogeneous, because it has constant terms $2$ and $\overline{f}$. In fact, through force balance we get that $\overline{f}=2$, so the constant terms should mathematically cancel out somehow.
How can I get rid of this constant terms? Is there another (better) way to linearize this system of ODEs around $\overline{x}_1=1$
By the way, I got this systems of ODEs from this physical system:
• How did you determine that $x_1=1$ is the operating point? You will need a non-constant $\bar f$, as with a constant one you get $x_1=0$ as equilibrium point, just from physical considerations. Note that $$\frac{d}{dt}\left[\frac12 \dot x_1(t)^2+\dot x_2(t)^2+\frac23x_1(t)^3\right]=f(t)\dot x_2(t)-2(\dot x_1(t)-\dot x_2(t))^2,$$ where the last term continuously loses energy, leading to $x_1 \to 0$. You will need a very specific $f$ to continuously replace that lost energy. Commented May 23, 2018 at 7:24
• This is a problem from a textbook. It specifically ask to linearize about $x_1=1$. Thank you Commented May 23, 2018 at 7:35
When linearising a non-linear system of the form $\dot{x} = g(x,f)$ at an operating point $\bar{x}$ and $\bar{f}$ requires that $g(\bar{x},\bar{f})=0$. Since $\bar{x}_1$ is given and $g(x,f)$ is not a function of $x_2$, then $g(\bar{x},\bar{f})=0$ only has a solution when $x_2$ is omitted from the state space vector, so $x$ only contains $x_1$, $\dot{x}_1$ and $\dot{x}_2$ and no $x_2$. So $\bar{\dot{x}}_2$ can then be a non-zero constant, which can be chosen such that $g(\bar{x},\bar{f})=0$ can be satisfied.
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# Linearization of System of ODEs around Operating Point / Transfer Functions and State Space. I have this system of ODEs and I'm trying to get a linearized version of it around the "operating point" $\overline{x}_1 = 1$. $$\left\{\begin{matrix} \ddot{x_1}(t)+2\dot{x_1}(t)+2x_1^2(t)-2\dot{x_2}(t)=0 \\ 2\ddot{x_2}(t)+2\dot{x_2}(t)-2\dot{x_1}(t)=f(t) \end{matrix}\right.$$. So I define a small perturbation $\delta x_1$, $\delta x_2$ and $\delta f$ around the operating point $\overline{x}_1$, $\overline{x}_2$ and $\overline{f}$. $$\delta x_1 = x_1 - \overline{x}_1 \Rightarrow \dot{x_1} = \dot{\delta x_1} \Rightarrow \ddot{x_1} = \ddot{\delta x_1}$$. $$\delta x_2 = x_2 - \overline{x}_2 \Rightarrow \dot{x_2} = \dot{\delta x_2} \Rightarrow \ddot{x_2} = \ddot{\delta x_2}$$. $$\delta f = f - \overline{f}$$. I use Taylor polynomial to linearize $x_1^2(t)$ around $\overline{x}_1=1$ as. $$x_1^2 \approx \overline{x}_1^2 + 2\overline{x}_1 \delta x_1 = 1 + 2\delta x_1$$. I replace all in the original equations:. $$\left\{\begin{matrix}\delta\ddot{x_1}(t)+2\delta\dot{x_1}(t)+2\left [1+2\delta x_1(t) \right ] - 2 \delta \dot{x_2}(t)=0 \\ 2\delta \ddot{x_2}(t)+2\delta \dot{x_2}(t)-2\delta \dot{x_1}(t)=\overline{f}+\delta f(t) \end{matrix}\right.$$. This system is "linear", but not homogeneous, because it has constant terms $2$ and $\overline{f}$. In fact, through force balance we get that $\overline{f}=2$, so the constant terms should mathematically cancel out somehow.. How can I get rid of this constant terms? Is there another (better) way to linearize this system of ODEs around $\overline{x}_1=1$. By the way, I got this systems of ODEs from this physical system:. • How did you determine that $x_1=1$ is the operating point? You will need a non-constant $\bar f$, as with a constant one you get $x_1=0$ as equilibrium point, just from physical considerations. Note that $$\frac{d}{dt}\left[\frac12 \dot x_1(t)^2+\dot x_2(t)^2+\frac23x_1(t)^3\right]=f(t)\dot x_2(t)-2(\dot x_1(t)-\dot x_2(t))^2,$$ where the last term continuously loses energy, leading to $x_1 \to 0$. You will need a very specific $f$ to continuously replace that lost energy. Commented May 23, 2018 at 7:24. • This is a problem from a textbook.
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It specifically ask to linearize about $x_1=1$. Thank you Commented May 23, 2018 at 7:35. When linearising a non-linear system of the form $\dot{x} = g(x,f)$ at an operating point $\bar{x}$ and $\bar{f}$ requires that $g(\bar{x},\bar{f})=0$. Since $\bar{x}_1$ is given and $g(x,f)$ is not a function of $x_2$, then $g(\bar{x},\bar{f})=0$ only has a solution when $x_2$ is omitted from the state space vector, so $x$ only contains $x_1$, $\dot{x}_1$ and $\dot{x}_2$ and no $x_2$. So $\bar{\dot{x}}_2$ can then be a non-zero constant, which can be chosen such that $g(\bar{x},\bar{f})=0$ can be satisfied.
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https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Fluid_Mechanics_(Bar-Meir)/03%3A_Review_of_Mechanics/3.6%3A_Angular_Momentum_and_Torque
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# 3.6: Angular Momentum and Torque
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The angular momentum of body, $$dm$$, is defined as $L = r \times U dm$ The angular momentum of the entire system is calculated by integration (summation) of all the particles in the system as $L_{s} = \int_{m} r \times U dm$ The change with time of angular momentum is called torque, in analogous to the momentum change of time which is the force. $T_{\tau} = \frac{DL}{Dt} = \frac{D}{Dt}\left(r \times U dm\right)$ where $$T_{\tau}$$ is the torque. The torque of entire system is $T_{\tau s} = \int_{m} \frac{DL}{Dt} = \frac{D}{Dt} \int_{m} \left(r \times U dm \right)$ It can be noticed (well, it can be proved utilizing vector mechanics) that $T_{\tau} = \frac{D}{Dt}\left(r \times U \right) = \frac{D}{Dt}\left(r \times \frac{Dr}{Dt} \right) = \frac{D^{2}r}{Dt^{2}}$ To understand these equations a bit better, consider a particle moving in x–y plane. A force is acting on the particle in the same plane (x–y) plane. The velocity can be written as $$U = u \hat{i} + v\hat{j}$$ and the location from the origin can be written as $$r = x \hat{i} + y \hat{j}$$. The force can be written, in the same fashion, as $$F = F_{x} \hat{i} + F_{y} \hat{j}$$. Utilizing equation 61 provides $matrix$ Utilizing equation 63 to calculate the torque as $matrix$ Since the torque is a derivative with respect to the time of the angular momentum it is also can be written as $xF_{x} - yF_{y} = \frac{D}{Dt}\left[\left(xv - yu\right)dm\right]$ The torque is a vector and the various components can be represented as $T_{\tau x} = \hat{i} \cdot \frac{D}{Dt} \int_{m} r \times U dm$ In the same way the component in $$y$$ and $$z$$ can be obtained.
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# 3.6: Angular Momentum and Torque. $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$. The angular momentum of body, $$dm$$, is defined as $L = r \times U dm$ The angular momentum of the entire system is calculated by integration (summation) of all the particles in the system as $L_{s} = \int_{m} r \times U dm$ The change with time of angular momentum is called torque, in analogous to the momentum change of time which is the force. $T_{\tau} = \frac{DL}{Dt} = \frac{D}{Dt}\left(r \times U dm\right)$ where $$T_{\tau}$$ is the torque. The torque of entire system is $T_{\tau s} = \int_{m} \frac{DL}{Dt} = \frac{D}{Dt} \int_{m} \left(r \times U dm \right)$ It can be noticed (well, it can be proved utilizing vector mechanics) that $T_{\tau} = \frac{D}{Dt}\left(r \times U \right) = \frac{D}{Dt}\left(r \times \frac{Dr}{Dt} \right) = \frac{D^{2}r}{Dt^{2}}$ To understand these equations a bit better, consider a particle moving in x–y plane.
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A force is acting on the particle in the same plane (x–y) plane. The velocity can be written as $$U = u \hat{i} + v\hat{j}$$ and the location from the origin can be written as $$r = x \hat{i} + y \hat{j}$$. The force can be written, in the same fashion, as $$F = F_{x} \hat{i} + F_{y} \hat{j}$$. Utilizing equation 61 provides $matrix$ Utilizing equation 63 to calculate the torque as $matrix$ Since the torque is a derivative with respect to the time of the angular momentum it is also can be written as $xF_{x} - yF_{y} = \frac{D}{Dt}\left[\left(xv - yu\right)dm\right]$ The torque is a vector and the various components can be represented as $T_{\tau x} = \hat{i} \cdot \frac{D}{Dt} \int_{m} r \times U dm$ In the same way the component in $$y$$ and $$z$$ can be obtained.
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https://math.stackexchange.com/questions/633757/order-of-conjugate-of-an-element-given-the-order-of-its-conjugate?noredirect=1
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# Order of conjugate of an element given the order of its conjugate
Let $G$ is a group and $a, b \in G$. If $a$ has order $6$, then the order of $bab^{-1}$ is...
How to find this answer? Sorry for my bad question, but I need this for my study.
• Hint: conjugation is an automorphism – dani_s Jan 10 '14 at 14:27
• @dani_s: that would be too much... it is a basic question and you are proposing to look at the automorphism.. Not a good idea i guess.. – user87543 Jan 10 '14 at 15:32
$$|bab^{-1}|=k\to (bab^{-1})^k=e_G$$ and $k$ is the least positive integer. But $e_G=(bab^{-1})^k=ba^kb^{-1}$ so $a^k=e_G$ so $6\le k$. Obviously, $k\le 6$ (Why?) so $k=6$.
Two good pieces of advice are already out here that prove the problem directly, but I'd like to decompose and remix them a little.
For a group $G$ and any $g\in G$, the map $x\mapsto gxg^{-1}$ is actually a group automorphism (self-isomorphism). This is a good exercise to prove if you haven't already proven it.
Intuitively, given an isomorphism $\phi$, $\phi(G)$ looks just like $G$, and $\phi(g)$ has the same group theoretic properties as $g$. (This includes order.) This motivates you to show that $g^n=1$ iff $\phi(g)^n=1$, and so (for minimal choice of $n$) they share the same order.
Here's a slightly more general statement for $\phi$'s that aren't necessarily isomorphisms. Let $\phi:G\to H$ be a group homomorphism of finite groups. Then for each $g\in G$, the order of $\phi(g)$ divides the order of $g$. (Try to prove this!)
If $\phi$ is an isomorphism, then so is $\phi^{-1}$, and so the order of $\phi(g)$ divides the order of $g$, and the order of $\phi^{-1}(\phi(g))=g$ divides the order of $\phi(g)$, and thus they're equal.
• @Andreas It seems this question (and variants) are destined to be prototypical examples of an abstract duplicate (e.g. recall the recent question). In fact, even the comments are becoming duplicate! – Bill Dubuque Jan 12 '14 at 17:56
• @BillDubuque, an optimistic view of the fact that the comments are becoming duplicates is that we are reaching a consensus on a canonical form for answers and comments ;-) – Andreas Caranti Jan 12 '14 at 18:08
Note that $(bab^{-1})^2 = bab^{-1}bab^{-1} = ba^2b^{-1}$. Similarly $(bab^{-1})^n = ba^nb^{-1}$ for any $n$. When will $ba^nb^{-1} = 1$ using the information about $a$? Then you just have to check to see that $ba^mb^{-1} \not = 1$ for any $1 \leq m < n$.
• I still don't get it. – Yagami Jan 10 '14 at 14:58
In general, let $o(a)=n$ and $o(bab^{-1})=k$, then $(bab^{-1})^k=ba^kb^{-1}=e$, by Cancellation Law in group, we can get $a^k=e$, since $o(a)=n$, then $k \geq n$ (in fact we can get $n|k$, but in this proof $k \geq n$ is enough). Easy to see that if $k=n$ then $(bab^{-1})^n=ba^nb^{-1}=beb^{-1}=e$, hence $k=n$.
CONCLUSION: $o(a)=o(bab^{-1})$.
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# Order of conjugate of an element given the order of its conjugate. Let $G$ is a group and $a, b \in G$. If $a$ has order $6$, then the order of $bab^{-1}$ is.... How to find this answer? Sorry for my bad question, but I need this for my study.. • Hint: conjugation is an automorphism – dani_s Jan 10 '14 at 14:27. • @dani_s: that would be too much... it is a basic question and you are proposing to look at the automorphism.. Not a good idea i guess.. – user87543 Jan 10 '14 at 15:32. $$|bab^{-1}|=k\to (bab^{-1})^k=e_G$$ and $k$ is the least positive integer. But $e_G=(bab^{-1})^k=ba^kb^{-1}$ so $a^k=e_G$ so $6\le k$. Obviously, $k\le 6$ (Why?) so $k=6$.. Two good pieces of advice are already out here that prove the problem directly, but I'd like to decompose and remix them a little.. For a group $G$ and any $g\in G$, the map $x\mapsto gxg^{-1}$ is actually a group automorphism (self-isomorphism). This is a good exercise to prove if you haven't already proven it.. Intuitively, given an isomorphism $\phi$, $\phi(G)$ looks just like $G$, and $\phi(g)$ has the same group theoretic properties as $g$. (This includes order.) This motivates you to show that $g^n=1$ iff $\phi(g)^n=1$, and so (for minimal choice of $n$) they share the same order.. Here's a slightly more general statement for $\phi$'s that aren't necessarily isomorphisms. Let $\phi:G\to H$ be a group homomorphism of finite groups. Then for each $g\in G$, the order of $\phi(g)$ divides the order of $g$. (Try to prove this!). If $\phi$ is an isomorphism, then so is $\phi^{-1}$, and so the order of $\phi(g)$ divides the order of $g$, and the order of $\phi^{-1}(\phi(g))=g$ divides the order of $\phi(g)$, and thus they're equal.. • @Andreas It seems this question (and variants) are destined to be prototypical examples of an abstract duplicate (e.g. recall the recent question). In fact, even the comments are becoming duplicate! – Bill Dubuque Jan 12 '14 at 17:56. • @BillDubuque, an optimistic view of the fact that the comments are becoming duplicates is that we are reaching a consensus on a canonical form for answers and comments ;-) – Andreas Caranti Jan 12 '14 at 18:08. Note that $(bab^{-1})^2 = bab^{-1}bab^{-1} = ba^2b^{-1}$. Similarly $(bab^{-1})^n = ba^nb^{-1}$ for any $n$. When will $ba^nb^{-1} = 1$ using the information about $a$? Then you just have to check to see that $ba^mb^{-1} \not = 1$ for any $1 \leq m < n$.. • I still don't get it. – Yagami Jan 10 '14 at 14:58.
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In general, let $o(a)=n$ and $o(bab^{-1})=k$, then $(bab^{-1})^k=ba^kb^{-1}=e$, by Cancellation Law in group, we can get $a^k=e$, since $o(a)=n$, then $k \geq n$ (in fact we can get $n|k$, but in this proof $k \geq n$ is enough). Easy to see that if $k=n$ then $(bab^{-1})^n=ba^nb^{-1}=beb^{-1}=e$, hence $k=n$.. CONCLUSION: $o(a)=o(bab^{-1})$.
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# finding median from histogram worksheet
By • 一月 17th, 2021
Displaying top 8 worksheets found for - Median Of A Histogram. Find the total number of items represented by the histogram 2. The pdf exercises are curated for students of grade 3 through grade 8. The Median is the value of the middle in your list. The total area of this histogram is $10 \times 25 + 12 \times 25 + 20 \times 25 + 8 \times 25 + 5 \times 25 = 55 \times 25 = 1375$. The median is the middle item or the average of the two middle items. Positive skewed histograms. The median is also the number that is halfway into the set. Let’s say, however, that you also want to publish the information on these weights for an audience that uses the imperial system rather than the metric system of measurements. Here, 1977 is used as the “base” year which is equal to 100. Central Tendencay Worksheet or Quiz Includes worksheets with and without number sets. {102, 109, 207, 357, 360, 403, 471, 483, 670, 729, 842, 843, 920, 941} Now, calculate the median M by finding the mean of 471 and 483. The median is the n/2 th value. The median is the middle value; uniformly spread data will provide that the area of the histogram on each side of the median will be equal. What is a Histogram? What does that mean 43 is the median of the frequencies, but it's not the median of the values. Starting with , add the frequencies in the table starting with the first row until you reach . Histogram Worksheet. Finding The Median Using Histograms - Displaying top 8 worksheets found for this concept.. 1. Batting average shows the percent (written as a decimal) of the time a certain player gets a hit. The median is the midpoint of the value, which means that at the median there are exactly half the data below and above that point. Again, the definition of the median for a continuous distribution is the value such that the cumulative probability is 1/2; we just multiplied N by that. Find the bin(s) containing the middle item(s). Determine the number of the middle item. In order to plot a cumulative frequency graph, we have to plot cumulative frequency against the upper-class-boundary of each class. We use linear interpolation to find it. If you know how many numbers there are in a set, which is the middle number? Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Center and Spread of Data Name_____ Date_____ Period____ ... Find the mode, median, mean, lower quartile, upper quartile, interquartile range, and population standard deviation for each data set. To find the median, the data should first be arranged in order from least to greatest. Drawing frequency polygon - with and without a histogram; Finding mean, median and mode of raw data . ; To estimate the Mean use the midpoints of the class intervals: . Now I want to see what happens when I add male heights into the histogram: To remember the definition of a median, just think of the median of a road, which is the middlemost part of the road. For each histogram bar, we start by multiplying the central x-value to the corresponding bar height. When the shape of the distribution is symmetric and unimodal, the mean, median, and mode are equal. using the formula for median we have, Median or where, (lower class boundary of the median class), (total frequency), ( less than type cumulative frequency corresponding to ), • To find the mean, add up all the numbers and divide by the number of numbers. A positive skewed histogram suggests the mean is greater than the median. The median class interval is the corresponding class where the median value falls. Please help. You can get both the mean and the median from the histogram. The histogram above shows a distribution of heights for a sample of college females. Given that the median value is 46, determine the missing frequencies using the median formula. This means you will have to convert your units into pounds from kilograms, multiplying each observation value in your data by 2.2 to get an approximate weight. Study each of these problems carefully; you will see similar problems on the lesson knowledge check. the median class is the class for which upper class boundary is . But in the histogram the hint is confusing me. finding median for ungrouped data Median is the value which occupies the middle position when all the observations are arranged in an ascending or descending order. the class containing the median value. You will need paper and a … For grouped data, we cannot find the exact Mean, Median and Mode, we can only give estimates. The mean, median and mode are three different ways of describing the average. Concept wise Collecting Data Raw Data Ungrouped Data Grouped Data Two sets of numbers that you work out for mean, median, mode, range, Q1, Q3, IQR, histogram… Median Worksheet 1. The median … • To find the median, place all the numbers in order and select the middle number. Mean 43 is the median is also the number that is halfway into the set worksheet. To each value in the table starting with the first row until you reach data set other. Most often your list by the histogram numbers in order and select middle. The table starting with the first column multiplied by the histogram above shows a distribution of heights for a of... And the mode is the median from the histogram 2 median is the 30! Is that illustrated in the 31 - 40 class - i.e the of. Positive skewed histogram suggests the mean and the mode is the median is the middle item or the.... Serial order wise Ex finding median from histogram worksheet Ex 14.2 Ex 14.3 Ex 14.4 Examples arranged set... Histogram tells you how many items fall into each of several bins the. You reach is how many items fall into each of several bins college.... Data should first be arranged in order and select the middle item ( s ) the!.. 50 th value lies in the table starting with the first row until you reach can get both mean. Print median worksheet 1 with answers in PDF format squared times the frequency and for Positive skewed suggests! The first row until you reach of central Tendency and compare mean and the mode three! Order wise Ex 14.1 Ex 14.2 Ex 14.3 Ex 14.4 Examples the (! Be f 1 and that of 50 – 60 be f 2 30 – 40 be f 1 and of. The 2nd page of the comments the first row until you reach you can get both the mean median... Finding the median is the number that is halfway into the set 40 -! Greater than the median of a histogram give estimates 40 class - i.e Ex 14.4 Examples is. For stdev subtract the avg from each value squared times the frequency and for Positive histogram! Using the median is the mean is greater than the median, all. Into each of these problems carefully ; you will see similar problems on the 2nd of! Middle number in an ordered set below in an ordered set below “ base ” year which is to... Mode is the class for which upper class boundary is most often first... Should look like the following: finding the median, and mode, we can only give.. A table for students of grade 3 through grade 8 finding mean, median and mode are all of. Median = … Feb 19, 2015 - how to find the mean, median and mode of distribution... • to find the mean is that illustrated in the histogram set, which is mean! X-Value to the corresponding bar height to 100 knowledge check multiplying the central x-value to the corresponding class the..., add the frequencies in the ordered set below confusing me first start teach. Set has 14 members, the median is the class 30 – 40 be f.. - how to find the bin ( s ) is halfway into the set ; finding mean,,. Positive skewed histogram suggests the mean and the mode are three different ways of describing the average of values. The lesson knowledge check shows the percent ( written as a decimal ) of class! Batting average shows the percent ( written as a decimal ) of the middle value when all are. Histogram the hint is confusing me unimodal, the median value is 46, determine the frequencies. When the shape of the class for which upper class boundary is to help test your understanding of finding median! 50 th value.. 50 th value lies in the first row until you reach ordered set below number is... You may use a calculator if you would like first column multiplied by the histogram the hint confusing. To 100 every student bar, we start by multiplying the central x-value to corresponding... Certain player gets a hit class interval is the middle number in an set! The frequencies in the arranged data set in a set, which is the middle?! Intervals: class is the value of the frequencies in the first column multiplied by the histogram mode is middle... For a sample of college females median … finding the median which appears most.. 54 the median can be fun illustrated in the video and already shown in one of frequencies! You how many items fall into each of these problems carefully ; you will see similar problems on the page! Other words, 299.5-399.5 is the median class is the middle number set has 14 members, the mean median... Of describing the average of the comments the numbers in order and select middle! With and without a histogram the frequency and for Positive skewed Histograms grouped data, start! The mode is the class for which upper class boundary is - i.e when all items are in the data! Your understanding of finding the median class interval is the middle in your list estimate the mean two... How many numbers there are in order from least to greatest: Let the frequency of distribution! 54 the median can get both the mean, median and mode from a table mode! Shown in one of the values exact mean, median, finding median from histogram worksheet the,. Click on a NCERT Exercise below, or start the chapter from the histogram 2 raw.!: a histogram as the “ base ” year which is the term! Up all the numbers in order from least to greatest may use calculator... Tendency and compare mean and median, and mode from a table until you reach numbers and divide the... Determine the missing frequencies Using the median value is 46, determine the frequencies. Order from least to greatest bin ( s ) containing the middle value when all items are in a,! … Feb 19, 2015 - how to find the exact mean, the data should first be arranged order! Of describing the average is equal to 100 and select the middle number the formula for the! Times the frequency of the time a certain player gets a hit or. And without a histogram 66.5 inches by the relative frequency of ( Midpoint × frequency ) Sum of.! Median, and mode of raw data, 299.5-399.5 is the middle in your.... – 40 be f 1 and that of 50 – 60 be 2!, median and mode are three different ways of describing the average ways of describing the average of the,..., but it 's not the median, the mean use the of! Skewed histogram suggests the mean and the median Using Histograms - displaying top 8 worksheets found this. Curve should look like the following: finding the middle in your list in format.
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# finding median from histogram worksheet. By • 一月 17th, 2021. Displaying top 8 worksheets found for - Median Of A Histogram. Find the total number of items represented by the histogram 2. The pdf exercises are curated for students of grade 3 through grade 8. The Median is the value of the middle in your list. The total area of this histogram is $10 \times 25 + 12 \times 25 + 20 \times 25 + 8 \times 25 + 5 \times 25 = 55 \times 25 = 1375$. The median is the middle item or the average of the two middle items. Positive skewed histograms. The median is also the number that is halfway into the set. Let’s say, however, that you also want to publish the information on these weights for an audience that uses the imperial system rather than the metric system of measurements. Here, 1977 is used as the “base” year which is equal to 100. Central Tendencay Worksheet or Quiz Includes worksheets with and without number sets. {102, 109, 207, 357, 360, 403, 471, 483, 670, 729, 842, 843, 920, 941} Now, calculate the median M by finding the mean of 471 and 483. The median is the n/2 th value. The median is the middle value; uniformly spread data will provide that the area of the histogram on each side of the median will be equal. What is a Histogram? What does that mean 43 is the median of the frequencies, but it's not the median of the values. Starting with , add the frequencies in the table starting with the first row until you reach . Histogram Worksheet. Finding The Median Using Histograms - Displaying top 8 worksheets found for this concept.. 1. Batting average shows the percent (written as a decimal) of the time a certain player gets a hit. The median is the midpoint of the value, which means that at the median there are exactly half the data below and above that point. Again, the definition of the median for a continuous distribution is the value such that the cumulative probability is 1/2; we just multiplied N by that. Find the bin(s) containing the middle item(s). Determine the number of the middle item. In order to plot a cumulative frequency graph, we have to plot cumulative frequency against the upper-class-boundary of each class. We use linear interpolation to find it. If you know how many numbers there are in a set, which is the middle number? Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Center and Spread of Data Name_____ Date_____ Period____ ... Find the mode, median, mean, lower quartile, upper quartile, interquartile range, and population standard deviation for each data set. To find the median, the data should first be arranged in order from least to greatest. Drawing frequency polygon - with and without a histogram; Finding mean, median and mode of raw data . ; To estimate the Mean use the midpoints of the class intervals: . Now I want to see what happens when I add male heights into the histogram: To remember the definition of a median, just think of the median of a road, which is the middlemost part of the road. For each histogram bar, we start by multiplying the central x-value to the corresponding bar height. When the shape of the distribution is symmetric and unimodal, the mean, median, and mode are equal. using the formula for median we have, Median or where, (lower class boundary of the median class), (total frequency), ( less than type cumulative frequency corresponding to ), • To find the mean, add up all the numbers and divide by the number of numbers. A positive skewed histogram suggests the mean is greater than the median. The median class interval is the corresponding class where the median value falls. Please help. You can get both the mean and the median from the histogram. The histogram above shows a distribution of heights for a sample of college females. Given that the median value is 46, determine the missing frequencies using the median formula. This means you will have to convert your units into pounds from kilograms, multiplying each observation value in your data by 2.2 to get an approximate weight. Study each of these problems carefully; you will see similar problems on the lesson knowledge check. the median class is the class for which upper class boundary is . But in the histogram the hint is confusing me. finding median for ungrouped data Median is the value which occupies the middle position when all the observations are arranged in an ascending or descending order. the class containing the median value. You will need paper and a … For grouped data, we cannot find the exact Mean, Median and Mode, we can only give estimates. 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And without a histogram 66.5 inches by the relative frequency of ( Midpoint × frequency ) Sum of.! Median, and mode of raw data, 299.5-399.5 is the middle in your.... – 40 be f 1 and that of 50 – 60 be 2!, median and mode are three different ways of describing the average ways of describing the average of the,..., but it 's not the median, the mean use the of! Skewed histogram suggests the mean and the median Using Histograms - displaying top 8 worksheets found this. Curve should look like the following: finding the middle in your list in format.
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# Tessellation and Miura Folds
## Mathematics
Say you’re going to launch a satellite into space. Once in orbit, it will be powered by an array of rigid solar panels that fan outward. But to launch the satellite, those panels have to be folded up and compact. How would you design them?
In 1985, Japanese astrophysicist Koryo Miura proposed a form of rigid origami, a style of folding paper (or other materials) that allows each section to remain flat—a necessary condition for stiff materials, like some solar panels. The Miura fold, or Miura-ori, was used in Japan’s Space Flyer Unit (a satellite launched in 1995), and has influenced the development of other folds that allow materials to be packed into a compact shape and then unfold in one continuous motion.
Miura folds are considered shape-memory origami because the fold can be “remembered”—that is, after unfolding, the sheet can easily be re-folded and returned to its compact shape. You are going to create a basic Miura fold out of a sheet of letter-size copy paper (8 ½ inches by 11 inches) and then use your observations of the pattern to reverse-engineer a Miura fold for a different sized sheet.
## Build a Basic Miura Fold
All you need is a sheet of plain letter-sized (8 ½” x 11″) copy paper.
1. Fold the paper into five evenly spaced sections, alternating mountain and valley folds to create an accordion fold, also called a concertina fold.
2. Keeping the paper folded—it should look like a tall, skinny rectangle—you will create seven more sections by alternating mountain and valley folds. To make the first fold, take the bottom of the paper and fold a section up and at an angle to the left, so that the top left corner is about 1 inch from the top of the rest of the paper and ¼ of an inch from the left edge of the rest of the paper (see blue arrow in the image below). Next, fold a portion of the paper that you just folded up back down so that its front right edge is parallel with the right edge of what remains of the original tall, skinny rectangle. You have created your first section. Continue this process until you’ve created seven sections. (You will have to turn the paper over at some point. See the diagram below and the video farther down for visual guides.)
3. Unfold the paper and lay it flat. You will notice seven zigzagging lines running across the page. You will again alternate mountain and valley folds using these lines as guides. If it helps, trace your zigzags with alternating colors to help you keep track of which lines should be mountain folds and which should be valley folds. You can also use your fingernail to reinforce the crease for each of the zigzags on either side of the paper, depending on whether it’s a mountain or a valley fold.
4. Push the zigzagging mountain folds together. As you do so, the structure should draw in on itself. This is the compressed structure of your Miura fold. Note: This may be difficult your first time, but keep at it!
5. Grabbing the two loose corners on either side of the “accordion,” stretch your paper in and out to work your Miura fold. Every time you do this, it becomes easier to return the structure to its compressed state.
## Stretch Your Paper Out Flat and Pick Out Some Patterns
### What kind of shapes do you see? Where are there parallel lines? Where are there diagonal lines? Where are there straight lines?
If you noticed a pattern of interlocking polygons (any closed figure with at least three straight sides) you have identified the tessellation in the Miura fold. A tessellation is an arrangement of polygons in a repeated pattern without gaps or overlap. You might have seen art that is inspired by tessellations, like pieces created by M.C. Escher. The type of polygon that comprises the Miura fold tessellation is a parallelogram.
### Examine the repeated polygon further by using a protractor to measure the angles at each vertex. What do you notice about the angles you found? Are there repeated angles?
Aside from the outer edges of your paper, you probably did not find any right angles. Miura did this on purpose—he proved that omitting right angles reduced the stress on the construction, thus increasing its durability. He was inspired by patterns he had observed in nature that don’t have right angles, like forehead wrinkles or the veins of a dragonfly’s wing. Since then, the Miura fold has been used as inspiration for foldable solar panels, surgical stents, and furniture that can be stowed away until needed.
You may have noted in your previous observations that opposite angles of each polygon in the tessellation are equal, which is a feature of parallelograms. The straight creases separate each parallelogram from its mirror image (reflection), while the zigzagging creases separate each parallelogram from an identical translation (copy) of itself.
## Reverse-Engineer a Miura Fold
In the previous exercise, you created a Miura fold by folding paper. In this activity, you’ll make a Miura fold by first creating a tessellation of parallelograms using reflection and translation.
1. Create a single parallelogram to use as a template for your tessellation. Be sure to adhere to these three basic traits of parallelograms:
— Opposite sides are parallel and equal in length.
— Opposite angles are equal.
— Two consecutive angles are always supplementary (add to 180°)
2. In a Miura fold, straight lines separate each parallelogram from its mirror reflection. Create a row consisting of five parallelograms across by tracing your parallelogram template you created in step one and then reflecting it across at the straight line.
3. Continue to add rows to the bottom until you have a total of seven rows.
4. Cut your reverse-engineered fold by following the outer lines of your drawing.
5. Create creases between each of your columns, alternating the creases on either side of the paper so you can create mountain and valley folds.
6. Crease the zigzags, again alternating on either side of the paper, then push the zigzagging mountain folds together.
You can create a Miura pattern on a coordinate plane to practice translation and reflection. (CCSS.MATH.CONTENT.8.G.A.3, CCSS.MATH.CONTENT.HSG.CO.A.5).
## Mess Around With the Basic Miura Fold
• Using a new sheet of 8 ½-inch by 11-inch paper, change the number of rows and/or columns, and then fold the paper using the same method as the original Miura fold. Document your results. What happens to the size and angles of your parallelograms? Fold and unfold your construction a few times. Which iteration of the Miura fold has the best shape memory?
• Calculate the decimal value of the ratio of zigzagging rows to columns, and calculate the decimal value of the ratio of length to width of your paper. Compare those values. Create new Miura fold constructions that bring those values closer together and then further apart by changing the length to width ratio of your paper. For each new structure, add the ratios for rows to columns and length to width of your paper to your Miura Fold Observation Sheet. For each iteration, calculate the area of your compressed structure (when the construction is folded up) and add this to your Miura Fold Observation Sheet. What changes do you notice in the tessellation? How does the area of the compressed structure change as the ratios get closer and further apart?
• Try making Miura folds with thicker and thinner paper. What effect does the paper’s thickness have on how it folds and compresses?
### Resources
— This program introduced students to Miura folds using thin wooden blocks and cloth tape to create their constructions.
— Nishiyama, Yutaka. “Miura Folding: Applying Origami to Space Exploration.” International Journal of Pure and Applied Mathematics 79.2 (2012): 269-79.
### Standards
CCSS.MATH.CONTENT.4.G.A.1: Draw points, lines, line segments, rays, angles (right, acute, obtuse), and perpendicular and parallel lines. Identify these in two-dimensional figures.
CCSS.MATH.CONTENT.4.G.A.2: Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size. Recognize right triangles as a category, and identify right triangles.
CCSS.MATH.CONTENT.8.G.A.2: Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations; given two congruent figures, describe a sequence that exhibits the congruence between them.
CCSS.MATH.CONTENT.8.G.A.4: Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations, and dilations; given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them.
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# Tessellation and Miura Folds. ## Mathematics. Say you’re going to launch a satellite into space. Once in orbit, it will be powered by an array of rigid solar panels that fan outward. But to launch the satellite, those panels have to be folded up and compact. How would you design them?. In 1985, Japanese astrophysicist Koryo Miura proposed a form of rigid origami, a style of folding paper (or other materials) that allows each section to remain flat—a necessary condition for stiff materials, like some solar panels. The Miura fold, or Miura-ori, was used in Japan’s Space Flyer Unit (a satellite launched in 1995), and has influenced the development of other folds that allow materials to be packed into a compact shape and then unfold in one continuous motion.. Miura folds are considered shape-memory origami because the fold can be “remembered”—that is, after unfolding, the sheet can easily be re-folded and returned to its compact shape. You are going to create a basic Miura fold out of a sheet of letter-size copy paper (8 ½ inches by 11 inches) and then use your observations of the pattern to reverse-engineer a Miura fold for a different sized sheet.. ## Build a Basic Miura Fold. All you need is a sheet of plain letter-sized (8 ½” x 11″) copy paper.. 1. Fold the paper into five evenly spaced sections, alternating mountain and valley folds to create an accordion fold, also called a concertina fold.. 2. Keeping the paper folded—it should look like a tall, skinny rectangle—you will create seven more sections by alternating mountain and valley folds. To make the first fold, take the bottom of the paper and fold a section up and at an angle to the left, so that the top left corner is about 1 inch from the top of the rest of the paper and ¼ of an inch from the left edge of the rest of the paper (see blue arrow in the image below). Next, fold a portion of the paper that you just folded up back down so that its front right edge is parallel with the right edge of what remains of the original tall, skinny rectangle. You have created your first section. Continue this process until you’ve created seven sections. (You will have to turn the paper over at some point. See the diagram below and the video farther down for visual guides.). 3. Unfold the paper and lay it flat. You will notice seven zigzagging lines running across the page. You will again alternate mountain and valley folds using these lines as guides. If it helps, trace your zigzags with alternating colors to help you keep track of which lines should be mountain folds and which should be valley folds. You can also use your fingernail to reinforce the crease for each of the zigzags on either side of the paper, depending on whether it’s a mountain or a valley fold.. 4. Push the zigzagging mountain folds together. As you do so, the structure should draw in on itself. This is the compressed structure of your Miura fold. Note: This may be difficult your first time, but keep at it!. 5. Grabbing the two loose corners on either side of the “accordion,” stretch your paper in and out to work your Miura fold. Every time you do this, it becomes easier to return the structure to its compressed state.. ## Stretch Your Paper Out Flat and Pick Out Some Patterns. ### What kind of shapes do you see? Where are there parallel lines? Where are there diagonal lines? Where are there straight lines?. If you noticed a pattern of interlocking polygons (any closed figure with at least three straight sides) you have identified the tessellation in the Miura fold. A tessellation is an arrangement of polygons in a repeated pattern without gaps or overlap. You might have seen art that is inspired by tessellations, like pieces created by M.C. Escher. The type of polygon that comprises the Miura fold tessellation is a parallelogram.. ### Examine the repeated polygon further by using a protractor to measure the angles at each vertex. What do you notice about the angles you found? Are there repeated angles?. Aside from the outer edges of your paper, you probably did not find any right angles. Miura did this on purpose—he proved that omitting right angles reduced the stress on the construction, thus increasing its durability. He was inspired by patterns he had observed in nature that don’t have right angles, like forehead wrinkles or the veins of a dragonfly’s wing. Since then, the Miura fold has been used as inspiration for foldable solar panels, surgical stents, and furniture that can be stowed away until needed.. You may have noted in your previous observations that opposite angles of each polygon in the tessellation are equal, which is a feature of parallelograms. The straight creases separate each parallelogram from its mirror image (reflection), while the zigzagging creases separate each parallelogram from an identical translation (copy) of itself.. ## Reverse-Engineer a Miura Fold. In the previous exercise, you created a Miura fold by folding paper. In this activity, you’ll make a Miura fold by first creating a tessellation of parallelograms using reflection and translation.. 1. Create a single parallelogram to use as a template for your tessellation. Be sure to adhere to these three basic traits of parallelograms:. — Opposite sides are parallel and equal in length.. — Opposite angles are equal.. — Two consecutive angles are always supplementary (add to 180°). 2. In a Miura fold, straight lines separate each parallelogram from its mirror reflection. Create a row consisting of five parallelograms across by tracing your parallelogram template you created in step one and then reflecting it across at the straight line.. 3. Continue to add rows to the bottom until you have a total of seven rows.. 4. Cut your reverse-engineered fold by following the outer lines of your drawing.. 5. Create creases between each of your columns, alternating the creases on either side of the paper so you can create mountain and valley folds.. 6. Crease the zigzags, again alternating on either side of the paper, then push the zigzagging mountain folds together.. You can create a Miura pattern on a coordinate plane to practice translation and reflection. (CCSS.MATH.CONTENT.8.G.A.3, CCSS.MATH.CONTENT.HSG.CO.A.5).. ## Mess Around With the Basic Miura Fold. • Using a new sheet of 8 ½-inch by 11-inch paper, change the number of rows and/or columns, and then fold the paper using the same method as the original Miura fold. Document your results. What happens to the size and angles of your parallelograms? Fold and unfold your construction a few times. Which iteration of the Miura fold has the best shape memory?. • Calculate the decimal value of the ratio of zigzagging rows to columns, and calculate the decimal value of the ratio of length to width of your paper. Compare those values. Create new Miura fold constructions that bring those values closer together and then further apart by changing the length to width ratio of your paper. For each new structure, add the ratios for rows to columns and length to width of your paper to your Miura Fold Observation Sheet. For each iteration, calculate the area of your compressed structure (when the construction is folded up) and add this to your Miura Fold Observation Sheet. What changes do you notice in the tessellation? How does the area of the compressed structure change as the ratios get closer and further apart?. • Try making Miura folds with thicker and thinner paper. What effect does the paper’s thickness have on how it folds and compresses?. ### Resources. — This program introduced students to Miura folds using thin wooden blocks and cloth tape to create their constructions.. — Nishiyama, Yutaka. “Miura Folding: Applying Origami to Space Exploration.” International Journal of Pure and Applied Mathematics 79.2 (2012): 269-79.. ### Standards. CCSS.MATH.CONTENT.4.G.A.1: Draw points, lines, line segments, rays, angles (right, acute, obtuse), and perpendicular and parallel lines. Identify these in two-dimensional figures.. CCSS.MATH.CONTENT.4.G.A.2: Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size. Recognize right triangles as a category, and identify right triangles.
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CCSS.MATH.CONTENT.8.G.A.2: Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations; given two congruent figures, describe a sequence that exhibits the congruence between them.. CCSS.MATH.CONTENT.8.G.A.4: Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations, and dilations; given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them.
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http://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.127233.html
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# SOLUTION: The sum of the first six terms of an A.P is 72 and the second term is seven times the fifth term. Form two equations in a' and d' and solve them to find the first term and the comm
Algebra -> Algebra -> Sequences-and-series -> SOLUTION: The sum of the first six terms of an A.P is 72 and the second term is seven times the fifth term. Form two equations in a' and d' and solve them to find the first term and the comm Log On
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Algebra: Sequences of numbers, series and how to sum them Solvers Lessons Answers archive Quiz In Depth
Question 127233: The sum of the first six terms of an A.P is 72 and the second term is seven times the fifth term. Form two equations in a' and d' and solve them to find the first term and the common diffrence. Thanks and pls help!Answer by solver91311(16885) (Show Source): You can put this solution on YOUR website!The first term is a, the second term is a + d, the third, a + 2d, etc. So the sum of the first 6 terms is . The 15 comes from So our first equation is The second term is and the fifth term is , so the second equation is: , which simplifies to Add the two equations, since the coefficients on the a terms are already additive inverses Substitute this value for d into the first equation: Check: Answer checks.
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# SOLUTION: The sum of the first six terms of an A.P is 72 and the second term is seven times the fifth term. Form two equations in a' and d' and solve them to find the first term and the comm. Algebra -> Algebra -> Sequences-and-series -> SOLUTION: The sum of the first six terms of an A.P is 72 and the second term is seven times the fifth term. Form two equations in a' and d' and solve them to find the first term and the comm Log On. Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!. Algebra: Sequences of numbers, series and how to sum them Solvers Lessons Answers archive Quiz In Depth. Question 127233: The sum of the first six terms of an A.P is 72 and the second term is seven times the fifth term. Form two equations in a' and d' and solve them to find the first term and the common diffrence.
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Thanks and pls help!Answer by solver91311(16885) (Show Source): You can put this solution on YOUR website!The first term is a, the second term is a + d, the third, a + 2d, etc. So the sum of the first 6 terms is . The 15 comes from So our first equation is The second term is and the fifth term is , so the second equation is: , which simplifies to Add the two equations, since the coefficients on the a terms are already additive inverses Substitute this value for d into the first equation: Check: Answer checks.
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# Partial Fractions
• Jun 11th 2009, 08:39 AM
coobe
Partial Fractions
hi again!
i have to make partialfractions out of the following integral:
$\int \frac{1-x}{x^3-2x^2+x-2} dx$
guessing a critical point for the denominator gives me $x_{0}=2$
then polynomial division gives me
$(x^3-2x^2+x-2) / (x-2) = x^2+1$
$\rightarrow \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}$
is this correct ? if so, i dont seem to get an equation system that is solvable...
could you give me a quick hint to do partial fractions in general ? like the different cases etc..., my professors script is a mess
thanks
• Jun 11th 2009, 09:28 AM
alexmahone
$\frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}$
Multiply throughout by (x-2)
$\frac {1-x}{x^2+1} = A + B\frac{x-2}{x^2+1}$
Sub x=2
$-\frac{1}{5}=A$
$A=-\frac{1}{5}$
• Jun 11th 2009, 09:32 AM
AMI
Quote:
Originally Posted by coobe
$\rightarrow \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}$
Actually, it is $\frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{Bx+C}{x^2+1}$
$\Rightarrow 1-x=A(x^2+1)+(Bx+C)(x-2)$ $\Rightarrow A+B=0,-2B+C=-1,A-2C=1$ $\Rightarrow A=-\frac{1}{5},B=\frac{1}{5},C=-\frac{3}{5}$.
Did you look here: Partial fraction - Wikipedia, the free encyclopedia ?! It seems pretty well written..
• Jun 12th 2009, 02:43 AM
coobe
sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither :(
• Jun 12th 2009, 02:59 AM
HallsofIvy
Quote:
Originally Posted by coobe
sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither :http://mathhelpforum.com/calculus/92565-partial-fractions.html#post328806\" rel=\"nofollow\">
sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither :(
(Wondering) Did you look at the link?? It's the case where you have an irreducible second degree factor in the denominator, just like HallsofIvy said..
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# Partial Fractions. • Jun 11th 2009, 08:39 AM. coobe. Partial Fractions. hi again!. i have to make partialfractions out of the following integral:. $\int \frac{1-x}{x^3-2x^2+x-2} dx$. guessing a critical point for the denominator gives me $x_{0}=2$. then polynomial division gives me. $(x^3-2x^2+x-2) / (x-2) = x^2+1$. $\rightarrow \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}$. is this correct ? if so, i dont seem to get an equation system that is solvable.... could you give me a quick hint to do partial fractions in general ? like the different cases etc..., my professors script is a mess. thanks. • Jun 11th 2009, 09:28 AM. alexmahone. $\frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}$. Multiply throughout by (x-2). $\frac {1-x}{x^2+1} = A + B\frac{x-2}{x^2+1}$. Sub x=2. $-\frac{1}{5}=A$. $A=-\frac{1}{5}$. • Jun 11th 2009, 09:32 AM. AMI. Quote:. Originally Posted by coobe. $\rightarrow \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}$. Actually, it is $\frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{Bx+C}{x^2+1}$. $\Rightarrow 1-x=A(x^2+1)+(Bx+C)(x-2)$ $\Rightarrow A+B=0,-2B+C=-1,A-2C=1$ $\Rightarrow A=-\frac{1}{5},B=\frac{1}{5},C=-\frac{3}{5}$.. Did you look here: Partial fraction - Wikipedia, the free encyclopedia ?! It seems pretty well written... • Jun 12th 2009, 02:43 AM. coobe. sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither :(. • Jun 12th 2009, 02:59 AM. HallsofIvy. Quote:. Originally Posted by coobe. sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither :http://mathhelpforum.com/calculus/92565-partial-fractions.html#post328806\" rel=\"nofollow\">.
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sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither :(. (Wondering) Did you look at the link?? It's the case where you have an irreducible second degree factor in the denominator, just like HallsofIvy said..
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# Intercept Theorem
Reviewed by:
Last updated date: 13th Sep 2024
Total views: 207.6k
Views today: 4.07k
## An Overview of the Theorem
The Intercept Theorem is a fundamental tool of Euclidean Geometry. The concept of parallel lines and transversal is of great importance in our day-to-day life. And, the Intercept theorem extends our understanding of parallel lines and transversal and we can apply these concepts in our day-to-day life.
A Transversal
In the above figure, we can see that there are 3 parallel lines \${L}_{1}\$,\${L}_{2}\$, \${L}_{3}\$ and then there is a transversal \$PR\$ which is intersecting all the 3 parallel lines at an equal distance. The intercept theorem, also known as Thale’s theorem, Basic Proportionality Theorem, or side splitter theorem, is an important theorem in elementary geometry about the ratios of various line segments that are created if two intersecting lines are intercepted by a pair of parallels.
### History of the Mathematician
Euclid
• Year of Birth: 325 BC
• Year of Death: 270 BC
• Contribution: He contributed significantly in the field of Mathematics and Physics by discovering the intercept theorem.
## Statement of the Theorem
If there are three or more parallel lines and the intercepts made by them on one transversal are equal, the corresponding intercepts of any transversal are also equal.
## Proof of the Theorem
Two Parallel Lines
Given:
\$l\$, \$m\$, \$n\$ are three parallel lines.
\$P\$ is a transversal intersecting the parallel lines such that \$AB=BC\$.
The transversal \$Q\$ has the intercepts \$DE\$ and \$HE\$ by the parallel lines \$l\$, \$m\$, \$n\$.
To prove:
\$DE=EF\$
Proof:
Draw a line \$E\$ parallel to the line \$P\$ which intersects the line \$n\$ at \$H\$ and line \$l\$ at \$G\$.
\$AG||BE\$ (Given)
\$GE||AB\$ (By construction)
From the information above, we can say that \$AGBE\$ is a parallelogram.
According to the properties of a parallelogram:
\$AB=GE\$ - (1)
Similarly, we can say that \$BEHC\$ is a parallelogram.
\$BC=HE\$ - (2)
From the given information, we know that \$AB=BC\$.
So, from equations (1) and (2), we can say that \$GE=HE\$.
In \$\Delta GED\$ and \$\Delta HEF\$,
\$GE=HE\$(Proved)
\$\angle GED=\angle FEH\$(Vertically Opposite Angles)
\$\angle DGE=\angle FHE\$(Alternate Interior Angles)
Hence, \$\Delta GED\cong \Delta HEF\$
As\$\Delta GED\cong \Delta HEF\$, the sides\$DE=EF\$.
Hence proved.
## Applications of the Theorem
The intercept theorem can be used to prove that a certain construction yields parallel line segments:
• If the midpoints of two triangle sides are connected, then the resulting line segment is parallel to the third triangle side (Mid point theorem of triangles).
• If the midpoints of the two non-parallel sides of a trapezoid are connected, then the resulting line segment is parallel to the other two sides of the trapezoid.
## Limitations of the Theorem
• The intercept theorem is not able to help us in finding the midpoint of the sides of the triangle.
• The basic proportionality theorem is an advanced version of the intercept theorem and it gives us a lot of information on the sides of the triangles.
## Solved Examples
1. In a \[\Delta ABC\], sides \[AB\] and \[AC\] are intersected by a line at \[D\] and \[E\], respectively, which is parallel to side \[BC\]. Prove that \[\dfrac{AD}{AB}=\dfrac{AE}{AC}\].
Ans:
Scalene Triangle
\[DE||BC\] (Given)
Interchanging the ratios,
Interchanging the ratios again,
Hence proved.
2. Find DE
Basic Proportionality Theorem
Ans: According to the basic proportionality theorem,
\[\dfrac{AE}{DE}=\dfrac{BE}{CE}\]
\[\dfrac{4}{DE}=\dfrac{6}{8.5}\]
\[\dfrac{4*8.5}{6}=DE\]
\[DE=5.66\]
So, \[DE=5.66\]
3. In \[\Delta ABC\], \[D\] and \[E\] are points on the sides \[AB\] and \[AC\], respectively, such that \[DE||BC\]. If \[\dfrac{AD}{DB}=\dfrac{3}{4}\] and \[AC=15cm\], find \[AE\].
Intercept Theorem
Ans:\[\dfrac{AD}{DB}=\dfrac{AE}{EC}\] (According to the intercept theorem)
Let \[AE=x\] and \[EC=15-x\]
So, \[\dfrac{3}{4}=\dfrac{x}{15-x}\]
\[3(15-x)=4x\]
\[45=7x\]
\[x=\dfrac{45}{7}\]
\[x=6.4cm\]
So, \[x=6.4cm\]
## Important Points
• The intercept theorem can only be applied when the lies are parallel, if the transversal is cutting lines that are not parallel, then the intercept theorem is not valid.
• The basic proportionality theorem and mid-point theorem are all applications of the intercept theorem but they are not the same theorems.
## Conclusion
In the above article, we have discussed the Equal intercept Theorem and its proof. We have also discussed the applications of the theorem. So, we can conclude that Intercept Theorem is a fundamental tool of Geometry and is based on applications of parallel lines and transversal and reduces our computational work based on its application as we have seen in the examples based on the theorem.
Competitive Exams after 12th Science
## FAQs on Intercept Theorem
1. Are the basic proportionality theorem, mid-point theorem, and intercept theorem the same?
No, basic proportionality theorem, intercept theorem, and mid-point theorem are 3 different kinds of theorems. The intercept theorem is a very broad theorem and it tells about how the properties of a transversal change when they interact with parallel lines. The basic proportionality theorem is covering about the interaction of lines parallel to one side of the triangle with the other two sides and the basic proportionality theorem is derived from the intercept theorem. The midpoint theorem talks about when a line bisects the other two sides of a triangle, then it is parallel to the third side, and the midpoint theorem is derived from the basic proportionality theorem.
2. What is the mid-point theorem?
The line segment of a triangle connecting the midpoint of two sides of the triangle is said to be parallel to its third side and is also half the length of the third side, according to the midpoint theorem. The mid-point theorem is derived from the basic proportionality theorem and it is a very useful theorem in solving the questions of geometry as it directly gives the value of the sides of a triangle and also the parallel line helps to get the values of angles.
3. Why can’t the intercept theorem be used for non-parallel lines?
The intercept theorem cannot be used for non-parallel lines because there are a lot of things in the derivation of the intercept theorem which we won’t get if the lines are not parallel. Like we won’t be able to use the angles such as the alternate interior angles and the vertically opposite angles and much more. The construction that we make in the derivation is only made because the lines are parallel, hence if the lines are not parallel the intercept theorem is not valid.
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Courses. Courses for Kids. Free study material. Offline Centres. More. Store. # Intercept Theorem. Reviewed by:. Last updated date: 13th Sep 2024. Total views: 207.6k. Views today: 4.07k. ## An Overview of the Theorem. The Intercept Theorem is a fundamental tool of Euclidean Geometry. The concept of parallel lines and transversal is of great importance in our day-to-day life. And, the Intercept theorem extends our understanding of parallel lines and transversal and we can apply these concepts in our day-to-day life.. A Transversal. In the above figure, we can see that there are 3 parallel lines \${L}_{1}\$,\${L}_{2}\$, \${L}_{3}\$ and then there is a transversal \$PR\$ which is intersecting all the 3 parallel lines at an equal distance. The intercept theorem, also known as Thale’s theorem, Basic Proportionality Theorem, or side splitter theorem, is an important theorem in elementary geometry about the ratios of various line segments that are created if two intersecting lines are intercepted by a pair of parallels.. ### History of the Mathematician. Euclid. • Year of Birth: 325 BC. • Year of Death: 270 BC. • Contribution: He contributed significantly in the field of Mathematics and Physics by discovering the intercept theorem.. ## Statement of the Theorem. If there are three or more parallel lines and the intercepts made by them on one transversal are equal, the corresponding intercepts of any transversal are also equal.. ## Proof of the Theorem. Two Parallel Lines. Given:. \$l\$, \$m\$, \$n\$ are three parallel lines.. \$P\$ is a transversal intersecting the parallel lines such that \$AB=BC\$.. The transversal \$Q\$ has the intercepts \$DE\$ and \$HE\$ by the parallel lines \$l\$, \$m\$, \$n\$.. To prove:. \$DE=EF\$. Proof:. Draw a line \$E\$ parallel to the line \$P\$ which intersects the line \$n\$ at \$H\$ and line \$l\$ at \$G\$.. \$AG||BE\$ (Given). \$GE||AB\$ (By construction). From the information above, we can say that \$AGBE\$ is a parallelogram.. According to the properties of a parallelogram:. \$AB=GE\$ - (1). Similarly, we can say that \$BEHC\$ is a parallelogram.. \$BC=HE\$ - (2). From the given information, we know that \$AB=BC\$.. So, from equations (1) and (2), we can say that \$GE=HE\$.. In \$\Delta GED\$ and \$\Delta HEF\$,. \$GE=HE\$(Proved). \$\angle GED=\angle FEH\$(Vertically Opposite Angles). \$\angle DGE=\angle FHE\$(Alternate Interior Angles). Hence, \$\Delta GED\cong \Delta HEF\$. As\$\Delta GED\cong \Delta HEF\$, the sides\$DE=EF\$.. Hence proved.. ## Applications of the Theorem. The intercept theorem can be used to prove that a certain construction yields parallel line segments:. • If the midpoints of two triangle sides are connected, then the resulting line segment is parallel to the third triangle side (Mid point theorem of triangles).. • If the midpoints of the two non-parallel sides of a trapezoid are connected, then the resulting line segment is parallel to the other two sides of the trapezoid.. ## Limitations of the Theorem. • The intercept theorem is not able to help us in finding the midpoint of the sides of the triangle.. • The basic proportionality theorem is an advanced version of the intercept theorem and it gives us a lot of information on the sides of the triangles.. ## Solved Examples. 1. In a \[\Delta ABC\], sides \[AB\] and \[AC\] are intersected by a line at \[D\] and \[E\], respectively, which is parallel to side \[BC\]. Prove that \[\dfrac{AD}{AB}=\dfrac{AE}{AC}\].. Ans:. Scalene Triangle. \[DE||BC\] (Given). Interchanging the ratios,. Interchanging the ratios again,. Hence proved.. 2. Find DE. Basic Proportionality Theorem. Ans: According to the basic proportionality theorem,. \[\dfrac{AE}{DE}=\dfrac{BE}{CE}\]. \[\dfrac{4}{DE}=\dfrac{6}{8.5}\]. \[\dfrac{4*8.5}{6}=DE\]. \[DE=5.66\]. So, \[DE=5.66\]. 3. In \[\Delta ABC\], \[D\] and \[E\] are points on the sides \[AB\] and \[AC\], respectively, such that \[DE||BC\]. If \[\dfrac{AD}{DB}=\dfrac{3}{4}\] and \[AC=15cm\], find \[AE\].. Intercept Theorem. Ans:\[\dfrac{AD}{DB}=\dfrac{AE}{EC}\] (According to the intercept theorem). Let \[AE=x\] and \[EC=15-x\]. So, \[\dfrac{3}{4}=\dfrac{x}{15-x}\]. \[3(15-x)=4x\]. \[45=7x\]. \[x=\dfrac{45}{7}\]. \[x=6.4cm\]. So, \[x=6.4cm\]. ## Important Points. • The intercept theorem can only be applied when the lies are parallel, if the transversal is cutting lines that are not parallel, then the intercept theorem is not valid.. • The basic proportionality theorem and mid-point theorem are all applications of the intercept theorem but they are not the same theorems.. ## Conclusion. In the above article, we have discussed the Equal intercept Theorem and its proof. We have also discussed the applications of the theorem. So, we can conclude that Intercept Theorem is a fundamental tool of Geometry and is based on applications of parallel lines and transversal and reduces our computational work based on its application as we have seen in the examples based on the theorem.. Competitive Exams after 12th Science. ## FAQs on Intercept Theorem. 1. Are the basic proportionality theorem, mid-point theorem, and intercept theorem the same?. No, basic proportionality theorem, intercept theorem, and mid-point theorem are 3 different kinds of theorems. The intercept theorem is a very broad theorem and it tells about how the properties of a transversal change when they interact with parallel lines. The basic proportionality theorem is covering about the interaction of lines parallel to one side of the triangle with the other two sides and the basic proportionality theorem is derived from the intercept theorem. The midpoint theorem talks about when a line bisects the other two sides of a triangle, then it is parallel to the third side, and the midpoint theorem is derived from the basic proportionality theorem.. 2. What is the mid-point theorem?. The line segment of a triangle connecting the midpoint of two sides of the triangle is said to be parallel to its third side and is also half the length of the third side, according to the midpoint theorem. The mid-point theorem is derived from the basic proportionality theorem and it is a very useful theorem in solving the questions of geometry as it directly gives the value of the sides of a triangle and also the parallel line helps to get the values of angles.. 3. Why can’t the intercept theorem be used for non-parallel lines?. The intercept theorem cannot be used for non-parallel lines because there are a lot of things in the derivation of the intercept theorem which we won’t get if the lines are not parallel.
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Like we won’t be able to use the angles such as the alternate interior angles and the vertically opposite angles and much more. The construction that we make in the derivation is only made because the lines are parallel, hence if the lines are not parallel the intercept theorem is not valid.
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# If $abc=1$ ,and $n$ is a natural number,prove $\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2}$
inequality
If $$a, b, c$$ are distinct positive real numbers such that $$abc=1$$,and $$n$$ is a natural number,prove
$$\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2}$$
I know for $$n=3$$,the answer is here,and how to go further?
As can be found here and here , we have that
$$P_n = \frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} = \displaystyle\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}$$ Now by AM-GM, it follows that
$$P_n = \displaystyle\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3} \geq Q [\displaystyle\prod_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3} ]^{1/Q}$$
where $$Q$$ is the number of terms in the sum $$\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}$$.
Now note that by symmetry,
$$P_n \geq Q [\displaystyle\prod_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3} ]^{1/Q} = Q [a b c ]^{\frac{Q(n-2)}{3 Q}} = Q$$
where the last one is true by the condition $$abc =1$$. So it remains to find $$Q$$. This derives from the sum's conditions: we can select $$n_1$$ from $$0$$ to $$n-2$$, i.e. $$n-1$$ possibilites. Then we have that $$n_2$$ can be taken from $$0$$ to $$n-2-n_1$$, i.e. $$n-1-n_1$$ possibilites. So $$Q = \sum_{n_1 =0}^{n-2} (n-1-n_1) = \frac12 n (n-1)$$ which proves the claim. $$\qquad \Box$$
Some examples, using AM-GM directly:
$$P_2 = 1 \geq 1 = \frac12 \cdot 2 \cdot (1)\\ P_3 = a + b +c \geq 3 [abc]^{\frac{1}{3}} = 3 = \frac12 \cdot 3 \cdot(2)\\ P_4 = a^2 + b^2 + c^2 + ab + bc + ac \geq 6 [abc]^{\frac{2}{3}} = 6 = \frac12 \cdot 4 \cdot(3)$$
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# If $abc=1$ ,and $n$ is a natural number,prove $\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2}$. inequality. If $$a, b, c$$ are distinct positive real numbers such that $$abc=1$$,and $$n$$ is a natural number,prove. $$\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2}$$. I know for $$n=3$$,the answer is here,and how to go further?. As can be found here and here , we have that. $$P_n = \frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} = \displaystyle\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}$$ Now by AM-GM, it follows that. $$P_n = \displaystyle\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3} \geq Q [\displaystyle\prod_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3} ]^{1/Q}$$. where $$Q$$ is the number of terms in the sum $$\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}$$.. Now note that by symmetry,. $$P_n \geq Q [\displaystyle\prod_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3} ]^{1/Q} = Q [a b c ]^{\frac{Q(n-2)}{3 Q}} = Q$$. where the last one is true by the condition $$abc =1$$. So it remains to find $$Q$$. This derives from the sum's conditions: we can select $$n_1$$ from $$0$$ to $$n-2$$, i.e. $$n-1$$ possibilites. Then we have that $$n_2$$ can be taken from $$0$$ to $$n-2-n_1$$, i.e. $$n-1-n_1$$ possibilites. So $$Q = \sum_{n_1 =0}^{n-2} (n-1-n_1) = \frac12 n (n-1)$$ which proves the claim. $$\qquad \Box$$.
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Some examples, using AM-GM directly:. $$P_2 = 1 \geq 1 = \frac12 \cdot 2 \cdot (1)\\ P_3 = a + b +c \geq 3 [abc]^{\frac{1}{3}} = 3 = \frac12 \cdot 3 \cdot(2)\\ P_4 = a^2 + b^2 + c^2 + ab + bc + ac \geq 6 [abc]^{\frac{2}{3}} = 6 = \frac12 \cdot 4 \cdot(3)$$.
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Home > Standard Error > Regression Equation And Standard Error
# Regression Equation And Standard Error
## Contents
We are working with a 99% confidence level. It is also possible to evaluate the properties under other assumptions, such as inhomogeneity, but this is discussed elsewhere.[clarification needed] Unbiasedness The estimators α ^ {\displaystyle {\hat {\alpha }}} and β From your table, it looks like you have 21 data points and are fitting 14 terms. Like us on: http://www.facebook.com/PartyMoreStud...Link to Playlist on Regression Analysishttp://www.youtube.com/course?list=EC...Created by David Longstreet, Professor of the Universe, MyBookSuckshttp://www.linkedin.com/in/davidlongs... Κατηγορία Εκπαίδευση Άδεια Τυπική άδεια YouTube Εμφάνιση περισσότερων Εμφάνιση λιγότερων Φόρτωση... Διαφήμιση Αυτόματη αναπαραγωγή http://wapgw.org/standard-error/regression-equation-standard-error.php
I actually haven't read a textbook for awhile. Suppose our requirement is that the predictions must be within +/- 5% of the actual value. It can be shown[citation needed] that at confidence level (1 − γ) the confidence band has hyperbolic form given by the equation y ^ | x = ξ ∈ [ α Estimation Requirements The approach described in this lesson is valid whenever the standard requirements for simple linear regression are met. http://onlinestatbook.com/lms/regression/accuracy.html
## Standard Error Of Regression Formula
The correlation between Y and X , denoted by rXY, is equal to the average product of their standardized values, i.e., the average of {the number of standard deviations by which It is common to make the additional hypothesis that the ordinary least squares method should be used to minimize the residuals. Under this assumption all formulas derived in the previous section remain valid, with the only exception that the quantile t*n−2 of Student's t distribution is replaced with the quantile q* of Mini-slump R2 = 0.98 DF SS F value Model 14 42070.4 20.8s Error 4 203.5 Total 20 42937.8 Name: Jim Frost • Thursday, July 3, 2014 Hi Nicholas, It appears like
Test Your Understanding Problem 1 The local utility company surveys 101 randomly selected customers. Standard Error Of The Slope Todd Grande 25.429 προβολές 9:33 What does r squared tell us? However, in the regression model the standard error of the mean also depends to some extent on the value of X, so the term is scaled up by a factor that my company I love the practical, intuitiveness of using the natural units of the response variable.
For each value of X, the probability distribution of Y has the same standard deviation σ. How To Calculate Standard Error Of Regression Coefficient So, for example, a 95% confidence interval for the forecast is given by In general, T.INV.2T(0.05, n-1) is fairly close to 2 except for very small samples, i.e., a 95% confidence Check out our Statistics Scholarship Page to apply! Thanks S!
## Standard Error Of The Slope
statisticsfun 157.040 προβολές 6:44 Calculating the Standard Error of the Mean in Excel - Διάρκεια: 9:33.
It can be computed in Excel using the T.INV.2T function. Standard Error Of Regression Formula Find the margin of error. Standard Error Of The Regression The following is based on assuming the validity of a model under which the estimates are optimal.
You can change this preference below. Κλείσιμο Ναι, θέλω να τη κρατήσω Αναίρεση Κλείσιμο Αυτό το βίντεο δεν είναι διαθέσιμο. Ουρά παρακολούθησηςΟυράΟυρά παρακολούθησηςΟυρά Κατάργηση όλωνΑποσύνδεση Φόρτωση... Ουρά παρακολούθησης Ουρά __count__/__total__ Standard check over here Smaller is better, other things being equal: we want the model to explain as much of the variation as possible. I was looking for something that would make my fundamentals crystal clear. Read more about how to obtain and use prediction intervals as well as my regression tutorial. Standard Error Of Regression Coefficient
For the model without the intercept term, y = βx, the OLS estimator for β simplifies to β ^ = ∑ i = 1 n x i y i ∑ i Minitab Inc. For example, type L1 and L2 if you entered your data into list L1 and list L2 in Step 1. http://wapgw.org/standard-error/relative-standard-error-equation.php Note, however, that the critical value is based on a t score with n - 2 degrees of freedom.
The standard error of the estimate is a measure of the accuracy of predictions. Linear Regression Standard Error Conversely, the unit-less R-squared doesn’t provide an intuitive feel for how close the predicted values are to the observed values. That for I need to find the standard deviation of a which I somehow just can't find out how to get it.
## Similarly, the confidence interval for the intercept coefficient α is given by α ∈ [ α ^ − s α ^ t n − 2 ∗ , α ^ +
and Keeping, E. You don′t need to memorize all these equations, but there is one important thing to note: the standard errors of the coefficients are directly proportional to the standard error of the Example data. Standard Error Of Regression Interpretation By taking square roots everywhere, the same equation can be rewritten in terms of standard deviations to show that the standard deviation of the errors is equal to the standard deviation
These authors apparently have a very similar textbook specifically for regression that sounds like it has content that is identical to the above book but only the content related to regression Princeton, NJ: Van Nostrand, pp. 252–285 External links Wolfram MathWorld's explanation of Least Squares Fitting, and how to calculate it Mathematics of simple regression (Robert Nau, Duke University) v t e Shashank Prasanna Shashank Prasanna (view profile) 0 questions 677 answers 269 accepted answers Reputation: 1,380 on 21 Jul 2014 Direct link to this comment: https://www.mathworks.com/matlabcentral/answers/142664#comment_226721 What do you mean by no weblink However, those formulas don't tell us how precise the estimates are, i.e., how much the estimators α ^ {\displaystyle {\hat {\alpha }}} and β ^ {\displaystyle {\hat {\beta }}} vary from
Play games and win prizes! Normality assumption Under the first assumption above, that of the normality of the error terms, the estimator of the slope coefficient will itself be normally distributed with mean β and variance Is there a textbook you'd recommend to get the basics of regression right (with the math involved)? In fact, you'll find the formula on the AP statistics formulas list given to you on the day of the exam.
Here the "best" will be understood as in the least-squares approach: a line that minimizes the sum of squared residuals of the linear regression model. We focus on the equation for simple linear regression, which is: ŷ = b0 + b1x where b0 is a constant, b1 is the slope (also called the regression coefficient), x You may need to scroll down with the arrow keys to see the result. statisticsfun 334.568 προβολές 8:29 Explanation of Regression Analysis Results - Διάρκεια: 6:14.
The Variability of the Slope Estimate To construct a confidence interval for the slope of the regression line, we need to know the standard error of the sampling distribution of the We look at various other statistics and charts that shed light on the validity of the model assumptions. If this is the case, then the mean model is clearly a better choice than the regression model. Smaller values are better because it indicates that the observations are closer to the fitted line.
von OehsenList Price: $49.95Buy Used:$0.52Buy New: $57.27Casio CFX-9850GC Plus Graphing Calculator (White)List Price:$139.99Buy Used: \$13.49Approved for AP Statistics and Calculus About Us Contact Us Privacy Terms of Use In the regression output for Minitab statistical software, you can find S in the Summary of Model section, right next to R-squared.
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Home > Standard Error > Regression Equation And Standard Error. # Regression Equation And Standard Error. ## Contents. We are working with a 99% confidence level. It is also possible to evaluate the properties under other assumptions, such as inhomogeneity, but this is discussed elsewhere.[clarification needed] Unbiasedness The estimators α ^ {\displaystyle {\hat {\alpha }}} and β From your table, it looks like you have 21 data points and are fitting 14 terms. Like us on: http://www.facebook.com/PartyMoreStud...Link to Playlist on Regression Analysishttp://www.youtube.com/course?list=EC...Created by David Longstreet, Professor of the Universe, MyBookSuckshttp://www.linkedin.com/in/davidlongs... Κατηγορία Εκπαίδευση Άδεια Τυπική άδεια YouTube Εμφάνιση περισσότερων Εμφάνιση λιγότερων Φόρτωση... Διαφήμιση Αυτόματη αναπαραγωγή http://wapgw.org/standard-error/regression-equation-standard-error.php. I actually haven't read a textbook for awhile. Suppose our requirement is that the predictions must be within +/- 5% of the actual value. It can be shown[citation needed] that at confidence level (1 − γ) the confidence band has hyperbolic form given by the equation y ^ | x = ξ ∈ [ α Estimation Requirements The approach described in this lesson is valid whenever the standard requirements for simple linear regression are met. http://onlinestatbook.com/lms/regression/accuracy.html. ## Standard Error Of Regression Formula. The correlation between Y and X , denoted by rXY, is equal to the average product of their standardized values, i.e., the average of {the number of standard deviations by which It is common to make the additional hypothesis that the ordinary least squares method should be used to minimize the residuals. Under this assumption all formulas derived in the previous section remain valid, with the only exception that the quantile t*n−2 of Student's t distribution is replaced with the quantile q* of Mini-slump R2 = 0.98 DF SS F value Model 14 42070.4 20.8s Error 4 203.5 Total 20 42937.8 Name: Jim Frost • Thursday, July 3, 2014 Hi Nicholas, It appears like. Test Your Understanding Problem 1 The local utility company surveys 101 randomly selected customers. Standard Error Of The Slope Todd Grande 25.429 προβολές 9:33 What does r squared tell us? However, in the regression model the standard error of the mean also depends to some extent on the value of X, so the term is scaled up by a factor that my company I love the practical, intuitiveness of using the natural units of the response variable.. For each value of X, the probability distribution of Y has the same standard deviation σ. How To Calculate Standard Error Of Regression Coefficient So, for example, a 95% confidence interval for the forecast is given by In general, T.INV.2T(0.05, n-1) is fairly close to 2 except for very small samples, i.e., a 95% confidence Check out our Statistics Scholarship Page to apply! Thanks S!. ## Standard Error Of The Slope. statisticsfun 157.040 προβολές 6:44 Calculating the Standard Error of the Mean in Excel - Διάρκεια: 9:33.. It can be computed in Excel using the T.INV.2T function. Standard Error Of Regression Formula Find the margin of error. Standard Error Of The Regression The following is based on assuming the validity of a model under which the estimates are optimal.. You can change this preference below. Κλείσιμο Ναι, θέλω να τη κρατήσω Αναίρεση Κλείσιμο Αυτό το βίντεο δεν είναι διαθέσιμο. Ουρά παρακολούθησηςΟυράΟυρά παρακολούθησηςΟυρά Κατάργηση όλωνΑποσύνδεση Φόρτωση... Ουρά παρακολούθησης Ουρά __count__/__total__ Standard check over here Smaller is better, other things being equal: we want the model to explain as much of the variation as possible. I was looking for something that would make my fundamentals crystal clear. Read more about how to obtain and use prediction intervals as well as my regression tutorial. Standard Error Of Regression Coefficient. For the model without the intercept term, y = βx, the OLS estimator for β simplifies to β ^ = ∑ i = 1 n x i y i ∑ i Minitab Inc. For example, type L1 and L2 if you entered your data into list L1 and list L2 in Step 1. http://wapgw.org/standard-error/relative-standard-error-equation.php Note, however, that the critical value is based on a t score with n - 2 degrees of freedom.. The standard error of the estimate is a measure of the accuracy of predictions. Linear Regression Standard Error Conversely, the unit-less R-squared doesn’t provide an intuitive feel for how close the predicted values are to the observed values. That for I need to find the standard deviation of a which I somehow just can't find out how to get it.. ## Similarly, the confidence interval for the intercept coefficient α is given by α ∈ [ α ^ − s α ^ t n − 2 ∗ , α ^ +. and Keeping, E. You don′t need to memorize all these equations, but there is one important thing to note: the standard errors of the coefficients are directly proportional to the standard error of the Example data. Standard Error Of Regression Interpretation By taking square roots everywhere, the same equation can be rewritten in terms of standard deviations to show that the standard deviation of the errors is equal to the standard deviation. These authors apparently have a very similar textbook specifically for regression that sounds like it has content that is identical to the above book but only the content related to regression Princeton, NJ: Van Nostrand, pp. 252–285 External links Wolfram MathWorld's explanation of Least Squares Fitting, and how to calculate it Mathematics of simple regression (Robert Nau, Duke University) v t e Shashank Prasanna Shashank Prasanna (view profile) 0 questions 677 answers 269 accepted answers Reputation: 1,380 on 21 Jul 2014 Direct link to this comment: https://www.mathworks.com/matlabcentral/answers/142664#comment_226721 What do you mean by no weblink However, those formulas don't tell us how precise the estimates are, i.e., how much the estimators α ^ {\displaystyle {\hat {\alpha }}} and β ^ {\displaystyle {\hat {\beta }}} vary from. Play games and win prizes! Normality assumption Under the first assumption above, that of the normality of the error terms, the estimator of the slope coefficient will itself be normally distributed with mean β and variance Is there a textbook you'd recommend to get the basics of regression right (with the math involved)? In fact, you'll find the formula on the AP statistics formulas list given to you on the day of the exam.. Here the "best" will be understood as in the least-squares approach: a line that minimizes the sum of squared residuals of the linear regression model. We focus on the equation for simple linear regression, which is: ŷ = b0 + b1x where b0 is a constant, b1 is the slope (also called the regression coefficient), x You may need to scroll down with the arrow keys to see the result. statisticsfun 334.568 προβολές 8:29 Explanation of Regression Analysis Results - Διάρκεια: 6:14.. The Variability of the Slope Estimate To construct a confidence interval for the slope of the regression line, we need to know the standard error of the sampling distribution of the We look at various other statistics and charts that shed light on the validity of the model assumptions. If this is the case, then the mean model is clearly a better choice than the regression model.
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Smaller values are better because it indicates that the observations are closer to the fitted line.. von OehsenList Price: $49.95Buy Used:$0.52Buy New: $57.27Casio CFX-9850GC Plus Graphing Calculator (White)List Price:$139.99Buy Used: \$13.49Approved for AP Statistics and Calculus About Us Contact Us Privacy Terms of Use In the regression output for Minitab statistical software, you can find S in the Summary of Model section, right next to R-squared.
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{br} STUCK with your assignment? {br} When is it due? {br} Get FREE assistance. Page Title: {title}{br} Page URL: {url}
UK: +44 748 007-0908, USA: +1 917 810-5386 [email protected]
Select Page
Q1: You are attending a championship soccer match with your family. In your pocket only \$40 to spend on Popcorns(S) and Cokes (C), supposed to allocate such amount to maximize satisfaction of the two goods, where the price of popcorn is \$5 and the price of coke is \$4, looking at the table below:
Various levels of S and C MUs MUs/Ps MUc MUc/Pc
1 40 120
2 30 80
3 25 40
4 20 32
5 15 16
6 10 8
MUs = marginal utility of popcorn, Ps = price of popcorn
MUc = marginal utility of coke, Pc = piece of coke
1. Fill up the blanks above?
2. Find the combination of popcorns and cokes yields and the level of utility, subject to the budget constraint?
3. Calculate the total utility of the optimal consumption of the two good?
4. What is the marginal rate of substitution(MRS) in such case?
5. State the budget equation in this case?
6. Graph the budget line and an indifferent curve, showing the optimal point?
7. What does the indifferent curve represent? How it relates to the indifference map?
8. What is the level of marginal utility per dollar spent, where you get the maximization or the optimal equation of the two goods?
9. The demand curve is derived from the marginal utility concept, explain?
10. Can you classify this case as a corner solution? Explain?
Q2: Decided on open tailor shop, the following table gives the level of outputs(shirts) produced daily, with the least cost of input combinations (labor wage(w) = \$300, and cost of capital, sewing machine(r) = \$200) per day:
output Labor # Capital # Long run total cost (LTC) Long run average cost(LAC) Long run marginal cost(LMC)
100 10 7
200 12 8
300 20 10
400 30 15
500 40 22
600 50 30
700 60 42
1. Fill up the blanks above?
2. Write down the cost equation?
3. State the total production function?
4. Draw up the graph for LAC and LMC in relation to the output?
5. Graph the Isoquant curve as well as the Isocost curve, with the optimal input combination?
6. In this case, you are operating in competitive market, what are the characteristics of such market?
7. At which price you are supposed to sell the shirt?
8. Draw up the expansion path for outputs and costs in the table above?
9. What is the approximate level of output(varies discretely by 100 units in the table), where profit is maximized and you are economically and technically efficient?
10. What is the Marginal Rate of Technical Substitution (MRTS) in this case?
11. Is the cost structure in the question long-run cost or short-run cost ones, what is the difference between the two, and how can they be related?
12. There are many factors reduce costs, and shift LAC downtrend, identify three of them?
13. At which level of output is the Minimum Efficient Scale(MES)?
14. What kind of demand curve you are facing in this market?
15. Is this competitive market, what is the economic profit, and should it be different in value from the opportunity cost? Explain?
Q3: Abdullah want to expand the capacity of his restaurant, but not sure of the 2020 economic growth (GDP) of the Saudi economy. He has the probability of 40% that the economy will maintain the expected growth rate in 2019 (1.8%) and the probability of 60% it will be little higher (2.1%) as the IMF forecasted. Accordingly, the table below is:
Growth rate 2.1% 1.8%
Probability Dist. 60% 40%
Profit Profit(\$Million) Profit(\$Million)
A. Decision maintain capacity 100 persons 3 2
B. Decision expand capacity by 20% 4 1
1. Compute the expected profits for both decisions?
2. Based on the expected profit only, which decision should Abdullah make?
3. Compute the Standard Deviation for decisions A and B, facing Abdullah?
4. Which decision would Abdullah make, using the coefficient of variation?
5. If Abdullah has no idea of the probability distribution of economic growth, operating in uncertainty world. Using the information above, what decision would Abdullah make, according to each of the following rules:
A. Maximax:
B. Maximin:
C. Minimax Regret:
D. Equal Probability:
6. Which decision is riskier, and how can you classify yourself as risk lover or averter or neutral?
Compelling correspondence is essential to the achievement all things considered but since of the changing idea of the present working environments, successful correspondence turns out to be more troublesome, and because of the numerous impediments that will permit beneficiaries to acknowledge the plan of the sender It is restricted. Misguided judgments.In spite of the fact that correspondence inside the association is rarely completely open, numerous straightforward arrangements can be executed to advance the effect of these hindrances.
Concerning specific contextual analysis, two significant correspondence standards, correspondence channel determination and commotion are self-evident. This course presents the standards of correspondence, the act of general correspondence, and different speculations to all the more likely comprehend the correspondence exchanges experienced in regular daily existence. The standards and practices that you learn in this course give the premise to additionally learning and correspondence.
This course starts with an outline of the correspondence cycle, the method of reasoning and hypothesis. In resulting modules of the course, we will look at explicit use of relational connections in close to home and expert life. These incorporate relational correspondence, bunch correspondence and dynamic, authoritative correspondence in the work environment or relational correspondence. Rule of Business Communication In request to make correspondence viable, it is important to follow a few rules and standards. Seven of them are fundamental and applicable, and these are clear, finished, brief, obliging, right, thought to be, concrete. These standards are frequently called 7C for business correspondence. The subtleties of these correspondence standards are examined underneath: Politeness Principle: When conveying, we should build up a cordial relationship with every individual who sends data to us.
To be inviting and polite is indistinguishable, and politeness requires an insightful and amicable activity against others. Axioms are notable that gracious “pay of graciousness is the main thing to win everything”. Correspondence staff ought to consistently remember this. The accompanying standards may assist with improving courtesy:Preliminary considering correspondence with family All glad families have the mystery of progress. This achievement originates from a strong establishment of closeness and closeness. Indeed, through private correspondence these cozy family connections become all the more intently. Correspondence is the foundation of different affiliations, building solid partners of obedient devotion, improving family way of life, and assisting with accomplishing satisfaction (Gosche, p. 1). In any case, so as to keep up an amicable relationship, a few families experienced tumultuous encounters. Correspondence in the family is an intricate and alluring marvel. Correspondence between families isn’t restricted to single messages between families or verbal correspondence.
It is a unique cycle that oversees force, closeness and limits, cohesiveness and flexibility of route frameworks, and makes pictures, topics, stories, ceremonies, rules, jobs, making implications, making a feeling of family life An intelligent cycle that makes a model. This model has passed ages. Notwithstanding the view as a family and family automatic framework, one of the greatest exploration establishments in between family correspondence centers around a family correspondence model. Family correspondence model (FCP) hypothesis clarifies why families impart in their own specific manner dependent on one another ‘s psychological direction. Early FCP research established in media research is keen on how families handle broad communications data. Family correspondence was perceived as an exceptional scholastic exploration field by the National Communications Association in 1989. Family correspondence researchers were at first impacted by family research, social brain science, and relational hypothesis, before long built up the hypothesis and began research in a family framework zeroed in on a significant job. Until 2001, the primary issue of the Family Communication Research Journal, Family Communication Magazine, was given. Family correspondence is more than the field of correspondence analysts in the family. Examination on family correspondence is normally done by individuals in brain science, humanism, and family research, to give some examples models. However, as the popular family correspondence researcher Leslie Baxter stated, it is the focal point of this intelligent semantic creation measure making the grant of family correspondence special. In the field of in-home correspondence, correspondence is normally not founded on autonomous messages from one sender to one beneficiary, yet dependent on the dynamic interdependency of data shared among families It is conceptualized. The focal point of this methodology is on the shared trait of semantic development inside family frameworks. As such, producing doesn’t happen in vacuum, however it happens in a wide scope of ages and social exchange.
Standards are rules end up being followed when performing work to agree to a given objective. Hierarchical achievement relies significantly upon compelling correspondence. So as to successfully impart, it is important to follow a few standards and rules. Coming up next are rules to guarantee powerful correspondence: clearness: lucidity of data is a significant guideline of correspondence. For beneficiaries to know the message plainly, the messages ought to be sorted out in a basic language. To guarantee that beneficiaries can without much of a stretch comprehend the importance of the message, the sender needs to impart unmistakably and unhesitatingly so the beneficiary can plainly and unquestionably comprehend the data.>
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{br} STUCK with your assignment? {br} When is it due? {br} Get FREE assistance. Page Title: {title}{br} Page URL: {url}. UK: +44 748 007-0908, USA: +1 917 810-5386 [email protected]. Select Page. Q1: You are attending a championship soccer match with your family. In your pocket only \$40 to spend on Popcorns(S) and Cokes (C), supposed to allocate such amount to maximize satisfaction of the two goods, where the price of popcorn is \$5 and the price of coke is \$4, looking at the table below:. Various levels of S and C MUs MUs/Ps MUc MUc/Pc. 1 40 120. 2 30 80. 3 25 40. 4 20 32. 5 15 16. 6 10 8. MUs = marginal utility of popcorn, Ps = price of popcorn. MUc = marginal utility of coke, Pc = piece of coke. 1. Fill up the blanks above?. 2. Find the combination of popcorns and cokes yields and the level of utility, subject to the budget constraint?. 3. Calculate the total utility of the optimal consumption of the two good?. 4. What is the marginal rate of substitution(MRS) in such case?. 5. State the budget equation in this case?. 6. Graph the budget line and an indifferent curve, showing the optimal point?. 7. What does the indifferent curve represent? How it relates to the indifference map?. 8. What is the level of marginal utility per dollar spent, where you get the maximization or the optimal equation of the two goods?. 9. The demand curve is derived from the marginal utility concept, explain?. 10. Can you classify this case as a corner solution? Explain?. Q2: Decided on open tailor shop, the following table gives the level of outputs(shirts) produced daily, with the least cost of input combinations (labor wage(w) = \$300, and cost of capital, sewing machine(r) = \$200) per day:. output Labor # Capital # Long run total cost (LTC) Long run average cost(LAC) Long run marginal cost(LMC). 100 10 7. 200 12 8. 300 20 10. 400 30 15. 500 40 22. 600 50 30. 700 60 42. 1. Fill up the blanks above?. 2. Write down the cost equation?. 3. State the total production function?. 4. Draw up the graph for LAC and LMC in relation to the output?. 5. Graph the Isoquant curve as well as the Isocost curve, with the optimal input combination?. 6. In this case, you are operating in competitive market, what are the characteristics of such market?. 7. At which price you are supposed to sell the shirt?. 8. Draw up the expansion path for outputs and costs in the table above?. 9. What is the approximate level of output(varies discretely by 100 units in the table), where profit is maximized and you are economically and technically efficient?. 10. What is the Marginal Rate of Technical Substitution (MRTS) in this case?. 11. Is the cost structure in the question long-run cost or short-run cost ones, what is the difference between the two, and how can they be related?. 12. There are many factors reduce costs, and shift LAC downtrend, identify three of them?. 13. At which level of output is the Minimum Efficient Scale(MES)?. 14. What kind of demand curve you are facing in this market?. 15. Is this competitive market, what is the economic profit, and should it be different in value from the opportunity cost? Explain?. Q3: Abdullah want to expand the capacity of his restaurant, but not sure of the 2020 economic growth (GDP) of the Saudi economy. He has the probability of 40% that the economy will maintain the expected growth rate in 2019 (1.8%) and the probability of 60% it will be little higher (2.1%) as the IMF forecasted. Accordingly, the table below is:. Growth rate 2.1% 1.8%. Probability Dist. 60% 40%. Profit Profit(\$Million) Profit(\$Million). A. Decision maintain capacity 100 persons 3 2. B. Decision expand capacity by 20% 4 1. 1. Compute the expected profits for both decisions?. 2. Based on the expected profit only, which decision should Abdullah make?. 3. Compute the Standard Deviation for decisions A and B, facing Abdullah?. 4. Which decision would Abdullah make, using the coefficient of variation?. 5. If Abdullah has no idea of the probability distribution of economic growth, operating in uncertainty world. Using the information above, what decision would Abdullah make, according to each of the following rules:. A. Maximax:. B. Maximin:. C. Minimax Regret:. D. Equal Probability:. 6. Which decision is riskier, and how can you classify yourself as risk lover or averter or neutral?. Compelling correspondence is essential to the achievement all things considered but since of the changing idea of the present working environments, successful correspondence turns out to be more troublesome, and because of the numerous impediments that will permit beneficiaries to acknowledge the plan of the sender It is restricted. Misguided judgments.In spite of the fact that correspondence inside the association is rarely completely open, numerous straightforward arrangements can be executed to advance the effect of these hindrances.. Concerning specific contextual analysis, two significant correspondence standards, correspondence channel determination and commotion are self-evident. This course presents the standards of correspondence, the act of general correspondence, and different speculations to all the more likely comprehend the correspondence exchanges experienced in regular daily existence. The standards and practices that you learn in this course give the premise to additionally learning and correspondence.. This course starts with an outline of the correspondence cycle, the method of reasoning and hypothesis. In resulting modules of the course, we will look at explicit use of relational connections in close to home and expert life. These incorporate relational correspondence, bunch correspondence and dynamic, authoritative correspondence in the work environment or relational correspondence. Rule of Business Communication In request to make correspondence viable, it is important to follow a few rules and standards. Seven of them are fundamental and applicable, and these are clear, finished, brief, obliging, right, thought to be, concrete. These standards are frequently called 7C for business correspondence. The subtleties of these correspondence standards are examined underneath: Politeness Principle: When conveying, we should build up a cordial relationship with every individual who sends data to us.. To be inviting and polite is indistinguishable, and politeness requires an insightful and amicable activity against others. Axioms are notable that gracious “pay of graciousness is the main thing to win everything”. Correspondence staff ought to consistently remember this. The accompanying standards may assist with improving courtesy:Preliminary considering correspondence with family All glad families have the mystery of progress. This achievement originates from a strong establishment of closeness and closeness. Indeed, through private correspondence these cozy family connections become all the more intently. Correspondence is the foundation of different affiliations, building solid partners of obedient devotion, improving family way of life, and assisting with accomplishing satisfaction (Gosche, p. 1). In any case, so as to keep up an amicable relationship, a few families experienced tumultuous encounters. Correspondence in the family is an intricate and alluring marvel. Correspondence between families isn’t restricted to single messages between families or verbal correspondence.. It is a unique cycle that oversees force, closeness and limits, cohesiveness and flexibility of route frameworks, and makes pictures, topics, stories, ceremonies, rules, jobs, making implications, making a feeling of family life An intelligent cycle that makes a model. This model has passed ages. Notwithstanding the view as a family and family automatic framework, one of the greatest exploration establishments in between family correspondence centers around a family correspondence model. Family correspondence model (FCP) hypothesis clarifies why families impart in their own specific manner dependent on one another ‘s psychological direction. Early FCP research established in media research is keen on how families handle broad communications data. Family correspondence was perceived as an exceptional scholastic exploration field by the National Communications Association in 1989. Family correspondence researchers were at first impacted by family research, social brain science, and relational hypothesis, before long built up the hypothesis and began research in a family framework zeroed in on a significant job. Until 2001, the primary issue of the Family Communication Research Journal, Family Communication Magazine, was given. Family correspondence is more than the field of correspondence analysts in the family. Examination on family correspondence is normally done by individuals in brain science, humanism, and family research, to give some examples models. However, as the popular family correspondence researcher Leslie Baxter stated, it is the focal point of this intelligent semantic creation measure making the grant of family correspondence special. In the field of in-home correspondence, correspondence is normally not founded on autonomous messages from one sender to one beneficiary, yet dependent on the dynamic interdependency of data shared among families It is conceptualized. The focal point of this methodology is on the shared trait of semantic development inside family frameworks. As such, producing doesn’t happen in vacuum, however it happens in a wide scope of ages and social exchange.. Standards are rules end up being followed when performing work to agree to a given objective. Hierarchical achievement relies significantly upon compelling correspondence. So as to successfully impart, it is important to follow a few standards and rules. Coming up next are rules to guarantee powerful correspondence: clearness: lucidity of data is a significant guideline of correspondence.
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For beneficiaries to know the message plainly, the messages ought to be sorted out in a basic language. To guarantee that beneficiaries can without much of a stretch comprehend the importance of the message, the sender needs to impart unmistakably and unhesitatingly so the beneficiary can plainly and unquestionably comprehend the data.>.
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# Will linear voltage regulator step up current?
I have a regulated 9 volt 300mA power supply I want to step it down to 5 volt using Linear Voltage Regulator LM7805 , I want to know how much current can I can draw at 5 volts, will it be 300mA or will it be close to 540mA, since power = voltage * current.
• With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching. – Sunnyskyguy EE75 Feb 20 at 20:32
No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.
$$\P_{in} = P_{out}\$$
but for a linear regulator it looks like this:
$$\V_{in} \times I_{in} = (V_{out} \times I_{out}) + [(V_{in} - V_{out}) \times I_{out}]\$$
The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:
$$\V_{in} \times I_{in} = V_{in} \times I_{out}\$$
Therefore:
$$\I_{in} = I_{out}\$$
No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.
However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.
A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)
A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)
They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.
• "drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook – quetzalcoatl Feb 21 at 10:16
since power = voltage * current.
It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.
In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.
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# Will linear voltage regulator step up current?. I have a regulated 9 volt 300mA power supply I want to step it down to 5 volt using Linear Voltage Regulator LM7805 , I want to know how much current can I can draw at 5 volts, will it be 300mA or will it be close to 540mA, since power = voltage * current.. • With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching. – Sunnyskyguy EE75 Feb 20 at 20:32. No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.. $$\P_{in} = P_{out}\$$. but for a linear regulator it looks like this:. $$\V_{in} \times I_{in} = (V_{out} \times I_{out}) + [(V_{in} - V_{out}) \times I_{out}]\$$. The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:. $$\V_{in} \times I_{in} = V_{in} \times I_{out}\$$. Therefore:. $$\I_{in} = I_{out}\$$. No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.. However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.. A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power). A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss). They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.. • "drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook – quetzalcoatl Feb 21 at 10:16. since power = voltage * current.. It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.
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In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.
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# Ramanujan results for middle school?
Pls I wonder what Ramanujan's results could be explained to middle school level audience, ie without using integral etc that is up to university curriculum?
For example Ramanujan's infinite radicals could be explained easily
$$3=\sqrt {1+2{\sqrt {1+3{\sqrt {1+\cdots }}}}}$$
• Why Ramanujan specifically? Jul 17, 2020 at 22:52
• just his results have many infinite forms, which sound fun! @ChrisCunningham Jul 18, 2020 at 12:54
## 1 Answer
You can try the Rogers–Ramanujan identities:
• The number of partitions of $$n$$ in which adjacent parts are at least 2 apart is the same as the number of partitions of $$n$$ in which each part ends with 1,4,6,9.
• The number of partitions of $$n$$ without 1 in which adjacent parts are at least 2 apart is the same as the number of partitions of $$n$$ in which each part ends with 2,3,7,8.
For example, taking $$n=10$$:
• Partitions in which adjacent parts are at least 2 apart: $$10 = 10 = 9 + 1 = 8 + 2 = 7 + 3 = 6 + 4 = 6 + 3 + 1$$
• Partitions in which each part ends with 1,4,6,9: $$10 = 9 + 1 = 6 + 4 = 6 + 1 + 1 + 1 + 1 = 4 + 4 + 1 + 1 = 4 + 6\times 1 = 10 \times 1$$
• Partitions without 1 in which adjacent parts are at least 2 apart: $$10 = 10 = 8 + 2 = 7 + 3 = 6 + 4$$
• Partitions in which each part ends with 2,3,5,8: $$10 = 8 + 2 = 7 + 3 = 3 + 3 + 2 + 2 = 2 + 2 + 2 + 2 + 2$$
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# Ramanujan results for middle school?. Pls I wonder what Ramanujan's results could be explained to middle school level audience, ie without using integral etc that is up to university curriculum?. For example Ramanujan's infinite radicals could be explained easily. $$3=\sqrt {1+2{\sqrt {1+3{\sqrt {1+\cdots }}}}}$$. • Why Ramanujan specifically? Jul 17, 2020 at 22:52. • just his results have many infinite forms, which sound fun! @ChrisCunningham Jul 18, 2020 at 12:54. ## 1 Answer. You can try the Rogers–Ramanujan identities:. • The number of partitions of $$n$$ in which adjacent parts are at least 2 apart is the same as the number of partitions of $$n$$ in which each part ends with 1,4,6,9.. • The number of partitions of $$n$$ without 1 in which adjacent parts are at least 2 apart is the same as the number of partitions of $$n$$ in which each part ends with 2,3,7,8.. For example, taking $$n=10$$:. • Partitions in which adjacent parts are at least 2 apart: $$10 = 10 = 9 + 1 = 8 + 2 = 7 + 3 = 6 + 4 = 6 + 3 + 1$$. • Partitions in which each part ends with 1,4,6,9: $$10 = 9 + 1 = 6 + 4 = 6 + 1 + 1 + 1 + 1 = 4 + 4 + 1 + 1 = 4 + 6\times 1 = 10 \times 1$$.
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• Partitions without 1 in which adjacent parts are at least 2 apart: $$10 = 10 = 8 + 2 = 7 + 3 = 6 + 4$$. • Partitions in which each part ends with 2,3,5,8: $$10 = 8 + 2 = 7 + 3 = 3 + 3 + 2 + 2 = 2 + 2 + 2 + 2 + 2$$.
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# matrix
Hi All, What's delta R and how to get it?
2. ### Transition matrix problem
I have the following problem that asks me what is the transition matrix from B to B'.In the second image you can view the answer but I don't understand how that answer appeared. Can someone explain me the steps to solve this problem? I am also confused because the answer is a 3x3 matrix and...
3. ### A function out of the matrix?
Hi everyone. I have a table of relations between various foods (each with each). Maybe that could be defined as a matrix. Here is a small 4 x 4 example: Rice Soup Ice cream Salad Rice 0 1 2 1 Soup 1 0 2 2 Ice cream 2 2 0 2 Salad 1 2 2 0 (The higher the...
4. ### determinant of symmetric matrix of size (n+1)(n+1)
Let for j = 0,. . . n aj = a0 + jd, where a0, d are fixed real numbers. Calculate the determinant of the matrix A of size (n + 1) × (n + 1)
5. ### Calculate the determinant of the matrix
Let for j = 0,. . ., n aj = a0 + jd, where a0 and d are fixed real numbers. Calculate the determinant of the matrix A of size (n + 1) × (n + 1). Matrix A is below
6. ### Rotation matrix
is rotation about OX by A° followed by rotation about OY by B° followed by rotation about OZ by C° equivalent to rotation about OX by A° followed by rotation about OY' by B° followed by rotation about OZ'' by C° ?
7. ### Finding square root of matrix without eigenvector
A is a 2x2 matrix. (0 -2) (2 0) Find the square root of this matrix. I managed to find one of the square root, because I know A is a 90-degree anti-clockwise rotation stretched by a factor of 2. So I just have to "half" that operation and I get (1 -1) (1 1) However, there...
8. ### Mathematical notation of matrix
Hi, I would like to know how is the mathematical notation to describe the selection of a part of a Matrix, eg. in Matlab B=A(1:3,1:2) I appreciate your help
9. ### Matrix operations for RBF solver
Hi All! This is my first post, I hope it is the right session. I will be describing the goal first, then the mathematical step of the current implementation I was given and then ask my question. GOAL: I got from someone the python code for an RBF solver. The solver stores 9...
10. ### Transformation matrix of permutation
I need help with the following: Let \sigma \in S_n be a permutation of numbers 1, \dots, n and f:K^n \rightarrow K^n, \begin{pmatrix} x^1 \\ \vdots \\ x^n \end{pmatrix} \mapsto \begin{pmatrix} x^{\sigma(1)} \\ \vdots \\ x^{\sigma(n)} \end{pmatrix} a linear mapping. (i) Calculate the linear...
11. ### Can somebody see whether my intuition re diagonalizability is correct?
Hi guys, can anybody see whether my intuition is correct? True or False? Let A and B be matrices of n x n. 1. If A and B are diagonalizable and they have the same characteristic polynomial, then A and B are similar. 2. If A and B are row equivalent and A is diagonalizable, then B is...
12. ### Matrix
Hi all, can someone please tell me how they got this equation from this matrix. I seem to get two products with X2 instead of one. Please see attachment. Thank you!
13. ### Solving this Matrix quickly in an Exam
Hi all, Does anyone know the best way to answer this maths question quickly? Get from the determinant 1 to the quadratic equation 2. Preferably on a casio calculator, Please see the attachments. Thank you!
14. ### Finding Convexivity or Convavity With 3X3 matrix
Question: https://i.postimg.cc/cCLWx65S/1b.png Can u help me pls? :)
15. ### Matrix Problem
A is 2 x 3 B is x x 4 C is y x z Find the values of x,y, and z such that: i)The matrix operation ABC is a square matrix. ii)The operation AB + C is possible.
16. ### Matrix in relation to solution subspace
Hi guys, I am having serious trouble analysing the following question: I have been thinking about it for an hour, but I am not progressing… Can anybody help out? Let A be a matrix over R of order 4x4, with rank 2. Suppose that the vectors u=(2,1,2,0) v=(1,-1,2,4) w=(1,0,2,-1) are...
17. ### Proving a matrix invertible
given that AB^2 - A is invertible (A,B square matrices of the same size) prove BA - A is invertible been thinking about it for hours. could anybody please help?
18. ### matrix construction
Hi guys, How can a matrix be constructed given 2 row spaces and a null space? any idea? I will work on solid exampled given recipe by you. thanks
19. ### Inverse of a matrix of decimal or floating numbers.
Can we find the Inverse of a 3*3 matrix having the following decimal number elements? 1.2 2.3 4.5 6.1 4.2 2.9 7.4 3.5 8.7 Thanks & Regards, Prashant S Akerkar
20. ### Geometric representation of matrix transformation
I want to sketch the 2-vector \textbf{u}=\left( \begin{array}{c} 2\\ 3 \end{array} \right) and its image under the matrix transformation f: R^2\rightarrow R^3 defined by f(u)=\left( \begin{array}{cc} 1 & 0\\ 1 & -1\\ 0 & 1 \end{array} \right)\textbf{u}. Can I sketch both of \textbf{u} and...
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# matrix. Hi All, What's delta R and how to get it?. 2. ### Transition matrix problem. I have the following problem that asks me what is the transition matrix from B to B'.In the second image you can view the answer but I don't understand how that answer appeared. Can someone explain me the steps to solve this problem? I am also confused because the answer is a 3x3 matrix and.... 3. ### A function out of the matrix?. Hi everyone. I have a table of relations between various foods (each with each). Maybe that could be defined as a matrix. Here is a small 4 x 4 example: Rice Soup Ice cream Salad Rice 0 1 2 1 Soup 1 0 2 2 Ice cream 2 2 0 2 Salad 1 2 2 0 (The higher the.... 4. ### determinant of symmetric matrix of size (n+1)(n+1). Let for j = 0,. . . n aj = a0 + jd, where a0, d are fixed real numbers. Calculate the determinant of the matrix A of size (n + 1) × (n + 1). 5. ### Calculate the determinant of the matrix. Let for j = 0,. . ., n aj = a0 + jd, where a0 and d are fixed real numbers. Calculate the determinant of the matrix A of size (n + 1) × (n + 1). Matrix A is below. 6. ### Rotation matrix. is rotation about OX by A° followed by rotation about OY by B° followed by rotation about OZ by C° equivalent to rotation about OX by A° followed by rotation about OY' by B° followed by rotation about OZ'' by C° ?. 7. ### Finding square root of matrix without eigenvector. A is a 2x2 matrix. (0 -2) (2 0) Find the square root of this matrix. I managed to find one of the square root, because I know A is a 90-degree anti-clockwise rotation stretched by a factor of 2. So I just have to "half" that operation and I get (1 -1) (1 1) However, there.... 8. ### Mathematical notation of matrix. Hi, I would like to know how is the mathematical notation to describe the selection of a part of a Matrix, eg. in Matlab B=A(1:3,1:2) I appreciate your help. 9. ### Matrix operations for RBF solver. Hi All! This is my first post, I hope it is the right session. I will be describing the goal first, then the mathematical step of the current implementation I was given and then ask my question. GOAL: I got from someone the python code for an RBF solver. The solver stores 9.... 10. ### Transformation matrix of permutation. I need help with the following: Let \sigma \in S_n be a permutation of numbers 1, \dots, n and f:K^n \rightarrow K^n, \begin{pmatrix} x^1 \\ \vdots \\ x^n \end{pmatrix} \mapsto \begin{pmatrix} x^{\sigma(1)} \\ \vdots \\ x^{\sigma(n)} \end{pmatrix} a linear mapping. (i) Calculate the linear.... 11. ### Can somebody see whether my intuition re diagonalizability is correct?. Hi guys, can anybody see whether my intuition is correct? True or False? Let A and B be matrices of n x n. 1. If A and B are diagonalizable and they have the same characteristic polynomial, then A and B are similar. 2. If A and B are row equivalent and A is diagonalizable, then B is.... 12. ### Matrix. Hi all, can someone please tell me how they got this equation from this matrix. I seem to get two products with X2 instead of one. Please see attachment. Thank you!. 13. ### Solving this Matrix quickly in an Exam. Hi all, Does anyone know the best way to answer this maths question quickly? Get from the determinant 1 to the quadratic equation 2. Preferably on a casio calculator, Please see the attachments. Thank you!. 14. ### Finding Convexivity or Convavity With 3X3 matrix. Question: https://i.postimg.cc/cCLWx65S/1b.png Can u help me pls? :). 15. ### Matrix Problem. A is 2 x 3 B is x x 4 C is y x z Find the values of x,y, and z such that: i)The matrix operation ABC is a square matrix. ii)The operation AB + C is possible.. 16. ### Matrix in relation to solution subspace. Hi guys, I am having serious trouble analysing the following question: I have been thinking about it for an hour, but I am not progressing… Can anybody help out? Let A be a matrix over R of order 4x4, with rank 2. Suppose that the vectors u=(2,1,2,0) v=(1,-1,2,4) w=(1,0,2,-1) are.... 17. ### Proving a matrix invertible. given that AB^2 - A is invertible (A,B square matrices of the same size) prove BA - A is invertible been thinking about it for hours. could anybody please help?. 18. ### matrix construction. Hi guys, How can a matrix be constructed given 2 row spaces and a null space? any idea? I will work on solid exampled given recipe by you. thanks. 19. ### Inverse of a matrix of decimal or floating numbers.. Can we find the Inverse of a 3*3 matrix having the following decimal number elements? 1.2 2.3 4.5 6.1 4.2 2.9 7.4 3.5 8.7 Thanks & Regards, Prashant S Akerkar. 20.
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### Geometric representation of matrix transformation. I want to sketch the 2-vector \textbf{u}=\left( \begin{array}{c} 2\\ 3 \end{array} \right) and its image under the matrix transformation f: R^2\rightarrow R^3 defined by f(u)=\left( \begin{array}{cc} 1 & 0\\ 1 & -1\\ 0 & 1 \end{array} \right)\textbf{u}. Can I sketch both of \textbf{u} and...
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From
Equation 19 we have the formula that the resisting moment at any point in the beam equals the area of the steel, times the unit tensile stress in the steel, times the distance from the steel to the centroid of compression of the steel, which is the distance d - x. We may compute the moment in the beam at two points at a unit-distance apart. The area of the steel is the same in each equation, and d - x is substantially the same in each case; and therefore the difference of moment, divided by (d - x), will evidently equal the difference in the unit-stress in the steel, times the area of the steel. To express this in an equation, we may say, denoting the difference in the moment by d M, and the difference in the unit-stress in the steel by d s: dM = A x ds.
(d-x)
But A X d s is evidently equal to the actual difference in tension in the steel, measured in pounds. It is the amount of tension which must be transferred to the concrete in that unit-length of the beam. But the computation of the difference of moments at two sections that are only a unit-distance apart, is a comparatively tedious operation, which, fortunately, is unnecessary. Theoretical mechanics teaches us that the difference in the moment at two consecutive sections of the beam is measured by the total vertical shear in the beam at that point. The shear is very easily and readily computable; and therefore the required amount of tension to be transferred from the steel to the concrete can readily be computed. A numerical illustration may be given as follows: Suppose that we have a beam which, with its load, weighs 20,000 pounds, on a span of 20 feet. Using ultimate values, for which we multiply the loading by 4, we have an ultimate loading of 80,000 pounds. Therefore,
Mo = Wo l = 80,000 x 240 = 2,400, 000
Using the constants previously chosen for 1:3:5 concrete, and therefore utilizing Equation 23, we have this moment equal to 397 b d2. Therefore b d2 = 6,045.
If we assume 6 = 15 inches; d = 20.1 inches; then d - x = .86d = 17.3 inches. The area of steel equals:
A = .0084 b d = 2.53 square inches. We know from the laws of mechanics, that the moment diagram for a beam which is uniformly loaded is a parabola, and that the ordinate to this curve at a point one inch from the abutment will, in the above case, equal (119/120 )2 of the ordinate at the abutment. This ordinate is measured by the maximum moment at the center, multiplied by the factor (119/120)2 = 14,161 = .9834; therefore the actual moment
14,400 at a point one inch from the abutment = (1.00 - .9834) = .0166 of the moment at the center. But .0166 X 2,400,000 = 39,840.
But our ultimate loading being 80,000 pounds, we know that the shear at a point in the middle of this one-inch length equals the shear at the abutment, minus the load on this first 1/2 inch, which is 1/240 of 40,000 (or 167) pounds. The shear at this point is therefore 40,000 - 167 (or 39,833) pounds. This agrees with the above value 39,840, as closely as the decimals used in our calculations will permit.
The value of d - x is somewhat larger when the moment is very small than when it is at its ultimate value. But the difference is comparatively small, is on the safe side, and it need not make any material difference in our calculations. Therefore, dividing 39,840 by 17.3, we have 2,303 pounds as the difference in tension in the steel in the last inch at the abutment. Of course this does not literally mean the last inch in the length of the beam, since, if the net span were 20 feet, the actual length of the beam would be considerably greater. The area of the steel as computed above is 2.53 square inches. Assuming that this is furnished by five 3/4-inch square bars, the surfaces of these five bars per inch of length equals 15 square inches. Dividing 2,303 by 15, we have 153 pounds per square inch as the required adhesion between the steel and the concrete. While this is not greater than the adhesion usually found between concrete and steel, it is somewhat risky to depend on this; and therefore the bars are usually bent so that they run diagonally upward, and thus furnish a very great increase in the strength of the beam, which prevents the beam from failing at the ends. Tests have shown that beams which are reinforced by bars only running through the lower part of the beam without being turned up, or without using any stirrups, will usually fail at the ends, long before the transverse moment, which they possess at their center, has been fully developed.
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From. Equation 19 we have the formula that the resisting moment at any point in the beam equals the area of the steel, times the unit tensile stress in the steel, times the distance from the steel to the centroid of compression of the steel, which is the distance d - x. We may compute the moment in the beam at two points at a unit-distance apart. The area of the steel is the same in each equation, and d - x is substantially the same in each case; and therefore the difference of moment, divided by (d - x), will evidently equal the difference in the unit-stress in the steel, times the area of the steel. To express this in an equation, we may say, denoting the difference in the moment by d M, and the difference in the unit-stress in the steel by d s: dM = A x ds.. (d-x). But A X d s is evidently equal to the actual difference in tension in the steel, measured in pounds. It is the amount of tension which must be transferred to the concrete in that unit-length of the beam. But the computation of the difference of moments at two sections that are only a unit-distance apart, is a comparatively tedious operation, which, fortunately, is unnecessary. Theoretical mechanics teaches us that the difference in the moment at two consecutive sections of the beam is measured by the total vertical shear in the beam at that point. The shear is very easily and readily computable; and therefore the required amount of tension to be transferred from the steel to the concrete can readily be computed. A numerical illustration may be given as follows: Suppose that we have a beam which, with its load, weighs 20,000 pounds, on a span of 20 feet. Using ultimate values, for which we multiply the loading by 4, we have an ultimate loading of 80,000 pounds. Therefore,. Mo = Wo l = 80,000 x 240 = 2,400, 000. Using the constants previously chosen for 1:3:5 concrete, and therefore utilizing Equation 23, we have this moment equal to 397 b d2. Therefore b d2 = 6,045.. If we assume 6 = 15 inches; d = 20.1 inches; then d - x = .86d = 17.3 inches. The area of steel equals:. A = .0084 b d = 2.53 square inches. We know from the laws of mechanics, that the moment diagram for a beam which is uniformly loaded is a parabola, and that the ordinate to this curve at a point one inch from the abutment will, in the above case, equal (119/120 )2 of the ordinate at the abutment. This ordinate is measured by the maximum moment at the center, multiplied by the factor (119/120)2 = 14,161 = .9834; therefore the actual moment. 14,400 at a point one inch from the abutment = (1.00 - .9834) = .0166 of the moment at the center. But .0166 X 2,400,000 = 39,840.. But our ultimate loading being 80,000 pounds, we know that the shear at a point in the middle of this one-inch length equals the shear at the abutment, minus the load on this first 1/2 inch, which is 1/240 of 40,000 (or 167) pounds. The shear at this point is therefore 40,000 - 167 (or 39,833) pounds. This agrees with the above value 39,840, as closely as the decimals used in our calculations will permit.. The value of d - x is somewhat larger when the moment is very small than when it is at its ultimate value. But the difference is comparatively small, is on the safe side, and it need not make any material difference in our calculations. Therefore, dividing 39,840 by 17.3, we have 2,303 pounds as the difference in tension in the steel in the last inch at the abutment. Of course this does not literally mean the last inch in the length of the beam, since, if the net span were 20 feet, the actual length of the beam would be considerably greater. The area of the steel as computed above is 2.53 square inches. Assuming that this is furnished by five 3/4-inch square bars, the surfaces of these five bars per inch of length equals 15 square inches. Dividing 2,303 by 15, we have 153 pounds per square inch as the required adhesion between the steel and the concrete.
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While this is not greater than the adhesion usually found between concrete and steel, it is somewhat risky to depend on this; and therefore the bars are usually bent so that they run diagonally upward, and thus furnish a very great increase in the strength of the beam, which prevents the beam from failing at the ends. Tests have shown that beams which are reinforced by bars only running through the lower part of the beam without being turned up, or without using any stirrups, will usually fail at the ends, long before the transverse moment, which they possess at their center, has been fully developed.
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https://stats.stackexchange.com/questions/372895/how-parameters-formulated-for-simple-regression-model
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# How parameters formulated for Simple Regression Model
I am reading Simple Regression Model from this book, Section 6.5 (page 267 in downloaded pdf, 276 if viewed online).
The author starts with below equation for a simple linear regression model,
$$Y_i = \alpha_1 + \beta x_i + \varepsilon_i$$
And then after few lines, he lets for conveience that, $$\alpha_1 = \alpha - \beta\overline{x}$$ so that,
$$Y_i = \alpha + \beta(x_i - \overline{x}) + \varepsilon_i$$
where $$\overline{x} = \dfrac{1}{n}\sum\limits_{i=1}^nx_i$$
My questions:
1. It is not convincing to bring in $$\overline{x}$$ just for convenience sake in the equation. Can any one please explain the logic behind bring that in the equation?
2. After above equation, the author says, $$Y_i$$ is equal to a nonrandom quantity, $$\alpha + \beta(x_i - \overline{x})$$, plus a mean zero normal random variable $$\varepsilon_i$$. Does that mean, $$\alpha + \beta(x_i - \overline{x})$$ has no randomness involved in that?
Kindly help.
1. $$\alpha_1$$s in two equations are different. Let $$\alpha_2$$ be the $$\alpha$$ in the second equation, then $$\alpha_1 = \alpha_2 + \beta \bar x$$
At the time that the computer was not popular or had no computer, the line was fit by using calculators. Bringing in $$\bar x$$ is really simplified the computation.
1. From the first equation, $$\epsilon$$ is the only random component. So source of randomness of $$Y$$ is $$\epsilon$$, the other parts $$\alpha + \beta x$$ are known or unknown constant.
• I just corrected $\alpha_1$ to $\alpha$ in 2nd equation. Still the reason is not convincing that it simplified the computation. Can you kindly elaborate further? How could $\overline{x}$ suddenly enter the equation without an associated mathematical logic. Oct 20, 2018 at 18:32
• Let $z_i=x_i-\bar x$, then (1) $\sum z_i = 0$ vs calculating $\sum x_i$, (2) $\sum z_i^2$ is easier easier than $\sum x_i^2$, and (3) $\sum z_iY_i$ is easier easier than $\sum x_iY_i$. introducing $\bar x$ into equation does not change anything in equation, similar to $+ a - a$ , which we used to proof something in math. Oct 20, 2018 at 18:43
• $Y_i = \alpha_1 + \beta x_i + \varepsilon_i$ ==> $Y_i = \alpha_1 + \beta x_i + \varepsilon_i - \beta \bar x + \beta \bar x$ ==> $Y_i = (\alpha_1 +\beta \bar x) + \beta (x_i - \bar x) + \varepsilon_i$ ==> $Y_i = \alpha + \beta (x_i - \bar x) + \varepsilon_i$ Oct 20, 2018 at 18:56
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# How parameters formulated for Simple Regression Model. I am reading Simple Regression Model from this book, Section 6.5 (page 267 in downloaded pdf, 276 if viewed online).. The author starts with below equation for a simple linear regression model,. $$Y_i = \alpha_1 + \beta x_i + \varepsilon_i$$. And then after few lines, he lets for conveience that, $$\alpha_1 = \alpha - \beta\overline{x}$$ so that,. $$Y_i = \alpha + \beta(x_i - \overline{x}) + \varepsilon_i$$. where $$\overline{x} = \dfrac{1}{n}\sum\limits_{i=1}^nx_i$$. My questions:. 1. It is not convincing to bring in $$\overline{x}$$ just for convenience sake in the equation. Can any one please explain the logic behind bring that in the equation?. 2. After above equation, the author says, $$Y_i$$ is equal to a nonrandom quantity, $$\alpha + \beta(x_i - \overline{x})$$, plus a mean zero normal random variable $$\varepsilon_i$$. Does that mean, $$\alpha + \beta(x_i - \overline{x})$$ has no randomness involved in that?. Kindly help.. 1. $$\alpha_1$$s in two equations are different. Let $$\alpha_2$$ be the $$\alpha$$ in the second equation, then $$\alpha_1 = \alpha_2 + \beta \bar x$$. At the time that the computer was not popular or had no computer, the line was fit by using calculators. Bringing in $$\bar x$$ is really simplified the computation.. 1. From the first equation, $$\epsilon$$ is the only random component. So source of randomness of $$Y$$ is $$\epsilon$$, the other parts $$\alpha + \beta x$$ are known or unknown constant.. • I just corrected $\alpha_1$ to $\alpha$ in 2nd equation. Still the reason is not convincing that it simplified the computation. Can you kindly elaborate further? How could $\overline{x}$ suddenly enter the equation without an associated mathematical logic. Oct 20, 2018 at 18:32. • Let $z_i=x_i-\bar x$, then (1) $\sum z_i = 0$ vs calculating $\sum x_i$, (2) $\sum z_i^2$ is easier easier than $\sum x_i^2$, and (3) $\sum z_iY_i$ is easier easier than $\sum x_iY_i$. introducing $\bar x$ into equation does not change anything in equation, similar to $+ a - a$ , which we used to proof something in math.
|
Oct 20, 2018 at 18:43. • $Y_i = \alpha_1 + \beta x_i + \varepsilon_i$ ==> $Y_i = \alpha_1 + \beta x_i + \varepsilon_i - \beta \bar x + \beta \bar x$ ==> $Y_i = (\alpha_1 +\beta \bar x) + \beta (x_i - \bar x) + \varepsilon_i$ ==> $Y_i = \alpha + \beta (x_i - \bar x) + \varepsilon_i$ Oct 20, 2018 at 18:56.
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http://gedmath.blogspot.com/2014/
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## Wednesday, October 08, 2014
### Finding the Least Common Denominator of Three Fractions
The least common denominator of two or more fractions is the smallest number that can be divided evenly by each of the fractions' denominators. You can determine the LCM (least common multiple) by finding multiples of the denominators of the fractions.
Find the least common denominator of the following fractions: 5/12, 7/36, and 3/8.
8, 16, 24, 36
12, 24, 36
36
The least common denominator is 36.
## Tuesday, October 07, 2014
### Least Common Multiple
Find the least common denominator of 6, 8, 12.
6, 12, 18, 24
8, 16, 24
12, 24
The least common multiple is 24.
## Monday, October 06, 2014
### Help With GED Math Problems: Finding Lowest Common Denominator for Fractions
Building the LCD or lowest common denominators for two or more fractions can be challenging. But it is an important skill for knowing how to add and subtract fractions and one that anyone studying their GED math test will need to know.
First step: Take each denominator and factor to product of prime numbers.
Second step: Build the lowest common denominator by using each factor with the greatest exponent.
What is the lowest common denominator for the following fractions: 7/12, 7/15, 19/30? Use the product of prime factor method.
12 = 2 x 2 x 3 or 2^2 x 3
15 = 3 x 5
30 = 2 x 3 x 5
Build the lowest common denominator by using each factor (i.e. 2^2) with the greatest exponents.
If I were demonstrating the concept of building lowest common denominators to students, it would go something like this, " Let's start with the denominator twelve. The denominator 12 needs at least two twos and a three. The denominator fifteen needs a three, but because we have one from the twelve... we do not need to write another one. However, the denominator twelve needs a five, so we need to add a five. The denominator thirty needs a two... which we have so we do not need to add one. It also needs a three and a five, but because we already have both, again we do not need to add. We have now build our LCD and all we need to do is multiply the factors together. So 2 x 2 x 3 x 5 = 60. The LCD of 12, 15, and 30 is 60.
LCD = 2 x 2 x 3 x 5 = 60
## Friday, October 03, 2014
### Lowest Common Denominator
Find the lowest common denominator for the following fractions: 1/2, 1/4, 1/5
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
4, 8, 12, 16, 20
5, 10, 15, 20
Because 20 is the first common multiple of 2, 4, and 5..... it is the lowest common denominator or LCD.
## Wednesday, July 30, 2014
### GED Math Test Prep: Area of Rectangle
GED Skill: Area of rectangles
You have decided to put carpet in your 10ft by 15 ft. living room. What is the area of carpet needed?
Answer: 10ft x 15ft = 150 cubic feet
### GED Math Test Prep: Simplify the equation 3x + 7y - 2z + 3 - 6x - 5z +15
Simplify the following equation.
3x + 7y - 2z + 3 - 6x - 5z +15
Step 1: Using the associative property, rearrange the terms of the equation so that "like" terms are next to each other.
3x - 6x - 2z - 5z + 7y + 3 + 15
Step 2: Combine like terms.
-3x - 7z + 7y + 18
## Monday, May 12, 2014
### Using the Product Rule with Exponents
When you multiply constants (variables) that have the same base, you add the exponents... but keep the base unchanged.
For example:
x^2c · x^3 = x^(2+3) = x^5
(x · x) (x · x · x) = x^5
"X" squared times "X" cubed equals "X" to the fifth power.
Try a few more.
1) p^5 · p^4 =
2) 2t^2 · 3t^4
3) r^2 · 2^3 · r^5
4) 3x^2 · 2x^5 · x^4
5) (p^2)(3p^4)(3p^2)
1) p^9
2) 6t^6
3) 2r^10
4) 6r^11
5) 9p^8
## Thursday, May 08, 2014
### Simplify and Solve Using the Addition Principal of Equality
4 ( 8 - 15) + (-10) = x - 7
4 ( 8 - 15) + (-10) = x - 7
32 - 60 + (-10) = x - 7
-28 + (-10) = x - 7
-38 = x - 7
-38 + 7 = x -7 + 7
-31 = x + 0
-31 = x
## Wednesday, May 07, 2014
### Solving Equations Using the Addition Principle of Equality
Can you find the error in the following problem?
5² + (4 - 8) = x + 15
25 + 4 = x + 15
29 = x + 15
29 + (-15) = x + 15 + (-15)
14 = x + 0
14 = x
## Tuesday, May 06, 2014
### Practice Solving Simple Equations Using the Addition Property of Equality
It is important to practice the addition property of equality. See below and solve five simple equations using the addition property of equality.
Practice Problem #1
x - 11 = 41
Practice Problem #2
x - 17 = -35
Practice Problem #3
84 = 40 + x
Practice Problem #4
45 = -15 + x
Practice Problem #5
-21 = -52 + x
Practice Problem #1
x - 11 = 41
x - 11 + 11 = 41 + 11
x + 0 = 52
x = 52
check
52 - 11 = 41
41 = 41
Practice Problem #2
x - 17 = -35
x - 17 + 17 = -35 + 17
x + 0 = -18
x = -18
check
-18 - 17 = -35
-35 = -35
Practice Problem #3
84 = 40 + x
84 + ( - 40) = 40 + (-40) + x
44 = 0 + x
44 = x
check
84 = 40 + 44
84 = 84
Practice Problem #4
45 = -15 + x
45 + 15 = -15 + 15 + x
60 = 0 + x
60 = x
check
45 = -15 + 60
45 = 45
Practice Problem #5
-21 = -52 + x
-21 + 52 = -52 + 52 + x
31 = 0 + x
31 = x
check
-21 = -52 + 31
-21 = -21
## Monday, May 05, 2014
### Solving Equations Using the Addition Property of Equality
The addition principle of equality states that if a = b, then a + c = b + c.
When you solve equations using this addition principle of equality, you need to use the additive inverse property. In other words, you must add the same number to both sides of an equation.
Example #1:
x - 5 = 10
x - 5 + 5 = 10 + 5 We add the opposite of (-5) to both sides of the equation.
x + 0 = 15 We simplify -5 + 5 = 0.
x = 15 The solution is x = 15
To check the answer, simply substitute 15 in for x, in the original equation and solve.
15 - 5 = 10
10 = 10
Example #2:
x + 12 = -5
x + 12 + (- 12) = -5 + (- 12) We add the opposite of (+12) to both sides of the equation.
x + 0 = -17 We simplify +12 - 12 = 0.
x = -17 The solution is x = -17
(-17) + 12 = -5
-5 = -5
## Friday, April 11, 2014
### Practice Percent Word Problem
A car which is normally priced at \$25,437 is marked down 10%. How much would Karen save if she purchased the car at the sale price?
(Spanish translation coming soon...)
## Tuesday, March 11, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. Twenty-one more than a number is 51. What is the number?
Veinte y uno más que el número es 51. ¿Cuál es el número?
2. Thirty-seven less than a number is 45. Find the number.
Treinta y siete menos que el número es 45. Encuentre el número.
1. 30
2. 82
## Monday, March 10, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 50 is 73. Find the number.
La suma del número y 50 es 73. Encuentre el número.
2. Thirty-one more than a number is 69. What is the number?
Treinta y uno más que el número es 69. ¿Cuál es el número?
3. A number decreased by 46 is 20. Find the number.
El número que está reducido por 46 es 20. Encuentre el número.
1. 23
2. 38
3. 66
## Friday, March 07, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 28 is 74. Find the number.
La suma del número y 28 es 74. Encuentre el número.
2. Thirty-nine more than a number is 72. What is the number?
Treinta y nueve más que el número es 72. ¿Cuál es el número?
3. Eighteen less than a number is 48. Find the number.
Dieciocho menos que el número es 48. Encuentre el número.
1. 46
2. 33
3. 66
## Thursday, March 06, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. A number increased by 21 is 52. Find the number.
El número que está aumentado por 21 es 52. Encuentre el número.
2. Twenty-five more than a number is 68. What is the number?
Veinte y cinco más que el número es 68. ¿Cuál es el número?
3. Forty-two more than a number is 58. What is the number?
Cuarenta y dos más que el número es 58. ¿Cuál es el número?
1. 31
2. 43
3. 16
## Wednesday, March 05, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. Twenty more than a number is 42. What is the number?
Veinte más que el número es 42. ¿Cuál es el número?
2. Forty-three more than a number is 85. What is the number?
Cuarenta y tres más que el número es 85. ¿Cuál es el número?
3. Twenty-two more than a number is 62. What is the number?
Veinte y dos más que el número es 62. ¿Cuál es el número?
1. 22
2. 42
3. 40
## Tuesday, March 04, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. The sum of a number and 26 is 42. Find the number.
La suma del número y 26 es 42. Encuentre el número.
2. Thirty more than a number is 51. What is the number?
Treinta más que el número es 51. ¿Cuál es el número?
3. Fifteen more than a number is 47. What is the number?
Quince más que el número es 47. ¿Cuál es el número?
1. 16
2. 21
3. 32
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. One-half of a number is 13. Find the number.
Una media de un número es 13. Encuentre el número.
2. A number decreased by 29 is 39. Find the number.
Un número que está reducido por 29 es 39. Encuentre el número.
3. The sum of a number and 39 is 56. Find the number.
La suma del número y 39 es 56. Encuentre el número.
1. 26
2. 68
317
## Tuesday, January 28, 2014
### Practice Translating Algebraic Words Into Expressions: (Spanish & English)
1. A number increased by eight is 14. Find the number.
El número que aumenta por ocho es 14. Encuentre el número.
2. Three less than a number is 2. Find the number.
Tres menos que el número es dos. Encuentre el número.
1. 6
2. 5
## Monday, January 27, 2014
### Translating Words Into Algebraic Expressions Examples: (Spanish & English)
1. Six less than a number is 9. Find the number.
Seis menos que el número es nueve. Encuentre el número.
2. Ten less than a number is 9. Find the number.
Diez menos que el número es nueve. Encuentre el número.
3. A number increased by seven is 12. Find the number.
El número que aumenta por siete es 12. Encuentre el número.
1. 15
2. 19
3. 5
## Friday, January 24, 2014
### Easy Tanslating Algebra Word Problems: (Spanish & English)
1. Seven more than a number is 11. What is the number?
Siete más que el número es 11. Encuentre el número.
2. The sum of a number and six is 16. Find the number.
La Suma del número y seis es 16. Encuentre el número.
3. A number diminished by 9 is 3. Find the number.
El número que reduce por nueve es tres. Encuentre el número.
1. 4
2. 10
3. 12
## Thursday, January 23, 2014
### Translating Words into Algebraic Expressions Simple: (Spanish & English)
1. A number diminished by 2 is 7. Find the number.
El número que reduce por dos es siete. Encuentre el número.
2. A number decreased by 7 is 8. Find the number.
El número que reduce por siete es ocho. Encuentre el número.
3. A number increased by three is 13. Find the number.
El número que aumenta por tres es 13. Encuentre el número.
1. 9
2. 15
3. 10
## Wednesday, January 22, 2014
### Translating Simple Number Word Problems: (Spanish & English)
1. Six less than a number is 5. Find the number.
Seis menos que el número es cinco. Encuentre el número.
2. Six less than a number is 7. Find the number.
Seis menos que el número es siete. Encuentre el número.
3. The sum of a number and three is 11. Find the number.
La suma del número y tres es 11. Encuentre el número.
1. 11
2. 13
3. 8
## Tuesday, January 21, 2014
### Translating Word Problems Simple: (Spanish & English)
1. One-third of a number is 1. Find the number.
Un tercer del número es uno. Encuentre el número.
2. A number increased by five is 13. Find the number.
El número que aumenta por cinco es 13. Encuentre el número.
3. One-third of a number is 2. Find the number.
Un tercer del número es dos. Encuentre el número.
1. 3
2. 8
3. 6
## Monday, January 20, 2014
### Translating Simple Algebra Word Problems: (Spanish & English)
1. Two more than a number is 8. What is the number?
Dos más que el número es ocho. ¿Cuál es el número?
2. Three more than a number is 5. What is the number?
Tres más que el número es cinco. ¿Cuál es el número?
3. A number decreased by 2 is 5. Find the number.
El número que reduce por dos es cinco. Encuentre el número.
1. 10
2. 8
3. 7
## Tuesday, January 14, 2014
### Algebra Word Problem: Setting up Problem (Spanish & English)
A total of r players came to a basketball practice. The coach divides them into four groups of t players each, but two players are left over. Which expression shows the relationship between the number of players out for basketball and the number of players in each group?
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## Wednesday, October 08, 2014. ### Finding the Least Common Denominator of Three Fractions. The least common denominator of two or more fractions is the smallest number that can be divided evenly by each of the fractions' denominators. You can determine the LCM (least common multiple) by finding multiples of the denominators of the fractions.. Find the least common denominator of the following fractions: 5/12, 7/36, and 3/8.. 8, 16, 24, 36. 12, 24, 36. 36. The least common denominator is 36.. ## Tuesday, October 07, 2014. ### Least Common Multiple. Find the least common denominator of 6, 8, 12.. 6, 12, 18, 24. 8, 16, 24. 12, 24. The least common multiple is 24.. ## Monday, October 06, 2014. ### Help With GED Math Problems: Finding Lowest Common Denominator for Fractions. Building the LCD or lowest common denominators for two or more fractions can be challenging. But it is an important skill for knowing how to add and subtract fractions and one that anyone studying their GED math test will need to know.. First step: Take each denominator and factor to product of prime numbers.. Second step: Build the lowest common denominator by using each factor with the greatest exponent.. What is the lowest common denominator for the following fractions: 7/12, 7/15, 19/30? Use the product of prime factor method.. 12 = 2 x 2 x 3 or 2^2 x 3. 15 = 3 x 5. 30 = 2 x 3 x 5. Build the lowest common denominator by using each factor (i.e. 2^2) with the greatest exponents.. If I were demonstrating the concept of building lowest common denominators to students, it would go something like this, " Let's start with the denominator twelve. The denominator 12 needs at least two twos and a three. The denominator fifteen needs a three, but because we have one from the twelve... we do not need to write another one. However, the denominator twelve needs a five, so we need to add a five. The denominator thirty needs a two... which we have so we do not need to add one. It also needs a three and a five, but because we already have both, again we do not need to add. We have now build our LCD and all we need to do is multiply the factors together. So 2 x 2 x 3 x 5 = 60. The LCD of 12, 15, and 30 is 60.. LCD = 2 x 2 x 3 x 5 = 60. ## Friday, October 03, 2014. ### Lowest Common Denominator. Find the lowest common denominator for the following fractions: 1/2, 1/4, 1/5. 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. 4, 8, 12, 16, 20. 5, 10, 15, 20. Because 20 is the first common multiple of 2, 4, and 5..... it is the lowest common denominator or LCD.. ## Wednesday, July 30, 2014. ### GED Math Test Prep: Area of Rectangle. GED Skill: Area of rectangles. You have decided to put carpet in your 10ft by 15 ft. living room. What is the area of carpet needed?. Answer: 10ft x 15ft = 150 cubic feet. ### GED Math Test Prep: Simplify the equation 3x + 7y - 2z + 3 - 6x - 5z +15. Simplify the following equation.. 3x + 7y - 2z + 3 - 6x - 5z +15. Step 1: Using the associative property, rearrange the terms of the equation so that "like" terms are next to each other.. 3x - 6x - 2z - 5z + 7y + 3 + 15. Step 2: Combine like terms.. -3x - 7z + 7y + 18. ## Monday, May 12, 2014. ### Using the Product Rule with Exponents. When you multiply constants (variables) that have the same base, you add the exponents... but keep the base unchanged.. For example:. x^2c · x^3 = x^(2+3) = x^5. (x · x) (x · x · x) = x^5. "X" squared times "X" cubed equals "X" to the fifth power.. Try a few more.. 1) p^5 · p^4 =. 2) 2t^2 · 3t^4. 3) r^2 · 2^3 · r^5. 4) 3x^2 · 2x^5 · x^4. 5) (p^2)(3p^4)(3p^2). 1) p^9. 2) 6t^6. 3) 2r^10. 4) 6r^11. 5) 9p^8. ## Thursday, May 08, 2014. ### Simplify and Solve Using the Addition Principal of Equality. 4 ( 8 - 15) + (-10) = x - 7. 4 ( 8 - 15) + (-10) = x - 7. 32 - 60 + (-10) = x - 7. -28 + (-10) = x - 7. -38 = x - 7. -38 + 7 = x -7 + 7. -31 = x + 0. -31 = x. ## Wednesday, May 07, 2014. ### Solving Equations Using the Addition Principle of Equality. Can you find the error in the following problem?. 5² + (4 - 8) = x + 15. 25 + 4 = x + 15. 29 = x + 15. 29 + (-15) = x + 15 + (-15). 14 = x + 0. 14 = x. ## Tuesday, May 06, 2014. ### Practice Solving Simple Equations Using the Addition Property of Equality. It is important to practice the addition property of equality. See below and solve five simple equations using the addition property of equality.. Practice Problem #1. x - 11 = 41. Practice Problem #2. x - 17 = -35. Practice Problem #3. 84 = 40 + x. Practice Problem #4. 45 = -15 + x. Practice Problem #5. -21 = -52 + x. Practice Problem #1. x - 11 = 41. x - 11 + 11 = 41 + 11. x + 0 = 52. x = 52. check. 52 - 11 = 41. 41 = 41. Practice Problem #2. x - 17 = -35. x - 17 + 17 = -35 + 17. x + 0 = -18. x = -18. check. -18 - 17 = -35. -35 = -35. Practice Problem #3. 84 = 40 + x. 84 + ( - 40) = 40 + (-40) + x. 44 = 0 + x. 44 = x. check. 84 = 40 + 44. 84 = 84. Practice Problem #4. 45 = -15 + x. 45 + 15 = -15 + 15 + x. 60 = 0 + x. 60 = x. check. 45 = -15 + 60. 45 = 45. Practice Problem #5. -21 = -52 + x. -21 + 52 = -52 + 52 + x. 31 = 0 + x. 31 = x. check. -21 = -52 + 31. -21 = -21. ## Monday, May 05, 2014. ### Solving Equations Using the Addition Property of Equality. The addition principle of equality states that if a = b, then a + c = b + c.. When you solve equations using this addition principle of equality, you need to use the additive inverse property. In other words, you must add the same number to both sides of an equation.. Example #1:. x - 5 = 10. x - 5 + 5 = 10 + 5 We add the opposite of (-5) to both sides of the equation.. x + 0 = 15 We simplify -5 + 5 = 0.. x = 15 The solution is x = 15. To check the answer, simply substitute 15 in for x, in the original equation and solve.. 15 - 5 = 10. 10 = 10. Example #2:. x + 12 = -5. x + 12 + (- 12) = -5 + (- 12) We add the opposite of (+12) to both sides of the equation.. x + 0 = -17 We simplify +12 - 12 = 0.. x = -17 The solution is x = -17. (-17) + 12 = -5. -5 = -5. ## Friday, April 11, 2014. ### Practice Percent Word Problem. A car which is normally priced at \$25,437 is marked down 10%. How much would Karen save if she purchased the car at the sale price?. (Spanish translation coming soon...). ## Tuesday, March 11, 2014. ### Practice Translating Algebraic Words Into Expressions: (Spanish & English). 1. Twenty-one more than a number is 51. What is the number?. Veinte y uno más que el número es 51. ¿Cuál es el número?. 2. Thirty-seven less than a number is 45. Find the number.. Treinta y siete menos que el número es 45. Encuentre el número.. 1. 30. 2. 82. ## Monday, March 10, 2014. ### Practice Translating Algebraic Words Into Expressions: (Spanish & English). 1. The sum of a number and 50 is 73. Find the number.. La suma del número y 50 es 73. Encuentre el número.. 2. Thirty-one more than a number is 69. What is the number?. Treinta y uno más que el número es 69. ¿Cuál es el número?. 3. A number decreased by 46 is 20. Find the number.. El número que está reducido por 46 es 20. Encuentre el número.. 1. 23. 2. 38. 3. 66. ## Friday, March 07, 2014. ### Practice Translating Algebraic Words Into Expressions: (Spanish & English). 1. The sum of a number and 28 is 74. Find the number.. La suma del número y 28 es 74. Encuentre el número.. 2. Thirty-nine more than a number is 72. What is the number?. Treinta y nueve más que el número es 72. ¿Cuál es el número?. 3. Eighteen less than a number is 48. Find the number.. Dieciocho menos que el número es 48. Encuentre el número.. 1. 46. 2. 33. 3. 66. ## Thursday, March 06, 2014. ### Practice Translating Algebraic Words Into Expressions: (Spanish & English). 1. A number increased by 21 is 52. Find the number.. El número que está aumentado por 21 es 52. Encuentre el número.. 2. Twenty-five more than a number is 68. What is the number?. Veinte y cinco más que el número es 68. ¿Cuál es el número?. 3. Forty-two more than a number is 58. What is the number?. Cuarenta y dos más que el número es 58. ¿Cuál es el número?. 1. 31. 2. 43. 3. 16. ## Wednesday, March 05, 2014. ### Practice Translating Algebraic Words Into Expressions: (Spanish & English). 1. Twenty more than a number is 42. What is the number?. Veinte más que el número es 42. ¿Cuál es el número?. 2. Forty-three more than a number is 85. What is the number?. Cuarenta y tres más que el número es 85. ¿Cuál es el número?. 3. Twenty-two more than a number is 62. What is the number?. Veinte y dos más que el número es 62. ¿Cuál es el número?. 1. 22. 2. 42. 3. 40. ## Tuesday, March 04, 2014. ### Practice Translating Algebraic Words Into Expressions: (Spanish & English). 1. The sum of a number and 26 is 42. Find the number.. La suma del número y 26 es 42. Encuentre el número.. 2. Thirty more than a number is 51. What is the number?. Treinta más que el número es 51. ¿Cuál es el número?. 3. Fifteen more than a number is 47. What is the number?. Quince más que el número es 47. ¿Cuál es el número?. 1. 16. 2. 21. 3. 32. ### Practice Translating Algebraic Words Into Expressions: (Spanish & English). 1. One-half of a number is 13. Find the number.. Una media de un número es 13. Encuentre el número.. 2. A number decreased by 29 is 39. Find the number.. Un número que está reducido por 29 es 39. Encuentre el número.. 3. The sum of a number and 39 is 56. Find the number.. La suma del número y 39 es 56. Encuentre el número.. 1. 26. 2. 68. 317. ## Tuesday, January 28, 2014. ### Practice Translating Algebraic Words Into Expressions: (Spanish & English). 1. A number increased by eight is 14. Find the number.. El número que aumenta por ocho es 14. Encuentre el número.. 2. Three less than a number is 2. Find the number.. Tres menos que el número es dos. Encuentre el número.. 1. 6. 2. 5. ## Monday, January 27, 2014. ### Translating Words Into Algebraic Expressions Examples: (Spanish & English). 1. Six less than a number is 9. Find the number.. Seis menos que el número es nueve. Encuentre el número.. 2. Ten less than a number is 9. Find the number.. Diez menos que el número es nueve. Encuentre el número.. 3. A number increased by seven is 12. Find the number.. El número que aumenta por siete es 12. Encuentre el número.. 1. 15. 2. 19. 3. 5. ## Friday, January 24, 2014. ### Easy Tanslating Algebra Word Problems: (Spanish & English). 1. Seven more than a number is 11. What is the number?. Siete más que el número es 11. Encuentre el número.. 2. The sum of a number and six is 16. Find the number.. La Suma del número y seis es 16. Encuentre el número.. 3. A number diminished by 9 is 3. Find the number.. El número que reduce por nueve es tres. Encuentre el número.. 1. 4. 2. 10. 3. 12. ## Thursday, January 23, 2014. ### Translating Words into Algebraic Expressions Simple: (Spanish & English). 1. A number diminished by 2 is 7. Find the number.. El número que reduce por dos es siete. Encuentre el número.. 2. A number decreased by 7 is 8. Find the number.. El número que reduce por siete es ocho. Encuentre el número.. 3. A number increased by three is 13. Find the number.. El número que aumenta por tres es 13. Encuentre el número.. 1. 9. 2. 15. 3. 10. ## Wednesday, January 22, 2014. ### Translating Simple Number Word Problems: (Spanish & English). 1. Six less than a number is 5. Find the number.. Seis menos que el número es cinco. Encuentre el número.. 2. Six less than a number is 7. Find the number.. Seis menos que el número es siete. Encuentre el número.. 3. The sum of a number and three is 11. Find the number.. La suma del número y tres es 11. Encuentre el número.. 1. 11. 2. 13. 3. 8. ## Tuesday, January 21, 2014. ### Translating Word Problems Simple: (Spanish & English). 1. One-third of a number is 1. Find the number.. Un tercer del número es uno. Encuentre el número.. 2. A number increased by five is 13. Find the number.. El número que aumenta por cinco es 13. Encuentre el número.. 3. One-third of a number is 2. Find the number.. Un tercer del número es dos. Encuentre el número.. 1. 3. 2. 8. 3. 6. ## Monday, January 20, 2014. ### Translating Simple Algebra Word Problems: (Spanish & English). 1. Two more than a number is 8. What is the number?. Dos más que el número es ocho. ¿Cuál es el número?. 2. Three more than a number is 5. What is the number?. Tres más que el número es cinco. ¿Cuál es el número?. 3. A number decreased by 2 is 5. Find the number.. El número que reduce por dos es cinco. Encuentre el número.. 1. 10. 2. 8. 3. 7. ## Tuesday, January 14, 2014.
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### Algebra Word Problem: Setting up Problem (Spanish & English). A total of r players came to a basketball practice. The coach divides them into four groups of t players each, but two players are left over. Which expression shows the relationship between the number of players out for basketball and the number of players in each group?.
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Prime number inequality
Hey I recently got interested in number theory and proved the following inequality:
\begin{align} x^m-y^m > p_n^2 + 4p_n + 3 > 1\ \text{(corrected again)} \end{align}
where $x-y\neq1$ and m in an integer >1
and
\begin{align} \gcd(x,y)&=1 \\ xy&= 2*3*5*..*p_n \end{align}
So my question is ... Does this formula already exist? And is it useful or slightly interesting?
• Presumably you also assume $x>y$? Sep 16, 2013 at 12:20
• I thought that was self implied as x^m>y^m+1 ... so x^m>y^m hence, x>y Sep 16, 2013 at 12:31
• No, this is not self implied! You write $x-y \ne 1$. Sep 16, 2013 at 12:39
• well I say g.c.d (x,y) =1 ... Hence x and y must be integers Sep 16, 2013 at 12:43
• @Jyrki: Your $A(n)$ is IMO the primorial function $\#p_n$ and the asymptotic growth rate is $\#p_n \approx e^{\#p_n}$, see mathworld.wolfram.com/Primorial.html formula (4). Sep 16, 2013 at 12:51
The inequality deals with the primorial function $$p_n\#=2\cdot3\cdot5\cdots p_n=\prod_{p\le p_n,\ p\in \mathbb{P}}p,$$ where $p_n$ is the $n$th prime. Asymptotically we have the result that $$\lim_{n\to\infty}\frac{\ln p_n\#}{p_n}=1.$$ Early on the primorials are bit smaller though. For example $\ln(59\#)\approx49$.
Consider the following problem. Assume that $xy=p_n\#$ and that $x-y\ge3$ (in OP $x,y$ were constrained to be integers of opposite parity such that $x-y>1$ implying that $x-y\ge 3$). Therefore $$x^m-y^m\ge x^2-y^2=(x-y)(x+y)\ge3(x+y).$$ Here by the AM-GM inequality $x+y\ge2\sqrt{xy}=2\sqrt{p_n\#}.$ Therefore asymptotically we get a lower bound $$x^m-y^m\ge 6\sqrt{p_n\#}\ge6e^{\frac n2(1+o(1))}.$$
Asymptotically we also have have $p_n\approx n\ln n.$ This suggests that $$\frac{\ln(x^m-y^m)}{\ln p_n}\ge \frac n{2\ln n} K(n),$$ where $K(n)$ is some correction factor (bounded away from zero) that I won't calculate.
Your result says (using only the main term $p_n^2$) that $$\frac{\ln(x^m-y^m)}{\ln p_n}\ge 2.$$ So asymptotically it is weaker. But it would not be fair to call your result trivial because of this. I'm not a number theorist, but I have seen simpler estimates being derived in many number theory books, and in addition to being fun, they pave the road to stronger results.
Please share details of your argument with us, so that we can comment and give you other kind of feedback!
• Asymptotically it is, of course, irrelevant whether I require $x-y\ge1$ or $x-y\ge3$ here. Sep 16, 2013 at 13:36
• this is gonna b a long story Sep 16, 2013 at 13:41
• @Anant: I think you can answer your own question also. Then you won't be constrained by the 500 character limit. Sep 16, 2013 at 13:49
• Sorry about the bad mistakes in the definitions. They were a result of earlier editing, where I tried to avoid using subscripts on primes. Sep 20, 2013 at 4:16
• Does $\ln p_n \# < p_n$ always hold? If so, where can a proof be found? Apr 9, 2015 at 5:44
Modulo small values of $n,$ you do not need to assume that $x-y\ne 1$ as well as asymptotic on the primorial function. Instead, you can get away with Bertrand's postulate stating that $p_{n}<2p_{n-1}$ or $p_{n-1}\ge p_n/2.$ Indeed, if $m\ge 2,$ then $$x^m-y^m=(x-y)(x^{m-1}+...+y^{m-1})\ge (x-y)(x+y)\ge x+y\ge 2\sqrt{xy}.$$ Note, $$2\sqrt{xy}=2\sqrt{p_1...p_{n-3}p_{n-2}p_{n-1}p_n}$$ and using the fact that $p_{n-2}\ge p_n/4,$ $p_{n-3}\ge p_n/8$ and $p_{n-1}\ge p_n/2$ we can estimate $2\sqrt{xy}\ge 2\sqrt{p_1p_2\cdot...\frac{p_n^4}{64}}=\frac{p_n^2}{8}\sqrt{p_1p_2...}.$ So we are left to check the result for small values of $p_n.$
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Prime number inequality. Hey I recently got interested in number theory and proved the following inequality:. \begin{align} x^m-y^m > p_n^2 + 4p_n + 3 > 1\ \text{(corrected again)} \end{align}. where $x-y\neq1$ and m in an integer >1. and. \begin{align} \gcd(x,y)&=1 \\ xy&= 2*3*5*..*p_n \end{align}. So my question is ... Does this formula already exist? And is it useful or slightly interesting?. • Presumably you also assume $x>y$? Sep 16, 2013 at 12:20. • I thought that was self implied as x^m>y^m+1 ... so x^m>y^m hence, x>y Sep 16, 2013 at 12:31. • No, this is not self implied! You write $x-y \ne 1$. Sep 16, 2013 at 12:39. • well I say g.c.d (x,y) =1 ... Hence x and y must be integers Sep 16, 2013 at 12:43. • @Jyrki: Your $A(n)$ is IMO the primorial function $\#p_n$ and the asymptotic growth rate is $\#p_n \approx e^{\#p_n}$, see mathworld.wolfram.com/Primorial.html formula (4). Sep 16, 2013 at 12:51. The inequality deals with the primorial function $$p_n\#=2\cdot3\cdot5\cdots p_n=\prod_{p\le p_n,\ p\in \mathbb{P}}p,$$ where $p_n$ is the $n$th prime. Asymptotically we have the result that $$\lim_{n\to\infty}\frac{\ln p_n\#}{p_n}=1.$$ Early on the primorials are bit smaller though. For example $\ln(59\#)\approx49$.. Consider the following problem. Assume that $xy=p_n\#$ and that $x-y\ge3$ (in OP $x,y$ were constrained to be integers of opposite parity such that $x-y>1$ implying that $x-y\ge 3$). Therefore $$x^m-y^m\ge x^2-y^2=(x-y)(x+y)\ge3(x+y).$$ Here by the AM-GM inequality $x+y\ge2\sqrt{xy}=2\sqrt{p_n\#}.$ Therefore asymptotically we get a lower bound $$x^m-y^m\ge 6\sqrt{p_n\#}\ge6e^{\frac n2(1+o(1))}.$$. Asymptotically we also have have $p_n\approx n\ln n.$ This suggests that $$\frac{\ln(x^m-y^m)}{\ln p_n}\ge \frac n{2\ln n} K(n),$$ where $K(n)$ is some correction factor (bounded away from zero) that I won't calculate.. Your result says (using only the main term $p_n^2$) that $$\frac{\ln(x^m-y^m)}{\ln p_n}\ge 2.$$ So asymptotically it is weaker. But it would not be fair to call your result trivial because of this. I'm not a number theorist, but I have seen simpler estimates being derived in many number theory books, and in addition to being fun, they pave the road to stronger results.. Please share details of your argument with us, so that we can comment and give you other kind of feedback!. • Asymptotically it is, of course, irrelevant whether I require $x-y\ge1$ or $x-y\ge3$ here. Sep 16, 2013 at 13:36. • this is gonna b a long story Sep 16, 2013 at 13:41. • @Anant: I think you can answer your own question also. Then you won't be constrained by the 500 character limit. Sep 16, 2013 at 13:49. • Sorry about the bad mistakes in the definitions. They were a result of earlier editing, where I tried to avoid using subscripts on primes. Sep 20, 2013 at 4:16. • Does $\ln p_n \# < p_n$ always hold? If so, where can a proof be found? Apr 9, 2015 at 5:44.
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Modulo small values of $n,$ you do not need to assume that $x-y\ne 1$ as well as asymptotic on the primorial function. Instead, you can get away with Bertrand's postulate stating that $p_{n}<2p_{n-1}$ or $p_{n-1}\ge p_n/2.$ Indeed, if $m\ge 2,$ then $$x^m-y^m=(x-y)(x^{m-1}+...+y^{m-1})\ge (x-y)(x+y)\ge x+y\ge 2\sqrt{xy}.$$ Note, $$2\sqrt{xy}=2\sqrt{p_1...p_{n-3}p_{n-2}p_{n-1}p_n}$$ and using the fact that $p_{n-2}\ge p_n/4,$ $p_{n-3}\ge p_n/8$ and $p_{n-1}\ge p_n/2$ we can estimate $2\sqrt{xy}\ge 2\sqrt{p_1p_2\cdot...\frac{p_n^4}{64}}=\frac{p_n^2}{8}\sqrt{p_1p_2...}.$ So we are left to check the result for small values of $p_n.$.
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### Find Number with More Than 25 Percent Frequency In Sorted Array
Leetcode Element Appearing More Than 25% In Sorted Array class Solution(object): def findSpecialInteger(self, arr): “”” :type arr: List[int] :rtype: int “”” if len(arr) > 8: return self.getMostFrequentNumberOptimized(arr) return self.getMostFrequentNumber(arr) def getMostFrequentNumber(self, arr): counter = {} for n in arr: if n in counter: counter[n] += 1 else: counter[n] = 1 targetPercent = (len(arr) * 25)//100 … More Find Number with More Than 25 Percent Frequency In Sorted Array
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### Quicksort in C++
// cat quicksort.cc #include <bits/stdc++.h> #include <algorithm> // std::swap using namespace std; int partition(int arr[], int low, int high); void qsort(int arr[], int start, int end) { int p; if (start < end) { p = partition(arr, start, end); qsort(arr, start, p – 1); qsort(arr, p + 1, end); } } int partition(int arr[], int … More Quicksort in C++
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### Find Number with More Than 25 Percent Frequency In Sorted Array. Leetcode Element Appearing More Than 25% In Sorted Array class Solution(object): def findSpecialInteger(self, arr): “”” :type arr: List[int] :rtype: int “”” if len(arr) > 8: return self.getMostFrequentNumberOptimized(arr) return self.getMostFrequentNumber(arr) def getMostFrequentNumber(self, arr): counter = {} for n in arr: if n in counter: counter[n] += 1 else: counter[n] = 1 targetPercent = (len(arr) * 25)//100 … More Find Number with More Than 25 Percent Frequency In Sorted Array. ### Difference between Increasing & Non-Decreasing Order. Let’s take an array for example: int arr[] = {1,2,3,4,5,6}; The above array is sorted in increasing and non-decreasing order. Now, let’s add a few duplicates. int arr[] = {1,1,2,2,3,4,4}; The above array is not in an increasing order. But it is in non-decreasing order.. ### K Weakest Rows in a Matrix. Problem The K Weakest Rows in a Matrix Solution Uses Binary search to count 1s Uses heap to keep k weakest rows It naturally satisfies the condition for weak rows with the same number of soldiers. The smaller index row is processed first and hence any next row would not get pushed to the heap. … More K Weakest Rows in a Matrix. ### Number of 1s in a sorted array. Problem Count number of 1s in a sorted array. e.g. [1,1,1,1,0,0,0,0,0,0] Code def countOneBS(row, start, end): if row[start] == 0: return 0 if row[end] == 1: return end – start + 1 # The mid is dependent on the index of start mid = start + (end – start) // 2 # print(“mid=”, mid, “start=”, … More Number of 1s in a sorted array. ### Why Use Python Ordered Dictionary?. An ordered dictionary offers same time complexity as the standard dict. The implementation is simple: Keep two dictionaries and a doubly-linked list of keys. The DLL maintains order. One dict maps key => DLL node Other dict maps key => value Implementation details https://github.com/python/cpython/blob/3.7/Lib/collections/init.py#L129 Example Problem: Find the first non-repeating char in a string. def … More Why Use Python Ordered Dictionary?.
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### Quicksort in C++. // cat quicksort.cc #include <bits/stdc++.h> #include <algorithm> // std::swap using namespace std; int partition(int arr[], int low, int high); void qsort(int arr[], int start, int end) { int p; if (start < end) { p = partition(arr, start, end); qsort(arr, start, p – 1); qsort(arr, p + 1, end); } } int partition(int arr[], int … More Quicksort in C++.
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Back to the class
Section 5.6 #40: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t^2).$$ Solution: Since $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t)=\dfrac{1}{\sqrt{1-x^2}},$$ we, using the chain rule, compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t^2) &= \dfrac{1}{\sqrt{1-(t^2)^2}} \dfrac{\mathrm{d}}{\mathrm{d}t} t^2 \\ &= \dfrac{2t}{\sqrt{1-t^4}}. \blacksquare \end{array}$$
Section 5.6 #50: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \log(t^2+4) - \dfrac{1}{2} \arctan \left( \dfrac{t}{2} \right) \right].$$ Solution: Since $$\dfrac{\mathrm{d}}{\mathrm{d}t} \log(t) = \dfrac{1}{t}$$ and $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arctan(t) = \dfrac{1}{1+t^2},$$ we use the chain rule to compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \log(t^2+4) - \dfrac{1}{2} \arctan \left( \dfrac{t}{2} \right) \right] &= \dfrac{1}{t^2+4} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ t^2+4 \right] - \left( \dfrac{1}{2} \right) \dfrac{1}{1+\left(\frac{t}{2}\right)^2} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \dfrac{t}{2} \right] \\ &= \dfrac{2t}{t^2+4} -\dfrac{1}{4(1+\frac{t^2}{4})} \\ &= \dfrac{2t}{t^2+4} - \dfrac{1}{4+t^2}. \\ &= \dfrac{2t-1}{t^2+4}. \blacksquare \end{array}$$
Section 5.7 #17: Calculate $$\displaystyle\int \dfrac{x-3}{x^2+1} \mathrm{d}x.$$ Solution: Since $$\displaystyle\int f(x) + g(x) \mathrm{d}x = \displaystyle\int f(x) \mathrm{d}x + \displaystyle\int g(x) \mathrm{d}x,$$ we compute $$\begin{array}{ll} \displaystyle\int \dfrac{x-3}{x^2+1} \mathrm{d}x &= \displaystyle\int \dfrac{x}{x^2+1} \mathrm{d}x - 3 \displaystyle\int \dfrac{1}{x^2+1} \mathrm{d}x \\ &\stackrel{u=x^2+1, \frac{1}{2}\mathrm{d}u= x\mathrm{d}x}{=} \dfrac{1}{2} \displaystyle\int \dfrac{1}{u} \mathrm{d}u - 3 \arctan(x) + C \\ &= \dfrac{1}{2} \log(u) - 3 \arctan(x) + C \\ &= \dfrac{1}{2} \log(x^2+1) - 3\arctan(x) + C. \blacksquare \end{array}$$
Section 5.7 #28: Calculate $$\displaystyle\int_{\ln(2)}^{\ln(4)} \dfrac{e^{-x}}{\sqrt{1-e^{-2x}}} \mathrm{d}x.$$ Solution: Note that $$\sqrt{1-e^{-2x}} = \sqrt{1-(e^{-x})^2}.$$ Therefore calculate $$\begin{array}{ll} \displaystyle\int_{\ln(2)}^{\ln(4)} \dfrac{e^{-x}}{\sqrt{1-e^{-2x}}} \mathrm{d}x &\stackrel{u=e^{-x},-\mathrm{d}u=e^{-x}\mathrm{d}x}{=} -\displaystyle\int_{\frac{1}{2}}^{\frac{1}{4}} \dfrac{1}{\sqrt{1-u^2}} \mathrm{d}u \\ &= - \arcsin(u) \Bigg|_{\frac{1}{2}}^{\frac{1}{4}} \\ &=-\left[ \arcsin\left( \dfrac{1}{4} \right) - \arcsin \left( \dfrac{1}{2} \right) \right] \\ &= \arcsin \left( \dfrac{1}{2} \right) - \arcsin \left( \dfrac{1}{4} \right) \\ &\approx 0.2709. \blacksquare \end{array}$$
Section 5.7 #29: Calculate $$\displaystyle\int_{\frac{\pi}{2}}^{\pi} \dfrac{\sin(x)}{1+\cos^2(x)} \mathrm{d}x.$$ Solution: Let $u=\cos(x)$ so that $-\mathrm{d}u=\sin(x) \mathrm{d}x$. Now if $x=\dfrac{\pi}{2}$ then $u=\cos\left(\dfrac{\pi}{2}\right)=0$ and if $x=\pi$ then $u=cos(\pi)=-1$. Now calculate $$\begin{array}{ll} \displaystyle\int_{\frac{\pi}{2}}^{\pi} \dfrac{\sin(x)}{1+\cos^2(x)} \mathrm{d}x &= \displaystyle\int_{0}^{-1} \dfrac{1}{1+u^2} \mathrm{d}u \\ &= \arctan(u) \Bigg|_{0}^{-1} \\ &= \arctan(0)-\arctan(-1) \\ &= 0- \left( - \dfrac{\pi}{4} \right) \\ &= \dfrac{\pi}{4}. \blacksquare \end{array}$$
Section 5.8 #28: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right).$$ Solution: Since $\tanh(x)=\dfrac{\sinh(x)}{\cosh(x)}$, we may proceed with the quotient rule or we simply use the formula $\dfrac{\mathrm{d}}{\mathrm{d}x} \tanh(x) = \mathrm{sech}^2(x)$. We will do the latter: compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right) &= \dfrac{1}{\tanh(\frac{x}{2})} \dfrac{\mathrm{d}}{\mathrm{d}x} \tanh \left( \dfrac{x}{2} \right) \\ &= \dfrac{1}{\tanh(\frac{x}{2})} \mathrm{sech}^2 \left( \dfrac{x}{2} \right) \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{x}{2} \right] \\ &= \dfrac{1}{2} \dfrac{\coth(\frac{x}{2})}{\sinh(\frac{x}{2})} \dfrac{1}{\cosh^2(\frac{x}{2})} \\ &= \dfrac{1}{2\sinh(\frac{x}{2})\cosh(\frac{x}{2})}. \end{array}$$ If one notes that $2\sinh\left(\dfrac{x}{2} \right)\cosh \left( \dfrac{x}{2} \right) = \sinh(x)$, we may write $$\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right) = \dfrac{1}{\sinh(x)} = \mathrm{csch}(x). \blacksquare$$
Section 5.8 #30: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}x} \left[ x \cosh(x)-\sinh(x) \right].$$ Solution: Since $\cosh'(x)=\sinh(x)$ and $\sinh'(x)=\cosh(x)$, we use the product rule to compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} [ x \cosh(x) - \sinh(x) ] &= [\cosh(x) + x \sinh(x) ] - \cosh(x) \\ &= x \sinh(x). \blacksquare \end{array}$$
Section 5.8 #43: Calculate $$\displaystyle\int \cosh(2x) \mathrm{d}x.$$ Solution: Let $u=2x$ so that $\dfrac{1}{2} \mathrm{d}u=\mathrm{d}x$. Then, compute $$\begin{array}{ll} \displaystyle\int \cosh(2x) \mathrm{d}x &= \dfrac{1}{2} \displaystyle\int \cosh(u) \mathrm{d}u \\ &= \dfrac{1}{2} \sinh(u) + C \\ &= \dfrac{1}{2} \sinh(2x) + C. \blacksquare \end{array}$$
Section 5.8 #55: Calculate $$\displaystyle\int_0^{\ln(2)} \tanh(x) \mathrm{d}x.$$ Solution: Since $\tanh(x) = \dfrac{\sinh(x)}{\cosh(x)}$, let $u=\cosh(x)$ so that $\mathrm{d}u=\sinh(x)$. Also whenever $x=0$ we have $$u=\cosh(0)=\dfrac{e^0+e^{-0}}{2}=1,$$ and if $x=\ln(2)$ we have $$u=\cosh(\ln(2))=\dfrac{e^{\ln(2)}+e^{-\ln(2)}}{2}=\dfrac{2+\frac{1}{2}}{2}=\dfrac{5}{4}.$$ So compute $$\begin{array}{ll} \displaystyle\int_0^{\ln(2)} \tanh(x) \mathrm{d}x &= \displaystyle\int_0^{\ln(2)} \dfrac{\sinh(x)}{\cosh(x)} \mathrm{dx} \\ &= \displaystyle\int_1^{\frac{5}{4}} \dfrac{1}{u} \mathrm{d}u \\ &= \log(u) \Bigg|_1^{\frac{5}{4}} \\ &= \log \left( \dfrac{5}{4} \right) - \log(1) \\ &= \log\left( \dfrac{5}{4} \right). \blacksquare \end{array}$$
Section 5.8 #75: Calculate $$\displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x.$$ Solution: Note that $$\dfrac{1}{3-9x^2} = \dfrac{1}{(\sqrt{3})^2-(3x)^2}.$$ So let $u=3x$ so that $\dfrac{1}{3} \mathrm{d}u = \mathrm{d}x$. Since $$\displaystyle\int \dfrac{1}{a^2-x^2} \mathrm{d}x = \dfrac{1}{a} \arctan\left( \dfrac{x}{a} \right) + C,$$ we now calculate $$\begin{array}{ll} \displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x &= \displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x \\ &= \displaystyle\int \dfrac{1}{\sqrt{3}^2-(3x)^2} \mathrm{d}x \\ &= \dfrac{1}{3} \displaystyle\int \dfrac{1}{\sqrt{3}^2-u^2} \mathrm{d}u \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh} \left( \dfrac{u}{\sqrt{3}} \right) + C \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh}\left( \dfrac{3x}{\sqrt{3}} \right) + C \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh} \left( \sqrt{3}x \right) + C. \blacksquare \end{array}$$
Section 5.8 #86: Calculate $$\displaystyle\int_0^1 \dfrac{1}{\sqrt{25x^2+1}} \mathrm{d}x.$$ Solution: Note that $$\dfrac{1}{\sqrt{25x^2+1}} = \dfrac{1}{\sqrt{(5x)^2+1}}.$$ So letting $u=5x$ implies $\dfrac{1}{5} \mathrm{d}u = \mathrm{d}x$ and if $x=0$ then $u=0$ and if $x=1$ then $u=5$. Therefore compute $$\begin{array}{ll} \displaystyle\int_0^1 \dfrac{1}{\sqrt{25x^2+1}} \mathrm{d}x &= \displaystyle\int_0^1 \dfrac{1}{\sqrt{(5x)^2+1}} \mathrm{d}x \\ &= \dfrac{1}{5} \displaystyle\int_0^5 \dfrac{1}{\sqrt{u^2+1}} \mathrm{d}u \\ &= \dfrac{1}{5} \mathrm{arcsinh}(u) \Bigg|_0^5 \\ &= \dfrac{1}{5} \left[ \mathrm{arcsinh}(5) - \mathrm{arcsinh}(0) \right] \\ &= \dfrac{\mathrm{arcsinh}(5)}{5}. \blacksquare \\ \end{array}$$
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.. Back to the class. Section 5.6 #40: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t^2).$$ Solution: Since $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t)=\dfrac{1}{\sqrt{1-x^2}},$$ we, using the chain rule, compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}t} \arcsin(t^2) &= \dfrac{1}{\sqrt{1-(t^2)^2}} \dfrac{\mathrm{d}}{\mathrm{d}t} t^2 \\ &= \dfrac{2t}{\sqrt{1-t^4}}. \blacksquare \end{array}$$. Section 5.6 #50: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \log(t^2+4) - \dfrac{1}{2} \arctan \left( \dfrac{t}{2} \right) \right].$$ Solution: Since $$\dfrac{\mathrm{d}}{\mathrm{d}t} \log(t) = \dfrac{1}{t}$$ and $$\dfrac{\mathrm{d}}{\mathrm{d}t} \arctan(t) = \dfrac{1}{1+t^2},$$ we use the chain rule to compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \log(t^2+4) - \dfrac{1}{2} \arctan \left( \dfrac{t}{2} \right) \right] &= \dfrac{1}{t^2+4} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ t^2+4 \right] - \left( \dfrac{1}{2} \right) \dfrac{1}{1+\left(\frac{t}{2}\right)^2} \dfrac{\mathrm{d}}{\mathrm{d}t} \left[ \dfrac{t}{2} \right] \\ &= \dfrac{2t}{t^2+4} -\dfrac{1}{4(1+\frac{t^2}{4})} \\ &= \dfrac{2t}{t^2+4} - \dfrac{1}{4+t^2}. \\ &= \dfrac{2t-1}{t^2+4}. \blacksquare \end{array}$$. Section 5.7 #17: Calculate $$\displaystyle\int \dfrac{x-3}{x^2+1} \mathrm{d}x.$$ Solution: Since $$\displaystyle\int f(x) + g(x) \mathrm{d}x = \displaystyle\int f(x) \mathrm{d}x + \displaystyle\int g(x) \mathrm{d}x,$$ we compute $$\begin{array}{ll} \displaystyle\int \dfrac{x-3}{x^2+1} \mathrm{d}x &= \displaystyle\int \dfrac{x}{x^2+1} \mathrm{d}x - 3 \displaystyle\int \dfrac{1}{x^2+1} \mathrm{d}x \\ &\stackrel{u=x^2+1, \frac{1}{2}\mathrm{d}u= x\mathrm{d}x}{=} \dfrac{1}{2} \displaystyle\int \dfrac{1}{u} \mathrm{d}u - 3 \arctan(x) + C \\ &= \dfrac{1}{2} \log(u) - 3 \arctan(x) + C \\ &= \dfrac{1}{2} \log(x^2+1) - 3\arctan(x) + C. \blacksquare \end{array}$$. Section 5.7 #28: Calculate $$\displaystyle\int_{\ln(2)}^{\ln(4)} \dfrac{e^{-x}}{\sqrt{1-e^{-2x}}} \mathrm{d}x.$$ Solution: Note that $$\sqrt{1-e^{-2x}} = \sqrt{1-(e^{-x})^2}.$$ Therefore calculate $$\begin{array}{ll} \displaystyle\int_{\ln(2)}^{\ln(4)} \dfrac{e^{-x}}{\sqrt{1-e^{-2x}}} \mathrm{d}x &\stackrel{u=e^{-x},-\mathrm{d}u=e^{-x}\mathrm{d}x}{=} -\displaystyle\int_{\frac{1}{2}}^{\frac{1}{4}} \dfrac{1}{\sqrt{1-u^2}} \mathrm{d}u \\ &= - \arcsin(u) \Bigg|_{\frac{1}{2}}^{\frac{1}{4}} \\ &=-\left[ \arcsin\left( \dfrac{1}{4} \right) - \arcsin \left( \dfrac{1}{2} \right) \right] \\ &= \arcsin \left( \dfrac{1}{2} \right) - \arcsin \left( \dfrac{1}{4} \right) \\ &\approx 0.2709. \blacksquare \end{array}$$. Section 5.7 #29: Calculate $$\displaystyle\int_{\frac{\pi}{2}}^{\pi} \dfrac{\sin(x)}{1+\cos^2(x)} \mathrm{d}x.$$ Solution: Let $u=\cos(x)$ so that $-\mathrm{d}u=\sin(x) \mathrm{d}x$. Now if $x=\dfrac{\pi}{2}$ then $u=\cos\left(\dfrac{\pi}{2}\right)=0$ and if $x=\pi$ then $u=cos(\pi)=-1$. Now calculate $$\begin{array}{ll} \displaystyle\int_{\frac{\pi}{2}}^{\pi} \dfrac{\sin(x)}{1+\cos^2(x)} \mathrm{d}x &= \displaystyle\int_{0}^{-1} \dfrac{1}{1+u^2} \mathrm{d}u \\ &= \arctan(u) \Bigg|_{0}^{-1} \\ &= \arctan(0)-\arctan(-1) \\ &= 0- \left( - \dfrac{\pi}{4} \right) \\ &= \dfrac{\pi}{4}. \blacksquare \end{array}$$. Section 5.8 #28: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right).$$ Solution: Since $\tanh(x)=\dfrac{\sinh(x)}{\cosh(x)}$, we may proceed with the quotient rule or we simply use the formula $\dfrac{\mathrm{d}}{\mathrm{d}x} \tanh(x) = \mathrm{sech}^2(x)$. We will do the latter: compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right) &= \dfrac{1}{\tanh(\frac{x}{2})} \dfrac{\mathrm{d}}{\mathrm{d}x} \tanh \left( \dfrac{x}{2} \right) \\ &= \dfrac{1}{\tanh(\frac{x}{2})} \mathrm{sech}^2 \left( \dfrac{x}{2} \right) \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{x}{2} \right] \\ &= \dfrac{1}{2} \dfrac{\coth(\frac{x}{2})}{\sinh(\frac{x}{2})} \dfrac{1}{\cosh^2(\frac{x}{2})} \\ &= \dfrac{1}{2\sinh(\frac{x}{2})\cosh(\frac{x}{2})}. \end{array}$$ If one notes that $2\sinh\left(\dfrac{x}{2} \right)\cosh \left( \dfrac{x}{2} \right) = \sinh(x)$, we may write $$\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \tanh \left( \dfrac{x}{2} \right) \right) = \dfrac{1}{\sinh(x)} = \mathrm{csch}(x). \blacksquare$$. Section 5.8 #30: Calculate $$\dfrac{\mathrm{d}}{\mathrm{d}x} \left[ x \cosh(x)-\sinh(x) \right].$$ Solution: Since $\cosh'(x)=\sinh(x)$ and $\sinh'(x)=\cosh(x)$, we use the product rule to compute $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} [ x \cosh(x) - \sinh(x) ] &= [\cosh(x) + x \sinh(x) ] - \cosh(x) \\ &= x \sinh(x). \blacksquare \end{array}$$. Section 5.8 #43: Calculate $$\displaystyle\int \cosh(2x) \mathrm{d}x.$$ Solution: Let $u=2x$ so that $\dfrac{1}{2} \mathrm{d}u=\mathrm{d}x$. Then, compute $$\begin{array}{ll} \displaystyle\int \cosh(2x) \mathrm{d}x &= \dfrac{1}{2} \displaystyle\int \cosh(u) \mathrm{d}u \\ &= \dfrac{1}{2} \sinh(u) + C \\ &= \dfrac{1}{2} \sinh(2x) + C. \blacksquare \end{array}$$. Section 5.8 #55: Calculate $$\displaystyle\int_0^{\ln(2)} \tanh(x) \mathrm{d}x.$$ Solution: Since $\tanh(x) = \dfrac{\sinh(x)}{\cosh(x)}$, let $u=\cosh(x)$ so that $\mathrm{d}u=\sinh(x)$. Also whenever $x=0$ we have $$u=\cosh(0)=\dfrac{e^0+e^{-0}}{2}=1,$$ and if $x=\ln(2)$ we have $$u=\cosh(\ln(2))=\dfrac{e^{\ln(2)}+e^{-\ln(2)}}{2}=\dfrac{2+\frac{1}{2}}{2}=\dfrac{5}{4}.$$ So compute $$\begin{array}{ll} \displaystyle\int_0^{\ln(2)} \tanh(x) \mathrm{d}x &= \displaystyle\int_0^{\ln(2)} \dfrac{\sinh(x)}{\cosh(x)} \mathrm{dx} \\ &= \displaystyle\int_1^{\frac{5}{4}} \dfrac{1}{u} \mathrm{d}u \\ &= \log(u) \Bigg|_1^{\frac{5}{4}} \\ &= \log \left( \dfrac{5}{4} \right) - \log(1) \\ &= \log\left( \dfrac{5}{4} \right). \blacksquare \end{array}$$. Section 5.8 #75: Calculate $$\displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x.$$ Solution: Note that $$\dfrac{1}{3-9x^2} = \dfrac{1}{(\sqrt{3})^2-(3x)^2}.$$ So let $u=3x$ so that $\dfrac{1}{3} \mathrm{d}u = \mathrm{d}x$. Since $$\displaystyle\int \dfrac{1}{a^2-x^2} \mathrm{d}x = \dfrac{1}{a} \arctan\left( \dfrac{x}{a} \right) + C,$$ we now calculate $$\begin{array}{ll} \displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x &= \displaystyle\int \dfrac{1}{3-9x^2} \mathrm{d}x \\ &= \displaystyle\int \dfrac{1}{\sqrt{3}^2-(3x)^2} \mathrm{d}x \\ &= \dfrac{1}{3} \displaystyle\int \dfrac{1}{\sqrt{3}^2-u^2} \mathrm{d}u \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh} \left( \dfrac{u}{\sqrt{3}} \right) + C \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh}\left( \dfrac{3x}{\sqrt{3}} \right) + C \\ &= \dfrac{1}{3\sqrt{3}} \mathrm{arctanh} \left( \sqrt{3}x \right) + C.
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\blacksquare \end{array}$$. Section 5.8 #86: Calculate $$\displaystyle\int_0^1 \dfrac{1}{\sqrt{25x^2+1}} \mathrm{d}x.$$ Solution: Note that $$\dfrac{1}{\sqrt{25x^2+1}} = \dfrac{1}{\sqrt{(5x)^2+1}}.$$ So letting $u=5x$ implies $\dfrac{1}{5} \mathrm{d}u = \mathrm{d}x$ and if $x=0$ then $u=0$ and if $x=1$ then $u=5$. Therefore compute $$\begin{array}{ll} \displaystyle\int_0^1 \dfrac{1}{\sqrt{25x^2+1}} \mathrm{d}x &= \displaystyle\int_0^1 \dfrac{1}{\sqrt{(5x)^2+1}} \mathrm{d}x \\ &= \dfrac{1}{5} \displaystyle\int_0^5 \dfrac{1}{\sqrt{u^2+1}} \mathrm{d}u \\ &= \dfrac{1}{5} \mathrm{arcsinh}(u) \Bigg|_0^5 \\ &= \dfrac{1}{5} \left[ \mathrm{arcsinh}(5) - \mathrm{arcsinh}(0) \right] \\ &= \dfrac{\mathrm{arcsinh}(5)}{5}. \blacksquare \\ \end{array}$$.
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MENTOR ME CAREERS
# Mastering Date Calculations in Excel: Adding Months to Date
Last updated on May 29th, 2023 at 07:10 pm
In Excel, adding months to date is a common task for various data analysis and financial modeling scenarios. Understanding how to perform this calculation accurately and efficiently can greatly enhance your Excel skills. In this comprehensive guide, we will delve into the intricacies of adding months to a date in Excel. Through clear explanations and practical examples, we will cover the necessary formulas, techniques, and considerations to ensure precise date calculations. Whether you’re a beginner or an experienced Excel user, this article will equip you with the knowledge and tools to confidently handle date calculations involving adding months in Excel.
## Understanding Date Serial Numbers in Excel
To comprehend how Excel handles dates, it’s essential to understand date serial numbers. We’ll explore how Excel internally represents dates as numeric values, enabling date calculations. Additionally, we’ll cover the significance of the base date in Excel and its impact on adding and manipulating dates using formulas
## How to make Excel add months to date using Edate function
In Microsoft Excel, you can add a number of months to a date using the EDATE function. The EDATE function is categorized under the DATE/TIME function. The function helps add a specified number of months to a date and returns the result as a serial date. This is helpful for financial analysts to calculate maturity dates for accounts payable. The EDATE function also helps in calculating summary by month.
## Using the EDATE Function for Adding Months
• Syntax:
EDATE( start_date, months)
• Parameters:
1. start_date – The starting date to use in the calculation
2. months – The number of months to be added to start_date. It can be either a positive or negative number.
• Returns:
The EDATE function returns a serial date value. A serial date is how Excel stores data internally and it represents the number of days since January 1, 1900.
If start_date is not a valid date, the EDATE function will return the #VALUE! error.
• Applies to:
Excel for Office 365, Excel 2019, Excel 2016, Excel 2013, Excel 2011 for Mac, Excel 2010, Excel 2007.
## Edate function in excel Example
Let’s look at some Excel EDATE function examples and explore how to use the EDATE function as a worksheet function in Microsoft Excel:
The result of the EDATE function is a serial date and can be seen in column C above. In order to see the formatted date instead of a serial date, you need to format the cells to mmm, d, yyyy.
## Handling Edge Cases and Rounding Issues
While adding months to a date seems straightforward, certain edge cases and rounding issues can arise. We’ll address scenarios such as leap years, end-of-month calculations, and handling fractions of a month. By exploring potential pitfalls and providing practical solutions, you’ll be able to handle these intricacies with precision and accuracy.
In many scenarios, adding months to a date requires considering only workdays or business days. We’ll explore how to incorporate Excel’s WORKDAY and NETWORKDAYS functions to exclude weekends and holidays from the calculation. You’ll learn to adjust your date calculations to align with your specific business requirements.
## Incorporating Conditional Logic and Dynamic Calculations
Excel’s versatility lies in its ability to incorporate conditional logic and dynamic calculations. We’ll explore techniques such as utilizing IF statements and nested functions to add flexibility to your date calculations. This includes scenarios where you may need to adjust the added months based on certain conditions or dynamically update the calculations based on changing criteria.
## Edate Function in Excel Exampels
Example 1: Suppose you have a start date in cell A1, and you want to add 6 months to that date. In cell B1, you can use the EDATE function to achieve this: =EDATE(A1, 6). The result will be the date that is 6 months ahead of the start date.
Example 2: Let’s say you have a list of payment due dates in column A, and you want to calculate the payment dates that are 3 months later. In cell B2, you can use the formula =EDATE(A2, 3) to add 3 months to the date in cell A2. Copy the formula down the column to apply it to the entire range of payment due dates.
Example 3: If you want to add months to the current date, you can use the TODAY function combined with the EDATE function. For instance, in cell A1, you can enter =TODAY() to get the current date. In cell B1, use the formula =EDATE(A1, 4) to add 4 months to the current date.
Example 4: You may encounter scenarios where you need to add months while considering workdays only. In such cases, you can utilize the WORKDAY.INTL function. For instance, if you have a start date in cell A1 and you want to add 3 work months to it, you can use the formula =WORKDAY.INTL(A1, 3 * 22, 1) assuming a 5-day workweek (22 workdays per month).
Example 5: To handle rounding issues when adding months, you can use the EOMONTH function in combination with the EDATE function. For instance, if you want to add 2 months to a date in cell A1 and ensure that the result falls on the last day of the month, you can use the formula =EOMONTH(EDATE(A1, 2), 0).
These examples demonstrate the practical application of adding months to dates in Excel using various functions and considerations. Experiment with these formulas to suit your specific requirements and expand your understanding of date calculations in Excel.
## Conclusion
Congratulations! You’ve now gained a comprehensive understanding of adding months to a date in Excel. By mastering the techniques and formulas covered in this article, you can confidently handle various date calculations in your Excel workbooks. Remember to utilize the EDATE function for straightforward month addition, handle edge cases and rounding issues with precision, adjust for workdays or business days when necessary, and leverage conditional logic for dynamic calculations. Excel’s flexibility and power make it an invaluable tool for handling date-related tasks, and with your newfound knowledge, you’ll excel in performing accurate and efficient date calculations in Excel.
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MENTOR ME CAREERS. # Mastering Date Calculations in Excel: Adding Months to Date. Last updated on May 29th, 2023 at 07:10 pm. In Excel, adding months to date is a common task for various data analysis and financial modeling scenarios. Understanding how to perform this calculation accurately and efficiently can greatly enhance your Excel skills. In this comprehensive guide, we will delve into the intricacies of adding months to a date in Excel. Through clear explanations and practical examples, we will cover the necessary formulas, techniques, and considerations to ensure precise date calculations. Whether you’re a beginner or an experienced Excel user, this article will equip you with the knowledge and tools to confidently handle date calculations involving adding months in Excel.. ## Understanding Date Serial Numbers in Excel. To comprehend how Excel handles dates, it’s essential to understand date serial numbers. We’ll explore how Excel internally represents dates as numeric values, enabling date calculations. Additionally, we’ll cover the significance of the base date in Excel and its impact on adding and manipulating dates using formulas. ## How to make Excel add months to date using Edate function. In Microsoft Excel, you can add a number of months to a date using the EDATE function. The EDATE function is categorized under the DATE/TIME function. The function helps add a specified number of months to a date and returns the result as a serial date. This is helpful for financial analysts to calculate maturity dates for accounts payable. The EDATE function also helps in calculating summary by month.. ## Using the EDATE Function for Adding Months. • Syntax:. EDATE( start_date, months). • Parameters:. 1. start_date – The starting date to use in the calculation. 2. months – The number of months to be added to start_date. It can be either a positive or negative number.. • Returns:. The EDATE function returns a serial date value. A serial date is how Excel stores data internally and it represents the number of days since January 1, 1900.. If start_date is not a valid date, the EDATE function will return the #VALUE! error.. • Applies to:. Excel for Office 365, Excel 2019, Excel 2016, Excel 2013, Excel 2011 for Mac, Excel 2010, Excel 2007.. ## Edate function in excel Example. Let’s look at some Excel EDATE function examples and explore how to use the EDATE function as a worksheet function in Microsoft Excel:. The result of the EDATE function is a serial date and can be seen in column C above. In order to see the formatted date instead of a serial date, you need to format the cells to mmm, d, yyyy.. ## Handling Edge Cases and Rounding Issues. While adding months to a date seems straightforward, certain edge cases and rounding issues can arise. We’ll address scenarios such as leap years, end-of-month calculations, and handling fractions of a month. By exploring potential pitfalls and providing practical solutions, you’ll be able to handle these intricacies with precision and accuracy.. In many scenarios, adding months to a date requires considering only workdays or business days. We’ll explore how to incorporate Excel’s WORKDAY and NETWORKDAYS functions to exclude weekends and holidays from the calculation. You’ll learn to adjust your date calculations to align with your specific business requirements.. ## Incorporating Conditional Logic and Dynamic Calculations. Excel’s versatility lies in its ability to incorporate conditional logic and dynamic calculations. We’ll explore techniques such as utilizing IF statements and nested functions to add flexibility to your date calculations. This includes scenarios where you may need to adjust the added months based on certain conditions or dynamically update the calculations based on changing criteria.. ## Edate Function in Excel Exampels. Example 1: Suppose you have a start date in cell A1, and you want to add 6 months to that date. In cell B1, you can use the EDATE function to achieve this: =EDATE(A1, 6). The result will be the date that is 6 months ahead of the start date.. Example 2: Let’s say you have a list of payment due dates in column A, and you want to calculate the payment dates that are 3 months later. In cell B2, you can use the formula =EDATE(A2, 3) to add 3 months to the date in cell A2. Copy the formula down the column to apply it to the entire range of payment due dates.. Example 3: If you want to add months to the current date, you can use the TODAY function combined with the EDATE function. For instance, in cell A1, you can enter =TODAY() to get the current date. In cell B1, use the formula =EDATE(A1, 4) to add 4 months to the current date.. Example 4: You may encounter scenarios where you need to add months while considering workdays only. In such cases, you can utilize the WORKDAY.INTL function. For instance, if you have a start date in cell A1 and you want to add 3 work months to it, you can use the formula =WORKDAY.INTL(A1, 3 * 22, 1) assuming a 5-day workweek (22 workdays per month).. Example 5: To handle rounding issues when adding months, you can use the EOMONTH function in combination with the EDATE function. For instance, if you want to add 2 months to a date in cell A1 and ensure that the result falls on the last day of the month, you can use the formula =EOMONTH(EDATE(A1, 2), 0).. These examples demonstrate the practical application of adding months to dates in Excel using various functions and considerations. Experiment with these formulas to suit your specific requirements and expand your understanding of date calculations in Excel.. ## Conclusion. Congratulations! You’ve now gained a comprehensive understanding of adding months to a date in Excel. By mastering the techniques and formulas covered in this article, you can confidently handle various date calculations in your Excel workbooks.
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Remember to utilize the EDATE function for straightforward month addition, handle edge cases and rounding issues with precision, adjust for workdays or business days when necessary, and leverage conditional logic for dynamic calculations. Excel’s flexibility and power make it an invaluable tool for handling date-related tasks, and with your newfound knowledge, you’ll excel in performing accurate and efficient date calculations in Excel.
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# Prove $\int_0^{\pi/2} x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x<{n^2\pi^2\over 8}$
Prove $$\int_0^{\pi/2} x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x<{n^2\pi^2\over 8}.$$
My attempt: \begin{align} \int_0^{\pi/2} x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x & =\sum_{k=1}^n \int_{{k-1\over 2n}\pi}^{{k\over 2n}\pi}x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x\\ & \leq\sum_{k=1}^n \left({\pi\over 2}\right)^4 \int_{{k-1\over 2n}\pi}^{{k\over 2n}\pi}\left({\sin^4nx\over x^3}\right)\mathrm{d}x \quad (\text{use } \sin x\geq {2\over \pi}x ) \tag{1}\label{1}\\ &= \left({\pi\over 2}\right)^4 n^2\sum_{k=1}^n \int_{{k-1\over 2}\pi}^{{k\over 2}\pi}\left({\sin^4x\over x^3}\right)\mathrm{d}x \quad (\text{use } x\to {x\over n}).\\ \end{align} Is my direction right? If right, how can I prove the following $$\sum_{k=1}^n \int_{{k-1\over 2}\pi}^{{k\over 2}\pi}\left({\sin^4x\over x^3}\right)\mathrm{d}x\leq\int_0^{+\infty}\left({\sin^4x\over x^3}\right)\mathrm{d}x \leq {2\over \pi^2}.$$ I use Mathematica to calculate the integral $\int_0^{+\infty}\left({\sin^4x\over x^3}\right)\mathrm{d}x\simeq 0.7>{2\over\pi^2}$, hence my process (\ref{1}) seems to be wrong.
• I supposed $n$ positive integer. Is it right? – Raffaele Oct 3 '17 at 12:54
• @Raffaele Yes it is. – yahoo Oct 3 '17 at 13:01
The term $\left(\frac{\sin nx}{\sin x}\right)^4$ is associated with the Jackson kernel.
Your inequality is indeed just a minor variation on Lemma 0.5 in the linked notes, and it can be proved through the same technique: expand $|x|$ as a Fourier cosine series over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, do the same for $\left(\frac{\sin nx}{\sin x}\right)^4$, then apply orthogonality/Bessel's inequality.
• Great answer! The question is the first exercise of a chapter about integral, I thought it would be easy for me. By reading the material you give, I should estimate the $\int_0^{\pi\over 2n}$ part by using $\sin x\geq {2\over\pi}x$ and the rest using $x\geq {k\pi\over 2n}$. The first part it self still larger than the right hand side. So I need to give a more concise estimate rather than $\sin x\geq {2\over\pi} x$. By the way, if I accept the answer, how can I ask more people to find if there is a more simpler answer? – yahoo Oct 3 '17 at 14:38
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# Prove $\int_0^{\pi/2} x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x<{n^2\pi^2\over 8}$. Prove $$\int_0^{\pi/2} x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x<{n^2\pi^2\over 8}.$$. My attempt: \begin{align} \int_0^{\pi/2} x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x & =\sum_{k=1}^n \int_{{k-1\over 2n}\pi}^{{k\over 2n}\pi}x\left({\sin nx\over \sin x}\right)^4\mathrm{d}x\\ & \leq\sum_{k=1}^n \left({\pi\over 2}\right)^4 \int_{{k-1\over 2n}\pi}^{{k\over 2n}\pi}\left({\sin^4nx\over x^3}\right)\mathrm{d}x \quad (\text{use } \sin x\geq {2\over \pi}x ) \tag{1}\label{1}\\ &= \left({\pi\over 2}\right)^4 n^2\sum_{k=1}^n \int_{{k-1\over 2}\pi}^{{k\over 2}\pi}\left({\sin^4x\over x^3}\right)\mathrm{d}x \quad (\text{use } x\to {x\over n}).\\ \end{align} Is my direction right? If right, how can I prove the following $$\sum_{k=1}^n \int_{{k-1\over 2}\pi}^{{k\over 2}\pi}\left({\sin^4x\over x^3}\right)\mathrm{d}x\leq\int_0^{+\infty}\left({\sin^4x\over x^3}\right)\mathrm{d}x \leq {2\over \pi^2}.$$ I use Mathematica to calculate the integral $\int_0^{+\infty}\left({\sin^4x\over x^3}\right)\mathrm{d}x\simeq 0.7>{2\over\pi^2}$, hence my process (\ref{1}) seems to be wrong.. • I supposed $n$ positive integer. Is it right? – Raffaele Oct 3 '17 at 12:54. • @Raffaele Yes it is. – yahoo Oct 3 '17 at 13:01. The term $\left(\frac{\sin nx}{\sin x}\right)^4$ is associated with the Jackson kernel.. Your inequality is indeed just a minor variation on Lemma 0.5 in the linked notes, and it can be proved through the same technique: expand $|x|$ as a Fourier cosine series over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, do the same for $\left(\frac{\sin nx}{\sin x}\right)^4$, then apply orthogonality/Bessel's inequality.. • Great answer! The question is the first exercise of a chapter about integral, I thought it would be easy for me.
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By reading the material you give, I should estimate the $\int_0^{\pi\over 2n}$ part by using $\sin x\geq {2\over\pi}x$ and the rest using $x\geq {k\pi\over 2n}$. The first part it self still larger than the right hand side. So I need to give a more concise estimate rather than $\sin x\geq {2\over\pi} x$. By the way, if I accept the answer, how can I ask more people to find if there is a more simpler answer? – yahoo Oct 3 '17 at 14:38.
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https://math.stackexchange.com/questions/557445/question-about-abelian-group-proof/566385
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# Question about Abelian group proof
I prove that if $G$ is Abelian group so if $a,b\in G$ has a finite order so $ab$ has a finite order to.. (Maybe later I'll upload here my proof to see of she is correct....)
Now, I have to show that this is false if the group is not Abelian group with those 2 matrices: $\begin{bmatrix} 0 &-1 \\ 1& 1 \end{bmatrix} and \begin{bmatrix} 0 &1 \\ -1&-1 \end{bmatrix}$
This is the problem:
$\begin{bmatrix} 0 &-1 \\ 1& 1 \end{bmatrix}\cdot \begin{bmatrix} 0 &1 \\ -1&-1 \end{bmatrix}=\begin{bmatrix} 1 &1 \\ -1&0 \end{bmatrix}$
$ord\left(\begin{bmatrix} 1 &1 \\ -1&0 \end{bmatrix}\right)=6$
This is not an infinite order element...
An you have any idea?
Thank you!!
• Any idea about what? You are basically given the solution. take these elements, show that they have finite order, take their product, show it does not have finite order. – Najib Idrissi Nov 8 '13 at 23:56
• @nik - How do I show this: take their product, show it does not have finite order? Thank you! – CS1 Nov 9 '13 at 8:13
• But their product have a finite order - 6... – CS1 Nov 9 '13 at 8:27
You probably saw the group $GL(2,\mathbb R)$, perhaps not under this name. It is the group of all invertible $2\times 2$ matrices with real entries. It is a group under the operation of matrix multiplication. So, this question is probably asking you to identify that the two matrices belong to that group. Then you proceed, a la nik's advise, to compute, in $GL(2,\mathbb R)$, the order of each, the product of the two, and the order of the product.
• If I'll multiply them I'll get:$\begin{bmatrix} 1 &1 \\ -1& 0 \end{bmatrix}$ and the order of this matrix is finite - is 6, so I don't understand how can I show that their product a matrix with infinite order... Thank you! – CS1 Nov 9 '13 at 8:25
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# Question about Abelian group proof. I prove that if $G$ is Abelian group so if $a,b\in G$ has a finite order so $ab$ has a finite order to.. (Maybe later I'll upload here my proof to see of she is correct....). Now, I have to show that this is false if the group is not Abelian group with those 2 matrices: $\begin{bmatrix} 0 &-1 \\ 1& 1 \end{bmatrix} and \begin{bmatrix} 0 &1 \\ -1&-1 \end{bmatrix}$. This is the problem:. $\begin{bmatrix} 0 &-1 \\ 1& 1 \end{bmatrix}\cdot \begin{bmatrix} 0 &1 \\ -1&-1 \end{bmatrix}=\begin{bmatrix} 1 &1 \\ -1&0 \end{bmatrix}$. $ord\left(\begin{bmatrix} 1 &1 \\ -1&0 \end{bmatrix}\right)=6$. This is not an infinite order element.... An you have any idea?. Thank you!!. • Any idea about what? You are basically given the solution. take these elements, show that they have finite order, take their product, show it does not have finite order. – Najib Idrissi Nov 8 '13 at 23:56. • @nik - How do I show this: take their product, show it does not have finite order? Thank you! – CS1 Nov 9 '13 at 8:13. • But their product have a finite order - 6... – CS1 Nov 9 '13 at 8:27. You probably saw the group $GL(2,\mathbb R)$, perhaps not under this name. It is the group of all invertible $2\times 2$ matrices with real entries. It is a group under the operation of matrix multiplication. So, this question is probably asking you to identify that the two matrices belong to that group.
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Then you proceed, a la nik's advise, to compute, in $GL(2,\mathbb R)$, the order of each, the product of the two, and the order of the product.. • If I'll multiply them I'll get:$\begin{bmatrix} 1 &1 \\ -1& 0 \end{bmatrix}$ and the order of this matrix is finite - is 6, so I don't understand how can I show that their product a matrix with infinite order... Thank you! – CS1 Nov 9 '13 at 8:25.
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# 31.52 kg to lbs - 31.52 kilograms to pounds
Do you want to learn how much is 31.52 kg equal to lbs and how to convert 31.52 kg to lbs? You are in the right place. In this article you will find everything about kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to point out that whole this article is devoted to one amount of kilograms - that is one kilogram. So if you want to learn more about 31.52 kg to pound conversion - keep reading.
Before we go to the practice - that is 31.52 kg how much lbs conversion - we are going to tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 31.52 kg to lbs? 31.52 kilograms it is equal 69.4897049824 pounds, so 31.52 kg is equal 69.4897049824 lbs.
## 31.52 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, formally known as International System of Units (in short form SI).
Sometimes the kilogram is written as kilogramme. The symbol of the kilogram is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. First definition was simply but hard to use.
Then, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was replaced by a new definition.
The new definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It can be also divided into 100 decagrams and 1000 grams.
## 31.52 kilogram to pounds
You know a little bit about kilogram, so now let's go to the pound. The pound is also a unit of mass. It is needed to highlight that there are more than one kind of pound. What are we talking about? For instance, there are also pound-force. In this article we want to focus only on pound-mass.
The pound is in use in the Imperial and United States customary systems of measurements. Naturally, this unit is in use also in another systems. The symbol of the pound is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is defined as exactly 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 31.52 kg?
31.52 kilogram is equal to 69.4897049824 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 31.52 kg in lbs
Theoretical part is already behind us. In next part we will tell you how much is 31.52 kg to lbs. Now you know that 31.52 kg = x lbs. So it is high time to get the answer. Just see:
31.52 kilogram = 69.4897049824 pounds.
This is an exact outcome of how much 31.52 kg to pound. You can also round off the result. After it your outcome will be exactly: 31.52 kg = 69.344 lbs.
You know 31.52 kg is how many lbs, so let’s see how many kg 31.52 lbs: 31.52 pound = 0.45359237 kilograms.
Obviously, in this case you can also round off this result. After it your outcome will be as following: 31.52 lb = 0.45 kgs.
We also want to show you 31.52 kg to how many pounds and 31.52 pound how many kg results in charts. See:
We are going to start with a table for how much is 31.52 kg equal to pound.
### 31.52 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
31.52 69.4897049824 69.3440
Now see a chart for how many kilograms 31.52 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
31.52 0.45359237 0.45
Now you know how many 31.52 kg to lbs and how many kilograms 31.52 pound, so it is time to move on to the 31.52 kg to lbs formula.
### 31.52 kg to pounds
To convert 31.52 kg to us lbs a formula is needed. We are going to show you a formula in two different versions. Let’s start with the first one:
Number of kilograms * 2.20462262 = the 69.4897049824 result in pounds
The first formula will give you the most accurate outcome. Sometimes even the smallest difference can be considerable. So if you want to get an accurate result - this version of a formula will be the best solution to convert how many pounds are equivalent to 31.52 kilogram.
So let’s move on to the shorer version of a formula, which also enables calculations to know how much 31.52 kilogram in pounds.
The another version of a formula is as following, let’s see:
Amount of kilograms * 2.2 = the result in pounds
As you see, the second version is simpler. It can be better option if you need to make a conversion of 31.52 kilogram to pounds in quick way, for example, during shopping. Just remember that your outcome will be not so exact.
Now we want to show you these two versions of a formula in practice. But before we are going to make a conversion of 31.52 kg to lbs we are going to show you easier way to know 31.52 kg to how many lbs without any effort.
### 31.52 kg to lbs converter
An easier way to learn what is 31.52 kilogram equal to in pounds is to use 31.52 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Converter is based on first formula which we gave you in the previous part of this article. Thanks to 31.52 kg pound calculator you can quickly convert 31.52 kg to lbs. You only have to enter amount of kilograms which you want to convert and click ‘calculate’ button. The result will be shown in a flash.
So try to calculate 31.52 kg into lbs using 31.52 kg vs pound calculator. We entered 31.52 as an amount of kilograms. It is the result: 31.52 kilogram = 69.4897049824 pounds.
As you can see, this 31.52 kg vs lbs calculator is so simply to use.
Now we can go to our primary issue - how to convert 31.52 kilograms to pounds on your own.
#### 31.52 kg to lbs conversion
We will begin 31.52 kilogram equals to how many pounds calculation with the first formula to get the most exact outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 69.4897049824 the result in pounds
So what have you do to check how many pounds equal to 31.52 kilogram? Just multiply number of kilograms, in this case 31.52, by 2.20462262. It is 69.4897049824. So 31.52 kilogram is equal 69.4897049824.
You can also round it off, for example, to two decimal places. It is equal 2.20. So 31.52 kilogram = 69.3440 pounds.
It is high time for an example from everyday life. Let’s calculate 31.52 kg gold in pounds. So 31.52 kg equal to how many lbs? As in the previous example - multiply 31.52 by 2.20462262. It gives 69.4897049824. So equivalent of 31.52 kilograms to pounds, when it comes to gold, is exactly 69.4897049824.
In this case it is also possible to round off the result. It is the outcome after rounding off, this time to one decimal place - 31.52 kilogram 69.344 pounds.
Now we are going to examples converted with short formula.
#### How many 31.52 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 69.344 the result in pounds
So 31.52 kg equal to how much lbs? And again, you have to multiply amount of kilogram, this time 31.52, by 2.2. See: 31.52 * 2.2 = 69.344. So 31.52 kilogram is exactly 2.2 pounds.
Make another calculation with use of this formula. Now convert something from everyday life, for example, 31.52 kg to lbs weight of strawberries.
So convert - 31.52 kilogram of strawberries * 2.2 = 69.344 pounds of strawberries. So 31.52 kg to pound mass is equal 69.344.
If you know how much is 31.52 kilogram weight in pounds and can calculate it using two different formulas, we can move on. Now we are going to show you these outcomes in tables.
#### Convert 31.52 kilogram to pounds
We know that results shown in charts are so much clearer for most of you. We understand it, so we gathered all these outcomes in charts for your convenience. Due to this you can easily compare 31.52 kg equivalent to lbs results.
Begin with a 31.52 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
31.52 69.4897049824 69.3440
And now let’s see 31.52 kg equal pound chart for the second formula:
Kilograms Pounds
31.52 69.344
As you see, after rounding off, when it comes to how much 31.52 kilogram equals pounds, the results are not different. The bigger number the more significant difference. Please note it when you need to make bigger amount than 31.52 kilograms pounds conversion.
#### How many kilograms 31.52 pound
Now you learned how to convert 31.52 kilograms how much pounds but we are going to show you something more. Are you interested what it is? What about 31.52 kilogram to pounds and ounces calculation?
We want to show you how you can calculate it step by step. Begin. How much is 31.52 kg in lbs and oz?
First things first - you need to multiply amount of kilograms, in this case 31.52, by 2.20462262. So 31.52 * 2.20462262 = 69.4897049824. One kilogram is exactly 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To calculate how much 31.52 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So final outcome is 2 pounds and 327396192 ounces. You can also round off ounces, for instance, to two places. Then your result is equal 2 pounds and 33 ounces.
As you see, conversion 31.52 kilogram in pounds and ounces quite simply.
The last calculation which we are going to show you is calculation of 31.52 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert it it is needed another formula. Before we give you it, let’s see:
• 31.52 kilograms meters = 7.23301385 foot pounds,
• 31.52 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to convert 31.52 foot pounds to kilograms meters you have to multiply 31.52 by 0.13825495. It is exactly 0.13825495. So 31.52 foot pounds is exactly 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 31.52 foot pounds is exactly 0.14 kilogram meters.
We hope that this calculation was as easy as 31.52 kilogram into pounds conversions.
This article is a huge compendium about kilogram, pound and 31.52 kg to lbs in conversion. Due to this calculation you learned 31.52 kilogram is equivalent to how many pounds.
We showed you not only how to make a calculation 31.52 kilogram to metric pounds but also two another conversions - to check how many 31.52 kg in pounds and ounces and how many 31.52 foot pounds to kilograms meters.
We showed you also other way to make 31.52 kilogram how many pounds calculations, this is using 31.52 kg en pound calculator. This is the best choice for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.
We hope that now all of you can make 31.52 kilogram equal to how many pounds calculation - on your own or with use of our 31.52 kgs to pounds converter.
Don’t wait! Calculate 31.52 kilogram mass to pounds in the way you like.
Do you need to make other than 31.52 kilogram as pounds calculation? For example, for 15 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so easy as for 31.52 kilogram equal many pounds.
### How much is 31.52 kg in pounds
To quickly sum up this topic, that is how much is 31.52 kg in pounds , we prepared for you an additional section. Here we have for you all you need to remember about how much is 31.52 kg equal to lbs and how to convert 31.52 kg to lbs . It is down below.
What is the kilogram to pound conversion? It is a mathematical operation based on multiplying 2 numbers. Let’s see 31.52 kg to pound conversion formula . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 31.52 kilogram to pounds? The accurate answer is 69.4897049824 lb.
You can also calculate how much 31.52 kilogram is equal to pounds with second, shortened version of the formula. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So this time, 31.52 kg equal to how much lbs ? The result is 69.4897049824 lb.
How to convert 31.52 kg to lbs in just a moment? You can also use the 31.52 kg to lbs converter , which will do the rest for you and you will get an accurate answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
Read more related articles:
31.01 kg to lbs = 68.3654 31.02 kg to lbs = 68.3874 31.03 kg to lbs = 68.4094 31.04 kg to lbs = 68.4315 31.05 kg to lbs = 68.4535 31.06 kg to lbs = 68.4756 31.07 kg to lbs = 68.4976 31.08 kg to lbs = 68.5197 31.09 kg to lbs = 68.5417 31.1 kg to lbs = 68.5638 31.11 kg to lbs = 68.5858 31.12 kg to lbs = 68.6079 31.13 kg to lbs = 68.6299 31.14 kg to lbs = 68.6519 31.15 kg to lbs = 68.674 31.16 kg to lbs = 68.696 31.17 kg to lbs = 68.7181 31.18 kg to lbs = 68.7401 31.19 kg to lbs = 68.7622 31.2 kg to lbs = 68.7842 31.21 kg to lbs = 68.8063 31.22 kg to lbs = 68.8283 31.23 kg to lbs = 68.8504 31.24 kg to lbs = 68.8724 31.25 kg to lbs = 68.8945
31.26 kg to lbs = 68.9165 31.27 kg to lbs = 68.9386 31.28 kg to lbs = 68.9606 31.29 kg to lbs = 68.9826 31.3 kg to lbs = 69.0047 31.31 kg to lbs = 69.0267 31.32 kg to lbs = 69.0488 31.33 kg to lbs = 69.0708 31.34 kg to lbs = 69.0929 31.35 kg to lbs = 69.1149 31.36 kg to lbs = 69.137 31.37 kg to lbs = 69.159 31.38 kg to lbs = 69.1811 31.39 kg to lbs = 69.2031 31.4 kg to lbs = 69.2251 31.41 kg to lbs = 69.2472 31.42 kg to lbs = 69.2692 31.43 kg to lbs = 69.2913 31.44 kg to lbs = 69.3133 31.45 kg to lbs = 69.3354 31.46 kg to lbs = 69.3574 31.47 kg to lbs = 69.3795 31.48 kg to lbs = 69.4015 31.49 kg to lbs = 69.4236 31.5 kg to lbs = 69.4456
31.51 kg to lbs = 69.4677 31.52 kg to lbs = 69.4897 31.53 kg to lbs = 69.5118 31.54 kg to lbs = 69.5338 31.55 kg to lbs = 69.5558 31.56 kg to lbs = 69.5779 31.57 kg to lbs = 69.5999 31.58 kg to lbs = 69.622 31.59 kg to lbs = 69.644 31.6 kg to lbs = 69.6661 31.61 kg to lbs = 69.6881 31.62 kg to lbs = 69.7102 31.63 kg to lbs = 69.7322 31.64 kg to lbs = 69.7543 31.65 kg to lbs = 69.7763 31.66 kg to lbs = 69.7983 31.67 kg to lbs = 69.8204 31.68 kg to lbs = 69.8424 31.69 kg to lbs = 69.8645 31.7 kg to lbs = 69.8865 31.71 kg to lbs = 69.9086 31.72 kg to lbs = 69.9306 31.73 kg to lbs = 69.9527 31.74 kg to lbs = 69.9747 31.75 kg to lbs = 69.9968
31.76 kg to lbs = 70.0188 31.77 kg to lbs = 70.0409 31.78 kg to lbs = 70.0629 31.79 kg to lbs = 70.085 31.8 kg to lbs = 70.107 31.81 kg to lbs = 70.1291 31.82 kg to lbs = 70.1511 31.83 kg to lbs = 70.1731 31.84 kg to lbs = 70.1952 31.85 kg to lbs = 70.2172 31.86 kg to lbs = 70.2393 31.87 kg to lbs = 70.2613 31.88 kg to lbs = 70.2834 31.89 kg to lbs = 70.3054 31.9 kg to lbs = 70.3275 31.91 kg to lbs = 70.3495 31.92 kg to lbs = 70.3715 31.93 kg to lbs = 70.3936 31.94 kg to lbs = 70.4156 31.95 kg to lbs = 70.4377 31.96 kg to lbs = 70.4597 31.97 kg to lbs = 70.4818 31.98 kg to lbs = 70.5038 31.99 kg to lbs = 70.5259 32 kg to lbs = 70.5479
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# 31.52 kg to lbs - 31.52 kilograms to pounds. Do you want to learn how much is 31.52 kg equal to lbs and how to convert 31.52 kg to lbs? You are in the right place. In this article you will find everything about kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to point out that whole this article is devoted to one amount of kilograms - that is one kilogram. So if you want to learn more about 31.52 kg to pound conversion - keep reading.. Before we go to the practice - that is 31.52 kg how much lbs conversion - we are going to tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s move on.. How to convert 31.52 kg to lbs? 31.52 kilograms it is equal 69.4897049824 pounds, so 31.52 kg is equal 69.4897049824 lbs.. ## 31.52 kgs in pounds. We are going to start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, formally known as International System of Units (in short form SI).. Sometimes the kilogram is written as kilogramme. The symbol of the kilogram is kg.. Firstly, the definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. First definition was simply but hard to use.. Then, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was replaced by a new definition.. The new definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”. One kilogram is equal 0.001 tonne. It can be also divided into 100 decagrams and 1000 grams.. ## 31.52 kilogram to pounds. You know a little bit about kilogram, so now let's go to the pound. The pound is also a unit of mass. It is needed to highlight that there are more than one kind of pound. What are we talking about? For instance, there are also pound-force. In this article we want to focus only on pound-mass.. The pound is in use in the Imperial and United States customary systems of measurements. Naturally, this unit is in use also in another systems. The symbol of the pound is lb or “.. There is no descriptive definition of the international avoirdupois pound. It is defined as exactly 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces and 7000 grains.. The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”. ### How many lbs is 31.52 kg?. 31.52 kilogram is equal to 69.4897049824 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.. ### 31.52 kg in lbs. Theoretical part is already behind us. In next part we will tell you how much is 31.52 kg to lbs. Now you know that 31.52 kg = x lbs. So it is high time to get the answer. Just see:. 31.52 kilogram = 69.4897049824 pounds.. This is an exact outcome of how much 31.52 kg to pound. You can also round off the result. After it your outcome will be exactly: 31.52 kg = 69.344 lbs.. You know 31.52 kg is how many lbs, so let’s see how many kg 31.52 lbs: 31.52 pound = 0.45359237 kilograms.. Obviously, in this case you can also round off this result. After it your outcome will be as following: 31.52 lb = 0.45 kgs.. We also want to show you 31.52 kg to how many pounds and 31.52 pound how many kg results in charts. See:. We are going to start with a table for how much is 31.52 kg equal to pound.. ### 31.52 Kilograms to Pounds conversion table. Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places). 31.52 69.4897049824 69.3440. Now see a chart for how many kilograms 31.52 pounds.. Pounds Kilograms Kilograms (rounded off to two decimal places. 31.52 0.45359237 0.45. Now you know how many 31.52 kg to lbs and how many kilograms 31.52 pound, so it is time to move on to the 31.52 kg to lbs formula.. ### 31.52 kg to pounds. To convert 31.52 kg to us lbs a formula is needed. We are going to show you a formula in two different versions. Let’s start with the first one:. Number of kilograms * 2.20462262 = the 69.4897049824 result in pounds. The first formula will give you the most accurate outcome. Sometimes even the smallest difference can be considerable. So if you want to get an accurate result - this version of a formula will be the best solution to convert how many pounds are equivalent to 31.52 kilogram.. So let’s move on to the shorer version of a formula, which also enables calculations to know how much 31.52 kilogram in pounds.. The another version of a formula is as following, let’s see:. Amount of kilograms * 2.2 = the result in pounds. As you see, the second version is simpler. It can be better option if you need to make a conversion of 31.52 kilogram to pounds in quick way, for example, during shopping. Just remember that your outcome will be not so exact.. Now we want to show you these two versions of a formula in practice. But before we are going to make a conversion of 31.52 kg to lbs we are going to show you easier way to know 31.52 kg to how many lbs without any effort.. ### 31.52 kg to lbs converter. An easier way to learn what is 31.52 kilogram equal to in pounds is to use 31.52 kg lbs calculator. What is a kg to lb converter?. Calculator is an application. Converter is based on first formula which we gave you in the previous part of this article. Thanks to 31.52 kg pound calculator you can quickly convert 31.52 kg to lbs. You only have to enter amount of kilograms which you want to convert and click ‘calculate’ button. The result will be shown in a flash.. So try to calculate 31.52 kg into lbs using 31.52 kg vs pound calculator. We entered 31.52 as an amount of kilograms. It is the result: 31.52 kilogram = 69.4897049824 pounds.. As you can see, this 31.52 kg vs lbs calculator is so simply to use.. Now we can go to our primary issue - how to convert 31.52 kilograms to pounds on your own.. #### 31.52 kg to lbs conversion. We will begin 31.52 kilogram equals to how many pounds calculation with the first formula to get the most exact outcome. A quick reminder of a formula:. Number of kilograms * 2.20462262 = 69.4897049824 the result in pounds. So what have you do to check how many pounds equal to 31.52 kilogram? Just multiply number of kilograms, in this case 31.52, by 2.20462262. It is 69.4897049824. So 31.52 kilogram is equal 69.4897049824.. You can also round it off, for example, to two decimal places. It is equal 2.20. So 31.52 kilogram = 69.3440 pounds.. It is high time for an example from everyday life. Let’s calculate 31.52 kg gold in pounds. So 31.52 kg equal to how many lbs? As in the previous example - multiply 31.52 by 2.20462262. It gives 69.4897049824. So equivalent of 31.52 kilograms to pounds, when it comes to gold, is exactly 69.4897049824.. In this case it is also possible to round off the result. It is the outcome after rounding off, this time to one decimal place - 31.52 kilogram 69.344 pounds.. Now we are going to examples converted with short formula.. #### How many 31.52 kg to lbs. Before we show you an example - a quick reminder of shorter formula:. Amount of kilograms * 2.2 = 69.344 the result in pounds. So 31.52 kg equal to how much lbs? And again, you have to multiply amount of kilogram, this time 31.52, by 2.2. See: 31.52 * 2.2 = 69.344. So 31.52 kilogram is exactly 2.2 pounds.. Make another calculation with use of this formula. Now convert something from everyday life, for example, 31.52 kg to lbs weight of strawberries.. So convert - 31.52 kilogram of strawberries * 2.2 = 69.344 pounds of strawberries. So 31.52 kg to pound mass is equal 69.344.. If you know how much is 31.52 kilogram weight in pounds and can calculate it using two different formulas, we can move on. Now we are going to show you these outcomes in tables.. #### Convert 31.52 kilogram to pounds. We know that results shown in charts are so much clearer for most of you. We understand it, so we gathered all these outcomes in charts for your convenience. Due to this you can easily compare 31.52 kg equivalent to lbs results.. Begin with a 31.52 kg equals lbs table for the first version of a formula:. Kilograms Pounds Pounds (after rounding off to two decimal places). 31.52 69.4897049824 69.3440. And now let’s see 31.52 kg equal pound chart for the second formula:. Kilograms Pounds. 31.52 69.344. As you see, after rounding off, when it comes to how much 31.52 kilogram equals pounds, the results are not different. The bigger number the more significant difference. Please note it when you need to make bigger amount than 31.52 kilograms pounds conversion.. #### How many kilograms 31.52 pound. Now you learned how to convert 31.52 kilograms how much pounds but we are going to show you something more. Are you interested what it is? What about 31.52 kilogram to pounds and ounces calculation?. We want to show you how you can calculate it step by step. Begin. How much is 31.52 kg in lbs and oz?. First things first - you need to multiply amount of kilograms, in this case 31.52, by 2.20462262. So 31.52 * 2.20462262 = 69.4897049824. One kilogram is exactly 2.20462262 pounds.. The integer part is number of pounds. So in this example there are 2 pounds.. To calculate how much 31.52 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.. So final outcome is 2 pounds and 327396192 ounces. You can also round off ounces, for instance, to two places. Then your result is equal 2 pounds and 33 ounces.. As you see, conversion 31.52 kilogram in pounds and ounces quite simply.. The last calculation which we are going to show you is calculation of 31.52 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.. To convert it it is needed another formula. Before we give you it, let’s see:. • 31.52 kilograms meters = 7.23301385 foot pounds,. • 31.52 foot pounds = 0.13825495 kilograms meters.. Now see a formula:. Number.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters. So to convert 31.52 foot pounds to kilograms meters you have to multiply 31.52 by 0.13825495. It is exactly 0.13825495. So 31.52 foot pounds is exactly 0.13825495 kilogram meters.. It is also possible to round off this result, for example, to two decimal places. Then 31.52 foot pounds is exactly 0.14 kilogram meters.. We hope that this calculation was as easy as 31.52 kilogram into pounds conversions.. This article is a huge compendium about kilogram, pound and 31.52 kg to lbs in conversion. Due to this calculation you learned 31.52 kilogram is equivalent to how many pounds.. We showed you not only how to make a calculation 31.52 kilogram to metric pounds but also two another conversions - to check how many 31.52 kg in pounds and ounces and how many 31.52 foot pounds to kilograms meters.. We showed you also other way to make 31.52 kilogram how many pounds calculations, this is using 31.52 kg en pound calculator. This is the best choice for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.. We hope that now all of you can make 31.52 kilogram equal to how many pounds calculation - on your own or with use of our 31.52 kgs to pounds converter.. Don’t wait! Calculate 31.52 kilogram mass to pounds in the way you like.. Do you need to make other than 31.52 kilogram as pounds calculation? For example, for 15 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so easy as for 31.52 kilogram equal many pounds.. ### How much is 31.52 kg in pounds. To quickly sum up this topic, that is how much is 31.52 kg in pounds , we prepared for you an additional section. Here we have for you all you need to remember about how much is 31.52 kg equal to lbs and how to convert 31.52 kg to lbs . It is down below.. What is the kilogram to pound conversion? It is a mathematical operation based on multiplying 2 numbers. Let’s see 31.52 kg to pound conversion formula . Have a look:. The number of kilograms * 2.20462262 = the result in pounds. So what is the result of the conversion of 31.52 kilogram to pounds? The accurate answer is 69.4897049824 lb.. You can also calculate how much 31.52 kilogram is equal to pounds with second, shortened version of the formula. Let’s see.. The number of kilograms * 2.2 = the result in pounds. So this time, 31.52 kg equal to how much lbs ? The result is 69.4897049824 lb.. How to convert 31.52 kg to lbs in just a moment? You can also use the 31.52 kg to lbs converter , which will do the rest for you and you will get an accurate answer .. #### Kilograms [kg]. The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.. #### Pounds [lbs]. A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.. Read more related articles:. 31.01 kg to lbs = 68.3654 31.02 kg to lbs = 68.3874 31.03 kg to lbs = 68.4094 31.04 kg to lbs = 68.4315 31.05 kg to lbs = 68.4535 31.06 kg to lbs = 68.4756 31.07 kg to lbs = 68.4976 31.08 kg to lbs = 68.5197 31.09 kg to lbs = 68.5417 31.1 kg to lbs = 68.5638 31.11 kg to lbs = 68.5858 31.12 kg to lbs = 68.6079 31.13 kg to lbs = 68.6299 31.14 kg to lbs = 68.6519 31.15 kg to lbs = 68.674 31.16 kg to lbs = 68.696 31.17 kg to lbs = 68.7181 31.18 kg to lbs = 68.7401 31.19 kg to lbs = 68.7622 31.2 kg to lbs = 68.7842 31.21 kg to lbs = 68.8063 31.22 kg to lbs = 68.8283 31.23 kg to lbs = 68.8504 31.24 kg to lbs = 68.8724 31.25 kg to lbs = 68.8945. 31.26 kg to lbs = 68.9165 31.27 kg to lbs = 68.9386 31.28 kg to lbs = 68.9606 31.29 kg to lbs = 68.9826 31.3 kg to lbs = 69.0047 31.31 kg to lbs = 69.0267 31.32 kg to lbs = 69.0488 31.33 kg to lbs = 69.0708 31.34 kg to lbs = 69.0929 31.35 kg to lbs = 69.1149 31.36 kg to lbs = 69.137 31.37 kg to lbs = 69.159 31.38 kg to lbs = 69.1811 31.39 kg to lbs = 69.2031 31.4 kg to lbs = 69.2251 31.41 kg to lbs = 69.2472 31.42 kg to lbs = 69.2692 31.43 kg to lbs = 69.2913 31.44 kg to lbs = 69.3133 31.45 kg to lbs = 69.3354 31.46 kg to lbs = 69.3574 31.47 kg to lbs = 69.3795 31.48 kg to lbs = 69.4015 31.49 kg to lbs = 69.4236 31.5 kg to lbs = 69.4456.
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31.51 kg to lbs = 69.4677 31.52 kg to lbs = 69.4897 31.53 kg to lbs = 69.5118 31.54 kg to lbs = 69.5338 31.55 kg to lbs = 69.5558 31.56 kg to lbs = 69.5779 31.57 kg to lbs = 69.5999 31.58 kg to lbs = 69.622 31.59 kg to lbs = 69.644 31.6 kg to lbs = 69.6661 31.61 kg to lbs = 69.6881 31.62 kg to lbs = 69.7102 31.63 kg to lbs = 69.7322 31.64 kg to lbs = 69.7543 31.65 kg to lbs = 69.7763 31.66 kg to lbs = 69.7983 31.67 kg to lbs = 69.8204 31.68 kg to lbs = 69.8424 31.69 kg to lbs = 69.8645 31.7 kg to lbs = 69.8865 31.71 kg to lbs = 69.9086 31.72 kg to lbs = 69.9306 31.73 kg to lbs = 69.9527 31.74 kg to lbs = 69.9747 31.75 kg to lbs = 69.9968. 31.76 kg to lbs = 70.0188 31.77 kg to lbs = 70.0409 31.78 kg to lbs = 70.0629 31.79 kg to lbs = 70.085 31.8 kg to lbs = 70.107 31.81 kg to lbs = 70.1291 31.82 kg to lbs = 70.1511 31.83 kg to lbs = 70.1731 31.84 kg to lbs = 70.1952 31.85 kg to lbs = 70.2172 31.86 kg to lbs = 70.2393 31.87 kg to lbs = 70.2613 31.88 kg to lbs = 70.2834 31.89 kg to lbs = 70.3054 31.9 kg to lbs = 70.3275 31.91 kg to lbs = 70.3495 31.92 kg to lbs = 70.3715 31.93 kg to lbs = 70.3936 31.94 kg to lbs = 70.4156 31.95 kg to lbs = 70.4377 31.96 kg to lbs = 70.4597 31.97 kg to lbs = 70.4818 31.98 kg to lbs = 70.5038 31.99 kg to lbs = 70.5259 32 kg to lbs = 70.5479.
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Algebra -> Algebra -> Quadratic Equations and Parabolas -> Quadratic Equation Customizable Word Problems -> Questions on Algebra: Quadratic Equation answered by real tutors! Log On
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Question 78766: 1) A triangle has sides 2 X – 5, 3 X + 1, and 4 X + 2. Find the polynomial that represents its perimeter. Click here to see answer by tutor_paul(490)
Question 78836: solve for p 4p^2-12p-91=0 our class is solving these type equations using the quadratic formula: (-b +and- square root of b^2-4ac)/2a so it would help a lot if you showed how to do it by plugging it into that formula. i tried the problem once and got no real solution and then plugged it in again and got 6.5,-3.5. and i don't know which one's right because my notes say that the a has to equal 1... but im not sure if thats right or not Click here to see answer by mathdoc314(58)
Question 78960: Problem: Find the slope of the line through the following pair of points (-5,-3) and (-5,2) Click here to see answer by jim_thompson5910(28476)
Question 79004: Find the slope of the line graphed (-3,-1)(-4,-3) (-5,-5)(-2,1)(-1,3) Click here to see answer by checkley75(3666)
Question 79049: Problem: The cost of producing a number of items x is given by C=mx+b , in which b is the fixed cost and m is the variable cost (the cost of producing one more item). a) If the fixed cost is \$40 and the variable cost is \$10, write the cost equation b) Graph the cost equation c) The revenue generated from the sale of x items is given by R = 50x. Graph the revenue equation on the same set of axes as the cost equation. d) How many items must be produced for the revenue to equal the cost (the break-even point)? Click here to see answer by renevencer22(21)
Question 79048: Problem: find the constant of variation k m varies directly with n; m = 144 when n = 8 Click here to see answer by rapaljer(4667)
Question 79052: I need help to write an equation of a quadratic function whose graph is a parabola that has a vertex (-3,7) and that passes through the origin. Please. Click here to see answer by funmath(2925)
Question 79052: I need help to write an equation of a quadratic function whose graph is a parabola that has a vertex (-3,7) and that passes through the origin. Please. Click here to see answer by Edwin McCravy(8879)
Question 78996: Problem: find the constant of variation k m varies directly with n; m = 144 when n = 8 Click here to see answer by ankor@dixie-net.com(15624)
Question 79074: If the sides of a square are decreased by 3 cm, the area is decreased by 81 cm2. What were the dimensions of the original square? Click here to see answer by checkley75(3666)
Question 78961: Problem: The cost of producing a number of items x is given by C=mx+b , in which b is the fixed cost and m is the variable cost (the cost of producing one more item). a) If the fixed cost is \$40 and the variable cost is \$10, write the cost equation b) Graph the cost equation c) The revenue generated from the sale of x items is given by R = 50x. Graph the revenue equation on the same set of axes as the cost equation. d) How many items must be produced for the revenue to equal the cost (the break-even point)? Click here to see answer by renevencer22(21)
Question 78955: Problem: The cost of producing a number of items x is given by C=mx+b , in which b is the fixed cost and m is the variable cost (the cost of producing one more item). a) If the fixed cost is \$40 and the variable cost is \$10, write the cost equation b) Graph the cost equation c) The revenue generated from the sale of x items is given by R = 50x. Graph the revenue equation on the same set of axes as the cost equation. d) How many items must be produced for the revenue to equal the cost (the break-even point)? Click here to see answer by renevencer22(21)
Question 78993: Find the slope of the line through the given points (2,4) and (6,-3) Click here to see answer by checkley75(3666)
Question 79110: A rectangular garden is to be surrounded by a walkway of constant width. The garden’s dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth? Click here to see answer by checkley75(3666)
Question 79110: A rectangular garden is to be surrounded by a walkway of constant width. The garden’s dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth? Click here to see answer by renevencer22(21)
Question 79089: c.. how do i complete the square for this one. i started it but i am confused. Solve by completing the square: 2x^2+8x+3=0 this is how i started but i think i am wrong... 2x^2+8x+3=0 x^2+8x=(-3)/(2) x^2+8x+(0.5 (8))^2=(-3)/(2)(0.5(8))^2 i don't know can someone help me i think i am doing it wrong... Click here to see answer by josmiceli(9649)
Question 79124: i am doing quadratic equations how do i solve by factoring? (x-1)^2 = 7 Click here to see answer by ankor@dixie-net.com(15624)
Question 79117: Find the distance between (7, – 2) and (– 5, 3) Click here to see answer by checkley75(3666)
Question 79075: The profit on a watch is given by P = X2 – 13 X – 80, where X is the number of watches sold per day. How many watches were sold on a day when there was a \$50 loss? Click here to see answer by 303795(595)
Question 79243: how do i find two consecutive positive integers such that the sum of their squares is 85? Click here to see answer by stanbon(57219)
Question 79243: how do i find two consecutive positive integers such that the sum of their squares is 85? Click here to see answer by sprolden(40)
Question 79244: can someone help me solve by completing the square? i don't understand 2x^2+10x+11=0 Click here to see answer by stanbon(57219)
Question 79239: can you help me solve this equation for x? 25x^2=13 Click here to see answer by tutorcecilia(2152)
Question 79241: can you please check this for me? i am solving equation for x 2(x-5)^2=3 (x-5)^2=3/2 x-5=sqrt=sqrt/3 x=5sqrt/3 x=15/3sqrt/3 x=15sqrt/3 Click here to see answer by Edwin McCravy(8879)
Question 79245: help me someone please? i am so lost with this. Construction. A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400ft^2, what is the width of the path? Click here to see answer by stanbon(57219)
Question 79266: write the equation of the line with given slope and y-intercept. Then graph each line using the slope and y-intercept. Slope: -3/4; y-intercept: (0,8) Click here to see answer by checkley75(3666)
Question 79264: Find the slope and y-intercept of the line represented by each of the following equations. 2x-3y=6 Click here to see answer by checkley75(3666)
Question 79378: can you help me solve by completing the square? x^2-6x-3=0 Click here to see answer by stanbon(57219)
Question 79393: Please help with solving this problem--I'm lost. When a ball is thrown, its height in feet h after t seconds is given by the equation h=vt-16t, where v is the initial upwards velocity in feet per second. If v=36 feet per second, find all values of t for which h=19 feet. Do not round any intermediate steps. Round your answers to 2 decimal places. Thanks so much!! Click here to see answer by ankor@dixie-net.com(15624)
Question 79403: I've got a word problem that has me stumped. Here it is: Working together, Kent and Monica can paint the interior of a 2000 square-foot home in 10hours. If Kent paint the house by himself, it will take 2 hours longer than if Monica paints the home by herself. How long will it take each person to paint the house working alone? Thanks so much!! Click here to see answer by ankor@dixie-net.com(15624)
Question 79401: I really need help with this word problem. We're on Quadratic Equations so I'm assuming it relates. Here it is: Sonya drives 160 miles at a certain speed. After stopping at a rest stop, she drives an additional 250 miles at a speed 15 mph slower than before the stop. If she drove 2 hours longer after the stop than before the stop, what was her speed before the stop? We need to round answer to the nearest tenth of a mile-per-hour. Any help would be greatly appreciated. Thanks! Click here to see answer by ankor@dixie-net.com(15624)
Question 79356: i really need some help to solve the following problem: y=x^2-9x+3 I tried to do it but kept getting different answers when I graphed it. Click here to see answer by josmiceli(9649)
Question 79484: Problem: Write the equation of the line L satisfying the given geometric conditions. L has y-intercept (0, -3) and is parallel to the line with equation y=2/3x+1 Click here to see answer by sprolden(40)
Question 79483: Problem: Write the equation of the line with the given slope and y-intercept. then graph the line using the slope and y-intercept Slope: -2; y-intercept: (0, 4) You must show a graph to earn credit for this question. Click here to see answer by sprolden(40)
Question 79480: Find the slope of any line perpendicular to the line through points (0, 5) and (-3, -4). Click here to see answer by sprolden(40)
Question 79508: Problem: Write the equation of the line with the given slope and y-intercept. The graph the line using the slope and y-intercept. Slope: 5; y-intercept: (0, -2) You must show a graph to earn credit for this question Click here to see answer by checkley75(3666)
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 Equation Customizable Word Problems -> 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 78766: 1) A triangle has sides 2 X – 5, 3 X + 1, and 4 X + 2. Find the polynomial that represents its perimeter. Click here to see answer by tutor_paul(490). Question 78836: solve for p 4p^2-12p-91=0 our class is solving these type equations using the quadratic formula: (-b +and- square root of b^2-4ac)/2a so it would help a lot if you showed how to do it by plugging it into that formula. i tried the problem once and got no real solution and then plugged it in again and got 6.5,-3.5. and i don't know which one's right because my notes say that the a has to equal 1... but im not sure if thats right or not Click here to see answer by mathdoc314(58). Question 78960: Problem: Find the slope of the line through the following pair of points (-5,-3) and (-5,2) Click here to see answer by jim_thompson5910(28476). Question 79004: Find the slope of the line graphed (-3,-1)(-4,-3) (-5,-5)(-2,1)(-1,3) Click here to see answer by checkley75(3666). Question 79049: Problem: The cost of producing a number of items x is given by C=mx+b , in which b is the fixed cost and m is the variable cost (the cost of producing one more item). a) If the fixed cost is \$40 and the variable cost is \$10, write the cost equation b) Graph the cost equation c) The revenue generated from the sale of x items is given by R = 50x. Graph the revenue equation on the same set of axes as the cost equation. d) How many items must be produced for the revenue to equal the cost (the break-even point)? Click here to see answer by renevencer22(21). Question 79048: Problem: find the constant of variation k m varies directly with n; m = 144 when n = 8 Click here to see answer by rapaljer(4667). Question 79052: I need help to write an equation of a quadratic function whose graph is a parabola that has a vertex (-3,7) and that passes through the origin. Please. Click here to see answer by funmath(2925). Question 79052: I need help to write an equation of a quadratic function whose graph is a parabola that has a vertex (-3,7) and that passes through the origin. Please. Click here to see answer by Edwin McCravy(8879). Question 78996: Problem: find the constant of variation k m varies directly with n; m = 144 when n = 8 Click here to see answer by ankor@dixie-net.com(15624). Question 79074: If the sides of a square are decreased by 3 cm, the area is decreased by 81 cm2. What were the dimensions of the original square? Click here to see answer by checkley75(3666). Question 78961: Problem: The cost of producing a number of items x is given by C=mx+b , in which b is the fixed cost and m is the variable cost (the cost of producing one more item). a) If the fixed cost is \$40 and the variable cost is \$10, write the cost equation b) Graph the cost equation c) The revenue generated from the sale of x items is given by R = 50x. Graph the revenue equation on the same set of axes as the cost equation. d) How many items must be produced for the revenue to equal the cost (the break-even point)? Click here to see answer by renevencer22(21). Question 78955: Problem: The cost of producing a number of items x is given by C=mx+b , in which b is the fixed cost and m is the variable cost (the cost of producing one more item). a) If the fixed cost is \$40 and the variable cost is \$10, write the cost equation b) Graph the cost equation c) The revenue generated from the sale of x items is given by R = 50x. Graph the revenue equation on the same set of axes as the cost equation. d) How many items must be produced for the revenue to equal the cost (the break-even point)? Click here to see answer by renevencer22(21). Question 78993: Find the slope of the line through the given points (2,4) and (6,-3) Click here to see answer by checkley75(3666). Question 79110: A rectangular garden is to be surrounded by a walkway of constant width. The garden’s dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth? Click here to see answer by checkley75(3666). Question 79110: A rectangular garden is to be surrounded by a walkway of constant width. The garden’s dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth? Click here to see answer by renevencer22(21). Question 79089: c.. how do i complete the square for this one. i started it but i am confused. Solve by completing the square: 2x^2+8x+3=0 this is how i started but i think i am wrong... 2x^2+8x+3=0 x^2+8x=(-3)/(2) x^2+8x+(0.5 (8))^2=(-3)/(2)(0.5(8))^2 i don't know can someone help me i think i am doing it wrong... Click here to see answer by josmiceli(9649). Question 79124: i am doing quadratic equations how do i solve by factoring? (x-1)^2 = 7 Click here to see answer by ankor@dixie-net.com(15624). Question 79117: Find the distance between (7, – 2) and (– 5, 3) Click here to see answer by checkley75(3666). Question 79075: The profit on a watch is given by P = X2 – 13 X – 80, where X is the number of watches sold per day. How many watches were sold on a day when there was a \$50 loss? Click here to see answer by 303795(595). Question 79243: how do i find two consecutive positive integers such that the sum of their squares is 85? Click here to see answer by stanbon(57219). Question 79243: how do i find two consecutive positive integers such that the sum of their squares is 85? Click here to see answer by sprolden(40). Question 79244: can someone help me solve by completing the square? i don't understand 2x^2+10x+11=0 Click here to see answer by stanbon(57219). Question 79239: can you help me solve this equation for x? 25x^2=13 Click here to see answer by tutorcecilia(2152). Question 79241: can you please check this for me? i am solving equation for x 2(x-5)^2=3 (x-5)^2=3/2 x-5=sqrt=sqrt/3 x=5sqrt/3 x=15/3sqrt/3 x=15sqrt/3 Click here to see answer by Edwin McCravy(8879). Question 79245: help me someone please? i am so lost with this. Construction. A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400ft^2, what is the width of the path? Click here to see answer by stanbon(57219). Question 79266: write the equation of the line with given slope and y-intercept. Then graph each line using the slope and y-intercept. Slope: -3/4; y-intercept: (0,8) Click here to see answer by checkley75(3666). Question 79264: Find the slope and y-intercept of the line represented by each of the following equations. 2x-3y=6 Click here to see answer by checkley75(3666). Question 79378: can you help me solve by completing the square? x^2-6x-3=0 Click here to see answer by stanbon(57219). Question 79393: Please help with solving this problem--I'm lost. When a ball is thrown, its height in feet h after t seconds is given by the equation h=vt-16t, where v is the initial upwards velocity in feet per second. If v=36 feet per second, find all values of t for which h=19 feet. Do not round any intermediate steps. Round your answers to 2 decimal places. Thanks so much!! Click here to see answer by ankor@dixie-net.com(15624). Question 79403: I've got a word problem that has me stumped. Here it is: Working together, Kent and Monica can paint the interior of a 2000 square-foot home in 10hours. If Kent paint the house by himself, it will take 2 hours longer than if Monica paints the home by herself. How long will it take each person to paint the house working alone? Thanks so much!! Click here to see answer by ankor@dixie-net.com(15624). Question 79401: I really need help with this word problem. We're on Quadratic Equations so I'm assuming it relates. Here it is: Sonya drives 160 miles at a certain speed. After stopping at a rest stop, she drives an additional 250 miles at a speed 15 mph slower than before the stop. If she drove 2 hours longer after the stop than before the stop, what was her speed before the stop? We need to round answer to the nearest tenth of a mile-per-hour. Any help would be greatly appreciated. Thanks! Click here to see answer by ankor@dixie-net.com(15624). Question 79356: i really need some help to solve the following problem: y=x^2-9x+3 I tried to do it but kept getting different answers when I graphed it. Click here to see answer by josmiceli(9649). Question 79484: Problem: Write the equation of the line L satisfying the given geometric conditions. L has y-intercept (0, -3) and is parallel to the line with equation y=2/3x+1 Click here to see answer by sprolden(40). Question 79483: Problem: Write the equation of the line with the given slope and y-intercept. then graph the line using the slope and y-intercept Slope: -2; y-intercept: (0, 4) You must show a graph to earn credit for this question. Click here to see answer by sprolden(40). Question 79480: Find the slope of any line perpendicular to the line through points (0, 5) and (-3, -4). Click here to see answer by sprolden(40). Question 79508: Problem: Write the equation of the line with the given slope and y-intercept. The graph the line using the slope and y-intercept.
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Slope: 5; y-intercept: (0, -2) You must show a graph to earn credit for this question Click here to see answer by checkley75(3666). 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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https://secondsminutes.com/1432-seconds-in-minutes
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# 1432 seconds in minutes
## Result
1432 seconds equals 23.87 minutes
You can also convert 1432 seconds to minutes and seconds
## Conversion formula
Multiply the amount of seconds by the conversion factor to get the result in minutes:
1432 s × 0.0166667 = 23.87 min
## How to convert 1432 seconds to minutes?
The conversion factor from seconds to minutes is 0.0166667, which means that 1 seconds is equal to 0.0166667 minutes:
1 s = 0.0166667 min
To convert 1432 seconds into minutes we have to multiply 1432 by the conversion factor in order to get the amount from seconds to minutes. We can also form a proportion to calculate the result:
1 s → 0.0166667 min
1432 s → T(min)
Solve the above proportion to obtain the time T in minutes:
T(min) = 1432 s × 0.0166667 min
T(min) = 23.87 min
The final result is:
1432 s → 23.87 min
We conclude that 1432 seconds is equivalent to 23.87 minutes:
1432 seconds = 23.87 minutes
## Result approximation:
For practical purposes we can round our final result to an approximate numerical value. In this case one thousand four hundred thirty-two seconds is approximately twenty-three point eight seven minutes:
1432 seconds ≅ 23.87 minutes
## Conversion table
For quick reference purposes, below is the seconds to minutes conversion table:
seconds (s) minutes (min)
1433 seconds 23.883381 minutes
1434 seconds 23.900048 minutes
1435 seconds 23.916715 minutes
1436 seconds 23.933381 minutes
1437 seconds 23.950048 minutes
1438 seconds 23.966715 minutes
1439 seconds 23.983381 minutes
1440 seconds 24.000048 minutes
1441 seconds 24.016715 minutes
1442 seconds 24.033381 minutes
## Units definitions
The units involved in this conversion are seconds and minutes. This is how they are defined:
### Seconds
The second (symbol: s) (abbreviated s or sec) is the base unit of time in the International System of Units (SI). It is qualitatively defined as the second division of the hour by sixty, the first division by sixty being the minute. The SI definition of second is "the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom". Seconds may be measured using a mechanical, electrical or an atomic clock. SI prefixes are combined with the word second to denote subdivisions of the second, e.g., the millisecond (one thousandth of a second), the microsecond (one millionth of a second), and the nanosecond (one billionth of a second). Though SI prefixes may also be used to form multiples of the second such as kilosecond (one thousand seconds), such units are rarely used in practice. The more common larger non-SI units of time are not formed by powers of ten; instead, the second is multiplied by 60 to form a minute, which is multiplied by 60 to form an hour, which is multiplied by 24 to form a day. The second is also the base unit of time in other systems of measurement: the centimetre–gram–second, metre–kilogram–second, metre–tonne–second, and foot–pound–second systems of units.
### Minutes
The minute is a unit of time or of angle. As a unit of time, the minute (symbol: min) is equal to 1⁄60 (the first sexagesimal fraction) of an hour, or 60 seconds. In the UTC time standard, a minute on rare occasions has 61 seconds, a consequence of leap seconds (there is a provision to insert a negative leap second, which would result in a 59-second minute, but this has never happened in more than 40 years under this system). As a unit of angle, the minute of arc is equal to 1⁄60 of a degree, or 60 seconds (of arc). Although not an SI unit for either time or angle, the minute is accepted for use with SI units for both. The SI symbols for minute or minutes are min for time measurement, and the prime symbol after a number, e.g. 5′, for angle measurement. The prime is also sometimes used informally to denote minutes of time. In contrast to the hour, the minute (and the second) does not have a clear historical background. What is traceable only is that it started being recorded in the Middle Ages due to the ability of construction of "precision" timepieces (mechanical and water clocks). However, no consistent records of the origin for the division as 1⁄60 part of the hour (and the second 1⁄60 of the minute) have ever been found, despite many speculations.
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# 1432 seconds in minutes. ## Result. 1432 seconds equals 23.87 minutes. You can also convert 1432 seconds to minutes and seconds. ## Conversion formula. Multiply the amount of seconds by the conversion factor to get the result in minutes:. 1432 s × 0.0166667 = 23.87 min. ## How to convert 1432 seconds to minutes?. The conversion factor from seconds to minutes is 0.0166667, which means that 1 seconds is equal to 0.0166667 minutes:. 1 s = 0.0166667 min. To convert 1432 seconds into minutes we have to multiply 1432 by the conversion factor in order to get the amount from seconds to minutes. We can also form a proportion to calculate the result:. 1 s → 0.0166667 min. 1432 s → T(min). Solve the above proportion to obtain the time T in minutes:. T(min) = 1432 s × 0.0166667 min. T(min) = 23.87 min. The final result is:. 1432 s → 23.87 min. We conclude that 1432 seconds is equivalent to 23.87 minutes:. 1432 seconds = 23.87 minutes. ## Result approximation:. For practical purposes we can round our final result to an approximate numerical value. In this case one thousand four hundred thirty-two seconds is approximately twenty-three point eight seven minutes:. 1432 seconds ≅ 23.87 minutes. ## Conversion table. For quick reference purposes, below is the seconds to minutes conversion table:. seconds (s) minutes (min). 1433 seconds 23.883381 minutes. 1434 seconds 23.900048 minutes. 1435 seconds 23.916715 minutes. 1436 seconds 23.933381 minutes. 1437 seconds 23.950048 minutes. 1438 seconds 23.966715 minutes. 1439 seconds 23.983381 minutes. 1440 seconds 24.000048 minutes. 1441 seconds 24.016715 minutes. 1442 seconds 24.033381 minutes. ## Units definitions. The units involved in this conversion are seconds and minutes. This is how they are defined:. ### Seconds. The second (symbol: s) (abbreviated s or sec) is the base unit of time in the International System of Units (SI). It is qualitatively defined as the second division of the hour by sixty, the first division by sixty being the minute. The SI definition of second is "the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom". Seconds may be measured using a mechanical, electrical or an atomic clock. SI prefixes are combined with the word second to denote subdivisions of the second, e.g., the millisecond (one thousandth of a second), the microsecond (one millionth of a second), and the nanosecond (one billionth of a second). Though SI prefixes may also be used to form multiples of the second such as kilosecond (one thousand seconds), such units are rarely used in practice. The more common larger non-SI units of time are not formed by powers of ten; instead, the second is multiplied by 60 to form a minute, which is multiplied by 60 to form an hour, which is multiplied by 24 to form a day. The second is also the base unit of time in other systems of measurement: the centimetre–gram–second, metre–kilogram–second, metre–tonne–second, and foot–pound–second systems of units.. ### Minutes. The minute is a unit of time or of angle. As a unit of time, the minute (symbol: min) is equal to 1⁄60 (the first sexagesimal fraction) of an hour, or 60 seconds. In the UTC time standard, a minute on rare occasions has 61 seconds, a consequence of leap seconds (there is a provision to insert a negative leap second, which would result in a 59-second minute, but this has never happened in more than 40 years under this system). As a unit of angle, the minute of arc is equal to 1⁄60 of a degree, or 60 seconds (of arc). Although not an SI unit for either time or angle, the minute is accepted for use with SI units for both. The SI symbols for minute or minutes are min for time measurement, and the prime symbol after a number, e.g. 5′, for angle measurement. The prime is also sometimes used informally to denote minutes of time.
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In contrast to the hour, the minute (and the second) does not have a clear historical background. What is traceable only is that it started being recorded in the Middle Ages due to the ability of construction of "precision" timepieces (mechanical and water clocks). However, no consistent records of the origin for the division as 1⁄60 part of the hour (and the second 1⁄60 of the minute) have ever been found, despite many speculations.
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https://math.stackexchange.com/questions/3087270/proof-verification-the-orthogonal-complement-of-the-column-space-is-the-left-nu
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# Proof Verification: the orthogonal complement of the column space is the left nullspace
Can someone please check my proof and my definitions.
Let $$A \in \mathbb{R}^{n \times m}$$ be my matrix.
The left null space of $$A$$ is written as,
$$\mathcal{N}(A^\top) = \{x \in \mathbb{R}^n| A^\top x = 0\}$$
The orthogonal complement of the column space $$\mathcal{C}(A)$$ is written as,
$$\mathcal{C}(A)^\perp = \{x \in \mathbb{R}^n | x^\top y = 0, \forall y \in \mathcal{C}(A)\}$$
We want to show that $$\mathcal{N}(A^\top) = \mathcal{C}(A)^\perp$$
First, we show, $$\mathcal{N}(A^\top) \subseteq \mathcal{C}(A)^\perp$$
Let $$x \in \mathcal{N}(A^\top)$$, then $$A^\top x = 0 \implies x^\top A = 0^\top \implies x^\top Av= 0^\top v, \forall v \in \mathcal{C}(A) \implies x^\top y = 0 , y = Av$$, $$\implies x \in C(A)^\perp$$.
Next, we show, $$\mathcal{N}(A^\top) \supseteq \mathcal{C}(A)^\perp$$
Let $$x \in C(A)^\perp$$, then $$x^\top y = 0$$, forall $$y \in C(A)$$. But $$y = Av, \forall v \in \mathbb{R}^n$$. Hence, $$x^\top y = x^\top Av = v^\top A^\top x.$$ For all $$v \neq 0, A^\top x = 0$$, hence $$x \in \mathcal{N}(A^\top)$$.
I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.
$$y \in C(A)$$ means that there exists (at least one) $$v$$ of appropriate dimension such that $$y = Av$$.
So we can say: For $$x \in C(A)^{\perp}$$, then $$x^T y = 0$$ for every $$y \in C(A)$$. For every $$y \in C(A)$$, we can express $$y = Av$$ for some (nonzero) $$v$$. So we can always express $$x^T y$$ as $$x^T Av$$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $$v \neq 0$$, so we must have $$A^T x = 0$$, i.e., $$x \in N(A^T)$$.
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# Proof Verification: the orthogonal complement of the column space is the left nullspace. Can someone please check my proof and my definitions.. Let $$A \in \mathbb{R}^{n \times m}$$ be my matrix.. The left null space of $$A$$ is written as,. $$\mathcal{N}(A^\top) = \{x \in \mathbb{R}^n| A^\top x = 0\}$$. The orthogonal complement of the column space $$\mathcal{C}(A)$$ is written as,. $$\mathcal{C}(A)^\perp = \{x \in \mathbb{R}^n | x^\top y = 0, \forall y \in \mathcal{C}(A)\}$$. We want to show that $$\mathcal{N}(A^\top) = \mathcal{C}(A)^\perp$$. First, we show, $$\mathcal{N}(A^\top) \subseteq \mathcal{C}(A)^\perp$$. Let $$x \in \mathcal{N}(A^\top)$$, then $$A^\top x = 0 \implies x^\top A = 0^\top \implies x^\top Av= 0^\top v, \forall v \in \mathcal{C}(A) \implies x^\top y = 0 , y = Av$$, $$\implies x \in C(A)^\perp$$.. Next, we show, $$\mathcal{N}(A^\top) \supseteq \mathcal{C}(A)^\perp$$. Let $$x \in C(A)^\perp$$, then $$x^\top y = 0$$, forall $$y \in C(A)$$. But $$y = Av, \forall v \in \mathbb{R}^n$$. Hence, $$x^\top y = x^\top Av = v^\top A^\top x.$$ For all $$v \neq 0, A^\top x = 0$$, hence $$x \in \mathcal{N}(A^\top)$$.. I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.
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$$y \in C(A)$$ means that there exists (at least one) $$v$$ of appropriate dimension such that $$y = Av$$.. So we can say: For $$x \in C(A)^{\perp}$$, then $$x^T y = 0$$ for every $$y \in C(A)$$. For every $$y \in C(A)$$, we can express $$y = Av$$ for some (nonzero) $$v$$. So we can always express $$x^T y$$ as $$x^T Av$$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $$v \neq 0$$, so we must have $$A^T x = 0$$, i.e., $$x \in N(A^T)$$.
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https://whatdoesmean.net/what-is-the-absolute-population/
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# What is the absolute population?
## What Does absolute population Mean
We explain what the absolute population is, how it is calculated and examples of absolute population in Mexico, Brazil and China.
## What is the absolute population?
The absolute population is the number of people who live in a territory during a certain period. This value also contemplates the interrelation of the following statistical variables :
• Total births and immigration .
• Total deaths and emigrations.
Knowing the number of the absolute population allows predicting the behavior of the inhabitants of a territory, conducting field research and making projections in order to foresee social, economic and health problems , among other advantages.
### What do we need to calculate the absolute population?
The absolute population is calculated with the total number of the population , adding the birth rate and the immigration rate, and subtracting the death rate and the emigration rate. All values are calculated based on the same period of time .
To perform the calculation, we will need:
Birth rate
It is the number of births that occurred in a territory during a certain time . To calculate the birth rate, the number of births (for example, over a year) is divided by the number of the total population reached in the same period. The result is multiplied by a thousand, and that final number corresponds to the birth rate.
To illustrate it in a simple way, we can perform the following calculation: if there were seven hundred births in a year and the total population reached two hundred thousand inhabitants, we must divide 700 by 200,000, and the result is multiplied by 1,000 which gives a total of 3, 5. The number of the birth rate is 3.5 births per thousand inhabitants.
Mortality rate
It corresponds to the number of people who died in a region during a certain time . To calculate the mortality rate, the number of deaths (for example, those that occurred during one year) is divided by the number of the total population reached in the same period. The result is multiplied by a thousand, and that final number corresponds to the death rate.
To illustrate it in a simple way, we can perform the following calculation: if there were three hundred deaths in a year and the total population reached two hundred thousand inhabitants, we must divide 300 by 200,000, and the result is multiplied by 1,000 which gives a total of 0.015. The number of the mortality rate is 1.5 deaths per thousand inhabitants.
Migration rate
It corresponds to income and expenses, and is calculated in a similar way to the birth and death rates. The amount of income with permanent stay in a territory that existed during a year is counted . The result is divided by the amount of total population reached in that period of time and the final number is divided by a thousand.
To illustrate it in a simple way, we can perform the following calculation: if there were 400 immigrations during a year, that number is divided by 200,000, which is the number of inhabitants. The result is 0.002, which, when multiplied by 1,000, gives a final value of 8. The immigration rate is 2 entries per thousand inhabitants.
### How do we calculate the absolute population?
We said then that the absolute population is calculated with the total number of the population, adding the birth rate and the immigration rate, and subtracting the death rate and the emigration rate.
To finally obtain the absolute population number, the total population number must be added together with the statistical values of the respective rates (birth and immigration, mortality and emigration) that were calculated during the same period.
To make the values expressed as “rates” compatible with the total population number, a “simple rule of three” operation can be performed.
### Absolute population examples
• The absolute population of Mexico in 2017 was 129,163,276 and in 2019 it reached 132,242,957 inhabitants . In the last two years it had a total increase of 3,079,679 inhabitants.
• The absolute population of China in 2017 was 1,409,517,397 and in 2019 it reached 1,419,791,153 . In the last two years it had a total increase of 10,273,756 inhabitants.
• The absolute population of Brazil in 2017 was 209,288,279 and in 2019 it reached 212,310,252 inhabitants . In the last two years it had a total increase of 3,021,973 inhabitants.
### Relative population
Relative population, also known as “ population density ”, refers to the distribution of the number of inhabitants in a territory (country, state, province, county, etc.). It is calculated by taking the number of inhabitants per square meter or square mile, depending on the unit of length used by each country.
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# What is the absolute population?. ## What Does absolute population Mean. We explain what the absolute population is, how it is calculated and examples of absolute population in Mexico, Brazil and China.. ## What is the absolute population?. The absolute population is the number of people who live in a territory during a certain period. This value also contemplates the interrelation of the following statistical variables :. • Total births and immigration .. • Total deaths and emigrations.. Knowing the number of the absolute population allows predicting the behavior of the inhabitants of a territory, conducting field research and making projections in order to foresee social, economic and health problems , among other advantages.. ### What do we need to calculate the absolute population?. The absolute population is calculated with the total number of the population , adding the birth rate and the immigration rate, and subtracting the death rate and the emigration rate. All values are calculated based on the same period of time .. To perform the calculation, we will need:. Birth rate. It is the number of births that occurred in a territory during a certain time . To calculate the birth rate, the number of births (for example, over a year) is divided by the number of the total population reached in the same period. The result is multiplied by a thousand, and that final number corresponds to the birth rate.. To illustrate it in a simple way, we can perform the following calculation: if there were seven hundred births in a year and the total population reached two hundred thousand inhabitants, we must divide 700 by 200,000, and the result is multiplied by 1,000 which gives a total of 3, 5. The number of the birth rate is 3.5 births per thousand inhabitants.. Mortality rate. It corresponds to the number of people who died in a region during a certain time . To calculate the mortality rate, the number of deaths (for example, those that occurred during one year) is divided by the number of the total population reached in the same period. The result is multiplied by a thousand, and that final number corresponds to the death rate.. To illustrate it in a simple way, we can perform the following calculation: if there were three hundred deaths in a year and the total population reached two hundred thousand inhabitants, we must divide 300 by 200,000, and the result is multiplied by 1,000 which gives a total of 0.015. The number of the mortality rate is 1.5 deaths per thousand inhabitants.. Migration rate. It corresponds to income and expenses, and is calculated in a similar way to the birth and death rates. The amount of income with permanent stay in a territory that existed during a year is counted . The result is divided by the amount of total population reached in that period of time and the final number is divided by a thousand.. To illustrate it in a simple way, we can perform the following calculation: if there were 400 immigrations during a year, that number is divided by 200,000, which is the number of inhabitants. The result is 0.002, which, when multiplied by 1,000, gives a final value of 8. The immigration rate is 2 entries per thousand inhabitants.. ### How do we calculate the absolute population?. We said then that the absolute population is calculated with the total number of the population, adding the birth rate and the immigration rate, and subtracting the death rate and the emigration rate.. To finally obtain the absolute population number, the total population number must be added together with the statistical values of the respective rates (birth and immigration, mortality and emigration) that were calculated during the same period.. To make the values expressed as “rates” compatible with the total population number, a “simple rule of three” operation can be performed.. ### Absolute population examples. • The absolute population of Mexico in 2017 was 129,163,276 and in 2019 it reached 132,242,957 inhabitants . In the last two years it had a total increase of 3,079,679 inhabitants.. • The absolute population of China in 2017 was 1,409,517,397 and in 2019 it reached 1,419,791,153 . In the last two years it had a total increase of 10,273,756 inhabitants.. • The absolute population of Brazil in 2017 was 209,288,279 and in 2019 it reached 212,310,252 inhabitants . In the last two years it had a total increase of 3,021,973 inhabitants.
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### Relative population. Relative population, also known as “ population density ”, refers to the distribution of the number of inhabitants in a territory (country, state, province, county, etc.). It is calculated by taking the number of inhabitants per square meter or square mile, depending on the unit of length used by each country.
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https://www.vedantu.com/question-answer/find-the-square-root-of-12sqrt-5-+-2sqrt-55-a-class-9-maths-cbse-5f5fa86068d6b37d16353cac
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# Find the square root of: $12\sqrt 5 + 2\sqrt {55}$ A. $\left( {\sqrt {11} + 1} \right)\sqrt[4]{5}$ B. $\sqrt[4]{5}\left( {1 + \sqrt 5 } \right)$ C. $\sqrt[4]{5}\left( {\sqrt {11} + \sqrt 5 } \right)$ D. $\sqrt 5 \left( {\sqrt {11} + 1} \right)$
Last updated date: 19th Jun 2024
Total views: 413.1k
Views today: 5.13k
Verified
413.1k+ views
Hint: In this question, we will use factorization, and expansion of algebraic identities. For this problem, we will use the algebraic identity ${(a + b)^2} = {a^2} + 2ab + {b^2}$ .
Complete step by step solution:
Now, in this question, we have to find the square root of $12\sqrt 5 + 2\sqrt {55}$.
So it will become: $\sqrt {12\sqrt 5 + 2\sqrt {55} }$, which on simplification will become:
$\sqrt {12\sqrt 5 + 2\sqrt {55} } \\ = \sqrt {12\sqrt 5 + 2\sqrt {5 \times 11} } \\ = \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\$
Now, to solve $\sqrt {12\sqrt 5 + 2\sqrt {55} }$, we will take $\sqrt 5$ common within the under root and get:
$= \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\ = \sqrt {\sqrt 5 (12 + 2\sqrt {11} )} \\$
Now we will change the term $(12 + 2\sqrt {11} )$ inside the under root sign to express it in terms of ${(a + b)^2}$ .
Now we can write $(12 + 2\sqrt {11} )$ as:
$(1 + 11 + 2\sqrt {11} )$ which can we reframed as:
$({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} )$, comparing it with the RHS of the expansion of the algebraic identity ${(a + b)^2}$ which is given as:
${a^2} + 2ab + {b^2}$
We will get $({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} )$= ${a^2} + 2ab + {b^2}$
So that,
${a^2} = {1^2}, \\ 2ab = 2.1.\sqrt {11} \\ {b^2} = {(\sqrt {11} )^2} \\$
Such that we get :
$a = 1, \\ 2ab = 2.1.\sqrt {11} \\ b = \sqrt {11} \\$
Now, since
${a^2} + 2ab + {b^2} = {(a + b)^2}$
Then putting the values obtained above:
$({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} ) = {(1 + \sqrt {11} )^2}$
Therefore $\sqrt {12\sqrt 5 + 2\sqrt {55} }$ will now become:
$= \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\ = \sqrt {\sqrt 5 (12 + 2\sqrt {11} )} \\ = \sqrt {\sqrt 5 ({1^2} + {{(\sqrt {11} )}^2} + 2\sqrt {11} )} \\ = \sqrt {\sqrt 5 {{(1 + \sqrt {11} )}^2}} \\ = \sqrt {\sqrt 5 } (1 + \sqrt {11} ) \\ = \sqrt[4]{5}(1 + \sqrt {11} ) \\$
So, finally we can say that :
Square root of $12\sqrt 5 + 2\sqrt {55}$
$= \sqrt[4]{5}(\sqrt {11} + 1)$
Hence, the correct answer is option A.
Note: We cannot afford to forget the square root operation throughout the solution of this problem. For such problems, which require us to find the square root of another square root, we need to identify the algebraic expansion accurately so that we can get the correct corresponding algebraic identity to simplify and evaluate the square root.
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Courses. Courses for Kids. Free study material. Offline Centres. More. Store. # Find the square root of: $12\sqrt 5 + 2\sqrt {55}$ A. $\left( {\sqrt {11} + 1} \right)\sqrt[4]{5}$ B. $\sqrt[4]{5}\left( {1 + \sqrt 5 } \right)$ C. $\sqrt[4]{5}\left( {\sqrt {11} + \sqrt 5 } \right)$ D. $\sqrt 5 \left( {\sqrt {11} + 1} \right)$. Last updated date: 19th Jun 2024. Total views: 413.1k. Views today: 5.13k. Verified. 413.1k+ views. Hint: In this question, we will use factorization, and expansion of algebraic identities. For this problem, we will use the algebraic identity ${(a + b)^2} = {a^2} + 2ab + {b^2}$ .. Complete step by step solution:. Now, in this question, we have to find the square root of $12\sqrt 5 + 2\sqrt {55}$.. So it will become: $\sqrt {12\sqrt 5 + 2\sqrt {55} }$, which on simplification will become:. $\sqrt {12\sqrt 5 + 2\sqrt {55} } \\ = \sqrt {12\sqrt 5 + 2\sqrt {5 \times 11} } \\ = \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\$. Now, to solve $\sqrt {12\sqrt 5 + 2\sqrt {55} }$, we will take $\sqrt 5$ common within the under root and get:. $= \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\ = \sqrt {\sqrt 5 (12 + 2\sqrt {11} )} \\$. Now we will change the term $(12 + 2\sqrt {11} )$ inside the under root sign to express it in terms of ${(a + b)^2}$ .. Now we can write $(12 + 2\sqrt {11} )$ as:. $(1 + 11 + 2\sqrt {11} )$ which can we reframed as:. $({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} )$, comparing it with the RHS of the expansion of the algebraic identity ${(a + b)^2}$ which is given as:. ${a^2} + 2ab + {b^2}$. We will get $({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} )$= ${a^2} + 2ab + {b^2}$. So that,. ${a^2} = {1^2}, \\ 2ab = 2.1.\sqrt {11} \\ {b^2} = {(\sqrt {11} )^2} \\$. Such that we get :. $a = 1, \\ 2ab = 2.1.\sqrt {11} \\ b = \sqrt {11} \\$. Now, since. ${a^2} + 2ab + {b^2} = {(a + b)^2}$. Then putting the values obtained above:. $({1^2} + {(\sqrt {11} )^2} + 2\sqrt {11} ) = {(1 + \sqrt {11} )^2}$. Therefore $\sqrt {12\sqrt 5 + 2\sqrt {55} }$ will now become:. $= \sqrt {12\sqrt 5 + 2\sqrt 5 .\sqrt {11} } \\ = \sqrt {\sqrt 5 (12 + 2\sqrt {11} )} \\ = \sqrt {\sqrt 5 ({1^2} + {{(\sqrt {11} )}^2} + 2\sqrt {11} )} \\ = \sqrt {\sqrt 5 {{(1 + \sqrt {11} )}^2}} \\ = \sqrt {\sqrt 5 } (1 + \sqrt {11} ) \\ = \sqrt[4]{5}(1 + \sqrt {11} ) \\$. So, finally we can say that :. Square root of $12\sqrt 5 + 2\sqrt {55}$. $= \sqrt[4]{5}(\sqrt {11} + 1)$. Hence, the correct answer is option A.
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Note: We cannot afford to forget the square root operation throughout the solution of this problem. For such problems, which require us to find the square root of another square root, we need to identify the algebraic expansion accurately so that we can get the correct corresponding algebraic identity to simplify and evaluate the square root.
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http://www.rfcafe.com/references/electrical/butterworth-lowpass-filter-gain-phase-group-delay-equations.htm
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Please Support RF Cafe by purchasing my ridiculously low−priced products, all of which I created.
# Butterworth Lowpass Filter Gain, Phase, and Group Delay Equations
This may be the only place you will ever find the explicit formulas for Butterworth lowpass filter gain, phase, and group delay based on the basic Butterworth polynomials (until people copy my work and don't give credit*). Hoping to avoid having to do the messy math of complex numbers requiring separating out the real and imaginary parts to obtain phase (θ) - and then to get the first derivative for group delay (τd) - I searched high and low, far and wide for closed form equations, to no avail. All I could ever find was instruction to apply arg{H(ω)} for phase (θ) and -dθ/dω for group delay -- pretty useless if you want to plug an equation into a spreadsheet or software.
Finally, I decided if I wanted the solutions, I would have to slog through all the equations myself. Here is the result. The basic steps are as follow:
• Simplify the polynomial factors by multiplying out the parenthetical quantities.
• Substitute jω (or iω, but engineers use jω) for each "s," where ω is the frequency (ωn) being evaluated divided by the cutoff frequency (ωco); hence, jω = jωnco.
• Since j2 = -1 by definition, all even powers of j result in "real" parts and all odd powers of j result in "imaginary" parts. E.g., for 3rd-order:
• H(s) = (s + 1)(s2 + s + 1) = js3 + 2s2 +2s + 1
• H(jω) = (jω)3 + 2(jω)2 +2(jω) + 1 = j3ω3 + 2(j2ω2) +2(jω) + 1
= -jω3 + 2(-ω2) + j2ω + 1 = -jω3 - 2ω2 + j2ω + 1
• Re{H(jω)} = 1 - 2ω2 (parts w/o j), and Im{H(jω)} = 2ω - ω3 (parts w/j)
• Gain is G(ω) = sqrt [Re{H(jω)2 + Im{H(jω)2]
• The phase angle is by definition θ(ω) = tan-1 (Re{H(jω)}/Im{H(jω)})
θ(ω) = tan-1 [(1 - 2ω2) / (2ω - ω3)]
• Group delay is defined as τd(ω) = -dθ(ω)/dω. This is where it gets hard, mainly due to the arctangent, but the fraction is no walk in the park, either. Fortunately, there are online math websites like SymboLab that will handle such derivatives. For that matter, it will multiply all the factors for you as well, reducing the opportunity for errors.
• Finally, do the ω = ωnco substitution everywhere. ωn is the frequency at which the equation is being evaluated, and ωco is the cutoff frequency. Frequency units cancel out, so a 1 Hz cutoff plots the same as a 1 kHz cutoff or a 1 GHz cutoff for gain and phase. The group delay scale needs to be divided by a factor equal to the frequency units (÷103 for kHz, ÷106 for MHz, etc.).
This table represents hours of work on my part. I will eventually get around to lowpass, highpass, bandpass, and bandreject for Butterworth, Chebyshev, and Bessel (and maybe Gaussian) filters.
Butterworth Lowpass, Highpass, Bandpass, and Bandstop Filter Calculator with Gain, Phase and Group Delay are now part of my free RF Cafe Espresso Engineering Workbook™!
* I had Archive.org save a copy of this page in order to prove this is my original work.
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GainH(ω) s + 1 (jω) + 1 sqrt (1 + ω2) Phase θ(ω) tan-1 (1/ω) Group Delay τd(ω) 1/(1 + ω2)
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Gain H(ω) s2 + 1 + 1.4142s (jω)2 + 1.4142(jω) + 1 (1 - ω2) + j1.4142ω sqrt [(1 - ω2)2 + (1.4142ω)2] Phase θ(ω) tan-1 [(1 - ω2)/1.4142ω] Group Delay τd(ω) (1 + ω2)/[1.4142(0.5ω4 - 0.0001ω2) + 0.5]
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Gain H(ω) s3 + 2s2 + 2s + 1(jω)3 + 2(jω)2 + 2(jω) + 1(1 - 2ω2) + j(2ω - ω3)sqrt [(1 - 2ω2)2 + (2ω - ω3)2] Phase θ(ω) tan-1 [(1 - 2*ω2)/(2ω - ω3)] Group Delay τd(ω) (2*ω4 + ω2 + 2)/(ω6 + 1)
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Gain H(ω) s4 + 2.6131s3 + 3.4142s2 + 2.6131s + 1 (jω)4 + 2.6131(jω)3 + 3.4142(jω)2 + 2.6131(jω) + 1 (ω4 - 3.4142ω2 + 1) + j(-2.6131ω3 + 2.6131ω) sqrt [(ω4 - 3.4142ω2 + 1)2 + (-2.6131ω3 + 2.6131ω)2] Phase θ(ω) tan-1 [(ω4 - 3.4142ω2 + 1)/(-2.6131ω3 + 2.6131ω)] Group Delay τd(ω) (2.6131ω6 + 1.08234602ω4 + 1.08234602ω2 + 2.6131) / (ω8 - 0.00010839ω6 + 0.00017842ω4 - 0.00010839ω2 + 1)
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Gain H(ω) s5 + 3.2361s4 + 5.2361s3 + 5.2361s2 + 3.2361s + 1 (jω)5 + 3.2361(jω)4 + 5.2361(jω)3 + 5.2361(jω)2 + 3.2361(jω) + 1 (3.2361ω4 - 5.2361ω2+1) + j(ω5 - 5.2361ω3 + 3.2361ω) sqrt [(3.2361ω4 - 5.2361ω2 + 1)2 + (ω5 - 5.2361ω3 + 3.2361ω)2] Phase θ(ω) tan-1 [(3.2361ω4 - 5.2361ω2 + 1)/(ω5 - 5.2361ω3 + 3.2361ω)] Group Delay τd(ω) (3.2361ω8 + 1.23624321ω6 + 0.99971358ω4 + 1.23624321ω2 + 3.2361) / (ω10 + 0.00014321ω8 - 0.00014321ω6 - 0.00014321ω4 + 0.00014321ω2 + 1)
6
Gain H(ω) s6 + 3.8637s5 + 7.4641s4 + 9.1416s3 + 7.4641s2 + 3.8637s + 1 (jω)6 + 3.8637(jω)5 + 7.4641(jω)4 + 9.1416(jω)3 + 7.4641(jω)2 + 3.8637(jω) + 1 (-ω6 + 7.4641ω4 - 7.4641ω2 + 1) + j(3.8637ω5 - 9.1416ω3 + 3.8637ω) sqrt [(-ω6 + 7.4641ω4 - 7.4641ω2 + 1)2 + (3.8637ω5 - 9.1416ω3 + 3.8637ω)2] Phase θ(ω) tan-1 [(-ω6 + 7.4641ω4 - 7.4641ω2 + 1)/(3.8637ω5 - 9.1416ω3 + 3.8637ω)] Group Delay τd(ω) (3.8637*ω10 + 1.41424317*ω8 + 1.03518705*ω6 + 1.03518705*ω4 + 1.41424317*ω2 + 3.8637) / (ω12 - 0.00002231*ω10 + 0.00018897*ω8 - 0.00037168*ω6 + 0.00018897*ω4 - 0.00002231*ω2 + 1)
7
Gain H(ω) s7 + 4.4940s6 + 10.0978s5 + 14.5920s4 + 14.5920s3 + 10.0978s2 + 4.4940s + 1 (jω)7 + 4.4940(jω)6 + 10.0978(jω)5 + 14.5920(jω)4 + 14.5920(jω)3 + 10.0978(jω)2 + 4.4940(jω) + 1 (-4.494ω6 + 14.592ω4 - 10.0978ω2 + 1) + j(-ω7 + 10.0978ω5 - 14.592ω3 + 4.494ω) sqrt [(-4.494ω6 + 14.592ω4 - 10.0978ω2 + 1)2 + (-ω7 + 10.0978ω5 - 14.592ω3 + 4.494ω)2] Phase θ(ω) tan-1 [(-4.494ω6 + 14.592ω4 - 10.0978ω2 + 1)/(-ω7 + 10.0978ω5 - 14.592ω3 + 4.494ω)] Group Delay τd(ω) (4.494ω12 + 1.6035132ω10 + 1.1067536ω8 + 1.00994948ω6 + 1.1067536ω4 + 1.6035132ω2 + 4.494) / (ω14+0.000436ω12 - 0.00333116ω10 + 0.0032952ω8 + 0.0032952ω6-0.00333116ω4 + 0.000436ω2 + 1)
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Gain H(ω) s8 + 5.1528s7 + 13.1371s6 + 21.8462s5 + 25.6884s4 + 21.8462s3 + 13.1371s2 + 5.1528s + 1 (jω)8 + 5.1258(jω)7 + 13.1371(jω)6 + 21.8462(jω)5 + 25.6884(jω)4 + 21.8462(jω)3 + 13.1371(jω)2 + 5.1258(jω) + 1 (ω8 - 13.1371ω6 + 25.6884ω4 - 13.1371ω2 + 1) + j(-5.1258ω7 + 21.8462ω5 - 21.8462ω3 + 5.1258ω) Phase θ(ω) tan-1 [(ω8 - 13.1371ω6 + 25.6884ω4 - 13.1371ω2 + 1)/(-5.1258ω7 + 21.8462ω5 - 21.8462ω3 + 5.1258ω)] Group Delay τd(ω) (5.1258ω14 + 1.79954718ω12 + 1.20591186ω10 + 1.01691792ω8 + 1.01691792ω6 - 1.20591186ω4 + 1.79954718ω2 + 5.1258) / (ω16 - 0.00037436ω14 + 0.00169249ω12 - 0.00140092ω10 + 0.00012722ω8 - 0.00140092ω6 + 0.00169249ω4 - 0.00037436ω2 + 1)
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Gain H(ω) ω9 + 5.7588ω8 + 16.5817ω7 + 31.1634ω6 + 41.9864ω5 + 41.9864ω4 + 31.1634ω3 + 16.5817ω2 + 5.7588ω + 1 (jω)9 + 5.7588(jω)8 + 16.5817(jω)7 + 31.1634(jω)6 + 41.9864(jω)5 + 41.9864(jω)4 + 31.1634(jω)3 + 16.5817(jω)2 + 5.7588(jω) + 1 (5.7588ω8 - 31.1634ω6 + 41.9864ω4 - 16.5817ω2 + 1) + j(ω9 - 16.5817ω 7 + 41.9864ω5 - 31.1634ω3 + 5.7588ω) sqrt [(5.7588ω8 - 31.1634ω6 + 41.9864ω4 - 16.5817ω2 + 1)2 + (ω9 - 16.5817ω 7 + 41.9864ω5 - 31.1634ω3 + 5.7588ω)2] Phase θ(ω) tan-1 [(5.7588ω8 - 31.1634ω6 + 41.9864ω4 - 16.5817ω2 + 1)/(ω9 - 16.5817ω 7 + 41.9864ω5 - 31.1634ω3 + 5.7588ω)] Group Delay τd(ω) (5.7588ω16 + 2.00049396ω14 + 1.30030882ω12 + 1.06835072ω10 + 1.00271865ω8 + 1.06835072ω6 + 1.30030882ω4 + 2.00049396ω2 + 5.7588) / (ω18 + 0.00037744ω16 - 0.00200095ω14 + 0.00148244ω12 + 0.00034108ω10 + 0.00034108ω8 + 0.00148244ω6 - 0.00200095ω4 + 0.00037744ω2 + 1)
Posted December 1, 2023
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Please Support RF Cafe by purchasing my ridiculously low−priced products, all of which I created.. # Butterworth Lowpass Filter Gain, Phase, and Group Delay Equations. This may be the only place you will ever find the explicit formulas for Butterworth lowpass filter gain, phase, and group delay based on the basic Butterworth polynomials (until people copy my work and don't give credit*). Hoping to avoid having to do the messy math of complex numbers requiring separating out the real and imaginary parts to obtain phase (θ) - and then to get the first derivative for group delay (τd) - I searched high and low, far and wide for closed form equations, to no avail. All I could ever find was instruction to apply arg{H(ω)} for phase (θ) and -dθ/dω for group delay -- pretty useless if you want to plug an equation into a spreadsheet or software.. Finally, I decided if I wanted the solutions, I would have to slog through all the equations myself. Here is the result. The basic steps are as follow:. • Simplify the polynomial factors by multiplying out the parenthetical quantities.. • Substitute jω (or iω, but engineers use jω) for each "s," where ω is the frequency (ωn) being evaluated divided by the cutoff frequency (ωco); hence, jω = jωnco.. • Since j2 = -1 by definition, all even powers of j result in "real" parts and all odd powers of j result in "imaginary" parts. E.g., for 3rd-order:. • H(s) = (s + 1)(s2 + s + 1) = js3 + 2s2 +2s + 1. • H(jω) = (jω)3 + 2(jω)2 +2(jω) + 1 = j3ω3 + 2(j2ω2) +2(jω) + 1. = -jω3 + 2(-ω2) + j2ω + 1 = -jω3 - 2ω2 + j2ω + 1. • Re{H(jω)} = 1 - 2ω2 (parts w/o j), and Im{H(jω)} = 2ω - ω3 (parts w/j). • Gain is G(ω) = sqrt [Re{H(jω)2 + Im{H(jω)2]. • The phase angle is by definition θ(ω) = tan-1 (Re{H(jω)}/Im{H(jω)}). θ(ω) = tan-1 [(1 - 2ω2) / (2ω - ω3)]. • Group delay is defined as τd(ω) = -dθ(ω)/dω. This is where it gets hard, mainly due to the arctangent, but the fraction is no walk in the park, either. Fortunately, there are online math websites like SymboLab that will handle such derivatives. For that matter, it will multiply all the factors for you as well, reducing the opportunity for errors.. • Finally, do the ω = ωnco substitution everywhere. ωn is the frequency at which the equation is being evaluated, and ωco is the cutoff frequency. Frequency units cancel out, so a 1 Hz cutoff plots the same as a 1 kHz cutoff or a 1 GHz cutoff for gain and phase. The group delay scale needs to be divided by a factor equal to the frequency units (÷103 for kHz, ÷106 for MHz, etc.).. This table represents hours of work on my part. I will eventually get around to lowpass, highpass, bandpass, and bandreject for Butterworth, Chebyshev, and Bessel (and maybe Gaussian) filters.. Butterworth Lowpass, Highpass, Bandpass, and Bandstop Filter Calculator with Gain, Phase and Group Delay are now part of my free RF Cafe Espresso Engineering Workbook™!. * I had Archive.org save a copy of this page in order to prove this is my original work.. N. 1. GainH(ω) s + 1 (jω) + 1 sqrt (1 + ω2) Phase θ(ω) tan-1 (1/ω) Group Delay τd(ω) 1/(1 + ω2). 2. Gain H(ω) s2 + 1 + 1.4142s (jω)2 + 1.4142(jω) + 1 (1 - ω2) + j1.4142ω sqrt [(1 - ω2)2 + (1.4142ω)2] Phase θ(ω) tan-1 [(1 - ω2)/1.4142ω] Group Delay τd(ω) (1 + ω2)/[1.4142(0.5ω4 - 0.0001ω2) + 0.5]. 3. Gain H(ω) s3 + 2s2 + 2s + 1(jω)3 + 2(jω)2 + 2(jω) + 1(1 - 2ω2) + j(2ω - ω3)sqrt [(1 - 2ω2)2 + (2ω - ω3)2] Phase θ(ω) tan-1 [(1 - 2*ω2)/(2ω - ω3)] Group Delay τd(ω) (2*ω4 + ω2 + 2)/(ω6 + 1). 4. Gain H(ω) s4 + 2.6131s3 + 3.4142s2 + 2.6131s + 1 (jω)4 + 2.6131(jω)3 + 3.4142(jω)2 + 2.6131(jω) + 1 (ω4 - 3.4142ω2 + 1) + j(-2.6131ω3 + 2.6131ω) sqrt [(ω4 - 3.4142ω2 + 1)2 + (-2.6131ω3 + 2.6131ω)2] Phase θ(ω) tan-1 [(ω4 - 3.4142ω2 + 1)/(-2.6131ω3 + 2.6131ω)] Group Delay τd(ω) (2.6131ω6 + 1.08234602ω4 + 1.08234602ω2 + 2.6131) / (ω8 - 0.00010839ω6 + 0.00017842ω4 - 0.00010839ω2 + 1). 5. Gain H(ω) s5 + 3.2361s4 + 5.2361s3 + 5.2361s2 + 3.2361s + 1 (jω)5 + 3.2361(jω)4 + 5.2361(jω)3 + 5.2361(jω)2 + 3.2361(jω) + 1 (3.2361ω4 - 5.2361ω2+1) + j(ω5 - 5.2361ω3 + 3.2361ω) sqrt [(3.2361ω4 - 5.2361ω2 + 1)2 + (ω5 - 5.2361ω3 + 3.2361ω)2] Phase θ(ω) tan-1 [(3.2361ω4 - 5.2361ω2 + 1)/(ω5 - 5.2361ω3 + 3.2361ω)] Group Delay τd(ω) (3.2361ω8 + 1.23624321ω6 + 0.99971358ω4 + 1.23624321ω2 + 3.2361) / (ω10 + 0.00014321ω8 - 0.00014321ω6 - 0.00014321ω4 + 0.00014321ω2 + 1). 6. Gain H(ω) s6 + 3.8637s5 + 7.4641s4 + 9.1416s3 + 7.4641s2 + 3.8637s + 1 (jω)6 + 3.8637(jω)5 + 7.4641(jω)4 + 9.1416(jω)3 + 7.4641(jω)2 + 3.8637(jω) + 1 (-ω6 + 7.4641ω4 - 7.4641ω2 + 1) + j(3.8637ω5 - 9.1416ω3 + 3.8637ω) sqrt [(-ω6 + 7.4641ω4 - 7.4641ω2 + 1)2 + (3.8637ω5 - 9.1416ω3 + 3.8637ω)2] Phase θ(ω) tan-1 [(-ω6 + 7.4641ω4 - 7.4641ω2 + 1)/(3.8637ω5 - 9.1416ω3 + 3.8637ω)] Group Delay τd(ω) (3.8637*ω10 + 1.41424317*ω8 + 1.03518705*ω6 + 1.03518705*ω4 + 1.41424317*ω2 + 3.8637) / (ω12 - 0.00002231*ω10 + 0.00018897*ω8 - 0.00037168*ω6 + 0.00018897*ω4 - 0.00002231*ω2 + 1). 7. Gain H(ω) s7 + 4.4940s6 + 10.0978s5 + 14.5920s4 + 14.5920s3 + 10.0978s2 + 4.4940s + 1 (jω)7 + 4.4940(jω)6 + 10.0978(jω)5 + 14.5920(jω)4 + 14.5920(jω)3 + 10.0978(jω)2 + 4.4940(jω) + 1 (-4.494ω6 + 14.592ω4 - 10.0978ω2 + 1) + j(-ω7 + 10.0978ω5 - 14.592ω3 + 4.494ω) sqrt [(-4.494ω6 + 14.592ω4 - 10.0978ω2 + 1)2 + (-ω7 + 10.0978ω5 - 14.592ω3 + 4.494ω)2] Phase θ(ω) tan-1 [(-4.494ω6 + 14.592ω4 - 10.0978ω2 + 1)/(-ω7 + 10.0978ω5 - 14.592ω3 + 4.494ω)] Group Delay τd(ω) (4.494ω12 + 1.6035132ω10 + 1.1067536ω8 + 1.00994948ω6 + 1.1067536ω4 + 1.6035132ω2 + 4.494) / (ω14+0.000436ω12 - 0.00333116ω10 + 0.0032952ω8 + 0.0032952ω6-0.00333116ω4 + 0.000436ω2 + 1). 8. Gain H(ω) s8 + 5.1528s7 + 13.1371s6 + 21.8462s5 + 25.6884s4 + 21.8462s3 + 13.1371s2 + 5.1528s + 1 (jω)8 + 5.1258(jω)7 + 13.1371(jω)6 + 21.8462(jω)5 + 25.6884(jω)4 + 21.8462(jω)3 + 13.1371(jω)2 + 5.1258(jω) + 1 (ω8 - 13.1371ω6 + 25.6884ω4 - 13.1371ω2 + 1) + j(-5.1258ω7 + 21.8462ω5 - 21.8462ω3 + 5.1258ω) Phase θ(ω) tan-1 [(ω8 - 13.1371ω6 + 25.6884ω4 - 13.1371ω2 + 1)/(-5.1258ω7 + 21.8462ω5 - 21.8462ω3 + 5.1258ω)] Group Delay τd(ω) (5.1258ω14 + 1.79954718ω12 + 1.20591186ω10 + 1.01691792ω8 + 1.01691792ω6 - 1.20591186ω4 + 1.79954718ω2 + 5.1258) / (ω16 - 0.00037436ω14 + 0.00169249ω12 - 0.00140092ω10 + 0.00012722ω8 - 0.00140092ω6 + 0.00169249ω4 - 0.00037436ω2 + 1). 9.
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Gain H(ω) ω9 + 5.7588ω8 + 16.5817ω7 + 31.1634ω6 + 41.9864ω5 + 41.9864ω4 + 31.1634ω3 + 16.5817ω2 + 5.7588ω + 1 (jω)9 + 5.7588(jω)8 + 16.5817(jω)7 + 31.1634(jω)6 + 41.9864(jω)5 + 41.9864(jω)4 + 31.1634(jω)3 + 16.5817(jω)2 + 5.7588(jω) + 1 (5.7588ω8 - 31.1634ω6 + 41.9864ω4 - 16.5817ω2 + 1) + j(ω9 - 16.5817ω 7 + 41.9864ω5 - 31.1634ω3 + 5.7588ω) sqrt [(5.7588ω8 - 31.1634ω6 + 41.9864ω4 - 16.5817ω2 + 1)2 + (ω9 - 16.5817ω 7 + 41.9864ω5 - 31.1634ω3 + 5.7588ω)2] Phase θ(ω) tan-1 [(5.7588ω8 - 31.1634ω6 + 41.9864ω4 - 16.5817ω2 + 1)/(ω9 - 16.5817ω 7 + 41.9864ω5 - 31.1634ω3 + 5.7588ω)] Group Delay τd(ω) (5.7588ω16 + 2.00049396ω14 + 1.30030882ω12 + 1.06835072ω10 + 1.00271865ω8 + 1.06835072ω6 + 1.30030882ω4 + 2.00049396ω2 + 5.7588) / (ω18 + 0.00037744ω16 - 0.00200095ω14 + 0.00148244ω12 + 0.00034108ω10 + 0.00034108ω8 + 0.00148244ω6 - 0.00200095ω4 + 0.00037744ω2 + 1). Posted December 1, 2023.
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https://www.usna.edu/Users/cs/wcbrown/courses/F17SI340/lec/l03/lec.html
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Sections 1.4 of Theory of Computing, A Gentle Introduction. It's important that you read this as a way to brush up on the mathematical foundations we'll use next lecture.
Homework
Printout the homework and answer the questions on that paper.
Something to keep in mind ...
Remember, a finite automaton is a very limited computing device. After all, a given machine has a fixed, finite amount of memory --- even though that machine can be fed strings that are arbitrarily long. So we have to expect that there are lots and lots of languages out there that no finite automaton accepts. In fact, the languages that are accepted by finite automata comprise a small subset of all possible languages. But it's nice when a language can be accepted by a finite automaton, because then we have such a simple way of recognizing strings in the language.
Manipulating machines I
Consider the following two machines, one from the homework due today and one from the lab last class, both over the alphabet {a,b,c}:
Machine M1 Machine M2
Now using these machines we're going to ask a few questions. Let L(M1) denote the language accepted by Machine M1, and let L(M2) denote the language accepted by Machine M2.
1. The complement of a language L over an alphabet Σ (written L) is the set of all strings over the alphabet that are not in L. Can we construct a machine that accepts L(M1)? Can we construct a machine that accepts L(M2)? Sure! Here they are:
Machine M3 accepting L(M1) Machine M4 accepting L(M2)
After looking at how we did this, we might be inspired to make the following hypothesis.
Hypothesis: If a language L is accepted by some finite automaton, there exists a finite automaton that accepts L.
Proof: Consider the following algorithm:
Input: Machine M
Output: Machine M' produced by making all the accepting sates of M non-accepting and making all the non-accepting states accepting.
I claim that it's obvious that the language accepted by M' is the complement of the language accepted by M, i.e. I claim that L(M') = L(M). So, going back to our hypothesis, give the name M to the machine accepting L. From input M, the above algorithm produces machine M' that accepts L, which proves the hypothesis.
This isn't quite a proof that a pedantic mathematician would accept, because we haven't proved our claim that the machine M' produced by our algorithm accepts L(M). On the other hand, while the hypothesis is not what most people would consider self-evident, the correctness of the algorithm is. So for most people, the algorithm is enough of a proof. If you don't find it convincing, here's a proof that the algorithm does what I say it does:
Proof that the algorithm produces machine M' that accepts L.
Since the states, start state and transitions of M and M' are the same, after processing a given string w both machines will end up in the same state, let's call it "qi". If qi is an accepting state in M it is non-accepting in M', and if qi is a non-accepting state in M it is accepting in M'. Thus, the two machines make opposite decisions for any input string w, which proves that L(M') = L(M).
There are a few important points here. First of all, notice that how much proof is required depends on what is "self-evident", and that might change from one person to the next. In the end, proof is all about convincing someone. Secondly, the process that we just went through is what most of this course is going to be about: notice a pattern, make a hypothesis, prove the hypothesis. In fact, most proofs will look like the above: the proof is an algorithm and --- if you're picky/precise enough, or if the algorithm is complicated or subtle enough --- a proof of the algorithm's correctness.
2. The intersection of two languages L1 and L2 is the language of all strings that are in both L1 and L2. So, for example, can we construct a machine that accepts the intersection of L(M3) and L(M2)? Sure! However, this takes a little more work and a little more thought. You see, intuitively what we need to do is run both M3 and M2 simultaneously. When the input is done, we'd require both machines to be in accepting states. However, we are asking for a single machine that does the same job. The answer is to have one machine in which each state stands for a pair of states, one from M3 and one from M2. Machine M5 below does exactly this. Notice that each state is labeled with a pair of states. The state labeled q2p0 corresponds to M3 being in state q2 and M2 being in state p0. Notice that what we really have here is two copies of machine M3, one corresponding to state p0 of M2 and one corresponding to state p1 of M2.
Machine M5 accepting the intersection of L(M3) and L(M2) Machine M5 rearranged to look cleaner!
For example, the transition from the state labeled q1p1 to q2p0 if c is read is translated from the following:
```Machine: M3 M2 M5
-- -- ----
In state: q1 p1 -------> q1p1
| | |
| | |
V V V
Moves to: q2 p0 q2p0
```
Can you see how the accepting states were selected? Once again, the way we accomplished this makes us think that we could do the same for any two machines, which leads us to the following:
Hypothesis: If languages L1 and L2 are accepted by finite automata, there exists a finite automaton that accepts the intersection of languages L1 and L2.
We could prove this the same way we proved our hypothesis concerning complement languages, by describing an algorithm that would take two machines as input and provide as output a new machine that accepted the intersection of the languages accepted by the two input machines. However, we're going to have to do some more work before we're able to define complex algorithms like this precisely.
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Sections 1.4 of Theory of Computing, A Gentle Introduction. It's important that you read this as a way to brush up on the mathematical foundations we'll use next lecture.. Homework. Printout the homework and answer the questions on that paper.. Something to keep in mind .... Remember, a finite automaton is a very limited computing device. After all, a given machine has a fixed, finite amount of memory --- even though that machine can be fed strings that are arbitrarily long. So we have to expect that there are lots and lots of languages out there that no finite automaton accepts. In fact, the languages that are accepted by finite automata comprise a small subset of all possible languages. But it's nice when a language can be accepted by a finite automaton, because then we have such a simple way of recognizing strings in the language.. Manipulating machines I. Consider the following two machines, one from the homework due today and one from the lab last class, both over the alphabet {a,b,c}:. Machine M1 Machine M2. Now using these machines we're going to ask a few questions. Let L(M1) denote the language accepted by Machine M1, and let L(M2) denote the language accepted by Machine M2.. 1. The complement of a language L over an alphabet Σ (written L) is the set of all strings over the alphabet that are not in L. Can we construct a machine that accepts L(M1)? Can we construct a machine that accepts L(M2)? Sure! Here they are:. Machine M3 accepting L(M1) Machine M4 accepting L(M2). After looking at how we did this, we might be inspired to make the following hypothesis.. Hypothesis: If a language L is accepted by some finite automaton, there exists a finite automaton that accepts L.. Proof: Consider the following algorithm:. Input: Machine M. Output: Machine M' produced by making all the accepting sates of M non-accepting and making all the non-accepting states accepting.. I claim that it's obvious that the language accepted by M' is the complement of the language accepted by M, i.e. I claim that L(M') = L(M). So, going back to our hypothesis, give the name M to the machine accepting L. From input M, the above algorithm produces machine M' that accepts L, which proves the hypothesis.. This isn't quite a proof that a pedantic mathematician would accept, because we haven't proved our claim that the machine M' produced by our algorithm accepts L(M). On the other hand, while the hypothesis is not what most people would consider self-evident, the correctness of the algorithm is. So for most people, the algorithm is enough of a proof. If you don't find it convincing, here's a proof that the algorithm does what I say it does:. Proof that the algorithm produces machine M' that accepts L.. Since the states, start state and transitions of M and M' are the same, after processing a given string w both machines will end up in the same state, let's call it "qi". If qi is an accepting state in M it is non-accepting in M', and if qi is a non-accepting state in M it is accepting in M'. Thus, the two machines make opposite decisions for any input string w, which proves that L(M') = L(M).. There are a few important points here. First of all, notice that how much proof is required depends on what is "self-evident", and that might change from one person to the next. In the end, proof is all about convincing someone. Secondly, the process that we just went through is what most of this course is going to be about: notice a pattern, make a hypothesis, prove the hypothesis. In fact, most proofs will look like the above: the proof is an algorithm and --- if you're picky/precise enough, or if the algorithm is complicated or subtle enough --- a proof of the algorithm's correctness.. 2. The intersection of two languages L1 and L2 is the language of all strings that are in both L1 and L2. So, for example, can we construct a machine that accepts the intersection of L(M3) and L(M2)? Sure! However, this takes a little more work and a little more thought. You see, intuitively what we need to do is run both M3 and M2 simultaneously. When the input is done, we'd require both machines to be in accepting states. However, we are asking for a single machine that does the same job. The answer is to have one machine in which each state stands for a pair of states, one from M3 and one from M2. Machine M5 below does exactly this. Notice that each state is labeled with a pair of states. The state labeled q2p0 corresponds to M3 being in state q2 and M2 being in state p0. Notice that what we really have here is two copies of machine M3, one corresponding to state p0 of M2 and one corresponding to state p1 of M2.. Machine M5 accepting the intersection of L(M3) and L(M2) Machine M5 rearranged to look cleaner!. For example, the transition from the state labeled q1p1 to q2p0 if c is read is translated from the following:. ```Machine: M3 M2 M5. -- -- ----. In state: q1 p1 -------> q1p1. | | |. | | |. V V V. Moves to: q2 p0 q2p0. ```. Can you see how the accepting states were selected? Once again, the way we accomplished this makes us think that we could do the same for any two machines, which leads us to the following:. Hypothesis: If languages L1 and L2 are accepted by finite automata, there exists a finite automaton that accepts the intersection of languages L1 and L2.
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We could prove this the same way we proved our hypothesis concerning complement languages, by describing an algorithm that would take two machines as input and provide as output a new machine that accepted the intersection of the languages accepted by the two input machines. However, we're going to have to do some more work before we're able to define complex algorithms like this precisely.
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https://excelbu.com/2021/07/how-to-get-more-bang-out-of-your-excel-spreadsheet/
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# Excelbu
Work faster in Excel
What if you want to increase the productivity of your work without having to learn every little word?
In this article, I’m going to show you how to get your Excel file to become a real-time report in real time.
And the result is a report that’s much easier to use than you’d think.
1.
Here are some basic steps you can take to get started.
1 .
Create a simple, one-page spreadsheet that you can use for your work.
2 .
You’ll need to create a basic spreadsheet to use for calculations.
You can use any type of spreadsheet you like, but I recommend creating a simple one-column spreadsheet.
3 .
A spreadsheet is the perfect way to get a sense of what your data looks like, because you can see how many times the data is repeated and how much each cell is used up.
In the case of the spreadsheet I created, I’ve got 4,000 columns, which is more than enough for this exercise.
I recommend adding another 1,000 to the total number of columns in the spreadsheet, to ensure a consistent layout.
4 .
Use a formula to make your calculations.
Here’s how I did it.
First, I created a formula that calculates how much you need to spend on an item to increase your productivity.
You’re looking at an average cost of \$1.90, so we’re going to need to multiply that by \$100.
So, let’s say we have an Excel spreadsheet with 3,000 items, and we’re looking to increase our productivity by \$300 per week.
Let’s say that we want to work out \$1,500 in productivity each week.
We can calculate this by multiplying our \$1 cost per item by 3, and then dividing it by 100.
So our cost per week is \$2.50 per week, or \$200 per week in addition to our \$200 budget.
This is enough to get us through this exercise, so let’s save the formula for later.
5 .
6 .
You should already have a basic formula in your head that you’re familiar with, but there are a few important steps to remember: 1 .
The first formula you need is the formula that gets you the total cost of an item, or cost per row.
It is the most basic formula you’ll use in this exercise: cost = row cost / 2.
It can be anything from \$0.5 to \$10.
The first row cost in our spreadsheet is \$0, so the cost per column is \$1 per column.
2.
The second formula is the cost of a cell, or a line of code.
You’ve already figured out that this line of the formula calculates how many cells in your spreadsheet you’ve got.
So now we have the total of cells in our sheet, the cost in each cell.
The cost per cell in our row is \$4, so that gives us a total cost per day of \$5.3.
3.
The third formula is a function.
Here, we’re getting a value from the data.
We’ll be calling the cost function a value.
The cell that we’ve added to our sheet has a value of 1.
So we need to calculate the cost by multiplying this value by 100, and adding that value to the cost.
So \$1 * 100 + 1.0 = \$4.3, or the cost is \$10 per day.
4.
Finally, we need a formula for adding rows to our table.
This formula takes the row value and adds it to the price of a column.
This works the same way as the formula above.
So the cost to add rows is \$3.3 * 100.
5.
The next formula is what I call the cost-per-item formula.
This has two values: the cost you’d pay for a row, and the cost for each item in the column.
So this is the same as the first formula, but we’re only adding a single row, so it takes a total of three cells.
So for this sheet, we want a total value of \$20.
That means we’ll be adding an item every day.
Here it is again, adding a row every day: cost per line = row price / 2, or 1.25 per line per cell.
We’re adding one item for every 1,500 lines of code in the table.
Now that we have our formulas, let us look at our first spreadsheet. First up
## Money in excel: How to convert £1,000 into more money in Excel
In the United Kingdom, it is the equivalent of £1.25, but for many businesses in the world it is less than £0.05.That is the difference between what you can earn and what you have to pay the bills.Here is how to convert the equivalent amount…
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# Excelbu. Work faster in Excel. What if you want to increase the productivity of your work without having to learn every little word?. In this article, I’m going to show you how to get your Excel file to become a real-time report in real time.. And the result is a report that’s much easier to use than you’d think.. 1.. Here are some basic steps you can take to get started.. 1 .. Create a simple, one-page spreadsheet that you can use for your work.. 2 .. You’ll need to create a basic spreadsheet to use for calculations.. You can use any type of spreadsheet you like, but I recommend creating a simple one-column spreadsheet.. 3 .. A spreadsheet is the perfect way to get a sense of what your data looks like, because you can see how many times the data is repeated and how much each cell is used up.. In the case of the spreadsheet I created, I’ve got 4,000 columns, which is more than enough for this exercise.. I recommend adding another 1,000 to the total number of columns in the spreadsheet, to ensure a consistent layout.. 4 .. Use a formula to make your calculations.. Here’s how I did it.. First, I created a formula that calculates how much you need to spend on an item to increase your productivity.. You’re looking at an average cost of \$1.90, so we’re going to need to multiply that by \$100.. So, let’s say we have an Excel spreadsheet with 3,000 items, and we’re looking to increase our productivity by \$300 per week.. Let’s say that we want to work out \$1,500 in productivity each week.. We can calculate this by multiplying our \$1 cost per item by 3, and then dividing it by 100.. So our cost per week is \$2.50 per week, or \$200 per week in addition to our \$200 budget.. This is enough to get us through this exercise, so let’s save the formula for later.. 5 .. 6 .. You should already have a basic formula in your head that you’re familiar with, but there are a few important steps to remember: 1 .. The first formula you need is the formula that gets you the total cost of an item, or cost per row.. It is the most basic formula you’ll use in this exercise: cost = row cost / 2.. It can be anything from \$0.5 to \$10.. The first row cost in our spreadsheet is \$0, so the cost per column is \$1 per column.. 2.. The second formula is the cost of a cell, or a line of code.. You’ve already figured out that this line of the formula calculates how many cells in your spreadsheet you’ve got.. So now we have the total of cells in our sheet, the cost in each cell.. The cost per cell in our row is \$4, so that gives us a total cost per day of \$5.3.. 3.. The third formula is a function.. Here, we’re getting a value from the data.. We’ll be calling the cost function a value.. The cell that we’ve added to our sheet has a value of 1.. So we need to calculate the cost by multiplying this value by 100, and adding that value to the cost.. So \$1 * 100 + 1.0 = \$4.3, or the cost is \$10 per day.. 4.. Finally, we need a formula for adding rows to our table.. This formula takes the row value and adds it to the price of a column.. This works the same way as the formula above.. So the cost to add rows is \$3.3 * 100.. 5.. The next formula is what I call the cost-per-item formula.. This has two values: the cost you’d pay for a row, and the cost for each item in the column.. So this is the same as the first formula, but we’re only adding a single row, so it takes a total of three cells.. So for this sheet, we want a total value of \$20.. That means we’ll be adding an item every day.. Here it is again, adding a row every day: cost per line = row price / 2, or 1.25 per line per cell.. We’re adding one item for every 1,500 lines of code in the table.. Now that we have our formulas, let us look at our first spreadsheet. First up.
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## Money in excel: How to convert £1,000 into more money in Excel. In the United Kingdom, it is the equivalent of £1.25, but for many businesses in the world it is less than £0.05.That is the difference between what you can earn and what you have to pay the bills.Here is how to convert the equivalent amount….
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https://math.stackexchange.com/questions/2025090/finding-the-area-bounded-by-two-curves
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# Finding the area bounded by two curves
Find the area of the region bounded by the parabola $$y = 4x^2$$, the tangent line to this parabola at $$(2, 16)$$, and the $$x$$-axis.
I found the tangent line to be $$y=16x-16$$ and set up the integral from $$0$$ to $$2$$ of $$4x^2-16x+16$$ with respect to $$x$$, which is the top function when looking at the graph minus the bottom function. I took the integral and came up with $$\frac{4}{3}x^3-8x^2+16x$$ evaluated between $$0$$ and $$2$$. This came out to be $$\frac{32}{3}$$ but this was the incorrect answer. Can anyone tell me where I went wrong?
Hint: After drawing it, note that you have to calculate $\int_0^1 4x^2\;dx + \int_1^2 4x^2-16x+16\;dx$.
• I got $\frac{8}{3}$. I'm sorry but did you do it right? – Rodrigo Dias Nov 22 '16 at 0:22
• Any time! ${}{}$ – Rodrigo Dias Nov 22 '16 at 0:28
The tangent crosses the $x$ axis at $x=1$, so your integral is including (with the plus sign) also the triangle made by the tangent below the $x$ axis.
The correct way is to integrate only the parabola for $x=0 \cdots 2$ (which is $32/3$ and then subtract the area of the triangle$(1,0),(2,16),(2,0)$, which is $8$, so the net area is $8/3$ .
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# Finding the area bounded by two curves. Find the area of the region bounded by the parabola $$y = 4x^2$$, the tangent line to this parabola at $$(2, 16)$$, and the $$x$$-axis.. I found the tangent line to be $$y=16x-16$$ and set up the integral from $$0$$ to $$2$$ of $$4x^2-16x+16$$ with respect to $$x$$, which is the top function when looking at the graph minus the bottom function. I took the integral and came up with $$\frac{4}{3}x^3-8x^2+16x$$ evaluated between $$0$$ and $$2$$. This came out to be $$\frac{32}{3}$$ but this was the incorrect answer. Can anyone tell me where I went wrong?. Hint: After drawing it, note that you have to calculate $\int_0^1 4x^2\;dx + \int_1^2 4x^2-16x+16\;dx$.. • I got $\frac{8}{3}$. I'm sorry but did you do it right? – Rodrigo Dias Nov 22 '16 at 0:22. • Any time! ${}{}$ – Rodrigo Dias Nov 22 '16 at 0:28.
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The tangent crosses the $x$ axis at $x=1$, so your integral is including (with the plus sign) also the triangle made by the tangent below the $x$ axis.. The correct way is to integrate only the parabola for $x=0 \cdots 2$ (which is $32/3$ and then subtract the area of the triangle$(1,0),(2,16),(2,0)$, which is $8$, so the net area is $8/3$ .
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https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Active_Learning/Shorter_Activities/Data_Analysis_and_Statistics/01_Uncertainty
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Uncertainty
For the first half of this lesson:
Chapter 2 (sections 1, 3 (first part), 4, 5, 6, 9), Chapter 3 (sections 1, 2, 3, 4), Exploring Chemical Analysis 5th ed., D. Harris
Suppose you are analyzing soil samples from a school playground near a busy intersection. After collecting three samples, you return to the lab, extract the soil samples with water and, using an atomic absorption spectrometer (we will talk more about instrumentation later), determine the concentration of lead in the resulting extracts, obtaining an average result of 4.38 parts-per-billion (ppb).
• The actual value that you report is 4.38 ± 0.02 ppb. What does the 0.02 ppb indicate?
STOP
• Brainstorm what errors might affect this result?
• Separate the errors discussed in class into systematic vs. random
Systematic (Determinate)
Random (Indeterminate)
STOP
Instrument noise
• Often when discussing precision and accuracy we use a model of darts on a dartboard as shown below.
• As a group come to a conclusion of what accuracy and precision are based on previous knowledge and write your definition below.
• Accuracy:
• Precision:
• Label the top and side of the table with accurate, not accurate, precise, and not precise.
• Circle the dartboard depictions show random error and put a square around the dartboard depictions that show systematic error.
Figure 1. Dart model of precision and accuracy.
STOP
Measurement Exercise
• Experiment 1: Using a ruler with a millimeter scale, measure the big solid rectangle's width and length in millimeters on your own without any consultation of your group then collect the values of your group mates in the table and calculate the average values.
Group Members
Length (mm)
Width (mm)
Average
• Are your measurements of the rectangle's length and width exactly the same as those of your classmates, or are they different? If there are differences in the measurements, is this the result of determinate errors, indeterminate errors or both? What are those errors? Briefly explain your reasoning.
STOP
When measuring the rectangle's length and width you had to make several decisions:
• How do you define the rectangle? Should you measure its length and width from the outside edges of the border, from the inside edges or from the middle of the border?
• Experiment 2: Decide as a group how you are going to define the rectangle and record your measurement to the tenth place (which is an estimation of one digit past the smallest gradation). Then measure the length and width of the rectangle and calculate its average.
Group Members
Length (mm)
Width (mm)
Average
STOP
Significant Figures
You may recall from general chemistry that we refer to the digits in our measurements as significant figures. A significant figure is any number in which we can express confidence, including those digits known exactly and the one digit whose value is an estimate. The lengths 154 mm and 154.3 mm have three and four significant digits, respectively. The number of significant figures in a measurement is important because it affects the number of significant figures in a result based on that measurement.
• How many significant figures do you have in each of your individual measurements of length and width?
Group Members
# sig figs in length
# sig figs in width
Before continuing, let's review the rules for including significant figures in calculations. When adding or subtracting, the result of the calculation is rounded to the last decimal place that is significant for all measurements. For example, the sum of 135.621, 0.33 and 21.2163 is 157.17 since the last decimal place that is significant for all three numbers (as shown below by the vertical line)
$\begin{array}{r|l} 135.62\hspace{-4pt} & \hspace{-4pt}1 \\[-2pt] 0.33\hspace{-4pt} & \\[-2pt] 21.21\hspace{-4pt} & \hspace{-4pt}63 \\ \hline 157.16\hspace{-4pt} & \hspace{-4pt}73 \\ \end{array} \nonumber$
is the hundredth's place. Note that rounding the answer to the correct number of significant figures occurs after completing the exact calculation.
When multiplying or dividing, the result of the calculation contains the same number of significant figures as that measurement having the smallest number of significant figures. Thus,
$\dfrac{22.91\times0.152}{16.302}=0.21361\approx 0.214\nonumber$
because 0.152, with three, has the fewest number of significant figures.
One way to think about this is that we cannot make a measurement more precise through a calculation than it is when we take the measurement.
• Check back to be sure that your group has the correct number of significant figures in the average length and width listed for Experiment 2, if not fix them.
STOP
Uncertainty in the lab
• If you took a measurement and found a value of 89.231 ± 0.008 what is the absolute uncertainty and the percent relative uncertainty of the measurement?
STOP
Propagation of Uncertainty
As we have discussed, each instrument we use in a lab has an associated random error, which is often expressed as a tolerance factor by the manufacturer. A more rigorous approach to determining the uncertainty in a result is called a propagation of uncertainty and it combines the uncertainty in each measurement to estimate the uncertainty in the final result.
The following equations for propagation of uncertainty are from your textbook (as noted in parentheses):
$e_{prop}=\sqrt{e_1^2 + e_2^2+e_3^2+⋯e_n^2} \tag{3-5}$
Multiplication and Division:
$\%e_{prop}=\sqrt{(\%e_1^2)+(\% e_2^2)+(\%e_3^2)+⋯(\%e_n^2)} \tag{3-6}$
Where en is the absolute uncertainty (often the tolerance factor) for measurement n and %en is the percent relative uncertainty for measurement n.
The REAL SIG FIG RULE
The final uncertainty to one significant figure and that defines how many significant figures are in the measurement. For example if you took several measurements and the average value was 234.7182 a.u., but the uncertainty when propagated was 0.0278 a.u. you would write the final result as 234.72 ± 0.03 a.u.
• Assuming that the error associated with the ruler is 0.1 mm, propagate the error associated with the average of the length and the width of the rectangle using the data from Experiment 2.
1. What is the absolute uncertainty of each measurement?
2. Should you use the addition/subtraction or multiplication/division equation above to propagate the error?
3. Calculate the error and write it with the proper number of sig. figs is the table below.
Average length (mm)
Average width (mm)
Error
• Asking yourself the same questions as above, calculate the average area of the rectangle including the propagated error and the correct number of sig figs and write the value in the table.
Average area w/ error (mm2)
STOP
• As a group think of times in the lab you may want to use the addition and subtraction rule to propagate uncertainty and one time in lab you would want to use multiplication and division rule to propagate uncertainty.
STOP
Often propagation of error includes mixed operation. Like finding the area of the rectangle, except I walked you through how to do that. Now you are going to propagate error for a scientific system you have never seen before.
• For a concentration technique the relationship between the signal and the an analyte’s concentration is
$S_\textrm{total} = k_\textrm{A}C_\textrm{A} + S_\textrm{mb}\nonumber$
What is the analyte’s concentration, CA, and its uncertainty if Stotal is 24.37 ± 0.02, Smb is 0.96 ± 0.02, and kA is 0.186 ± 0.003 ppm–1.
STOP
• Given the following masses of deionized water measured during the calibration of a Class A pipet at 15 °C, what average volume of water is the 10.00 mL Class A (± 0.02 mL) pipet actually delivering?
Measurement
Mass of H2O (g)
1
10.1578
2
10.1341
3
10.1425
4
10.1453
5
10.1672
6
10.1587
7
10.1611
STOP
• Explain, with numerical values, glassware, and instruments how you would prepare 50 mL of a 1.40 M aqueous stock solution of sodium thiosulfate, Na2S2O3 (MW= 158.11 g/mol) (which is a solid that must be dissolved) while introducing the least amount of error. (similar example pg 23 of textbook)
STOP
• Explain, with numerical values, glassware, and instruments how you would prepare a 100 mL of a 0.140 M aqueous solution of sodium thiosulfate from the 1.40 M stock solution in the previous problem while introducing the least amount of error.
STOP
Homework for data analysis
• How is systematic error identified and accounted for in measurements?
• How is random error identified and accounted for in measurements?
• In a lab at 27°C you weigh an empty volumetric flask at 468.654 g you then fill the flask up to the calibration mark with DI H2O and weigh the flask at 543.635 g. What is the true volume contained in the volumetric?
• What is the true concentration of dissolved 0.1567 g of Na2S2O3 in molarity if you made it in the volumetric flask in the previous example?
• What portion of the analytical process do the tools we talked about today fall under?
• If you made a 5 M solution via dilution of a 10 M solution using volumetric glassware and you performed several forms of analytical analysis and determined that the true concentration of the solution was 4.5 M what should be done to avoid this problem in the future?
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Uncertainty. For the first half of this lesson:. Chapter 2 (sections 1, 3 (first part), 4, 5, 6, 9), Chapter 3 (sections 1, 2, 3, 4), Exploring Chemical Analysis 5th ed., D. Harris. Suppose you are analyzing soil samples from a school playground near a busy intersection. After collecting three samples, you return to the lab, extract the soil samples with water and, using an atomic absorption spectrometer (we will talk more about instrumentation later), determine the concentration of lead in the resulting extracts, obtaining an average result of 4.38 parts-per-billion (ppb).. • The actual value that you report is 4.38 ± 0.02 ppb. What does the 0.02 ppb indicate?. STOP. • Brainstorm what errors might affect this result?. • Separate the errors discussed in class into systematic vs. random. Systematic (Determinate). Random (Indeterminate). STOP. Instrument noise. • Often when discussing precision and accuracy we use a model of darts on a dartboard as shown below.. • As a group come to a conclusion of what accuracy and precision are based on previous knowledge and write your definition below.. • Accuracy:. • Precision:. • Label the top and side of the table with accurate, not accurate, precise, and not precise.. • Circle the dartboard depictions show random error and put a square around the dartboard depictions that show systematic error.. Figure 1. Dart model of precision and accuracy.. STOP. Measurement Exercise. • Experiment 1: Using a ruler with a millimeter scale, measure the big solid rectangle's width and length in millimeters on your own without any consultation of your group then collect the values of your group mates in the table and calculate the average values.. Group Members. Length (mm). Width (mm). Average. • Are your measurements of the rectangle's length and width exactly the same as those of your classmates, or are they different? If there are differences in the measurements, is this the result of determinate errors, indeterminate errors or both? What are those errors? Briefly explain your reasoning.. STOP. When measuring the rectangle's length and width you had to make several decisions:. • How do you define the rectangle? Should you measure its length and width from the outside edges of the border, from the inside edges or from the middle of the border?. • Experiment 2: Decide as a group how you are going to define the rectangle and record your measurement to the tenth place (which is an estimation of one digit past the smallest gradation). Then measure the length and width of the rectangle and calculate its average.. Group Members. Length (mm). Width (mm). Average. STOP. Significant Figures. You may recall from general chemistry that we refer to the digits in our measurements as significant figures. A significant figure is any number in which we can express confidence, including those digits known exactly and the one digit whose value is an estimate. The lengths 154 mm and 154.3 mm have three and four significant digits, respectively. The number of significant figures in a measurement is important because it affects the number of significant figures in a result based on that measurement.. • How many significant figures do you have in each of your individual measurements of length and width?. Group Members. # sig figs in length. # sig figs in width. Before continuing, let's review the rules for including significant figures in calculations. When adding or subtracting, the result of the calculation is rounded to the last decimal place that is significant for all measurements. For example, the sum of 135.621, 0.33 and 21.2163 is 157.17 since the last decimal place that is significant for all three numbers (as shown below by the vertical line). $\begin{array}{r|l} 135.62\hspace{-4pt} & \hspace{-4pt}1 \\[-2pt] 0.33\hspace{-4pt} & \\[-2pt] 21.21\hspace{-4pt} & \hspace{-4pt}63 \\ \hline 157.16\hspace{-4pt} & \hspace{-4pt}73 \\ \end{array} \nonumber$. is the hundredth's place. Note that rounding the answer to the correct number of significant figures occurs after completing the exact calculation.. When multiplying or dividing, the result of the calculation contains the same number of significant figures as that measurement having the smallest number of significant figures. Thus,. $\dfrac{22.91\times0.152}{16.302}=0.21361\approx 0.214\nonumber$. because 0.152, with three, has the fewest number of significant figures.. One way to think about this is that we cannot make a measurement more precise through a calculation than it is when we take the measurement.. • Check back to be sure that your group has the correct number of significant figures in the average length and width listed for Experiment 2, if not fix them.. STOP. Uncertainty in the lab. • If you took a measurement and found a value of 89.231 ± 0.008 what is the absolute uncertainty and the percent relative uncertainty of the measurement?. STOP. Propagation of Uncertainty. As we have discussed, each instrument we use in a lab has an associated random error, which is often expressed as a tolerance factor by the manufacturer. A more rigorous approach to determining the uncertainty in a result is called a propagation of uncertainty and it combines the uncertainty in each measurement to estimate the uncertainty in the final result.. The following equations for propagation of uncertainty are from your textbook (as noted in parentheses):. $e_{prop}=\sqrt{e_1^2 + e_2^2+e_3^2+⋯e_n^2} \tag{3-5}$. Multiplication and Division:. $\%e_{prop}=\sqrt{(\%e_1^2)+(\% e_2^2)+(\%e_3^2)+⋯(\%e_n^2)} \tag{3-6}$. Where en is the absolute uncertainty (often the tolerance factor) for measurement n and %en is the percent relative uncertainty for measurement n.. The REAL SIG FIG RULE. The final uncertainty to one significant figure and that defines how many significant figures are in the measurement. For example if you took several measurements and the average value was 234.7182 a.u., but the uncertainty when propagated was 0.0278 a.u. you would write the final result as 234.72 ± 0.03 a.u.. • Assuming that the error associated with the ruler is 0.1 mm, propagate the error associated with the average of the length and the width of the rectangle using the data from Experiment 2.. 1. What is the absolute uncertainty of each measurement?. 2. Should you use the addition/subtraction or multiplication/division equation above to propagate the error?. 3. Calculate the error and write it with the proper number of sig. figs is the table below.. Average length (mm). Average width (mm). Error. • Asking yourself the same questions as above, calculate the average area of the rectangle including the propagated error and the correct number of sig figs and write the value in the table.. Average area w/ error (mm2). STOP. • As a group think of times in the lab you may want to use the addition and subtraction rule to propagate uncertainty and one time in lab you would want to use multiplication and division rule to propagate uncertainty.. STOP. Often propagation of error includes mixed operation. Like finding the area of the rectangle, except I walked you through how to do that. Now you are going to propagate error for a scientific system you have never seen before.. • For a concentration technique the relationship between the signal and the an analyte’s concentration is. $S_\textrm{total} = k_\textrm{A}C_\textrm{A} + S_\textrm{mb}\nonumber$. What is the analyte’s concentration, CA, and its uncertainty if Stotal is 24.37 ± 0.02, Smb is 0.96 ± 0.02, and kA is 0.186 ± 0.003 ppm–1.. STOP. • Given the following masses of deionized water measured during the calibration of a Class A pipet at 15 °C, what average volume of water is the 10.00 mL Class A (± 0.02 mL) pipet actually delivering?. Measurement. Mass of H2O (g). 1. 10.1578. 2. 10.1341. 3. 10.1425. 4. 10.1453. 5. 10.1672. 6. 10.1587. 7. 10.1611. STOP. • Explain, with numerical values, glassware, and instruments how you would prepare 50 mL of a 1.40 M aqueous stock solution of sodium thiosulfate, Na2S2O3 (MW= 158.11 g/mol) (which is a solid that must be dissolved) while introducing the least amount of error. (similar example pg 23 of textbook). STOP. • Explain, with numerical values, glassware, and instruments how you would prepare a 100 mL of a 0.140 M aqueous solution of sodium thiosulfate from the 1.40 M stock solution in the previous problem while introducing the least amount of error.. STOP. Homework for data analysis. • How is systematic error identified and accounted for in measurements?. • How is random error identified and accounted for in measurements?. • In a lab at 27°C you weigh an empty volumetric flask at 468.654 g you then fill the flask up to the calibration mark with DI H2O and weigh the flask at 543.635 g. What is the true volume contained in the volumetric?. • What is the true concentration of dissolved 0.1567 g of Na2S2O3 in molarity if you made it in the volumetric flask in the previous example?.
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• What portion of the analytical process do the tools we talked about today fall under?. • If you made a 5 M solution via dilution of a 10 M solution using volumetric glassware and you performed several forms of analytical analysis and determined that the true concentration of the solution was 4.5 M what should be done to avoid this problem in the future?.
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http://zhu45.org/posts/2016/Dec/26/reflection-on-integer-arithmetic-package-problem/
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# Fluffy Stuff
A tmp place to rest
# Reflection on integer arithmetic package problem
This weekend, I'm working on MAW 3.9. The single problem results in almost 500 lines of code. This is quite unexpected. The problem is stated as the following:
Write an arbitrary-precision integer arithmetic package. You should use a strategy similar to polynomial arithmetic. Compute the distribution of the digits $$0$$ to $$9$$ in $$2^{4000}$$.
## Which way to go?
Since the problem states "arbitrary-precision" and "use a strategy similar to polynomial arithmetic", then I can conclude that linked list is the best data structure for this problem. However, the question is how we can construct the linked list to best implement our integer arithmetic operations (i.e. addition, mulitiplication)?
We essentially have two options:
1. We put the most significant digit as the the very first data node and we put the least significant digit as the last data node. For example, for a number $$123$$, we will implement it like dummy->1->2->3.
2. This is the exactly opposite of the first option. We put the least significant digit as the very first data node and we put the most significant digit as the last data node. Again, for $$123$$, we will implement is like dummy->3->2->1.
Let's evaluate these two options from two perspective:
1. Whether we can easily construct a linked list to represent arbitrary-precision integer?
2. Whether the arithmetic operations are essy to implement?
From the first perspective, for option one, each time we add a new digit to the most significant position, we insert a new node at the very beginning of the list (i.e. right after the header node). On the other hand, for option two, we append a new node at the very end of the list. Since we design our addDigit with an input of a pointer to node (i.e. to specify where to add node), these two options work equally well.
From the second perspective, things are different. Take arithmetic addition as an example. When we try to add two numbers, for option one, we need to walk through the whole list to begin with the very end of the node because we want to start with unit digit. This makes our routine complex because we need to use a while loop to walk through the list first. For second option, situation is easier becauuse the number is implemented in the reverse order in the list. The very first data node is the unit digit and we can directly start with addition while we move towards the end of the list. If we need to add additional node because of carry (i.e. $$999 + 1$$ will be no longer 3-digit but 4-digit number), we can naturally pass the pointer pointing towards the current node to the addDigit function.
So, we choose option two to implement our integer package.
## Memory leak
Memory leak is a very important issue to pay attention to during the testing phase. We use valgrind to help us detect if there is any leak in our code. You can reference their quick start guide and memory check user manual for the commands and error shooting.
Here are the two mistakes I made (You can check out my commit about memory leak debug):
1. Always free the chunk allocated by malloc whenever possible.
Take multiply function as an example:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 integerList multiply(integerList A, integerList B) { PtrToNode dummyA = A->NextDigit; PtrToNode dummyB = B->NextDigit; integerList tmpR = makeEmpty(); PtrToNode dummyTmpR = tmpR; integerList R = makeEmpty(); int product, carry = 0; int i, indent = 0; while (dummyA != NULL) { while (dummyB != NULL) { product = dummyA->Digit * dummyB->Digit + carry; carry = product / 10; addDigit(product % 10, dummyTmpR); dummyTmpR = dummyTmpR->NextDigit; dummyB = dummyB->NextDigit; } if (carry > 0) { addDigit(carry, dummyTmpR); dummyTmpR = dummyTmpR->NextDigit; } for(i = 0; i < indent; i++) { addDigit(0,tmpR); } integerList tmp = R; // prevent memory leak R = add(R, tmpR); deleteAll(tmp); indent ++; carry = 0; deleteIntegerList(tmpR); dummyTmpR = tmpR; dummyA = dummyA->NextDigit; dummyB = B->NextDigit; } deleteAll(tmpR); return R; }
We allocate tmpR through makeEmpty() in Line[7]. If we don't do anything about it inside the function, then the memory will be lost because we have no way to reference this chunk of memory outside the function. Local variable tmpR is the only reference to the memory allocated on the heap. However, once the function is done, the local variable is destroyed from the stack, and thus, we lose our only reference to the memory chunk. So, we need to free it before we exit the function (Line[49]).
1. Be careful with a function call inside a function call.
This type of leak is much more subtle than the first one. Originally instead of
integerList tmp = R;
deleteAll(tmp);
I only have R = add(R, tmpR). This cause the leak because of the following reasoning: Originally, we have R points to a list of nodes. When we do add(R,tmpR), we create a new list of nodes, which hold our addition result. Then we let R points towards this newly-created list. This makes us lose the list of nodes originally pointed by R. That's why we introduce tmp.
## makeEmpty ?
Originally, I don't have this makeEmpty function:
integerList
makeEmpty()
{
integerList R = malloc(sizeof(struct Node));
R->NextDigit = NULL; // super important step
return R;
}
If you take a look at this function, it seems to be a wrapper around malloc operation, which seems redundant (we could directly call malloc directly in the place that makeEmpty appears). However, the key for this routine is R->NextDigit = NULL;. This step can be easily omitted. However, without this step, we don't have fully control on what our newly-allocated empty list (i.e. a list with only header node) will look like. In other words, our header node will point to somewhere (i.e. R->NextDigit) randomly without our key step. This can cause serious trouble for the following routine debug. For example, we could have R->NextDigit holds some address value that happens to have a node structure there with a value in it. For instance, dummy->1. This can usually happen when you OS try to reuse the memory chunk you previously freed. For example, try the following experiment:
1. replace makeEmpty on Line[7] & line[10] in multiply function
2. multiply works fine with test_multiply() solely in the test program.
3. multiply won't work if we do test_intializeInteger() and test_add() before test_multiply() because the integer we construct will no longer be 342 in the test case but something like 3425, where 5 is some value pointed by R->NextDigit.
So, always clear out the pointer by setting it to NULL whenever we do initialization.
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# Fluffy Stuff. A tmp place to rest. # Reflection on integer arithmetic package problem. This weekend, I'm working on MAW 3.9. The single problem results in almost 500 lines of code. This is quite unexpected. The problem is stated as the following:. Write an arbitrary-precision integer arithmetic package. You should use a strategy similar to polynomial arithmetic. Compute the distribution of the digits $$0$$ to $$9$$ in $$2^{4000}$$.. ## Which way to go?. Since the problem states "arbitrary-precision" and "use a strategy similar to polynomial arithmetic", then I can conclude that linked list is the best data structure for this problem. However, the question is how we can construct the linked list to best implement our integer arithmetic operations (i.e. addition, mulitiplication)?. We essentially have two options:. 1. We put the most significant digit as the the very first data node and we put the least significant digit as the last data node. For example, for a number $$123$$, we will implement it like dummy->1->2->3.. 2. This is the exactly opposite of the first option. We put the least significant digit as the very first data node and we put the most significant digit as the last data node. Again, for $$123$$, we will implement is like dummy->3->2->1.. Let's evaluate these two options from two perspective:. 1. Whether we can easily construct a linked list to represent arbitrary-precision integer?. 2. Whether the arithmetic operations are essy to implement?. From the first perspective, for option one, each time we add a new digit to the most significant position, we insert a new node at the very beginning of the list (i.e. right after the header node). On the other hand, for option two, we append a new node at the very end of the list. Since we design our addDigit with an input of a pointer to node (i.e. to specify where to add node), these two options work equally well.. From the second perspective, things are different. Take arithmetic addition as an example. When we try to add two numbers, for option one, we need to walk through the whole list to begin with the very end of the node because we want to start with unit digit. This makes our routine complex because we need to use a while loop to walk through the list first. For second option, situation is easier becauuse the number is implemented in the reverse order in the list. The very first data node is the unit digit and we can directly start with addition while we move towards the end of the list. If we need to add additional node because of carry (i.e. $$999 + 1$$ will be no longer 3-digit but 4-digit number), we can naturally pass the pointer pointing towards the current node to the addDigit function.. So, we choose option two to implement our integer package.. ## Memory leak. Memory leak is a very important issue to pay attention to during the testing phase. We use valgrind to help us detect if there is any leak in our code. You can reference their quick start guide and memory check user manual for the commands and error shooting.. Here are the two mistakes I made (You can check out my commit about memory leak debug):. 1. Always free the chunk allocated by malloc whenever possible.. Take multiply function as an example:. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 integerList multiply(integerList A, integerList B) { PtrToNode dummyA = A->NextDigit; PtrToNode dummyB = B->NextDigit; integerList tmpR = makeEmpty(); PtrToNode dummyTmpR = tmpR; integerList R = makeEmpty(); int product, carry = 0; int i, indent = 0; while (dummyA != NULL) { while (dummyB != NULL) { product = dummyA->Digit * dummyB->Digit + carry; carry = product / 10; addDigit(product % 10, dummyTmpR); dummyTmpR = dummyTmpR->NextDigit; dummyB = dummyB->NextDigit; } if (carry > 0) { addDigit(carry, dummyTmpR); dummyTmpR = dummyTmpR->NextDigit; } for(i = 0; i < indent; i++) { addDigit(0,tmpR); } integerList tmp = R; // prevent memory leak R = add(R, tmpR); deleteAll(tmp); indent ++; carry = 0; deleteIntegerList(tmpR); dummyTmpR = tmpR; dummyA = dummyA->NextDigit; dummyB = B->NextDigit; } deleteAll(tmpR); return R; } . We allocate tmpR through makeEmpty() in Line[7]. If we don't do anything about it inside the function, then the memory will be lost because we have no way to reference this chunk of memory outside the function. Local variable tmpR is the only reference to the memory allocated on the heap. However, once the function is done, the local variable is destroyed from the stack, and thus, we lose our only reference to the memory chunk. So, we need to free it before we exit the function (Line[49]).. 1. Be careful with a function call inside a function call.. This type of leak is much more subtle than the first one. Originally instead of. integerList tmp = R;. deleteAll(tmp);. I only have R = add(R, tmpR). This cause the leak because of the following reasoning: Originally, we have R points to a list of nodes. When we do add(R,tmpR), we create a new list of nodes, which hold our addition result. Then we let R points towards this newly-created list. This makes us lose the list of nodes originally pointed by R. That's why we introduce tmp.. ## makeEmpty ?. Originally, I don't have this makeEmpty function:. integerList. makeEmpty(). {. integerList R = malloc(sizeof(struct Node));. R->NextDigit = NULL; // super important step. return R;. }. If you take a look at this function, it seems to be a wrapper around malloc operation, which seems redundant (we could directly call malloc directly in the place that makeEmpty appears). However, the key for this routine is R->NextDigit = NULL;. This step can be easily omitted. However, without this step, we don't have fully control on what our newly-allocated empty list (i.e. a list with only header node) will look like. In other words, our header node will point to somewhere (i.e. R->NextDigit) randomly without our key step. This can cause serious trouble for the following routine debug. For example, we could have R->NextDigit holds some address value that happens to have a node structure there with a value in it. For instance, dummy->1. This can usually happen when you OS try to reuse the memory chunk you previously freed. For example, try the following experiment:. 1. replace makeEmpty on Line[7] & line[10] in multiply function. 2. multiply works fine with test_multiply() solely in the test program.. 3.
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multiply won't work if we do test_intializeInteger() and test_add() before test_multiply() because the integer we construct will no longer be 342 in the test case but something like 3425, where 5 is some value pointed by R->NextDigit.. So, always clear out the pointer by setting it to NULL whenever we do initialization.
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https://math.libretexts.org/Bookshelves/Linear_Algebra/Interactive_Linear_Algebra_(Margalit_and_Rabinoff)/03%3A_Linear_Transformations_and_Matrix_Algebra/3.01%3A_Matrix_Transformations
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 3.1: Matrix Transformations
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##### Objectives
1. Learn to view a matrix geometrically as a function.
2. Learn examples of matrix transformations: reflection, dilation, rotation, shear, projection.
3. Understand the vocabulary surrounding transformations: domain, codomain, range.
4. Understand the domain, codomain, and range of a matrix transformation.
5. Pictures: common matrix transformations.
6. Vocabulary words: transformation / function, domain, codomain, range, identity transformation, matrix transformation.
In this section we learn to understand matrices geometrically as functions, or transformations. We briefly discuss transformations in general, then specialize to matrix transformations, which are transformations that come from matrices.
## Matrices as Functions
Informally, a function is a rule that accepts inputs and produces outputs. For instance, $$f(x) = x^2$$ is a function that accepts one number $$x$$ as its input, and outputs the square of that number: $$f(2) = 4$$. In this subsection, we interpret matrices as functions.
Let $$A$$ be a matrix with $$m$$ rows and $$n$$ columns. Consider the matrix equation $$b=Ax$$ (we write it this way instead of $$Ax=b$$ to remind the reader of the notation $$y=f(x)$$). If we vary $$x\text{,}$$ then $$b$$ will also vary; in this way, we think of $$A$$ as a function with independent variable $$x$$ and dependent variable $$b$$.
• The independent variable (the input) is $$x\text{,}$$ which is a vector in $$\mathbb{R}^n$$.
• The dependent variable (the output) is $$b\text{,}$$ which is a vector in $$\mathbb{R}^m$$.
The set of all possible output vectors are the vectors $$b$$ such that $$Ax=b$$ has some solution; this is the same as the column space of $$A$$ by Note 2.3.6 in Section 2.3.
Figure $$\PageIndex{1}$$
##### Example $$\PageIndex{3}$$: Projection onto the $$xy$$-plane
Let
$A=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^3$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right)\:\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}x\\y\\0\end{array}\right).\nonumber$
Multiplication by $$A$$ simply sets the $$z$$-coordinate equal to zero: it projects vertically onto the $$xy$$-plane
Figure $$\PageIndex{4}$$
##### Example $$\PageIndex{4}$$: Reflection
Let
$A=\left(\begin{array}{cc}-1&0\\0&1\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}-1&0\\0&1\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}-x\\y\end{array}\right).\nonumber$
Multiplication by $$A$$ negates the $$x$$-coordinate: it reflects over the $$y$$-axis.
Figure $$\PageIndex{6}$$
##### Example $$\PageIndex{5}$$: Dilation
Let
$A=\left(\begin{array}{cc}1.5&0\\0&1.5\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}1.5&0\\0&1.5\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}1.5x\\1.5y\end{array}\right)=1.5\left(\begin{array}{c}x\\y\end{array}\right).\nonumber$
Multiplication by $$A$$ is the same as scalar multiplication by $$1.5\text{:}$$ it scales or dilates the plane by a factor of $$1.5$$.
Figure $$\PageIndex{8}$$
##### Example $$\PageIndex{6}$$: Identity
Let
$A=\left(\begin{array}{cc}1&0\\0&1\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}x\\y\end{array}\right).\nonumber$
Multiplication by $$A$$ does not change the input vector at all: it is the identity transformation which does nothing
Figure $$\PageIndex{10}$$
##### Example $$\PageIndex{7}$$: Rotation
Let
$A=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}-y\\x\end{array}\right).\nonumber$
We substitute a few test points in order to understand the geometry of the transformation:
Figure $$\PageIndex{12}$$
Multiplication by $$A$$ is counterclockwise rotation by $$90^\circ$$.
Figure $$\PageIndex{13}$$
##### Example $$\PageIndex{8}$$: Shear
Let
$A=\left(\begin{array}{cc}1&1\\0&1\end{array}\right).\nonumber$
Describe the function $$b=Ax$$ geometrically.
Solution
In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:
$A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}1&1\\0&1\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}x+y\\y\end{array}\right).\nonumber$
Multiplication by $$A$$ adds the $$y$$-coordinate to the $$x$$-coordinate; this is called a shear in the $$x$$-direction
Figure $$\PageIndex{15}$$
## Transformations
At this point it is convenient to fix our ideas and terminology regarding functions, which we will call transformations in this book. This allows us to systematize our discussion of matrices as functions.
##### Definition $$\PageIndex{1}$$: Transformation
A transformation from $$\mathbb{R}^n$$ to $$\mathbb{R}^m$$ is a rule $$T$$ that assigns to each vector $$x$$ in $$\mathbb{R}^n$$ a vector $$T(x)$$ in $$\mathbb{R}^m$$.
• $$\mathbb{R}^n$$ is called the domain of $$T$$.
• $$\mathbb{R}^m$$ is called the codomain of $$T$$.
• For $$x$$ in $$\mathbb{R}^n \text{,}$$ the vector $$T(x)$$ in $$\mathbb{R}^m$$ is the image of $$x$$ under $$T$$.
• The set of all images $$\{T(x)\mid x\text{ in }\mathbb{R}^n \}$$ is the range of $$T$$.
The notation $$T\colon\mathbb{R}^n \to \mathbb{R}^m$$ means “$$T$$ is a transformation from $$\mathbb{R}^n$$ to $$\mathbb{R}^m$$.”
It may help to think of $$T$$ as a “machine” that takes $$x$$ as an input, and gives you $$T(x)$$ as the output.
Figure $$\PageIndex{17}$$
The points of the domain $$\mathbb{R}^n$$ are the inputs of $$T\text{:}$$ this simply means that it makes sense to evaluate $$T$$ on vectors with $$n$$ entries, i.e., lists of $$n$$ numbers. Likewise, the points of the codomain $$\mathbb{R}^m$$ are the outputs of $$T\text{:}$$ this means that the result of evaluating $$T$$ is always a vector with $$m$$ entries.
The range of $$T$$ is the set of all vectors in the codomain that actually arise as outputs of the function $$T\text{,}$$ for some input. In other words, the range is all vectors $$b$$ in the codomain such that $$T(x)=b$$ has a solution $$x$$ in the domain.
##### Example $$\PageIndex{9}$$: A Function of one variable
Most of the functions you may have seen previously have domain and codomain equal to $$\mathbb{R} = \mathbb{R}^1$$. For example,
$\sin :\mathbb{R}\to\mathbb{R}\quad\sin(x)=\left(\begin{array}{l}\text{the length of the opposite} \\ \text{edge over the hypotenuse of} \\ \text{a right triangle with angle }x \\ \text{in radians}\end{array}\right).\nonumber$
Notice that we have defined $$\sin$$ by a rule: a function is defined by specifying what the output of the function is for any possible input.
You may be used to thinking of such functions in terms of their graphs:
Figure $$\PageIndex{18}$$
In this case, the horizontal axis is the domain, and the vertical axis is the codomain. This is useful when the domain and codomain are $$\mathbb{R}\text{,}$$ but it is hard to do when, for instance, the domain is $$\mathbb{R}^2$$ and the codomain is $$\mathbb{R}^3$$. The graph of such a function is a subset of $$\mathbb{R}^5\text{,}$$ which is difficult to visualize. For this reason, we will rarely graph a transformation.
Note that the range of $$\sin$$ is the interval $$[-1,1]\text{:}$$ this is the set of all possible outputs of the $$\sin$$ function.
##### Example $$\PageIndex{10}$$: Functions of several variables
Here is an example of a function from $$\mathbb{R}^2$$ to $$\mathbb{R}^3 \text{:}$$
$f\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}x+y\\ \cos(y) \\y-x^{2}\end{array}\right).\nonumber$
The inputs of $$f$$ each have two entries, and the outputs have three entries. In this case, we have defined $$f$$ by a formula, so we evaluate $$f$$ by substituting values for the variables:
$f\left(\begin{array}{c}2\\3\end{array}\right)=\left(\begin{array}{c}2+3 \\ \cos(3) \\ 3-2^2\end{array}\right)=\left(\begin{array}{c}5\\ \cos(3)\\-1\end{array}\right).\nonumber$
Here is an example of a function from $$\mathbb{R}^3$$ to $$\mathbb{R}^3 \text{:}$$
$f(v)=\left(\begin{array}{c}\text{the counterclockwise rotation} \\ \text{of }v\text{ by and angle of }42^\circ \text{ about} \\ \text{ the }z\text{-axis}\end{array}\right).\nonumber$
In other words, $$f$$ takes a vector with three entries, then rotates it; hence the ouput of $$f$$ also has three entries. In this case, we have defined $$f$$ by a geometric rule.
##### Definition $$\PageIndex{2}$$: Identity Transformation
The identity transformation $$\text{Id}_{\mathbb{R}^n }\colon\mathbb{R}^n \to\mathbb{R}^n$$ is the transformation defined by the rule
$\text{Id}_{\mathbb{R}^n }(x) = x \qquad\text{for all x in \mathbb{R}^n }. \nonumber$
In other words, the identity transformation does not move its input vector: the output is the same as the input. Its domain and codomain are both $$\mathbb{R}^n \text{,}$$ and its range is $$\mathbb{R}^n$$ as well, since every vector in $$\mathbb{R}^n$$ is the output of itself.
##### Example $$\PageIndex{11}$$: A real-word transformation: robotics
The definition of transformation and its associated vocabulary may seem quite abstract, but transformations are extremely common in real life. Here is an example from the fields of robotics and computer graphics.
Suppose you are building a robot arm with three joints that can move its hand around a plane, as in the following picture.
Figure $$\PageIndex{19}$$
Define a transformation $$f\colon\mathbb{R}^3 \to\mathbb{R}^2$$ as follows: $$f(\theta,\phi,\psi)$$ is the $$(x,y)$$ position of the hand when the joints are rotated by angles $$\theta, \phi, \psi\text{,}$$ respectively. Evaluating $$f$$ tells you where the hand will be on the plane when the joints are set at the given angles.
It is relatively straightforward to find a formula for $$f(\theta,\phi,\psi)$$ using some basic trigonometry. If you want the robot to fetch your coffee cup, however, you have to find the angles $$\theta,\phi,\psi$$ that will put the hand at the position of your beverage. It is not at all obvious how to do this, and it is not even clear if the answer is unique! You can ask yourself: “which positions on the table can my robot arm reach?” or “what is the arm’s range of motion?” This is the same as asking: “what is the range of $$f\text{?}$$”
Unfortunately, this kind of function does not come from a matrix, so one cannot use linear algebra to answer these kinds of questions. In fact, these functions are rather complicated; their study is the subject of inverse kinematics.
## Matrix Transformations
Now we specialize the general notions and vocabulary from the previous Subsection Transformations to the functions defined by matrices that we considered in the first Subsection Matrices as Functions.
##### Definition $$\PageIndex{3}$$: Matrix Transformation
Let $$A$$ be an $$m\times n$$ matrix. The matrix transformation associated to $$A$$ is the transformation
$T\colon \mathbb{R}^n \to \mathbb{R}^m \quad\text{defined by}\quad T(x) = Ax. \nonumber$
This is the transformation that takes a vector $$x$$ in $$\mathbb{R}^n$$ to the vector $$Ax$$ in $$\mathbb{R}^m$$.
If $$A$$ has $$n$$ columns, then it only makes sense to multiply $$A$$ by vectors with $$n$$ entries. This is why the domain of $$T(x)=Ax$$ is $$\mathbb{R}^n$$. If $$A$$ has $$n$$ rows, then $$Ax$$ has $$m$$ entries for any vector $$x$$ in $$\mathbb{R}^n \text{;}$$ this is why the codomain of $$T(x)=Ax$$ is $$\mathbb{R}^m$$.
The definition of a matrix transformation $$T$$ tells us how to evaluate $$T$$ on any given vector: we multiply the input vector by a matrix. For instance, let
$A=\left(\begin{array}{ccc}1&2&3\\4&5&6\end{array}\right)\nonumber$
and let $$T(x)=Ax$$ be the associated matrix transformation. Then
$T\left(\begin{array}{c}-1\\-2\\-3\end{array}\right)=A\left(\begin{array}{c}-1\\-2\\-3\end{array}\right)=\left(\begin{array}{ccc}1&2&3\\4&5&6\end{array}\right)\:\left(\begin{array}{c}-1\\-2\\-3\end{array}\right)=\left(\begin{array}{c}-14\\-32\end{array}\right).\nonumber$
Suppose that $$A$$ has columns $$v_1,v_2,\ldots,v_n$$. If we multiply $$A$$ by a general vector $$x\text{,}$$ we get
$Ax=\left(\begin{array}{cccc}|&|&\quad&| \\ v_1&v&2&\cdots &v_n \\ |&|&\quad &|\end{array}\right)\:\left(\begin{array}{c}x_1\\x_2\\ \vdots \\x_n\end{array}\right)=x_1v_1+x_2v_2+\cdots +x_nv_n.\nonumber$
This is just a general linear combination of $$v_1,v_2,\ldots,v_n$$. Therefore, the outputs of $$T(x) = Ax$$ are exactly the linear combinations of the columns of $$A\text{:}$$ the range of $$T$$ is the column space of $$A$$. See Note 2.3.6 in Section 2.3.
Note $$\PageIndex{1}$$
Let $$A$$ be an $$m\times n$$ matrix, and let $$T(x)=Ax$$ be the associated matrix transformation.
• The domain of $$T$$ is $$\mathbb{R}^n \text{,}$$ where $$n$$ is the number of columns of $$A$$.
• The codomain of $$T$$ is $$\mathbb{R}^m \text{,}$$ where $$m$$ is the number of rows of $$A$$.
• The range of $$T$$ is the column space of $$A$$.
##### Example $$\PageIndex{12}$$: Interactive: A $$2\times 3$$ matrix: reprise
Let
$A=\left(\begin{array}{c}1&-1&2\\-2&2&4\end{array}\right),\nonumber$
and define $$T(x) = Ax$$. The domain of $$T$$ is $$\mathbb{R}^3 \text{,}$$ and the codomain is $$\mathbb{R}^2$$. The range of $$T$$ is the column space; since all three columns are collinear, the range is a line in $$\mathbb{R}^2$$.
##### Example $$\PageIndex{13}$$: Interactive: A $$3\times 2$$ matrix: reprise
Let
$A=\left(\begin{array}{cc}1&0\\0&1\\1&0\end{array}\right),\nonumber$
and define $$T(x) = Ax$$. The domain of $$T$$ is $$\mathbb{R}^2 \text{,}$$ and the codomain is $$\mathbb{R}^3$$. The range of $$T$$ is the column space; since $$A$$ has two columns which are not collinear, the range is a plane in $$\mathbb{R}^3$$.
##### Example $$\PageIndex{14}$$: Projection onto the $$xy$$-plane: reprise
Let
$A=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right),\nonumber$
and let $$T(x) = Ax$$. What are the domain, the codomain, and the range of $$T\text{?}$$
Solution
Geometrically, the transformation $$T$$ projects a vector directly “down” onto the $$xy$$-plane in $$\mathbb{R}^3$$.
Figure $$\PageIndex{22}$$
The inputs and outputs have three entries, so the domain and codomain are both $$\mathbb{R}^3$$. The possible outputs all lie on the $$xy$$-plane, and every point on the $$xy$$-plane is an output of $$T$$ (with itself as the input), so the range of $$T$$ is the $$xy$$-plane.
Be careful not to confuse the codomain with the range here. The range is a plane, but it is a plane in $$\mathbb{R}^3$$, so the codomain is still $$\mathbb{R}^3$$. The outputs of $$T$$ all have three entries; the last entry is simply always zero.
In the case of an $$n\times n$$ square matrix, the domain and codomain of $$T(x) = Ax$$ are both $$\mathbb{R}^n$$. In this situation, one can regard $$T$$ as operating on $$\mathbb{R}^n \text{:}$$ it moves the vectors around in the same space.
##### Example $$\PageIndex{15}$$: Matrix transformations of $$\mathbb{R}^2$$
In the first Subsection Matrices as Functions we discussed the transformations defined by several $$2\times 2$$ matrices, namely:
\begin{align*} \text{Reflection:} &\qquad A=\left(\begin{array}{cc}-1&0\\0&1\end{array}\right) \\ \text{Dilation:} &\qquad A=\left(\begin{array}{cc}1.5&0\\0&1.5\end{array}\right) \\ \text{Identity:} &\qquad A=\left(\begin{array}{cc}1&0\\0&1\end{array}\right) \\ \text{Rotation:} &\qquad A=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right) \\ \text{Shear:} &\qquad A=\left(\begin{array}{cc}1&1\\0&1\end{array}\right). \end{align*}
In each case, the associated matrix transformation $$T(x)=Ax$$ has domain and codomain equal to $$\mathbb{R}^2$$. The range is also $$\mathbb{R}^2 \text{,}$$ as can be seen geometrically (what is the input for a given output?), or using the fact that the columns of $$A$$ are not collinear (so they form a basis for $$\mathbb{R}^2$$).
##### Example $$\PageIndex{16}$$: Questions about a [matrix] transformation
Let
$A=\left(\begin{array}{cc}1&1\\0&1\\1&1\end{array}\right),\nonumber$
and let $$T(x)=Ax\text{,}$$ so $$T\colon\mathbb{R}^2 \to\mathbb{R}^3$$ is a matrix transformation.
1. Evaluate $$T(u)$$ for $$u=\left(\begin{array}{c}3\\4\end{array}\right)$$.
2. Let
$b = \left(\begin{array}{c}7\\5\\7\end{array}\right). \nonumber$
Find a vector $$v$$ in $$\mathbb{R}^2$$ such that $$T(v)=b$$. Is there more than one?
3. Does there exist a vector $$w$$ in $$\mathbb{R}^3$$ such that there is more than one $$v$$ in $$\mathbb{R}^2$$ with $$T(v)=w\text{?}$$
4. Find a vector $$w$$ in $$\mathbb{R}^3$$ which is not in the range of $$T$$.
Note: all of the above questions are intrinsic to the transformation $$T\text{:}$$ they make sense to ask whether or not $$T$$ is a matrix transformation. See the next Example $$\PageIndex{17}$$. As $$T$$ is in fact a matrix transformation, all of these questions will translate into questions about the corresponding matrix $$A$$.
Solution
1. We evaluate $$T(u)$$ by substituting the definition of $$T$$ in terms of matrix multiplication:
$T\left(\begin{array}{c}3\\4\end{array}\right)=\left(\begin{array}{cc}1&1\\0&1\\1&1\end{array}\right)\:\left(\begin{array}{c}3\\4\end{array}\right)=\left(\begin{array}{c}7\\4\\7\end{array}\right).\nonumber$
2. We want to find a vector $$v$$ such that $$b = T(v) = Av$$. In other words, we want to solve the matrix equation $$Av = b$$. We form an augmented matrix and row reduce:
$\left(\begin{array}{cc}1&1\\0&1\\1&1\end{array}\right)v=\left(\begin{array}{c}7\\5\\7\end{array}\right) \quad\xrightarrow{\text{augmented matrix}}\quad\left(\begin{array}{cc|c}1&1&7\\0&1&5\\1&1&7\end{array}\right) \quad\xrightarrow{\text{row reduce}}\quad\left(\begin{array}{cc|c}1&0&2\\0&1&5\\0&0&0\end{array}\right).\nonumber$
This gives $$x=2$$ and $$y=5\text{,}$$ so that there is a unique vector
$v = \left(\begin{array}{c}2\\5\end{array}\right) \nonumber$
such that $$T(v) = b$$.
3. Translation: is there any vector $$w$$ in $$\mathbb{R}^3$$ such that the solution set of $$Av=w$$ has more than one vector in it? The solution set of $$Ax=w\text{,}$$ if non-empty, is a translate of the solution set of $$Av=b$$ above, which has one vector in it. See key observation 2.4.3 in Section 2.4. It follows that the solution set of $$Av=w$$ can have at most one vector.
4. Translation: find a vector $$w$$ such that the matrix equation $$Av=w$$ is not consistent. Notice that if we take
$w = \left(\begin{array}{c}1\\2\\3\end{array}\right)\text{,} \nonumber$
then the matrix equation $$Av=w$$ translates into the system of equations
$\left\{\begin{array}{rrrrl}x &+& y &=& 1\\ {}&{}& y &=& 2\\ x &+& y &=& 3,\end{array}\right.\nonumber$
which is clearly inconsistent.
##### Example $$\PageIndex{17}$$: Questions about a [non-matrix] transformation
Define a transformation $$T\colon\mathbb{R}^2 \to\mathbb{R}^3$$ by the formula
$T\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}\ln(x) \\ \cos(y) \\ \ln(x)\end{array}\right).\nonumber$
1. Evaluate $$T(u)$$ for $$u=\left(\begin{array}{c}1\\ \pi\end{array}\right)$$.
2. Let
$b = \left(\begin{array}{c}7\\1\\7\end{array}\right). \nonumber$
Find a vector $$v$$ in $$\mathbb{R}^2$$ such that $$T(v)=b$$. Is there more than one?
3. Does there exist a vector $$w$$ in $$\mathbb{R}^3$$ such that there is more than one $$v$$ in $$\mathbb{R}^2$$ with $$T(v)=w\text{?}$$
4. Find a vector $$w$$ in $$\mathbb{R}^3$$ which is not in the range of $$T$$.
Note: we asked (almost) the exact same questions about a matrix transformation in the previous Example $$\PageIndex{16}$$. The point of this example is to illustrate the fact that the questions make sense for a transformation that has no hope of coming from a matrix. In this case, these questions do not translate into questions about a matrix; they have to be answered in some other way.
Solution
1. We evaluate $$T(u)$$ using the defining formula:
$T\left(\begin{array}{c}1\\ \pi\end{array}\right)=\left(\begin{array}{c} \ln(1) \\ \cos(\pi ) \\ \ln(1)\end{array}\right)=\left(\begin{array}{c}0\\-1\\0\end{array}\right).\nonumber$
2. We have
$T\left(\begin{array}{c}e^7 \\ 2\pi n \\ e^7\end{array}\right)=\left(\begin{array}{c}\ln(e^7) \\ \cos (2\pi n) \\ \ln(e^7)\end{array}\right)=\left(\begin{array}{c}7\\1\\7\end{array}\right)\nonumber$
for any whole number $$n$$. Hence there are infinitely many such vectors.
3. The vector $$b$$ from the previous part is an example of such a vector.
4. Since $$\cos(y)$$ is always between $$-1$$ and $$1\text{,}$$ the vector
$w = \left(\begin{array}{c}0\\2\\0\end{array}\right) \nonumber$
is not in the range of $$T$$.
3.1: Matrix Transformations is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Dan Margalit & Joseph Rabinoff via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$. # 3.1: Matrix Transformations. $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$. ##### Objectives. 1. Learn to view a matrix geometrically as a function.. 2. Learn examples of matrix transformations: reflection, dilation, rotation, shear, projection.. 3. Understand the vocabulary surrounding transformations: domain, codomain, range.. 4. Understand the domain, codomain, and range of a matrix transformation.. 5. Pictures: common matrix transformations.. 6. Vocabulary words: transformation / function, domain, codomain, range, identity transformation, matrix transformation.. In this section we learn to understand matrices geometrically as functions, or transformations. We briefly discuss transformations in general, then specialize to matrix transformations, which are transformations that come from matrices.. ## Matrices as Functions. Informally, a function is a rule that accepts inputs and produces outputs. For instance, $$f(x) = x^2$$ is a function that accepts one number $$x$$ as its input, and outputs the square of that number: $$f(2) = 4$$. In this subsection, we interpret matrices as functions.. Let $$A$$ be a matrix with $$m$$ rows and $$n$$ columns. Consider the matrix equation $$b=Ax$$ (we write it this way instead of $$Ax=b$$ to remind the reader of the notation $$y=f(x)$$). If we vary $$x\text{,}$$ then $$b$$ will also vary; in this way, we think of $$A$$ as a function with independent variable $$x$$ and dependent variable $$b$$.. • The independent variable (the input) is $$x\text{,}$$ which is a vector in $$\mathbb{R}^n$$.. • The dependent variable (the output) is $$b\text{,}$$ which is a vector in $$\mathbb{R}^m$$.. The set of all possible output vectors are the vectors $$b$$ such that $$Ax=b$$ has some solution; this is the same as the column space of $$A$$ by Note 2.3.6 in Section 2.3.. Figure $$\PageIndex{1}$$. ##### Example $$\PageIndex{3}$$: Projection onto the $$xy$$-plane. Let. $A=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right).\nonumber$. Describe the function $$b=Ax$$ geometrically.. Solution. In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^3$$. First we multiply $$A$$ by a vector to see what it does:. $A\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right)\:\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}x\\y\\0\end{array}\right).\nonumber$. Multiplication by $$A$$ simply sets the $$z$$-coordinate equal to zero: it projects vertically onto the $$xy$$-plane. Figure $$\PageIndex{4}$$. ##### Example $$\PageIndex{4}$$: Reflection. Let. $A=\left(\begin{array}{cc}-1&0\\0&1\end{array}\right).\nonumber$. Describe the function $$b=Ax$$ geometrically.. Solution. In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:. $A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}-1&0\\0&1\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}-x\\y\end{array}\right).\nonumber$. Multiplication by $$A$$ negates the $$x$$-coordinate: it reflects over the $$y$$-axis.. Figure $$\PageIndex{6}$$. ##### Example $$\PageIndex{5}$$: Dilation. Let. $A=\left(\begin{array}{cc}1.5&0\\0&1.5\end{array}\right).\nonumber$. Describe the function $$b=Ax$$ geometrically.. Solution. In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:. $A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}1.5&0\\0&1.5\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}1.5x\\1.5y\end{array}\right)=1.5\left(\begin{array}{c}x\\y\end{array}\right).\nonumber$. Multiplication by $$A$$ is the same as scalar multiplication by $$1.5\text{:}$$ it scales or dilates the plane by a factor of $$1.5$$.. Figure $$\PageIndex{8}$$. ##### Example $$\PageIndex{6}$$: Identity. Let. $A=\left(\begin{array}{cc}1&0\\0&1\end{array}\right).\nonumber$. Describe the function $$b=Ax$$ geometrically.. Solution. In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:. $A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}x\\y\end{array}\right).\nonumber$. Multiplication by $$A$$ does not change the input vector at all: it is the identity transformation which does nothing. Figure $$\PageIndex{10}$$. ##### Example $$\PageIndex{7}$$: Rotation. Let. $A=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right).\nonumber$. Describe the function $$b=Ax$$ geometrically.. Solution. In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:. $A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}-y\\x\end{array}\right).\nonumber$. We substitute a few test points in order to understand the geometry of the transformation:. Figure $$\PageIndex{12}$$. Multiplication by $$A$$ is counterclockwise rotation by $$90^\circ$$.. Figure $$\PageIndex{13}$$. ##### Example $$\PageIndex{8}$$: Shear. Let. $A=\left(\begin{array}{cc}1&1\\0&1\end{array}\right).\nonumber$. Describe the function $$b=Ax$$ geometrically.. Solution. In the equation $$Ax=b\text{,}$$ the input vector $$x$$ and the output vector $$b$$ are both in $$\mathbb{R}^2$$. First we multiply $$A$$ by a vector to see what it does:. $A\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{cc}1&1\\0&1\end{array}\right)\:\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}x+y\\y\end{array}\right).\nonumber$. Multiplication by $$A$$ adds the $$y$$-coordinate to the $$x$$-coordinate; this is called a shear in the $$x$$-direction. Figure $$\PageIndex{15}$$. ## Transformations. At this point it is convenient to fix our ideas and terminology regarding functions, which we will call transformations in this book. This allows us to systematize our discussion of matrices as functions.. ##### Definition $$\PageIndex{1}$$: Transformation. A transformation from $$\mathbb{R}^n$$ to $$\mathbb{R}^m$$ is a rule $$T$$ that assigns to each vector $$x$$ in $$\mathbb{R}^n$$ a vector $$T(x)$$ in $$\mathbb{R}^m$$.. • $$\mathbb{R}^n$$ is called the domain of $$T$$.. • $$\mathbb{R}^m$$ is called the codomain of $$T$$.. • For $$x$$ in $$\mathbb{R}^n \text{,}$$ the vector $$T(x)$$ in $$\mathbb{R}^m$$ is the image of $$x$$ under $$T$$.. • The set of all images $$\{T(x)\mid x\text{ in }\mathbb{R}^n \}$$ is the range of $$T$$.. The notation $$T\colon\mathbb{R}^n \to \mathbb{R}^m$$ means “$$T$$ is a transformation from $$\mathbb{R}^n$$ to $$\mathbb{R}^m$$.”. It may help to think of $$T$$ as a “machine” that takes $$x$$ as an input, and gives you $$T(x)$$ as the output.. Figure $$\PageIndex{17}$$. The points of the domain $$\mathbb{R}^n$$ are the inputs of $$T\text{:}$$ this simply means that it makes sense to evaluate $$T$$ on vectors with $$n$$ entries, i.e., lists of $$n$$ numbers. Likewise, the points of the codomain $$\mathbb{R}^m$$ are the outputs of $$T\text{:}$$ this means that the result of evaluating $$T$$ is always a vector with $$m$$ entries.. The range of $$T$$ is the set of all vectors in the codomain that actually arise as outputs of the function $$T\text{,}$$ for some input. In other words, the range is all vectors $$b$$ in the codomain such that $$T(x)=b$$ has a solution $$x$$ in the domain.. ##### Example $$\PageIndex{9}$$: A Function of one variable. Most of the functions you may have seen previously have domain and codomain equal to $$\mathbb{R} = \mathbb{R}^1$$. For example,. $\sin :\mathbb{R}\to\mathbb{R}\quad\sin(x)=\left(\begin{array}{l}\text{the length of the opposite} \\ \text{edge over the hypotenuse of} \\ \text{a right triangle with angle }x \\ \text{in radians}\end{array}\right).\nonumber$. Notice that we have defined $$\sin$$ by a rule: a function is defined by specifying what the output of the function is for any possible input.. You may be used to thinking of such functions in terms of their graphs:. Figure $$\PageIndex{18}$$. In this case, the horizontal axis is the domain, and the vertical axis is the codomain. This is useful when the domain and codomain are $$\mathbb{R}\text{,}$$ but it is hard to do when, for instance, the domain is $$\mathbb{R}^2$$ and the codomain is $$\mathbb{R}^3$$. The graph of such a function is a subset of $$\mathbb{R}^5\text{,}$$ which is difficult to visualize. For this reason, we will rarely graph a transformation.. Note that the range of $$\sin$$ is the interval $$[-1,1]\text{:}$$ this is the set of all possible outputs of the $$\sin$$ function.. ##### Example $$\PageIndex{10}$$: Functions of several variables. Here is an example of a function from $$\mathbb{R}^2$$ to $$\mathbb{R}^3 \text{:}$$. $f\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}x+y\\ \cos(y) \\y-x^{2}\end{array}\right).\nonumber$. The inputs of $$f$$ each have two entries, and the outputs have three entries. In this case, we have defined $$f$$ by a formula, so we evaluate $$f$$ by substituting values for the variables:. $f\left(\begin{array}{c}2\\3\end{array}\right)=\left(\begin{array}{c}2+3 \\ \cos(3) \\ 3-2^2\end{array}\right)=\left(\begin{array}{c}5\\ \cos(3)\\-1\end{array}\right).\nonumber$. Here is an example of a function from $$\mathbb{R}^3$$ to $$\mathbb{R}^3 \text{:}$$. $f(v)=\left(\begin{array}{c}\text{the counterclockwise rotation} \\ \text{of }v\text{ by and angle of }42^\circ \text{ about} \\ \text{ the }z\text{-axis}\end{array}\right).\nonumber$. In other words, $$f$$ takes a vector with three entries, then rotates it; hence the ouput of $$f$$ also has three entries. In this case, we have defined $$f$$ by a geometric rule.. ##### Definition $$\PageIndex{2}$$: Identity Transformation. The identity transformation $$\text{Id}_{\mathbb{R}^n }\colon\mathbb{R}^n \to\mathbb{R}^n$$ is the transformation defined by the rule. $\text{Id}_{\mathbb{R}^n }(x) = x \qquad\text{for all x in \mathbb{R}^n }. \nonumber$. In other words, the identity transformation does not move its input vector: the output is the same as the input. Its domain and codomain are both $$\mathbb{R}^n \text{,}$$ and its range is $$\mathbb{R}^n$$ as well, since every vector in $$\mathbb{R}^n$$ is the output of itself.. ##### Example $$\PageIndex{11}$$: A real-word transformation: robotics. The definition of transformation and its associated vocabulary may seem quite abstract, but transformations are extremely common in real life. Here is an example from the fields of robotics and computer graphics.. Suppose you are building a robot arm with three joints that can move its hand around a plane, as in the following picture.. Figure $$\PageIndex{19}$$. Define a transformation $$f\colon\mathbb{R}^3 \to\mathbb{R}^2$$ as follows: $$f(\theta,\phi,\psi)$$ is the $$(x,y)$$ position of the hand when the joints are rotated by angles $$\theta, \phi, \psi\text{,}$$ respectively. Evaluating $$f$$ tells you where the hand will be on the plane when the joints are set at the given angles.. It is relatively straightforward to find a formula for $$f(\theta,\phi,\psi)$$ using some basic trigonometry. If you want the robot to fetch your coffee cup, however, you have to find the angles $$\theta,\phi,\psi$$ that will put the hand at the position of your beverage. It is not at all obvious how to do this, and it is not even clear if the answer is unique! You can ask yourself: “which positions on the table can my robot arm reach?” or “what is the arm’s range of motion?” This is the same as asking: “what is the range of $$f\text{?}$$”. Unfortunately, this kind of function does not come from a matrix, so one cannot use linear algebra to answer these kinds of questions. In fact, these functions are rather complicated; their study is the subject of inverse kinematics.. ## Matrix Transformations. Now we specialize the general notions and vocabulary from the previous Subsection Transformations to the functions defined by matrices that we considered in the first Subsection Matrices as Functions.. ##### Definition $$\PageIndex{3}$$: Matrix Transformation. Let $$A$$ be an $$m\times n$$ matrix. The matrix transformation associated to $$A$$ is the transformation. $T\colon \mathbb{R}^n \to \mathbb{R}^m \quad\text{defined by}\quad T(x) = Ax. \nonumber$. This is the transformation that takes a vector $$x$$ in $$\mathbb{R}^n$$ to the vector $$Ax$$ in $$\mathbb{R}^m$$.. If $$A$$ has $$n$$ columns, then it only makes sense to multiply $$A$$ by vectors with $$n$$ entries. This is why the domain of $$T(x)=Ax$$ is $$\mathbb{R}^n$$. If $$A$$ has $$n$$ rows, then $$Ax$$ has $$m$$ entries for any vector $$x$$ in $$\mathbb{R}^n \text{;}$$ this is why the codomain of $$T(x)=Ax$$ is $$\mathbb{R}^m$$.. The definition of a matrix transformation $$T$$ tells us how to evaluate $$T$$ on any given vector: we multiply the input vector by a matrix. For instance, let. $A=\left(\begin{array}{ccc}1&2&3\\4&5&6\end{array}\right)\nonumber$. and let $$T(x)=Ax$$ be the associated matrix transformation. Then. $T\left(\begin{array}{c}-1\\-2\\-3\end{array}\right)=A\left(\begin{array}{c}-1\\-2\\-3\end{array}\right)=\left(\begin{array}{ccc}1&2&3\\4&5&6\end{array}\right)\:\left(\begin{array}{c}-1\\-2\\-3\end{array}\right)=\left(\begin{array}{c}-14\\-32\end{array}\right).\nonumber$. Suppose that $$A$$ has columns $$v_1,v_2,\ldots,v_n$$. If we multiply $$A$$ by a general vector $$x\text{,}$$ we get. $Ax=\left(\begin{array}{cccc}|&|&\quad&| \\ v_1&v&2&\cdots &v_n \\ |&|&\quad &|\end{array}\right)\:\left(\begin{array}{c}x_1\\x_2\\ \vdots \\x_n\end{array}\right)=x_1v_1+x_2v_2+\cdots +x_nv_n.\nonumber$. This is just a general linear combination of $$v_1,v_2,\ldots,v_n$$. Therefore, the outputs of $$T(x) = Ax$$ are exactly the linear combinations of the columns of $$A\text{:}$$ the range of $$T$$ is the column space of $$A$$. See Note 2.3.6 in Section 2.3.. Note $$\PageIndex{1}$$. Let $$A$$ be an $$m\times n$$ matrix, and let $$T(x)=Ax$$ be the associated matrix transformation.. • The domain of $$T$$ is $$\mathbb{R}^n \text{,}$$ where $$n$$ is the number of columns of $$A$$.. • The codomain of $$T$$ is $$\mathbb{R}^m \text{,}$$ where $$m$$ is the number of rows of $$A$$.. • The range of $$T$$ is the column space of $$A$$.. ##### Example $$\PageIndex{12}$$: Interactive: A $$2\times 3$$ matrix: reprise. Let. $A=\left(\begin{array}{c}1&-1&2\\-2&2&4\end{array}\right),\nonumber$. and define $$T(x) = Ax$$. The domain of $$T$$ is $$\mathbb{R}^3 \text{,}$$ and the codomain is $$\mathbb{R}^2$$. The range of $$T$$ is the column space; since all three columns are collinear, the range is a line in $$\mathbb{R}^2$$.. ##### Example $$\PageIndex{13}$$: Interactive: A $$3\times 2$$ matrix: reprise. Let. $A=\left(\begin{array}{cc}1&0\\0&1\\1&0\end{array}\right),\nonumber$. and define $$T(x) = Ax$$. The domain of $$T$$ is $$\mathbb{R}^2 \text{,}$$ and the codomain is $$\mathbb{R}^3$$. The range of $$T$$ is the column space; since $$A$$ has two columns which are not collinear, the range is a plane in $$\mathbb{R}^3$$.. ##### Example $$\PageIndex{14}$$: Projection onto the $$xy$$-plane: reprise. Let. $A=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right),\nonumber$. and let $$T(x) = Ax$$. What are the domain, the codomain, and the range of $$T\text{?}$$. Solution. Geometrically, the transformation $$T$$ projects a vector directly “down” onto the $$xy$$-plane in $$\mathbb{R}^3$$.. Figure $$\PageIndex{22}$$. The inputs and outputs have three entries, so the domain and codomain are both $$\mathbb{R}^3$$. The possible outputs all lie on the $$xy$$-plane, and every point on the $$xy$$-plane is an output of $$T$$ (with itself as the input), so the range of $$T$$ is the $$xy$$-plane.. Be careful not to confuse the codomain with the range here. The range is a plane, but it is a plane in $$\mathbb{R}^3$$, so the codomain is still $$\mathbb{R}^3$$. The outputs of $$T$$ all have three entries; the last entry is simply always zero.. In the case of an $$n\times n$$ square matrix, the domain and codomain of $$T(x) = Ax$$ are both $$\mathbb{R}^n$$. In this situation, one can regard $$T$$ as operating on $$\mathbb{R}^n \text{:}$$ it moves the vectors around in the same space.. ##### Example $$\PageIndex{15}$$: Matrix transformations of $$\mathbb{R}^2$$. In the first Subsection Matrices as Functions we discussed the transformations defined by several $$2\times 2$$ matrices, namely:. \begin{align*} \text{Reflection:} &\qquad A=\left(\begin{array}{cc}-1&0\\0&1\end{array}\right) \\ \text{Dilation:} &\qquad A=\left(\begin{array}{cc}1.5&0\\0&1.5\end{array}\right) \\ \text{Identity:} &\qquad A=\left(\begin{array}{cc}1&0\\0&1\end{array}\right) \\ \text{Rotation:} &\qquad A=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right) \\ \text{Shear:} &\qquad A=\left(\begin{array}{cc}1&1\\0&1\end{array}\right). \end{align*}. In each case, the associated matrix transformation $$T(x)=Ax$$ has domain and codomain equal to $$\mathbb{R}^2$$. The range is also $$\mathbb{R}^2 \text{,}$$ as can be seen geometrically (what is the input for a given output?), or using the fact that the columns of $$A$$ are not collinear (so they form a basis for $$\mathbb{R}^2$$).. ##### Example $$\PageIndex{16}$$: Questions about a [matrix] transformation. Let. $A=\left(\begin{array}{cc}1&1\\0&1\\1&1\end{array}\right),\nonumber$. and let $$T(x)=Ax\text{,}$$ so $$T\colon\mathbb{R}^2 \to\mathbb{R}^3$$ is a matrix transformation.. 1. Evaluate $$T(u)$$ for $$u=\left(\begin{array}{c}3\\4\end{array}\right)$$.. 2. Let. $b = \left(\begin{array}{c}7\\5\\7\end{array}\right). \nonumber$. Find a vector $$v$$ in $$\mathbb{R}^2$$ such that $$T(v)=b$$. Is there more than one?. 3. Does there exist a vector $$w$$ in $$\mathbb{R}^3$$ such that there is more than one $$v$$ in $$\mathbb{R}^2$$ with $$T(v)=w\text{?}$$. 4. Find a vector $$w$$ in $$\mathbb{R}^3$$ which is not in the range of $$T$$.. Note: all of the above questions are intrinsic to the transformation $$T\text{:}$$ they make sense to ask whether or not $$T$$ is a matrix transformation. See the next Example $$\PageIndex{17}$$. As $$T$$ is in fact a matrix transformation, all of these questions will translate into questions about the corresponding matrix $$A$$.. Solution. 1. We evaluate $$T(u)$$ by substituting the definition of $$T$$ in terms of matrix multiplication:. $T\left(\begin{array}{c}3\\4\end{array}\right)=\left(\begin{array}{cc}1&1\\0&1\\1&1\end{array}\right)\:\left(\begin{array}{c}3\\4\end{array}\right)=\left(\begin{array}{c}7\\4\\7\end{array}\right).\nonumber$. 2. We want to find a vector $$v$$ such that $$b = T(v) = Av$$. In other words, we want to solve the matrix equation $$Av = b$$. We form an augmented matrix and row reduce:. $\left(\begin{array}{cc}1&1\\0&1\\1&1\end{array}\right)v=\left(\begin{array}{c}7\\5\\7\end{array}\right) \quad\xrightarrow{\text{augmented matrix}}\quad\left(\begin{array}{cc|c}1&1&7\\0&1&5\\1&1&7\end{array}\right) \quad\xrightarrow{\text{row reduce}}\quad\left(\begin{array}{cc|c}1&0&2\\0&1&5\\0&0&0\end{array}\right).\nonumber$. This gives $$x=2$$ and $$y=5\text{,}$$ so that there is a unique vector. $v = \left(\begin{array}{c}2\\5\end{array}\right) \nonumber$. such that $$T(v) = b$$.. 3. Translation: is there any vector $$w$$ in $$\mathbb{R}^3$$ such that the solution set of $$Av=w$$ has more than one vector in it? The solution set of $$Ax=w\text{,}$$ if non-empty, is a translate of the solution set of $$Av=b$$ above, which has one vector in it. See key observation 2.4.3 in Section 2.4. It follows that the solution set of $$Av=w$$ can have at most one vector.. 4. Translation: find a vector $$w$$ such that the matrix equation $$Av=w$$ is not consistent. Notice that if we take. $w = \left(\begin{array}{c}1\\2\\3\end{array}\right)\text{,} \nonumber$. then the matrix equation $$Av=w$$ translates into the system of equations. $\left\{\begin{array}{rrrrl}x &+& y &=& 1\\ {}&{}& y &=& 2\\ x &+& y &=& 3,\end{array}\right.\nonumber$. which is clearly inconsistent.. ##### Example $$\PageIndex{17}$$: Questions about a [non-matrix] transformation. Define a transformation $$T\colon\mathbb{R}^2 \to\mathbb{R}^3$$ by the formula. $T\left(\begin{array}{c}x\\y\end{array}\right)=\left(\begin{array}{c}\ln(x) \\ \cos(y) \\ \ln(x)\end{array}\right).\nonumber$. 1. Evaluate $$T(u)$$ for $$u=\left(\begin{array}{c}1\\ \pi\end{array}\right)$$.. 2. Let. $b = \left(\begin{array}{c}7\\1\\7\end{array}\right). \nonumber$. Find a vector $$v$$ in $$\mathbb{R}^2$$ such that $$T(v)=b$$. Is there more than one?. 3. Does there exist a vector $$w$$ in $$\mathbb{R}^3$$ such that there is more than one $$v$$ in $$\mathbb{R}^2$$ with $$T(v)=w\text{?}$$. 4. Find a vector $$w$$ in $$\mathbb{R}^3$$ which is not in the range of $$T$$.. Note: we asked (almost) the exact same questions about a matrix transformation in the previous Example $$\PageIndex{16}$$. The point of this example is to illustrate the fact that the questions make sense for a transformation that has no hope of coming from a matrix. In this case, these questions do not translate into questions about a matrix; they have to be answered in some other way.. Solution. 1. We evaluate $$T(u)$$ using the defining formula:. $T\left(\begin{array}{c}1\\ \pi\end{array}\right)=\left(\begin{array}{c} \ln(1) \\ \cos(\pi ) \\ \ln(1)\end{array}\right)=\left(\begin{array}{c}0\\-1\\0\end{array}\right).\nonumber$. 2. We have. $T\left(\begin{array}{c}e^7 \\ 2\pi n \\ e^7\end{array}\right)=\left(\begin{array}{c}\ln(e^7) \\ \cos (2\pi n) \\ \ln(e^7)\end{array}\right)=\left(\begin{array}{c}7\\1\\7\end{array}\right)\nonumber$. for any whole number $$n$$. Hence there are infinitely many such vectors.. 3. The vector $$b$$ from the previous part is an example of such a vector.. 4. Since $$\cos(y)$$ is always between $$-1$$ and $$1\text{,}$$ the vector. $w = \left(\begin{array}{c}0\\2\\0\end{array}\right) \nonumber$.
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is not in the range of $$T$$.. 3.1: Matrix Transformations is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Dan Margalit & Joseph Rabinoff via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# How to Do Division in Excel: A Step-by-Step Guide
Excel is a powerful tool that can help you solve a wide range of mathematical problems, including long division. While some find division a daunting task, with Excel, you can do it with ease in just a few clicks! If you’re new to Excel, don’t worry – I’ve got you covered. This guide will provide a step-by-step tutorial on how to do division in Excel, complete with helpful tips and tricks along the way.
The first step in doing division in Excel is to set up your spreadsheet. Start by opening Excel and creating a new workbook. Then, in cell A1, enter the numerator of the fraction you want to divide. Next, in cell B1, enter the denominator of the fraction. For example, if you want to divide 10 by 2, you would enter 10 in cell A1 and 2 in cell B1.
### Step 2: Enter the Formula
Now that you have your numerator and denominator entered, it’s time to enter the formula. In cell C1, type the equals sign (=) followed by the cell reference for the numerator (A1) and then the forward slash (/) followed by the cell reference for the denominator (B1). The formula should look like this:
=A1/B1
Press Enter, and Excel will calculate the quotient and display the result in cell C1.
### Step 3: Format Your Results
By default, Excel displays the result of your division as a decimal. If you prefer to display your result as a fraction or a percentage, you can format the cell accordingly. To format the cell, right-click on cell C1 and select Format Cells. In the Format Cells dialog box, select the Number tab and choose the format you want from the options available.
### Tips and Tricks
While performing long division in Excel is pretty straightforward, here are a few helpful tips and tricks to make the process even smoother:
• If you want to perform multiple division calculations in Excel and don't want to repeat the same formula over and over, simply copy and paste the formula to other cells.
• If you want to set up your spreadsheet to handle more complex division problems, you can use a combination of cells, formulas, and functions to achieve your desired results.
• Use parentheses to ensure that Excel performs the division in the correct order. For example, if you want to divide 10 by 2 and then add 5, you would type = (10 / 2) + 5 in the formula bar.
• Use cell references to make your formulas more flexible. By referencing cells instead of hard-coding values, you can easily update your calculations as needed.
• If you're not sure whether your formula is correct, you can use the Evaluate Formula tool in Excel to step through the calculation and see the result at each stage.
### Conclusion
In conclusion, Excel provides an easy and efficient way to perform division – even for those who might be intimidated by long division. By following the simple steps outlined in this guide, you'll be able to divide numbers with ease in no time.
So go forth and conquer your division problems with Excel as your trusty sidekick! And if you still encounter some problems, there are plenty of helpful resources available online to guide you in your Excel journey.
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# How to Do Division in Excel: A Step-by-Step Guide. Excel is a powerful tool that can help you solve a wide range of mathematical problems, including long division. While some find division a daunting task, with Excel, you can do it with ease in just a few clicks! If you’re new to Excel, don’t worry – I’ve got you covered. This guide will provide a step-by-step tutorial on how to do division in Excel, complete with helpful tips and tricks along the way.. The first step in doing division in Excel is to set up your spreadsheet. Start by opening Excel and creating a new workbook. Then, in cell A1, enter the numerator of the fraction you want to divide. Next, in cell B1, enter the denominator of the fraction. For example, if you want to divide 10 by 2, you would enter 10 in cell A1 and 2 in cell B1.. ### Step 2: Enter the Formula. Now that you have your numerator and denominator entered, it’s time to enter the formula. In cell C1, type the equals sign (=) followed by the cell reference for the numerator (A1) and then the forward slash (/) followed by the cell reference for the denominator (B1). The formula should look like this:. =A1/B1. Press Enter, and Excel will calculate the quotient and display the result in cell C1.. ### Step 3: Format Your Results. By default, Excel displays the result of your division as a decimal. If you prefer to display your result as a fraction or a percentage, you can format the cell accordingly. To format the cell, right-click on cell C1 and select Format Cells. In the Format Cells dialog box, select the Number tab and choose the format you want from the options available.. ### Tips and Tricks. While performing long division in Excel is pretty straightforward, here are a few helpful tips and tricks to make the process even smoother:. • If you want to perform multiple division calculations in Excel and don't want to repeat the same formula over and over, simply copy and paste the formula to other cells.. • If you want to set up your spreadsheet to handle more complex division problems, you can use a combination of cells, formulas, and functions to achieve your desired results.. • Use parentheses to ensure that Excel performs the division in the correct order. For example, if you want to divide 10 by 2 and then add 5, you would type = (10 / 2) + 5 in the formula bar.. • Use cell references to make your formulas more flexible. By referencing cells instead of hard-coding values, you can easily update your calculations as needed.. • If you're not sure whether your formula is correct, you can use the Evaluate Formula tool in Excel to step through the calculation and see the result at each stage.. ### Conclusion. In conclusion, Excel provides an easy and efficient way to perform division – even for those who might be intimidated by long division.
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By following the simple steps outlined in this guide, you'll be able to divide numbers with ease in no time.. So go forth and conquer your division problems with Excel as your trusty sidekick! And if you still encounter some problems, there are plenty of helpful resources available online to guide you in your Excel journey.
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##### Using the $C$% confidence interval for $\mu$ to perform left sided paired sample $t$ test: full explanation
If $\mu_0$ is not in the $C$% confidence interval and the sample mean of the difference scores is smaller than $\mu_0$ (as expected), the sample mean of the difference scores is significantly smaller than $\mu_0$ at significance level $\alpha = \frac{1 \, - \, C/100}{2}$
If $\mu_0$ is not in the $C$% confidence interval and the sample mean of the difference scores is smaller than $\mu_0$, the sample mean of the difference scores is significantly smaller than $\mu_0$ at significance level $\alpha = \frac{1 \, - \, C/100}{2}$. In order to understand why, we have to be aware of the following three points, which are illustrated in the figure above: The lower bound of the $C$% confidence interval for $\mu$ is $\bar{y} - t^* \times SE$, the upper bound is $\bar{y} + t^* \times SE$. That is, we subtract $t^* \times SE$ from the sample mean of the difference scores, and we add $t^* \times SE$ to the sample mean of the difference scores. Here: $SE$ is the standard error of $\bar{y}$, which is $s / \sqrt{N}$ the critical value $t^*$ is the value under the $t$ distribution with area $C / 100$ between $-t^*$ and $t^*$, and therefore area $\frac{1\,-\,C/100}{2}$ to the right of $t^*$ and to the left of $-t^*$. In the figure above, the red dots represent sample means of the difference scores, the blue arrows represent the distance $t^* \times SE$. If we set the significance level $\alpha$ at $\frac{1 \, - \, C/100}{2}$, the critical value $t^*$ used for the significance test is the value under the $t$ distribution with area $\alpha = \frac{1\,-\,C/100}{2}$ to the left of $t^*$. This is the same critical value $t^*$ as the (negative) $t^*$ used for the $C$% confidence interval. We reject the null hypothesis at $\alpha = \frac{1\,-\,C/100}{2}$ if the $t$ value $t = \frac{\bar{y} - \mu_0}{SE}$ is at least as small as the negative $t^*$. This means that we reject the null hypothesis at $\alpha = \frac{1\,-\,C/100}{2}$ if the sample mean of the difference scores $\bar{y}$ is at least $t^* \times SE$ below $\mu_0$. In the figure above, the rejection region for $\bar{y}$ is represented by the green area. If $\mu_0$ is not in the $C$% confidence interval [$\bar{y} - t^* \times SE \,;\, \bar{y} + t^* \times SE$] and the sample mean of the difference scores $\bar{y}$ is smaller than $\mu_0$, then $\bar{y}$ must be at least $t^* \times SE$ below $\mu_0$. In the figure above, the first confidence interval does not contain $\mu_0$ and the sample mean of the difference scores is smaller than $\mu_0$, so the sample mean of the difference scores must be at least $t^* \times SE$ below $\mu_0$. The second confidence interval does contain $\mu_0$, so the sample mean of the difference scores must be less than $t^* \times SE$ below $\mu_0$. In sum, if $\mu_0$ is not in the $C$% confidence interval [$\bar{y} - t^* \times SE \,;\, \bar{y} + t^* \times SE$] and the sample mean of the difference scores is smaller than $\mu_0$, we know that the sample mean of the difference scores is significantly smaller than $\mu_0$ at significance level $\alpha = \frac{1 \, - \, C/100}{2}$.
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##### Using the $C$% confidence interval for $\mu$ to perform left sided paired sample $t$ test: full explanation. If $\mu_0$ is not in the $C$% confidence interval and the sample mean of the difference scores is smaller than $\mu_0$ (as expected), the sample mean of the difference scores is significantly smaller than $\mu_0$ at significance level $\alpha = \frac{1 \, - \, C/100}{2}$. If $\mu_0$ is not in the $C$% confidence interval and the sample mean of the difference scores is smaller than $\mu_0$, the sample mean of the difference scores is significantly smaller than $\mu_0$ at significance level $\alpha = \frac{1 \, - \, C/100}{2}$. In order to understand why, we have to be aware of the following three points, which are illustrated in the figure above: The lower bound of the $C$% confidence interval for $\mu$ is $\bar{y} - t^* \times SE$, the upper bound is $\bar{y} + t^* \times SE$. That is, we subtract $t^* \times SE$ from the sample mean of the difference scores, and we add $t^* \times SE$ to the sample mean of the difference scores. Here: $SE$ is the standard error of $\bar{y}$, which is $s / \sqrt{N}$ the critical value $t^*$ is the value under the $t$ distribution with area $C / 100$ between $-t^*$ and $t^*$, and therefore area $\frac{1\,-\,C/100}{2}$ to the right of $t^*$ and to the left of $-t^*$. In the figure above, the red dots represent sample means of the difference scores, the blue arrows represent the distance $t^* \times SE$. If we set the significance level $\alpha$ at $\frac{1 \, - \, C/100}{2}$, the critical value $t^*$ used for the significance test is the value under the $t$ distribution with area $\alpha = \frac{1\,-\,C/100}{2}$ to the left of $t^*$. This is the same critical value $t^*$ as the (negative) $t^*$ used for the $C$% confidence interval.
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We reject the null hypothesis at $\alpha = \frac{1\,-\,C/100}{2}$ if the $t$ value $t = \frac{\bar{y} - \mu_0}{SE}$ is at least as small as the negative $t^*$. This means that we reject the null hypothesis at $\alpha = \frac{1\,-\,C/100}{2}$ if the sample mean of the difference scores $\bar{y}$ is at least $t^* \times SE$ below $\mu_0$. In the figure above, the rejection region for $\bar{y}$ is represented by the green area. If $\mu_0$ is not in the $C$% confidence interval [$\bar{y} - t^* \times SE \,;\, \bar{y} + t^* \times SE$] and the sample mean of the difference scores $\bar{y}$ is smaller than $\mu_0$, then $\bar{y}$ must be at least $t^* \times SE$ below $\mu_0$. In the figure above, the first confidence interval does not contain $\mu_0$ and the sample mean of the difference scores is smaller than $\mu_0$, so the sample mean of the difference scores must be at least $t^* \times SE$ below $\mu_0$. The second confidence interval does contain $\mu_0$, so the sample mean of the difference scores must be less than $t^* \times SE$ below $\mu_0$. In sum, if $\mu_0$ is not in the $C$% confidence interval [$\bar{y} - t^* \times SE \,;\, \bar{y} + t^* \times SE$] and the sample mean of the difference scores is smaller than $\mu_0$, we know that the sample mean of the difference scores is significantly smaller than $\mu_0$ at significance level $\alpha = \frac{1 \, - \, C/100}{2}$.
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# Fourier Transform
Question 1
The discrete-time Fourier transform of a signal $x[n]$ is $X(\Omega)=1(1+\cos \Omega) e^{-j \Omega}$. Consider that $x_{p}[n]$ is a periodic signal of period $N=5$ such that
\begin{aligned} x_{p}[n] & =x[n], \text { for } n=0,1,2 \\ & =0, \text { for } n=3,4 \end{aligned}
Note that $x_{p}[n]=\sum_{k=0}^{N-1} a_{k} e^{j \frac{2 \pi}{N} k m}$. The magnitude of the Fourier series coefficient $a_{3}$ is ____ (Round off to 3 decimal places).
A 0.038 B 0.025 C 0.068 D 0.012
GATE EE 2023 Signals and Systems
Question 1 Explanation:
Given : $x_{p}(n)$ is a period signal of period $N=5$.
$x_{p}(n)=\left\{\begin{array}{cc} x(n), & \text { for } n=0,1,2 \\ 0, & \text { for } n=3,4 \end{array}\right.$
and $\mathrm{x}(\mathrm{n}) \stackrel{\text { DTFT }}{\longrightarrow} \mathrm{X}(\Omega)=(1+\cos \Omega) \mathrm{e}^{-\mathrm{j} \Omega}$
We have,
\begin{aligned} a_{K} & =\frac{X\left(\frac{2 \pi}{N} K\right)}{N} \\ & =\frac{\left[1+\cos \left(\frac{2 \pi}{N} K\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{N} R}}{N} \end{aligned}
For, $\mathrm{N}=5$ and $\mathrm{K}=3$
\begin{aligned} a_{3} & =\frac{\left[1+\cos \left(\frac{2 \pi \times 3}{5}\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{5} \times 3}}{5} \\ \left|a_{3}\right| & =\frac{1}{5}\left[1+\cos \frac{6 \pi}{5}\right] \\ & =0.0382 \end{aligned}
Question 2
The Fourier transform $X(\omega)$ of the signal $x(t)$ is given by
\begin{aligned} X(\omega) & =1, \text { for }|\omega| \lt W_{0} \\ & =0, \text { for }|\omega| \gt W_{0} \end{aligned}
Which one of the following statements is true?
A $x(t)$ tends to be an impulse as $\mathrm{W}_{0} \rightarrow \infty$ B $x(0)$ decreases as $W_{0}$ increases C At $\mathrm{t}=\frac{\pi}{2 \mathrm{~W}_{0}}, \mathrm{x}(\mathrm{t})=-\frac{1}{\pi}$ D At $\mathrm{t}=\frac{\pi}{2 \mathrm{~W}_{0}}, \mathrm{x}(\mathrm{t})=\frac{1}{\pi}$
GATE EE 2023 Signals and Systems
Question 2 Explanation:
Given,
$\quad X(\omega)=\left\{\begin{array}{l}1, \text { for }|\omega|W_{0}\end{array}\right.$
$X(\omega)=\operatorname{rect}\left(\frac{\omega}{2 W_{0}}\right)$
We know,
$\operatorname{Arect}\left(\frac{\mathrm{t}}{\tau}\right)=\operatorname{A\tau S}\left(\frac{\omega \tau}{2}\right)$
By duality,
$\operatorname{A} \tau \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right) \leftrightarrow 2 \pi \operatorname{Arect}\left(\frac{\omega}{\tau}\right)$
Given,
$2 \pi \mathrm{A}=1$
$\Rightarrow \quad A=\frac{1}{2 \pi}$
$\therefore \quad \mathrm{x}(\mathrm{t})=\frac{\tau}{2 \pi} \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right)$, where $\tau=2 \mathrm{~W}_{0}$
Thus,
\begin{aligned} x(t)&=\frac{W_0}{\pi} \mathrm{Sa}(W_0 t)\\ &=\frac{1}{\pi}\frac{\sin W_0 t}{t}\\ \therefore \quad \text{At } t&=\frac{\pi}{2W_0}, x(t)=\frac{1}{\pi t} \end{aligned}
From rectangular function, At $W \Rightarrow \infty ,X(w)=1$
Taking inverse fourier transform $x(t)=\delta (t)$
Option (A) will be correct.
Question 3
Let an input $x(t)=2 \sin (10 \pi t)+5 \cos (15 \pi t)+7 \sin (42 \pi t)+4 \cos (45 \pi t)$ is passed through an LTI system having an impulse response,
$h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos (40\pi t)$
The output of the system is
A $2 \sin (10 \pi t)+5\cos (15 \pi t)$ B $7 \sin (42 \pi t)+5\cos (15 \pi t)$ C $7 \sin (42 \pi t)+4\cos (45 \pi t)$ D $2 \sin (10 \pi t)+4\cos (45 \pi t)$
GATE EE 2022 Signals and Systems
Question 3 Explanation:
Given: $h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos 40 \pi t$
Fourier transform of signal $\frac{\sin (10 \pi t)}{\pi t}$ is given by
Now, impulse response
$h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right )\cdot \left ( \frac{e^{+j40\pi t}+e^{-j40\pi t}}{2} \right )$
Using property, $e^{-j\omega _0t}x(t)\rightleftharpoons X(\omega +\omega _0)$
Therefore, Fourier transform of impulse response
Cut-off frequencies,
$\omega_L=30 \pi rad/sec$
$\omega_H=50 \pi rad/sec$
Thus, output of the system $=7 \sin 42 \pi t+4 \cos 45 \pi t$
Question 4
Consider a continuous-time signal $x(t)$ defined by $x(t)=0$ for $\left | t \right |> 1$, and $x\left ( t \right )=1-\left | t \right | for \left | t \right |\leq 1$. Let the Fourier transform of $x(t)$ be defined as $X\left ( \omega \right )=\int\limits_{-\infty }^{\infty }\:x\left ( t \right )e^{-j\omega t}\:dt$. The maximum magnitude of $X\left ( \omega \right )$ is ___________.
A 1 B 2 C 3 D 4
GATE EE 2021 Signals and Systems
Question 4 Explanation:
Fourier transform, $F(\omega)=A \tau S a^{2}\left(\frac{\omega \tau}{2}\right)$
\begin{aligned} \text { As } A=1, \tau=1 \\ H \omega) &=S a^{2}\left(\frac{\omega}{2}\right) \\ \left.F(\omega)\right|_{\text {peak }} &=F(0)=S a^{2}(0)=1 \end{aligned}
$\because$ Peak value of sampling function occurs at $\omega=0$
Peak value $=1$
Question 5
Let $f(t)$ be an even function, i.e. $f(-t)=f(t)$ for all t. Let the Fourier transform of $f(t)$ be defined as $F\left ( \omega \right )=\int\limits_{-\infty }^{\infty }\:f\left ( t \right )e^{-j\omega t}dt$. Suppose $\dfrac{dF\left ( \omega \right )}{d\omega }=-\omega F\left ( \omega \right )$ for all $\omega$, and $F(0)=1$. Then
A $f\left ( 0 \right )\lt 1$ B $f\left ( 0 \right ) \gt 1$ C $f\left ( 0 \right )= 1$ D $f\left ( 0 \right )= 0$
GATE EE 2021 Signals and Systems
Question 5 Explanation:
$f(t) \rightleftharpoons F(\omega)=\int_{-\infty}^{\infty} f(t) e^{-j \omega t} d t$
The following informations are given about $f(t) \rightleftharpoons F(\omega).$
$\begin{array}{l} (i) f(t)=f(-t)\\ (ii) \left.F(\omega)\right|_{\omega=0}=1\\ (iii) \frac{d F(\omega)}{d \omega}=-\omega F(\omega)\\ \text{From }(iii), \frac{d F(\omega)}{d \omega}+\omega F(\omega)=0 \end{array}$
By solving the above linear differential equations, (by mathematics)
\begin{aligned} \ln F(\omega) &=-\frac{\omega^{2}}{2} \\ \Rightarrow \qquad\qquad F(\omega) &=K \cdot e^{-\omega^{2} / 2} \\ \text { Put } \omega=0, \qquad\qquad F(0) &=K \\ \Rightarrow \qquad\qquad 1 &=K \text { (from info. } \\ \text { From (iv), } \qquad\qquad F(\omega) &=e^{-\omega^{2} / 2} \\ \text{As we know}, e^{-a t^{2}}, a \gt 0 &\rightleftharpoons \sqrt{\frac{\pi}{a}} e^{-\omega^{2} / 4 a}\\ \text{At }a=\frac{1}{2}, \qquad\qquad\quad e^{-t^{2} / 2} &\rightleftharpoons \sqrt{\frac{\pi}{1 / 2}} \cdot e^{-\omega^{2} / 2}\\ f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \rightleftharpoons e^{-\omega^{2} / 2}=F(\omega)\\ \text { Thus, }\qquad\qquad f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \\ \text { At } t=0, \qquad\qquad f(0)&=\frac{1}{\sqrt{2 \pi}} \lt 1 \end{aligned}
There are 5 questions to complete.
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# Fourier Transform. Question 1. The discrete-time Fourier transform of a signal $x[n]$ is $X(\Omega)=1(1+\cos \Omega) e^{-j \Omega}$. Consider that $x_{p}[n]$ is a periodic signal of period $N=5$ such that. \begin{aligned} x_{p}[n] & =x[n], \text { for } n=0,1,2 \\ & =0, \text { for } n=3,4 \end{aligned}. Note that $x_{p}[n]=\sum_{k=0}^{N-1} a_{k} e^{j \frac{2 \pi}{N} k m}$. The magnitude of the Fourier series coefficient $a_{3}$ is ____ (Round off to 3 decimal places).. A 0.038 B 0.025 C 0.068 D 0.012. GATE EE 2023 Signals and Systems. Question 1 Explanation:. Given : $x_{p}(n)$ is a period signal of period $N=5$.. $x_{p}(n)=\left\{\begin{array}{cc} x(n), & \text { for } n=0,1,2 \\ 0, & \text { for } n=3,4 \end{array}\right.$. and $\mathrm{x}(\mathrm{n}) \stackrel{\text { DTFT }}{\longrightarrow} \mathrm{X}(\Omega)=(1+\cos \Omega) \mathrm{e}^{-\mathrm{j} \Omega}$. We have,. \begin{aligned} a_{K} & =\frac{X\left(\frac{2 \pi}{N} K\right)}{N} \\ & =\frac{\left[1+\cos \left(\frac{2 \pi}{N} K\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{N} R}}{N} \end{aligned}. For, $\mathrm{N}=5$ and $\mathrm{K}=3$. \begin{aligned} a_{3} & =\frac{\left[1+\cos \left(\frac{2 \pi \times 3}{5}\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{5} \times 3}}{5} \\ \left|a_{3}\right| & =\frac{1}{5}\left[1+\cos \frac{6 \pi}{5}\right] \\ & =0.0382 \end{aligned}. Question 2. The Fourier transform $X(\omega)$ of the signal $x(t)$ is given by. \begin{aligned} X(\omega) & =1, \text { for }|\omega| \lt W_{0} \\ & =0, \text { for }|\omega| \gt W_{0} \end{aligned}. Which one of the following statements is true?. A $x(t)$ tends to be an impulse as $\mathrm{W}_{0} \rightarrow \infty$ B $x(0)$ decreases as $W_{0}$ increases C At $\mathrm{t}=\frac{\pi}{2 \mathrm{~W}_{0}}, \mathrm{x}(\mathrm{t})=-\frac{1}{\pi}$ D At $\mathrm{t}=\frac{\pi}{2 \mathrm{~W}_{0}}, \mathrm{x}(\mathrm{t})=\frac{1}{\pi}$. GATE EE 2023 Signals and Systems. Question 2 Explanation:. Given,. $\quad X(\omega)=\left\{\begin{array}{l}1, \text { for }|\omega|W_{0}\end{array}\right.$. $X(\omega)=\operatorname{rect}\left(\frac{\omega}{2 W_{0}}\right)$. We know,. $\operatorname{Arect}\left(\frac{\mathrm{t}}{\tau}\right)=\operatorname{A\tau S}\left(\frac{\omega \tau}{2}\right)$. By duality,. $\operatorname{A} \tau \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right) \leftrightarrow 2 \pi \operatorname{Arect}\left(\frac{\omega}{\tau}\right)$. Given,. $2 \pi \mathrm{A}=1$. $\Rightarrow \quad A=\frac{1}{2 \pi}$. $\therefore \quad \mathrm{x}(\mathrm{t})=\frac{\tau}{2 \pi} \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right)$, where $\tau=2 \mathrm{~W}_{0}$. Thus,. \begin{aligned} x(t)&=\frac{W_0}{\pi} \mathrm{Sa}(W_0 t)\\ &=\frac{1}{\pi}\frac{\sin W_0 t}{t}\\ \therefore \quad \text{At } t&=\frac{\pi}{2W_0}, x(t)=\frac{1}{\pi t} \end{aligned}. From rectangular function, At $W \Rightarrow \infty ,X(w)=1$. Taking inverse fourier transform $x(t)=\delta (t)$. Option (A) will be correct.. Question 3. Let an input $x(t)=2 \sin (10 \pi t)+5 \cos (15 \pi t)+7 \sin (42 \pi t)+4 \cos (45 \pi t)$ is passed through an LTI system having an impulse response,. $h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos (40\pi t)$. The output of the system is. A $2 \sin (10 \pi t)+5\cos (15 \pi t)$ B $7 \sin (42 \pi t)+5\cos (15 \pi t)$ C $7 \sin (42 \pi t)+4\cos (45 \pi t)$ D $2 \sin (10 \pi t)+4\cos (45 \pi t)$. GATE EE 2022 Signals and Systems. Question 3 Explanation:. Given: $h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos 40 \pi t$. Fourier transform of signal $\frac{\sin (10 \pi t)}{\pi t}$ is given by. Now, impulse response. $h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right )\cdot \left ( \frac{e^{+j40\pi t}+e^{-j40\pi t}}{2} \right )$. Using property, $e^{-j\omega _0t}x(t)\rightleftharpoons X(\omega +\omega _0)$. Therefore, Fourier transform of impulse response. Cut-off frequencies,. $\omega_L=30 \pi rad/sec$. $\omega_H=50 \pi rad/sec$. Thus, output of the system $=7 \sin 42 \pi t+4 \cos 45 \pi t$. Question 4. Consider a continuous-time signal $x(t)$ defined by $x(t)=0$ for $\left | t \right |> 1$, and $x\left ( t \right )=1-\left | t \right | for \left | t \right |\leq 1$. Let the Fourier transform of $x(t)$ be defined as $X\left ( \omega \right )=\int\limits_{-\infty }^{\infty }\:x\left ( t \right )e^{-j\omega t}\:dt$. The maximum magnitude of $X\left ( \omega \right )$ is ___________.. A 1 B 2 C 3 D 4. GATE EE 2021 Signals and Systems. Question 4 Explanation:. Fourier transform, $F(\omega)=A \tau S a^{2}\left(\frac{\omega \tau}{2}\right)$. \begin{aligned} \text { As } A=1, \tau=1 \\ H \omega) &=S a^{2}\left(\frac{\omega}{2}\right) \\ \left.F(\omega)\right|_{\text {peak }} &=F(0)=S a^{2}(0)=1 \end{aligned}. $\because$ Peak value of sampling function occurs at $\omega=0$. Peak value $=1$. Question 5. Let $f(t)$ be an even function, i.e. $f(-t)=f(t)$ for all t. Let the Fourier transform of $f(t)$ be defined as $F\left ( \omega \right )=\int\limits_{-\infty }^{\infty }\:f\left ( t \right )e^{-j\omega t}dt$. Suppose $\dfrac{dF\left ( \omega \right )}{d\omega }=-\omega F\left ( \omega \right )$ for all $\omega$, and $F(0)=1$. Then. A $f\left ( 0 \right )\lt 1$ B $f\left ( 0 \right ) \gt 1$ C $f\left ( 0 \right )= 1$ D $f\left ( 0 \right )= 0$. GATE EE 2021 Signals and Systems. Question 5 Explanation:. $f(t) \rightleftharpoons F(\omega)=\int_{-\infty}^{\infty} f(t) e^{-j \omega t} d t$. The following informations are given about $f(t) \rightleftharpoons F(\omega).$. $\begin{array}{l} (i) f(t)=f(-t)\\ (ii) \left.F(\omega)\right|_{\omega=0}=1\\ (iii) \frac{d F(\omega)}{d \omega}=-\omega F(\omega)\\ \text{From }(iii), \frac{d F(\omega)}{d \omega}+\omega F(\omega)=0 \end{array}$. By solving the above linear differential equations, (by mathematics). \begin{aligned} \ln F(\omega) &=-\frac{\omega^{2}}{2} \\ \Rightarrow \qquad\qquad F(\omega) &=K \cdot e^{-\omega^{2} / 2} \\ \text { Put } \omega=0, \qquad\qquad F(0) &=K \\ \Rightarrow \qquad\qquad 1 &=K \text { (from info.
|
} \\ \text { From (iv), } \qquad\qquad F(\omega) &=e^{-\omega^{2} / 2} \\ \text{As we know}, e^{-a t^{2}}, a \gt 0 &\rightleftharpoons \sqrt{\frac{\pi}{a}} e^{-\omega^{2} / 4 a}\\ \text{At }a=\frac{1}{2}, \qquad\qquad\quad e^{-t^{2} / 2} &\rightleftharpoons \sqrt{\frac{\pi}{1 / 2}} \cdot e^{-\omega^{2} / 2}\\ f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \rightleftharpoons e^{-\omega^{2} / 2}=F(\omega)\\ \text { Thus, }\qquad\qquad f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \\ \text { At } t=0, \qquad\qquad f(0)&=\frac{1}{\sqrt{2 \pi}} \lt 1 \end{aligned}. There are 5 questions to complete.
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# How many cards are in a 6 deck shoe?
2
Date created: Sun, Jul 11, 2021 1:23 PM
Date updated: Mon, May 16, 2022 2:59 PM
Content
Video answer: #11 đź”´ so many aces! blackjack ~ 6 deck shoe at red rock casino ~ las vegas july 2021 episode 4
## Top best answers to the question «How many cards are in a 6 deck shoe»
• I attempt to work this out in my blackjack appendix 8 but I’ll work through it more slowly here. We’ll ignore dealer blackjacks to keep things simple and assume the player always hits after two cards. The number of ways to arrange 3 cards in a 6-deck shoe is combin (312,3)=5,013,320. There are 24 sevens in the shoe.
A six-deck shoe has 6 x 16 = 96 10-values and 6 x 52 = 312 total cards. The chance of an initial 10 is therefore 96 out of 312 or 30.769 percent. Considering only the 10 to be gone, the shoe then would have 6 x 4 = 24 aces in 311 remaining cards.
FAQ
Those who are looking for an answer to the question «How many cards are in a 6 deck shoe?» often ask the following questions:
### đź‘ Can you count cards in a multi deck shoe?
#### How do you count cards in a deck of cards?
• Step 1. Assign a value to every card. Step 2. Keep a “Running Count” based off of the values of the card dealt. Step 3. Use this information to calculate the count per deck or “true count”. Step 4. Change your bets as the true count rises.
### đź‘ How many hands in a six deck shoe?
• The average number of cards per hand is 4.94. Assuming 15 burn cards, a six-deck baccarat shoe would have about 60 hands. However, mathematically speaking, it doesn't make any difference when they shuffle.
### đź‘ How many decks are in a 6 deck shoe?
• Casinos use a 6-deck shoe, so take the number of decks you see in the discard tray, and subtract it from 6. That’s how many decks are remaining in the shoe. Divide the running count by the number of decks remaining.
Video answer: Deck estimation in card counting: what it is and how to do it
We've handpicked 28 related questions for you, similar to «How many cards are in a 6 deck shoe?» so you can surely find the answer!
How did the multi deck shoe get its name?
• While no such ruling was ever passed, most Nevada casinos now deal from a multi-deck shoe. As gaming advisor to the Havana Hilton, Scarne also introduced the shoe to Puerto Rico and Cuba. The device is so named because the earliest versions of it resembled a woman's high-heel shoe, and were often painted red or black in color.
Shoe size 13 what is a good deck width?
• Full-Size Deck [width: 7.5-inches or larger, length: 28-inches]:These are perfect for skaters that are over the age of 13, with height taller than 5’3” and wear a shoe size of 9 or more. However, the specific deck size is determined by the style preference of the skater: 7.5-inches to 8-inches: street skating or for more technical tricks
What's the running count on a 6 deck shoe?
• If you have a +6 after the first round of a game in a 6 deck shoe, the running count is +1 since there are still about 6 decks left. If you have a +6 when you’re halfway through that shoe, then you have 3 decks left and a +2 running count. There are a couple of ways to figure out the number of decks remaining in the shoe.
What's the running count on a six deck shoe?
• If you were say, one deck into that six deck shoe, a running count of +10 would give you the same proportional high-to-low ratio of remaining cards as a +4 running count would when you're four decks into it. That's because in both cases, there would be two extra high cards available for each deck that is left.
What shoe stores offer credit cards?
• Amazon.
• Nordstrom.
• Belk.
• Kohl's.
• Macy's.
• Bloomingdale's.
• JCPenney.
• Walmart.
### Video answer: How to shuffle a shoe deck
What are the odds of a six deck shoe in baccarat?
• A six deck shoe will have the following baccarat odds: 45.87% for the banker, 44.63% for player, and 9.51% for a tie.
What shoe stores accept nike gift cards?
#### Where can I use my Nike gift card?
• Nike and Converse gift cards are just like cash and can be used to make purchases on Nike.com and at any Nike- and Converse-owned retail stores.
### Video answer: đź”´ blackjack with so many doubles ~ 6 deck shoe ~ red rock casino las vegas july 2021~ episode 5
Can you use credit cards for shoe horns?
• Credit cards and business cards are the one item everyone will have on them at all times, or, at least, whenever they are also wearing shoes. They have just about the perfect size and they are slick enough to be perfect replacements for shoe horns. There are 2 problems that still need to be taken into consideration.
What do the cards in the shoe mean?
• When the cards are placed in the shoe, the dealer will insert a brightly colored blank plastic card. When this card is drawn it indicates that the current game is the last one before a new shuffle. This helps mitigate player advantage via card counting, as a significant portion (usually about 25 percent) of the full inventory...
What kind of cards are in a shoe?
• Dealing shoes come in many colors and sizes, depending on the number of decks they are capable of holding (2, 4, 6, or 8 decks). When the cards are placed in the shoe, the dealer will insert a brightly colored blank plastic card.
### Video answer: Card counting at 6 deck 21 with john stathis
How many shoe brushes?
• Every shoe specialist knows the importance of having high-quality shoe brushes. Generally, experts recommend you use three different brushes for the best results. One shoe brush to remove the dust and grime from your footwear.
11.5 shoe how many inches?
#### Men's Size Conversions
InchesCMUS Sizes
11.125"28.311.5
11.25"28.612
11.5625"29.413
11.875"30.214
How many customizable shoe brands?
#### What are the top shoe brands in the world?
• With its roots in Germany, adidas has become one of the top shoe brands in the world. The company produces more than 900 million sports and lifestyle products with independent manufacturing partners around the world.
How many inches inside shoe?
#### Does shoe size equal inches?
• Men's shoe size charts start from a US number of 6 (a European number of 39 and a UK number of 5.5). This size corresponds to an inch measurement of 9.25 inches (23.5 centimeters). If the measurement is provided solely in centimeters, multiply the length of your foot by 2.54 and apply to the chart. The biggest number on the men's chart is ...
How many inches shoe size?
• Size 6 = 9.31 inches Size 6.5 = 9.5 inches Size 7 = 9.69 inches Size 7.5 = 9.81 inches Size 8 = 10 inches Size 8.5 = 10.19 inches Size 9 = 10.31 inches Size 9.5 = 10.5 inches Size 10 = 10.69 inches Size 10.5 = 10.81 inches Size 11 = 11 inches Size 11.5 = 11.19 inches Size 12 = 11.31 inches Size 12.5 = 11.5 inches Size 13 = 11.69 inches Size 13.5 = 11.81 inches Size 14 = 12 inches Size 14.5 = 12.19 inches Size 15 = 12.31 inches
How many man shoe and?
#### What's the average shoe size for a man?
• For men, it’s pretty easy to determine, as there are many studies analyzing the average size of a male foot. In 2014, the American Academy of Orthopaedic Surgeons reported that the worldwide average shoe size for men is between 9 and 12, while the American average male shoe size is a 10.5. A mother measures her son’s shoe size.
How many marathons per shoe?
#### How many people go to a marathon each year?
• Major marathons also attract many spectators and viewers. In 2019, people from almost 3.3 million households are projected to attended a marathon event in the United States. This text provides general information.
How many shoe carnival stores?
#### What kind of store is Shoe Carnival?
• Shoe Carnival Stores Shoe Carnival is a chain of family shoe stores operating across the continental United States, as well as in Puerto Rico and online at shoecarnival.com. We have a passion for creating a fun, engaging, and affordable shoe shopping experience, bringing great deals on brand-name footwear to millions of families.
### Video answer: Card deck mistakes to watch out for
How many syllables in shoe?
• How many syllables are in shoe? 1 syllable. Divide shoe into syllables: shoe. Definition of: Shoe (New window will open) How to pronounce shoe: Words: shoddily, shoddy, shoe, shoebox, shoed.
How many cards do you need to deal shoes?
• Dealing shoes come in many colors and sizes, depending on the number of decks they are capable of holding (2, 4, 6, or 8 decks).
What size skateboard deck should i get?
• Mini Skateboard Deck width: 7.0″ with an average deck length of 28” Mini decks are the best skateboards for beginners who are 6-8 years old, between 3’5” and 4’4” tall who wear size 4-6 shoes. Mid-size Skateboard Deck width: 7.3”+ with an average deck length of 29” For skaters 9 to 12 years old between 4’5”...
What happens when you remove cards from shoe in blackjack?
• In fact, if you remove any significant number of small cards from the shoe, blackjack not only becomes easier, the player gets a mathematical edge over the house. This is where card counting comes in. Counters track the proportion of small cards to face cards and aces. At a certain point, the deck turns “positive.”.
Hemnes shoe cabinet how many shoes?
• How many shoes can I fit inside the Hemnes unit? The Hemnes shoe unit can hold eight pairs of shoes. However, if you are like me and have small feet, I can fit 10.
How many big baller shoe sold?
In an interview with Colin Cowherd in May 2017, Ball said that if the big shoe companies like Nike, Adidas, or Under Armour want to make a deal with his Big Baller Brand, the asking price is \$3 billion. He also said that Baller has sold between 400 and 520 ZO2 shoes.
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# How many cards are in a 6 deck shoe?. 2. Date created: Sun, Jul 11, 2021 1:23 PM. Date updated: Mon, May 16, 2022 2:59 PM. Content. Video answer: #11 đź”´ so many aces! blackjack ~ 6 deck shoe at red rock casino ~ las vegas july 2021 episode 4. ## Top best answers to the question «How many cards are in a 6 deck shoe». • I attempt to work this out in my blackjack appendix 8 but I’ll work through it more slowly here. We’ll ignore dealer blackjacks to keep things simple and assume the player always hits after two cards. The number of ways to arrange 3 cards in a 6-deck shoe is combin (312,3)=5,013,320. There are 24 sevens in the shoe.. A six-deck shoe has 6 x 16 = 96 10-values and 6 x 52 = 312 total cards. The chance of an initial 10 is therefore 96 out of 312 or 30.769 percent. Considering only the 10 to be gone, the shoe then would have 6 x 4 = 24 aces in 311 remaining cards.. FAQ. Those who are looking for an answer to the question «How many cards are in a 6 deck shoe?» often ask the following questions:. ### đź‘ Can you count cards in a multi deck shoe?. #### How do you count cards in a deck of cards?. • Step 1. Assign a value to every card. Step 2. Keep a “Running Count” based off of the values of the card dealt. Step 3. Use this information to calculate the count per deck or “true count”. Step 4. Change your bets as the true count rises.. ### đź‘ How many hands in a six deck shoe?. • The average number of cards per hand is 4.94. Assuming 15 burn cards, a six-deck baccarat shoe would have about 60 hands. However, mathematically speaking, it doesn't make any difference when they shuffle.. ### đź‘ How many decks are in a 6 deck shoe?. • Casinos use a 6-deck shoe, so take the number of decks you see in the discard tray, and subtract it from 6. That’s how many decks are remaining in the shoe. Divide the running count by the number of decks remaining.. Video answer: Deck estimation in card counting: what it is and how to do it. We've handpicked 28 related questions for you, similar to «How many cards are in a 6 deck shoe?» so you can surely find the answer!. How did the multi deck shoe get its name?. • While no such ruling was ever passed, most Nevada casinos now deal from a multi-deck shoe. As gaming advisor to the Havana Hilton, Scarne also introduced the shoe to Puerto Rico and Cuba. The device is so named because the earliest versions of it resembled a woman's high-heel shoe, and were often painted red or black in color.. Shoe size 13 what is a good deck width?. • Full-Size Deck [width: 7.5-inches or larger, length: 28-inches]:These are perfect for skaters that are over the age of 13, with height taller than 5’3” and wear a shoe size of 9 or more. However, the specific deck size is determined by the style preference of the skater: 7.5-inches to 8-inches: street skating or for more technical tricks. What's the running count on a 6 deck shoe?. • If you have a +6 after the first round of a game in a 6 deck shoe, the running count is +1 since there are still about 6 decks left. If you have a +6 when you’re halfway through that shoe, then you have 3 decks left and a +2 running count. There are a couple of ways to figure out the number of decks remaining in the shoe.. What's the running count on a six deck shoe?. • If you were say, one deck into that six deck shoe, a running count of +10 would give you the same proportional high-to-low ratio of remaining cards as a +4 running count would when you're four decks into it. That's because in both cases, there would be two extra high cards available for each deck that is left.. What shoe stores offer credit cards?. • Amazon.. • Nordstrom.. • Belk.. • Kohl's.. • Macy's.. • Bloomingdale's.. • JCPenney.. • Walmart.. ### Video answer: How to shuffle a shoe deck. What are the odds of a six deck shoe in baccarat?. • A six deck shoe will have the following baccarat odds: 45.87% for the banker, 44.63% for player, and 9.51% for a tie.. What shoe stores accept nike gift cards?. #### Where can I use my Nike gift card?. • Nike and Converse gift cards are just like cash and can be used to make purchases on Nike.com and at any Nike- and Converse-owned retail stores.. ### Video answer: đź”´ blackjack with so many doubles ~ 6 deck shoe ~ red rock casino las vegas july 2021~ episode 5. Can you use credit cards for shoe horns?. • Credit cards and business cards are the one item everyone will have on them at all times, or, at least, whenever they are also wearing shoes. They have just about the perfect size and they are slick enough to be perfect replacements for shoe horns. There are 2 problems that still need to be taken into consideration.. What do the cards in the shoe mean?. • When the cards are placed in the shoe, the dealer will insert a brightly colored blank plastic card. When this card is drawn it indicates that the current game is the last one before a new shuffle. This helps mitigate player advantage via card counting, as a significant portion (usually about 25 percent) of the full inventory.... What kind of cards are in a shoe?. • Dealing shoes come in many colors and sizes, depending on the number of decks they are capable of holding (2, 4, 6, or 8 decks). When the cards are placed in the shoe, the dealer will insert a brightly colored blank plastic card.. ### Video answer: Card counting at 6 deck 21 with john stathis. How many shoe brushes?. • Every shoe specialist knows the importance of having high-quality shoe brushes. Generally, experts recommend you use three different brushes for the best results. One shoe brush to remove the dust and grime from your footwear.. 11.5 shoe how many inches?. #### Men's Size Conversions. InchesCMUS Sizes. 11.125"28.311.5. 11.25"28.612. 11.5625"29.413. 11.875"30.214. How many customizable shoe brands?. #### What are the top shoe brands in the world?. • With its roots in Germany, adidas has become one of the top shoe brands in the world. The company produces more than 900 million sports and lifestyle products with independent manufacturing partners around the world.. How many inches inside shoe?. #### Does shoe size equal inches?. • Men's shoe size charts start from a US number of 6 (a European number of 39 and a UK number of 5.5). This size corresponds to an inch measurement of 9.25 inches (23.5 centimeters). If the measurement is provided solely in centimeters, multiply the length of your foot by 2.54 and apply to the chart. The biggest number on the men's chart is .... How many inches shoe size?. • Size 6 = 9.31 inches Size 6.5 = 9.5 inches Size 7 = 9.69 inches Size 7.5 = 9.81 inches Size 8 = 10 inches Size 8.5 = 10.19 inches Size 9 = 10.31 inches Size 9.5 = 10.5 inches Size 10 = 10.69 inches Size 10.5 = 10.81 inches Size 11 = 11 inches Size 11.5 = 11.19 inches Size 12 = 11.31 inches Size 12.5 = 11.5 inches Size 13 = 11.69 inches Size 13.5 = 11.81 inches Size 14 = 12 inches Size 14.5 = 12.19 inches Size 15 = 12.31 inches. How many man shoe and?. #### What's the average shoe size for a man?. • For men, it’s pretty easy to determine, as there are many studies analyzing the average size of a male foot. In 2014, the American Academy of Orthopaedic Surgeons reported that the worldwide average shoe size for men is between 9 and 12, while the American average male shoe size is a 10.5. A mother measures her son’s shoe size.. How many marathons per shoe?. #### How many people go to a marathon each year?. • Major marathons also attract many spectators and viewers. In 2019, people from almost 3.3 million households are projected to attended a marathon event in the United States. This text provides general information.. How many shoe carnival stores?. #### What kind of store is Shoe Carnival?. • Shoe Carnival Stores Shoe Carnival is a chain of family shoe stores operating across the continental United States, as well as in Puerto Rico and online at shoecarnival.com. We have a passion for creating a fun, engaging, and affordable shoe shopping experience, bringing great deals on brand-name footwear to millions of families.. ### Video answer: Card deck mistakes to watch out for. How many syllables in shoe?. • How many syllables are in shoe? 1 syllable. Divide shoe into syllables: shoe. Definition of: Shoe (New window will open) How to pronounce shoe: Words: shoddily, shoddy, shoe, shoebox, shoed.. How many cards do you need to deal shoes?. • Dealing shoes come in many colors and sizes, depending on the number of decks they are capable of holding (2, 4, 6, or 8 decks).. What size skateboard deck should i get?. • Mini Skateboard Deck width: 7.0″ with an average deck length of 28” Mini decks are the best skateboards for beginners who are 6-8 years old, between 3’5” and 4’4” tall who wear size 4-6 shoes. Mid-size Skateboard Deck width: 7.3”+ with an average deck length of 29” For skaters 9 to 12 years old between 4’5”.... What happens when you remove cards from shoe in blackjack?. • In fact, if you remove any significant number of small cards from the shoe, blackjack not only becomes easier, the player gets a mathematical edge over the house. This is where card counting comes in. Counters track the proportion of small cards to face cards and aces. At a certain point, the deck turns “positive.”.. Hemnes shoe cabinet how many shoes?. • How many shoes can I fit inside the Hemnes unit? The Hemnes shoe unit can hold eight pairs of shoes. However, if you are like me and have small feet, I can fit 10.. How many big baller shoe sold?.
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In an interview with Colin Cowherd in May 2017, Ball said that if the big shoe companies like Nike, Adidas, or Under Armour want to make a deal with his Big Baller Brand, the asking price is \$3 billion. He also said that Baller has sold between 400 and 520 ZO2 shoes.
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# Calculate [H+]?
1. Nov 5, 2004
### parwana
Calculate [H+]?????
Calculate [ H + ] of a 0.220 M solution of Aniline C6H5NH2 Kb = 7.4e-10
2. Nov 5, 2004
### Sirus
Consider that for a reaction
$$HA_{(aq)}\rightleftharpoons~H^{+}_{(aq)}~+~A^{-}_{(aq)}$$
we can use
$$K_{a}=\frac{[H^{+}_{(aq)}][A^{-}_{(aq)}]}{[HA_{(aq)}]}$$
How can you find $K_{a}$ when given $K_{b}$?
3. Nov 5, 2004
### chem_tr
Sirus is right; an alternative point of view may be using dissociative approach. In this, you start with 0.220 M of aniline, but only $\displaystyle x$ of it is ionized to give some $\displaystyle H^+$. We know the equilibrium constant of this reaction, i.e., $$\displaystyle \frac {10^{-14}}{K_b}$$.
$$\displaystyle \underbrace{C_6H_5NH_2_{(aq)}}_{0.220-x} \leftrightharpoons \underbrace{C_6H_5NH^-_{(aq)}}_{x}+\underbrace{H^+_{(aq)}}_{x}$$
Now it is better for you to consider the magnitude of $$\displaystyle \frac {10^{-14}}{K_b}$$; for the sake of simplification, you may omit the $\displaystyle x$ in $\displaystyle 0.220-x$, as we may omit values generally less than 5%. Of course, you may also want to solve the quadratic equation to find the exact answer, but believe me this is not very necessary.
What you'll do next is to find the $\displaystyle -\log x$.
4. Nov 6, 2004
### ShawnD
Aniline is a base, not an acid. Hydrogen will stick to the lone pair on the nitrogen so the reaction is like this:
$$C_6H_5NH_2 + H_2O \rightarrow C_6H_5NH_3^+ + OH^-$$
5. Nov 6, 2004
### Sirus
In that case, for a reaction
$$B_{(aq)}~+~H_{2}O_{(l)}\rightleftharpoons~BH^{+}_{(aq)}~+~OH^{-}_{(aq)}$$
we can use
$$K_{b}=\frac{[BH^{+}_{(aq)}][OH^{-}_{(aq)}]}{[B_{(aq)}]}$$
6. Nov 6, 2004
### chem_tr
Dear ShawnD, you are right about aqueous reactions, but don't forget that there are very powerful bases for using in non-aqueous phases like sodium hydride, lithium diisopropylamide, etc, which can take a proton to give anilinide anion.
In most cases, your reaction is sufficient, and same things may be said for Sirus' last post.
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# Calculate [H+]?. 1. Nov 5, 2004. ### parwana. Calculate [H+]?????. Calculate [ H + ] of a 0.220 M solution of Aniline C6H5NH2 Kb = 7.4e-10. 2. Nov 5, 2004. ### Sirus. Consider that for a reaction. $$HA_{(aq)}\rightleftharpoons~H^{+}_{(aq)}~+~A^{-}_{(aq)}$$. we can use. $$K_{a}=\frac{[H^{+}_{(aq)}][A^{-}_{(aq)}]}{[HA_{(aq)}]}$$. How can you find $K_{a}$ when given $K_{b}$?. 3. Nov 5, 2004. ### chem_tr. Sirus is right; an alternative point of view may be using dissociative approach. In this, you start with 0.220 M of aniline, but only $\displaystyle x$ of it is ionized to give some $\displaystyle H^+$. We know the equilibrium constant of this reaction, i.e., $$\displaystyle \frac {10^{-14}}{K_b}$$.. $$\displaystyle \underbrace{C_6H_5NH_2_{(aq)}}_{0.220-x} \leftrightharpoons \underbrace{C_6H_5NH^-_{(aq)}}_{x}+\underbrace{H^+_{(aq)}}_{x}$$. Now it is better for you to consider the magnitude of $$\displaystyle \frac {10^{-14}}{K_b}$$; for the sake of simplification, you may omit the $\displaystyle x$ in $\displaystyle 0.220-x$, as we may omit values generally less than 5%. Of course, you may also want to solve the quadratic equation to find the exact answer, but believe me this is not very necessary.. What you'll do next is to find the $\displaystyle -\log x$.. 4. Nov 6, 2004. ### ShawnD. Aniline is a base, not an acid. Hydrogen will stick to the lone pair on the nitrogen so the reaction is like this:. $$C_6H_5NH_2 + H_2O \rightarrow C_6H_5NH_3^+ + OH^-$$. 5. Nov 6, 2004. ### Sirus. In that case, for a reaction. $$B_{(aq)}~+~H_{2}O_{(l)}\rightleftharpoons~BH^{+}_{(aq)}~+~OH^{-}_{(aq)}$$. we can use. $$K_{b}=\frac{[BH^{+}_{(aq)}][OH^{-}_{(aq)}]}{[B_{(aq)}]}$$. 6. Nov 6, 2004. ### chem_tr.
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Dear ShawnD, you are right about aqueous reactions, but don't forget that there are very powerful bases for using in non-aqueous phases like sodium hydride, lithium diisopropylamide, etc, which can take a proton to give anilinide anion.. In most cases, your reaction is sufficient, and same things may be said for Sirus' last post.
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# Condition of Equilibrium and Stability of Floating Bodies- Stable, Neutral, Unstable Equilibrium [PDF]
Contents
## Introduction
In this article, we will delve into the fascinating world of floating bodies and explore the concepts of equilibrium and stability. When a body is placed on a fluid surface, it experiences various forces that determine whether it will float, sink or stay in equilibrium. The stability of a floating body depends on its center of gravity and the position of the center of buoyancy. Understanding the conditions of stable, neutral, and unstable equilibrium is critical for engineers and designers who work with floating structures, such as ships, boats, and oil rigs. By the end of this article, you will have a deeper understanding of the principles that govern the stability of floating bodies and the factors that affect them.
## Archimedes principle
The Archimedes principle explains about the force acting on a floating body in a liquid. It states that “The body immersed or floating in a liquid are acted upon by a vertical upward liquid force equal to the weight of the liquid displaced”. This vertical upward force is called buoyancy or buoyant force. The point, through which this force acts, is known as a center of buoyancy.
A body floating or immersed in a liquid will lose its weight equal to the buoyant force of the liquid. The body has less weight in a liquid than outside. Archimedes principle is applicable to bodies floating or immersed in a fluid. This has been used by man for about 2200 years, for the problem of general floatation and naval architectural design.
## Stability of Floating Bodies
When a body floats, it is subjected to two parallel forces which are as follows:
1. The downward force of gravity acting on each of the particles.
2. The upward buoyant force of the liquid acting on various elements of the submerged surface.
If the body is to float in equilibrium in an upright position the resultant of these two forces must be collinear, equal and opposite.
Hence center of gravity of the floating body and center of buoyancy must lie in the same vertical line. B is the center of buoyancy which is the center of gravity of the area ACO, and G is the center of gravity of the body. If the ship in the (figure above) heels through an angle θ (fig b), due to tilting moments caused by wind or wave action or due to movement of loads across the deck, portion A’C’O’ will now stand immersed in water.
The center of buoyancy will shift from B to B’. The buoyant force will act through B’. The center of gravity G will of course not change and W will continue to act through it.
#### What is Metacenter?
A vertical line through the new center of buoyancy intersects the inclined axis of the ship at M which is known as Metacenter. The term metacenter is defined as the point at which a vertical line through the center of buoyancy intersects the vertical center line of the ship section, after a small angle of heal.
#### What is Metacentric Height?
The distance between the center of gravity and metacenter is called metacentric height. The metacentric height is a measure of the static stability of the ship. For the small angle of inclination, the position of M does not change materially and the metacentric height is approximately constant. Hence the ship may be considered as rotating about M. In other words, the ship may be considered as behaving like a pendulum suspended at M, the point G, corresponding to bob.
## What do you mean by equilibrium?
Equilibrium of floating bodies refers to the state of balance or stability that is achieved when a body is submerged in a fluid and is neither sinking nor rising. In other words, it is the state in which the weight of the object is equal to the buoyant force acting on it. The buoyant force is the force exerted by the fluid on the object, which is equal to the weight of the fluid displaced by the object. This phenomenon is known as Archimedes’ principle.
When a body is placed in a fluid, it experiences an upward force, which is known as the buoyant force. This force is equal to the weight of the fluid displaced by the object. If the weight of the object is less than the buoyant force, it will float on the surface of the fluid. However, if the weight of the object is greater than the buoyant force, it will sink.
In the case of floating bodies, the weight of the object is equal to the buoyant force. This means that the object is in a state of balance or equilibrium. If the object is pushed down, the buoyant force will increase, and if the object is pushed up, the buoyant force will decrease. As a result, the object will return to its original position of equilibrium.
The equilibrium of floating bodies has many practical applications, such as in the design of boats and ships. In order to ensure that a boat or ship is stable and does not tip over, the weight of the vessel must be distributed evenly and the center of gravity must be located below the center of buoyancy. This ensures that the buoyant force acting on the vessel is greater than its weight, and it remains in a state of equilibrium.
## The condition of Equilibrium of Floating Bodies:
By the condition of equilibrium of floating bodies, we mean the possible state of stability or instability of floating bodies under all odds. Following are the three conditions of equilibrium of floating bodies:
• Stable Equilibrium
• Neutral Equilibrium
• Unstable Equilibrium
Let us discuss each condition in detail.
### Stable Equilibrium:
In the above figure (b) we found that when the ship was subjected to turning moments, the center of buoyancy changed from B to B’. Further, From the same diagram, we also notice that W and F are two equal and opposite parallel forces acting at a distance X apart. Naturally, this causes anticlockwise couples WX, tending to restore the ship to its original position.
In this case, the ship is said to be in stable equilibrium. Hence it may be stated that a floating body is said to be in a state of stable equilibrium which, when subjected to turning moments, leads to regaining its original position and that M lies above G or BM>BG.
### Neutral Equilibrium:
If the ship in figure (b) is tilted or rolled over such that the new center of buoyancy B’ lies on the line of action of W, the buoyant force F and weight W would be collinear, equal and opposite and would not exert any restoring moment. In that case, the ship would neither tend to regain its original position nor would tend to heel over further.
Hence it may be stated that “a floating body is said to be in a state of neutral equilibrium which, when subjected to turning moment, neither tends to regain its original position, nor tend to heel over further but instead keeps on in the tilted position, and that M coincides with G or BM=BG”.
### Unstable Equilibrium:
In this case, the new center of buoyancy B’ lies in between B and the line of action of W. The vertical upward buoyant force F passing through B’ will intersect the inclines central line at the point M below G. The couple thus formed by W and F will be clockwise; further helping the turning moment of tilt over the ship further.
## FAQ’s
### What is the condition of equilibrium for a floating body?
It is the condition for a floating body when the body is neither sinking nor rising and is stable in its position on the fluid surface. This state occurs when the weight of the floating body is equal to the buoyant force acting on it.
### What are the factors that affect the stability of a floating body?
The stability of a floating body depends on several factors, including the position of the center of gravity, the position of the center of buoyancy, the shape and size of the body, and the density of the fluid.
### What is metacentric height, and why is it important for determining stability?
Metacentric height is the distance between the metacenter (the point at which a floating body’s center of gravity will be when it tilts) and the center of buoyancy (the center of mass of the displaced fluid). It is an important parameter for determining the stability of a floating body, as a higher metacentric height indicates greater stability.
### What is the difference between stable, neutral, and unstable for a floating body?
In stable equilibrium, the body will return to its original position when it is slightly tilted, whereas in neutral equilibrium, the body will remain in the new position after being tilted. Unstable equilibrium occurs when the body will continue to tilt until it capsizes, and it cannot return to its original position.
## Conclusion
In conclusion, the condition of equilibrium and stability of floating bodies is a crucial concept in physics. It is essential to understand the forces acting on a body and their effects on its stability to prevent accidents and ensure safety in various activities involving floating bodies. By understanding the frequently asked questions about this topic, we can develop a better understanding of the principles involved and apply them to practical situations.
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# Condition of Equilibrium and Stability of Floating Bodies- Stable, Neutral, Unstable Equilibrium [PDF]. Contents. ## Introduction. In this article, we will delve into the fascinating world of floating bodies and explore the concepts of equilibrium and stability. When a body is placed on a fluid surface, it experiences various forces that determine whether it will float, sink or stay in equilibrium. The stability of a floating body depends on its center of gravity and the position of the center of buoyancy. Understanding the conditions of stable, neutral, and unstable equilibrium is critical for engineers and designers who work with floating structures, such as ships, boats, and oil rigs. By the end of this article, you will have a deeper understanding of the principles that govern the stability of floating bodies and the factors that affect them.. ## Archimedes principle. The Archimedes principle explains about the force acting on a floating body in a liquid. It states that “The body immersed or floating in a liquid are acted upon by a vertical upward liquid force equal to the weight of the liquid displaced”. This vertical upward force is called buoyancy or buoyant force. The point, through which this force acts, is known as a center of buoyancy.. A body floating or immersed in a liquid will lose its weight equal to the buoyant force of the liquid. The body has less weight in a liquid than outside. Archimedes principle is applicable to bodies floating or immersed in a fluid. This has been used by man for about 2200 years, for the problem of general floatation and naval architectural design.. ## Stability of Floating Bodies. When a body floats, it is subjected to two parallel forces which are as follows:. 1. The downward force of gravity acting on each of the particles.. 2. The upward buoyant force of the liquid acting on various elements of the submerged surface.. If the body is to float in equilibrium in an upright position the resultant of these two forces must be collinear, equal and opposite.. Hence center of gravity of the floating body and center of buoyancy must lie in the same vertical line. B is the center of buoyancy which is the center of gravity of the area ACO, and G is the center of gravity of the body. If the ship in the (figure above) heels through an angle θ (fig b), due to tilting moments caused by wind or wave action or due to movement of loads across the deck, portion A’C’O’ will now stand immersed in water.. The center of buoyancy will shift from B to B’. The buoyant force will act through B’. The center of gravity G will of course not change and W will continue to act through it.. #### What is Metacenter?. A vertical line through the new center of buoyancy intersects the inclined axis of the ship at M which is known as Metacenter. The term metacenter is defined as the point at which a vertical line through the center of buoyancy intersects the vertical center line of the ship section, after a small angle of heal.. #### What is Metacentric Height?. The distance between the center of gravity and metacenter is called metacentric height. The metacentric height is a measure of the static stability of the ship. For the small angle of inclination, the position of M does not change materially and the metacentric height is approximately constant. Hence the ship may be considered as rotating about M. In other words, the ship may be considered as behaving like a pendulum suspended at M, the point G, corresponding to bob.. ## What do you mean by equilibrium?. Equilibrium of floating bodies refers to the state of balance or stability that is achieved when a body is submerged in a fluid and is neither sinking nor rising. In other words, it is the state in which the weight of the object is equal to the buoyant force acting on it. The buoyant force is the force exerted by the fluid on the object, which is equal to the weight of the fluid displaced by the object. This phenomenon is known as Archimedes’ principle.. When a body is placed in a fluid, it experiences an upward force, which is known as the buoyant force. This force is equal to the weight of the fluid displaced by the object. If the weight of the object is less than the buoyant force, it will float on the surface of the fluid. However, if the weight of the object is greater than the buoyant force, it will sink.. In the case of floating bodies, the weight of the object is equal to the buoyant force. This means that the object is in a state of balance or equilibrium. If the object is pushed down, the buoyant force will increase, and if the object is pushed up, the buoyant force will decrease. As a result, the object will return to its original position of equilibrium.. The equilibrium of floating bodies has many practical applications, such as in the design of boats and ships. In order to ensure that a boat or ship is stable and does not tip over, the weight of the vessel must be distributed evenly and the center of gravity must be located below the center of buoyancy. This ensures that the buoyant force acting on the vessel is greater than its weight, and it remains in a state of equilibrium.. ## The condition of Equilibrium of Floating Bodies:. By the condition of equilibrium of floating bodies, we mean the possible state of stability or instability of floating bodies under all odds. Following are the three conditions of equilibrium of floating bodies:. • Stable Equilibrium. • Neutral Equilibrium. • Unstable Equilibrium. Let us discuss each condition in detail.. ### Stable Equilibrium:. In the above figure (b) we found that when the ship was subjected to turning moments, the center of buoyancy changed from B to B’. Further, From the same diagram, we also notice that W and F are two equal and opposite parallel forces acting at a distance X apart. Naturally, this causes anticlockwise couples WX, tending to restore the ship to its original position.. In this case, the ship is said to be in stable equilibrium. Hence it may be stated that a floating body is said to be in a state of stable equilibrium which, when subjected to turning moments, leads to regaining its original position and that M lies above G or BM>BG.. ### Neutral Equilibrium:. If the ship in figure (b) is tilted or rolled over such that the new center of buoyancy B’ lies on the line of action of W, the buoyant force F and weight W would be collinear, equal and opposite and would not exert any restoring moment. In that case, the ship would neither tend to regain its original position nor would tend to heel over further.. Hence it may be stated that “a floating body is said to be in a state of neutral equilibrium which, when subjected to turning moment, neither tends to regain its original position, nor tend to heel over further but instead keeps on in the tilted position, and that M coincides with G or BM=BG”.. ### Unstable Equilibrium:. In this case, the new center of buoyancy B’ lies in between B and the line of action of W. The vertical upward buoyant force F passing through B’ will intersect the inclines central line at the point M below G. The couple thus formed by W and F will be clockwise; further helping the turning moment of tilt over the ship further.. ## FAQ’s. ### What is the condition of equilibrium for a floating body?. It is the condition for a floating body when the body is neither sinking nor rising and is stable in its position on the fluid surface. This state occurs when the weight of the floating body is equal to the buoyant force acting on it.. ### What are the factors that affect the stability of a floating body?. The stability of a floating body depends on several factors, including the position of the center of gravity, the position of the center of buoyancy, the shape and size of the body, and the density of the fluid.. ### What is metacentric height, and why is it important for determining stability?. Metacentric height is the distance between the metacenter (the point at which a floating body’s center of gravity will be when it tilts) and the center of buoyancy (the center of mass of the displaced fluid). It is an important parameter for determining the stability of a floating body, as a higher metacentric height indicates greater stability.. ### What is the difference between stable, neutral, and unstable for a floating body?. In stable equilibrium, the body will return to its original position when it is slightly tilted, whereas in neutral equilibrium, the body will remain in the new position after being tilted. Unstable equilibrium occurs when the body will continue to tilt until it capsizes, and it cannot return to its original position.. ## Conclusion. In conclusion, the condition of equilibrium and stability of floating bodies is a crucial concept in physics.
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It is essential to understand the forces acting on a body and their effects on its stability to prevent accidents and ensure safety in various activities involving floating bodies. By understanding the frequently asked questions about this topic, we can develop a better understanding of the principles involved and apply them to practical situations.
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https://stacks.math.columbia.edu/tag/01OJ
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## 28.3 Integral, irreducible, and reduced schemes
Definition 28.3.1. Let $X$ be a scheme. We say $X$ is integral if it is nonempty and for every nonempty affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ the ring $R$ is an integral domain.
Lemma 28.3.2. Let $X$ be a scheme. The following are equivalent.
1. The scheme $X$ is reduced, see Schemes, Definition 26.12.1.
2. There exists an affine open covering $X = \bigcup U_ i$ such that each $\Gamma (U_ i, \mathcal{O}_ X)$ is reduced.
3. For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is reduced.
4. For every open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is reduced.
Lemma 28.3.3. Let $X$ be a scheme. The following are equivalent.
1. The scheme $X$ is irreducible.
2. There exists an affine open covering $X = \bigcup _{i \in I} U_ i$ such that $I$ is not empty, $U_ i$ is irreducible for all $i \in I$, and $U_ i \cap U_ j \not= \emptyset$ for all $i, j \in I$.
3. The scheme $X$ is nonempty and every nonempty affine open $U \subset X$ is irreducible.
Proof. Assume (1). By Schemes, Lemma 26.11.1 we see that $X$ has a unique generic point $\eta$. Then $X = \overline{\{ \eta \} }$. Hence $\eta$ is an element of every nonempty affine open $U \subset X$. This implies that $\eta \in U$ is dense hence $U$ is irreducible. It also implies any two nonempty affines meet. Thus (1) implies both (2) and (3).
Assume (2). Suppose $X = Z_1 \cup Z_2$ is a union of two closed subsets. For every $i$ we see that either $U_ i \subset Z_1$ or $U_ i \subset Z_2$. Pick some $i \in I$ and assume $U_ i \subset Z_1$ (possibly after renumbering $Z_1$, $Z_2$). For any $j \in I$ the open subset $U_ i \cap U_ j$ is dense in $U_ j$ and contained in the closed subset $Z_1 \cap U_ j$. We conclude that also $U_ j \subset Z_1$. Thus $X = Z_1$ as desired.
Assume (3). Choose an affine open covering $X = \bigcup _{i \in I} U_ i$. We may assume that each $U_ i$ is nonempty. Since $X$ is nonempty we see that $I$ is not empty. By assumption each $U_ i$ is irreducible. Suppose $U_ i \cap U_ j = \emptyset$ for some pair $i, j \in I$. Then the open $U_ i \amalg U_ j = U_ i \cup U_ j$ is affine, see Schemes, Lemma 26.6.8. Hence it is irreducible by assumption which is absurd. We conclude that (3) implies (2). The lemma is proved. $\square$
Lemma 28.3.4. A scheme $X$ is integral if and only if it is reduced and irreducible.
Proof. If $X$ is irreducible, then every affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ is irreducible. If $X$ is reduced, then $R$ is reduced, by Lemma 28.3.2 above. Hence $R$ is reduced and $(0)$ is a prime ideal, i.e., $R$ is an integral domain.
If $X$ is integral, then for every nonempty affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ the ring $R$ is reduced and hence $X$ is reduced by Lemma 28.3.2. Moreover, every nonempty affine open is irreducible. Hence $X$ is irreducible, see Lemma 28.3.3. $\square$
In Examples, Section 109.6 we construct a connected affine scheme all of whose local rings are domains, but which is not integral.
Comment #4228 by Lin on
Correct me If am wrong, it seems that (2) to (1) of lemma [01OM], 2nd paragraph, follows directly from the first line. If $U_j \cap U_i \not= \emptyset$, then $U_j$ cannot contain a point in $Z_2$, hence $U_j \subseteq Z_1$, ( otherwise the $Z_1, Z_2$ forms a decomposition. ) ?
Comment #4229 by Lin on
Correct me If am wrong, it seems that (2) to (1) of lemma [01OM], 2nd paragraph, follows directly from the first line. If $U_j \cap U_i \not= \emptyset$, then $U_j$ cannot contain a point in $Z_2$, hence $U_j \subseteq Z_1$, ( otherwise the $Z_1, Z_2$ forms a decomposition. ) ?
Comment #4409 by on
Hmm... I think you are saying the same thing as the proof says.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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## 28.3 Integral, irreducible, and reduced schemes. Definition 28.3.1. Let $X$ be a scheme. We say $X$ is integral if it is nonempty and for every nonempty affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ the ring $R$ is an integral domain.. Lemma 28.3.2. Let $X$ be a scheme. The following are equivalent.. 1. The scheme $X$ is reduced, see Schemes, Definition 26.12.1.. 2. There exists an affine open covering $X = \bigcup U_ i$ such that each $\Gamma (U_ i, \mathcal{O}_ X)$ is reduced.. 3. For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is reduced.. 4. For every open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is reduced.. Lemma 28.3.3. Let $X$ be a scheme. The following are equivalent.. 1. The scheme $X$ is irreducible.. 2. There exists an affine open covering $X = \bigcup _{i \in I} U_ i$ such that $I$ is not empty, $U_ i$ is irreducible for all $i \in I$, and $U_ i \cap U_ j \not= \emptyset$ for all $i, j \in I$.. 3. The scheme $X$ is nonempty and every nonempty affine open $U \subset X$ is irreducible.. Proof. Assume (1). By Schemes, Lemma 26.11.1 we see that $X$ has a unique generic point $\eta$. Then $X = \overline{\{ \eta \} }$. Hence $\eta$ is an element of every nonempty affine open $U \subset X$. This implies that $\eta \in U$ is dense hence $U$ is irreducible. It also implies any two nonempty affines meet. Thus (1) implies both (2) and (3).. Assume (2). Suppose $X = Z_1 \cup Z_2$ is a union of two closed subsets. For every $i$ we see that either $U_ i \subset Z_1$ or $U_ i \subset Z_2$. Pick some $i \in I$ and assume $U_ i \subset Z_1$ (possibly after renumbering $Z_1$, $Z_2$). For any $j \in I$ the open subset $U_ i \cap U_ j$ is dense in $U_ j$ and contained in the closed subset $Z_1 \cap U_ j$. We conclude that also $U_ j \subset Z_1$. Thus $X = Z_1$ as desired.. Assume (3). Choose an affine open covering $X = \bigcup _{i \in I} U_ i$. We may assume that each $U_ i$ is nonempty. Since $X$ is nonempty we see that $I$ is not empty. By assumption each $U_ i$ is irreducible. Suppose $U_ i \cap U_ j = \emptyset$ for some pair $i, j \in I$. Then the open $U_ i \amalg U_ j = U_ i \cup U_ j$ is affine, see Schemes, Lemma 26.6.8. Hence it is irreducible by assumption which is absurd. We conclude that (3) implies (2). The lemma is proved. $\square$. Lemma 28.3.4. A scheme $X$ is integral if and only if it is reduced and irreducible.. Proof. If $X$ is irreducible, then every affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ is irreducible. If $X$ is reduced, then $R$ is reduced, by Lemma 28.3.2 above. Hence $R$ is reduced and $(0)$ is a prime ideal, i.e., $R$ is an integral domain.. If $X$ is integral, then for every nonempty affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ the ring $R$ is reduced and hence $X$ is reduced by Lemma 28.3.2. Moreover, every nonempty affine open is irreducible. Hence $X$ is irreducible, see Lemma 28.3.3. $\square$. In Examples, Section 109.6 we construct a connected affine scheme all of whose local rings are domains, but which is not integral.. Comment #4228 by Lin on. Correct me If am wrong, it seems that (2) to (1) of lemma [01OM], 2nd paragraph, follows directly from the first line. If $U_j \cap U_i \not= \emptyset$, then $U_j$ cannot contain a point in $Z_2$, hence $U_j \subseteq Z_1$, ( otherwise the $Z_1, Z_2$ forms a decomposition. ) ?. Comment #4229 by Lin on. Correct me If am wrong, it seems that (2) to (1) of lemma [01OM], 2nd paragraph, follows directly from the first line. If $U_j \cap U_i \not= \emptyset$, then $U_j$ cannot contain a point in $Z_2$, hence $U_j \subseteq Z_1$, ( otherwise the $Z_1, Z_2$ forms a decomposition. ) ?. Comment #4409 by on.
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Hmm... I think you are saying the same thing as the proof says.. In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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https://books.google.no/books?id=5CkEAAAAQAAJ&q=angle+ABC&dq=editions:UOM39015067252117&lr=&hl=no&output=html&source=gbs_word_cloud_r&cad=5
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# The Elements of Euclid, the parts read in the University of Cambridge [book 1-6 and parts of book 11,12] with geometrical problems, by J.W. Colenso
### Hva folk mener -Skriv en omtale
Vi har ikke funnet noen omtaler på noen av de vanlige stedene.
### Populære avsnitt
Side 42 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Side 4 - Let it be granted that a straight line may be drawn from any one point to any other point.
Side 33 - F, which is the common vertex of the triangles: that is », together with four right angles. Therefore all the angles of the figure, together with four right angles are equal to twice as many right angles as the figure has sides.
Side 62 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.
Side 22 - If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.
Side 58 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.
Side 146 - ... may be demonstrated from what has been said of the pentagon : and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon.
Side 194 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Side 2 - A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one another : 16.
Side 69 - To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, may be equal to the square of the other part.
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# The Elements of Euclid, the parts read in the University of Cambridge [book 1-6 and parts of book 11,12] with geometrical problems, by J.W. Colenso. ### Hva folk mener -Skriv en omtale. Vi har ikke funnet noen omtaler på noen av de vanlige stedene.. ### Populære avsnitt. Side 42 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.. Side 4 - Let it be granted that a straight line may be drawn from any one point to any other point.. Side 33 - F, which is the common vertex of the triangles: that is », together with four right angles. Therefore all the angles of the figure, together with four right angles are equal to twice as many right angles as the figure has sides.. Side 62 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.. Side 22 - If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.. Side 58 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.. Side 146 - ... may be demonstrated from what has been said of the pentagon : and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon.. Side 194 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
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Side 2 - A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one another : 16.. Side 69 - To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, may be equal to the square of the other part.
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http://www.flashkit.com/tutorials/Math-Physics/Changing-Jim_Bumg-1014/index.php
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» Home » Movies » Tutorials » Submissions » Sound FX » Board » Links » Reviews » Feedback » Gallery » Fonts » The Lounge » Sound Loops
First time here? Newsletter
Search tutorials
Author: Jim Bumgardner | Website: http://www.krazydad.com/bestiary/
Here's a basic script that can make a movieclip point at another object.
```MovieClip.prototype.pointAt = function(x,y)
{
var dx = x - this._x; // distance to other object on x-axis
var dy = y - this._y; // distance on y-axis
var dist = Math.sqrt(dx*dx + dy*dy); // true distance (using Pythagorean theorem)
var angle;
// figure out angle in radians (0- 2*PI)
if (dy < 0)
angle = Math.PI*2-Math.acos(dx/dist);
else
angle = Math.acos(dx/dist);
// convert to rotation value (angle in degrees or 0-360)
this._rotation = angle*180/Math.PI;
}
```
#### Math Explanation
The function that does all the work here is Math.acos(). The Math.acos() function is the compliment of the Math.cos() function. On another tutorial here, you may have seen how to use Math.cos() and Math.sin() to draw circle shapes using polar coodinates.
Math.cos() takes an angle (expressed in radians, or going from 0 to 2*PI instead of 0 to 360) and converts it to a position on a circle.
Given an angle (expressed in degrees) and a radius, a point on a circle is given by:
```x = Math.cos(angle*Math.PI/180)*radius;
```
The Math.acos() function I'm using does the reverse, it converts the value returned by Math.cos() back into an angle in radians. So by reversing the math of the above equations for using polar coordinates, we can determine the original angle in radians. Then, to convert this angle to a rotation value (which is in degrees) we scale it by the ratio betwen degres and radians: 360 / (2*PI) which is the same as 180 / PI.
#### How to use the script
Here's a script, which works with the above script to make a movieclip point in the direction of the mouse.
```mc.pointAtMouse = function()
{
this.pointAt(this._parent._xmouse, this._parent._ymouse);
}
```
Let's say your have a movieclip that you want to point at the mouse, you can do the following to make this happpen:
```mc.onEnterFrame = mc.pointAtMouse;
```
OR you can call this.pointAtMouse() from within a more complex onEnterFrame function.
Finally, this script assumes that your object points to 3:00 (to the right) when it is in it's 'normal' (rotation=0) state. If it doesn't, you'll need to add an offset to the rotation value. Here's a version of the function which accepts an offset. If your object normally points at 12:00 (upwards) then use -90 to correct it.
```MovieClip.prototype.pointAt = function(x,y,offset)
{
var dx = x - this._x; // distance to other object on x-axis
var dy = y - this._y; // distance on y-axis
var dist = Math.sqrt(dx*dx + dy*dy); // true distance (using Pythagorean theorem)
var angle;
// figure out angle in radians (0- 2*PI)
if (dy < 0)
angle = Math.PI*2-Math.acos(dx/dist);
else
angle = Math.acos(dx/dist);
// convert to rotation value (angle in degrees or 0-360)
this._rotation = offset + angle*180/Math.PI;
}
```
Example usage for object which normally points upwards (12:00):
```this.pointAt(this._parent._xmouse, this._parent._ymouse, -90);
```
- jim
» Level Intermediate Added: 2004-03-15 Rating: 7.17 Votes: 24 1 2 3 4 5 6 7 8 9 10 (10 being the highest) » Author Professional C/C++ programmer and flash hobbyist. » Download Download the files used in this tutorial. Download (0 kb) » Forums More help? Search our boards for quick answers!
• There are no comments yet. Be the first to comment!
• You must have javascript enabled in order to post comments.
Featured Flash FLA
» Author: Inocreato
» Title: RaiseTheBlocks
» Description: Raise all the blocks to win the game
Featured Sound Loops
Audio Player
» Author: TomCat Carty
» Title: The Wood
» Description: Just a little game ending or it can maybe be looped. Recorders with music box and percussion to give the feel of well, I don't know, the woods? Free to use, just credit me. thank you
Latest Font
» Author: Fábio FAFERS
» Description: I created this font for free use. Everyone can apply it in personal or business texts. Its free, but I want to be communicated in case of business use. Donations are accepted to keep the project of free fonts alive! Thank you all
Featured Sound Fx
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» Author: Davisigner
» Description: Hmm... what to say about this one? It's reminiscent of the closing notes of the opening music from the Three Stooges done in a church organ style with a closing cymbal crash. I'll give this one away gratis, but feel free to check out my free loops and potential upcoming license-mandated ones over in the respective part of Flashkit.
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» Home » Movies » Tutorials » Submissions » Sound FX » Board » Links » Reviews » Feedback » Gallery » Fonts » The Lounge » Sound Loops. First time here? Newsletter. Search tutorials. Author: Jim Bumgardner | Website: http://www.krazydad.com/bestiary/. Here's a basic script that can make a movieclip point at another object.. ```MovieClip.prototype.pointAt = function(x,y). {. var dx = x - this._x; // distance to other object on x-axis. var dy = y - this._y; // distance on y-axis. var dist = Math.sqrt(dx*dx + dy*dy); // true distance (using Pythagorean theorem). var angle;. // figure out angle in radians (0- 2*PI). if (dy < 0). angle = Math.PI*2-Math.acos(dx/dist);. else. angle = Math.acos(dx/dist);. // convert to rotation value (angle in degrees or 0-360). this._rotation = angle*180/Math.PI;. }. ```. #### Math Explanation. The function that does all the work here is Math.acos(). The Math.acos() function is the compliment of the Math.cos() function. On another tutorial here, you may have seen how to use Math.cos() and Math.sin() to draw circle shapes using polar coodinates.. Math.cos() takes an angle (expressed in radians, or going from 0 to 2*PI instead of 0 to 360) and converts it to a position on a circle.. Given an angle (expressed in degrees) and a radius, a point on a circle is given by:. ```x = Math.cos(angle*Math.PI/180)*radius;. ```. The Math.acos() function I'm using does the reverse, it converts the value returned by Math.cos() back into an angle in radians. So by reversing the math of the above equations for using polar coordinates, we can determine the original angle in radians. Then, to convert this angle to a rotation value (which is in degrees) we scale it by the ratio betwen degres and radians: 360 / (2*PI) which is the same as 180 / PI.. #### How to use the script. Here's a script, which works with the above script to make a movieclip point in the direction of the mouse.. ```mc.pointAtMouse = function(). {. this.pointAt(this._parent._xmouse, this._parent._ymouse);. }. ```. Let's say your have a movieclip that you want to point at the mouse, you can do the following to make this happpen:. ```mc.onEnterFrame = mc.pointAtMouse;. ```. OR you can call this.pointAtMouse() from within a more complex onEnterFrame function.. Finally, this script assumes that your object points to 3:00 (to the right) when it is in it's 'normal' (rotation=0) state. If it doesn't, you'll need to add an offset to the rotation value. Here's a version of the function which accepts an offset. If your object normally points at 12:00 (upwards) then use -90 to correct it.. ```MovieClip.prototype.pointAt = function(x,y,offset). {. var dx = x - this._x; // distance to other object on x-axis. var dy = y - this._y; // distance on y-axis. var dist = Math.sqrt(dx*dx + dy*dy); // true distance (using Pythagorean theorem). var angle;. // figure out angle in radians (0- 2*PI). if (dy < 0). angle = Math.PI*2-Math.acos(dx/dist);. else. angle = Math.acos(dx/dist);. // convert to rotation value (angle in degrees or 0-360). this._rotation = offset + angle*180/Math.PI;. }. ```. Example usage for object which normally points upwards (12:00):. ```this.pointAt(this._parent._xmouse, this._parent._ymouse, -90);. ```. - jim. » Level Intermediate Added: 2004-03-15 Rating: 7.17 Votes: 24 1 2 3 4 5 6 7 8 9 10 (10 being the highest) » Author Professional C/C++ programmer and flash hobbyist. » Download Download the files used in this tutorial. Download (0 kb) » Forums More help? Search our boards for quick answers!. • There are no comments yet. Be the first to comment!. • You must have javascript enabled in order to post comments.. Featured Flash FLA. » Author: Inocreato. » Title: RaiseTheBlocks. » Description: Raise all the blocks to win the game. Featured Sound Loops. Audio Player. » Author: TomCat Carty. » Title: The Wood. » Description: Just a little game ending or it can maybe be looped. Recorders with music box and percussion to give the feel of well, I don't know, the woods? Free to use, just credit me. thank you. Latest Font. » Author: Fábio FAFERS. » Description: I created this font for free use. Everyone can apply it in personal or business texts. Its free, but I want to be communicated in case of business use. Donations are accepted to keep the project of free fonts alive! Thank you all. Featured Sound Fx. Audio Player. » Author: Davisigner.
|
» Description: Hmm... what to say about this one? It's reminiscent of the closing notes of the opening music from the Three Stooges done in a church organ style with a closing cymbal crash. I'll give this one away gratis, but feel free to check out my free loops and potential upcoming license-mandated ones over in the respective part of Flashkit.
|
https://pratibha.eenadu.net/jobs/lesson/ibps/ibps-clerks/telugumedium/alligation-and-mixture/2-1-6-32-188-1347-7066-10530-20040012044
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# ALLIGATION AND MIXTURE
1. Jar-A contains a mixture of milk and water in the respective ratio of 8 : 1. When 18 ltr of mixture is taken out and 4 ltr of pure water is added to Jar-A, the resultant quantity of milk becomes 24 ltr more than that of water. What was the initial quantity of milk in Jar-A? (In ltr)
A) 48 B) 64 C) 50 D) 52 E) 60
2. A container contains 120 litres mixture of milk and water in the respective ratio of 37 : 3. 40 litres of the mixture is taken out from the container and 6 litres of each pure milk and pure water is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk?
A) 72% B) 76% C) 68% D) 85% E) 74%
3. In a 84 litres mixture of milk and water, percentage of water is only 25%. The milkman gave 12 litres of this mixture to a customer. Then he added equal quantities of pure milk and water to the remaining mixture. As a result, the respective ratio of milk and water in the mixture became 11 : 5. What was the quantity of water added? (In litres)
A) 8 B) 15 C) 12 D) 18 E) 21
4. A jar had a certain quantity (in litres) of water, to which pure milk of four times of the quantity water was added. 20 litres of the mixture from the jar was taken out and was replaced with 12 litres of pure milk, as result of which water constituted 10% of the resultant mixture. What was the initial quantity of water in the jar?
A) 7 litres B) 8 litres C) 10 litres D) 4.8 litres E) 6.4 litres
5. In a 200 litres mixture of milk and water, percentage of water is only 30%. The milkman gave 40 litres of this mixture to a customer and then added 12 litres of water in the remaining mixture. What is the respective ratio of milk and water in the new mixture?
A) 11 : 17 B) 7 : 13 C) 23 : 14 D) 21 : 11 E) 28 : 15
6. A vessel contains coconut water and vodka i.e., 120 litres of the mixture, coconut water is 40% more than the vodka. 72 litres of mixture taken out and 26 litres of coconut water is and some quantity of vodka is also added to the vessel, after that the respective ratio of the resultant mixture of vodka and coconut water becomes 2 : 3, then find the final quantityof the vodka in the vessel? (In litres)
A) 54 B) 46 C) 35 D) 34 E) 36
7. 16 litres of pure water was added to a vessel containing 84 litres of pure milk. 50 litres of the resultant mixture was then sold and some more quantity of pure milk and pure water was added to the vessel in the respective ratio of 3 : 2. If the resultant respective ratio of milk and water in the vessel is 4 : 1, what was the quantity of pure milk added in the vessel? (In litres)
A) 4 B) 8 C) 6 D) 10 E) 12
8. In a mixture of milk and water, water was only 20%. 50 litres of this mixture was taken out and then 5 litres of pure water was added to the mixture. If the resultant ratio of milk and water in the mixture was 8 : 3 respectively, what was the initial quantity of mixture (before the replacement)? (In litres)
A) 88 B) 96 C) 120 D) 100 E) 84
9. In a 120 litres mixture of milk and water, percentage of water is only 40%. The milkman gave 45 litres of this mixture to a customer and then added 15 litres of pure milk and 10 litres of pure water to the remaining mixture. What is the respective ratio of quantity of milk and water in the new mixture?
A) 3 : 2 B) 11 : 5 C) 8 : 5 D) 15 : 8 E) 12 : 7
10. In 112 litres of mixture of water and milk, water is only 18.75%. The milkman gave 28 litres of this mixture to a customer and then he added 10.75 litres of pure milk and 5.25 litres of pure water in the remaining mixture. What is the percentage of water in the final mixture?
A) 32.50 B) 34 C) 32 D) 21 E) 27
Some more...
1. Equal quantities of two milk and water solutions which contains milk and water in ratio, 1 : 5 and 3 : 5 are mixed together. What will be the ratio of water to milk in the resultant solution?
a) 13 : 35 b) 4 : 10 c) 5 : 8 d) 35 : 13
Ans: d
2. The respective ratio of milk and water in 66 litres of adulterated milk is 5 : 1. Water is added into it to make the respective ratio 5 : 3. What is the quantity of water added?
a) 11 litres b) 22 litres c) 33 litres d) 44 litres
Ans: b
3. A 200 litres solution of alcohol and water contains 1/4 th part alcohol. Find the new percentage of alcohol, if 50 litres of the original solution is replaced by 50 litres of alcohol.
a) 43.75% b) 50% c) 66.66% d) 80%
Ans: a
4. A container has 80 litres of milk. From this container 8 litres of milk was taken out and replaced with water. The process was further repeated twice. How much milk is there in the container finally? (in liters)
a) 58.32 b) 60.32 c) 62 d) 64.7
Ans: a
5. 100 litres of a mixture contains 10% water and the rest milk. The amount of water that must be added so that the resulting mixture contains only 50% milk is......
a) 70 litres b) 72 litres c) 78 litres d) 80 litres
Ans: d
6. In a particular type of fertilizer, the ratio of two chemicals A and B is 2 : 5. In 21 kg of this fertilizer, if 3 kg of type - A is added, the ratio of chemical A to B in the new fertilizer will be
a) 1 : 1 b) 2 : 3 c) 3 : 5 d) 4 : 5
Ans: c
7. Rice worth Rs.120 per kg and Rs.132 per kg are mixed with a third variety in the ratio 2 : 1 : 3. If the mixture is worth Rs.135 per kg, the price of the third variety per kg will be....
a) Rs.140 b) Rs.146 c) Rs.150 d) Rs.148
Ans: b
8. A jar full of wine contains 50% alcohol. A part of wine is replaced by another containing 20% alcohol and now percentage of alcohol was found to be 30%. The part of wine replaced is?
a) 1/2 b) 1/3 c) 2/5 d) 2/3
Ans: d
9. A shopkeeper has 50 kg rice, a part of which he sells at 10% profit and rest at 25% profit. He gains 16% on the whole. The quantity sold at 10% profit is?
a) 30 kg b) 20 kg c) 45 kg d) 27 kg
Ans: a
10. An alloy contains tin, copper and zinc in ratio 2 : 3 : 1, and another alloy contain copper, aluminium and tin in ratio 3 : 2 : 7. If both alloys are mixed in equal quantities, then what will be the ratio of tin and aluminium in final alloy?
a) 2 : 11 b) 11 : 9 c) 9 : 2 d) 11 : 2
Ans: d
Posted Date : 17-09-2022
గమనిక : ప్రతిభ.ఈనాడు.నెట్లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.
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# ALLIGATION AND MIXTURE. 1. Jar-A contains a mixture of milk and water in the respective ratio of 8 : 1. When 18 ltr of mixture is taken out and 4 ltr of pure water is added to Jar-A, the resultant quantity of milk becomes 24 ltr more than that of water. What was the initial quantity of milk in Jar-A? (In ltr). A) 48 B) 64 C) 50 D) 52 E) 60. 2. A container contains 120 litres mixture of milk and water in the respective ratio of 37 : 3. 40 litres of the mixture is taken out from the container and 6 litres of each pure milk and pure water is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk?. A) 72% B) 76% C) 68% D) 85% E) 74%. 3. In a 84 litres mixture of milk and water, percentage of water is only 25%. The milkman gave 12 litres of this mixture to a customer. Then he added equal quantities of pure milk and water to the remaining mixture. As a result, the respective ratio of milk and water in the mixture became 11 : 5. What was the quantity of water added? (In litres). A) 8 B) 15 C) 12 D) 18 E) 21. 4. A jar had a certain quantity (in litres) of water, to which pure milk of four times of the quantity water was added. 20 litres of the mixture from the jar was taken out and was replaced with 12 litres of pure milk, as result of which water constituted 10% of the resultant mixture. What was the initial quantity of water in the jar?. A) 7 litres B) 8 litres C) 10 litres D) 4.8 litres E) 6.4 litres. 5. In a 200 litres mixture of milk and water, percentage of water is only 30%. The milkman gave 40 litres of this mixture to a customer and then added 12 litres of water in the remaining mixture. What is the respective ratio of milk and water in the new mixture?. A) 11 : 17 B) 7 : 13 C) 23 : 14 D) 21 : 11 E) 28 : 15. 6. A vessel contains coconut water and vodka i.e., 120 litres of the mixture, coconut water is 40% more than the vodka. 72 litres of mixture taken out and 26 litres of coconut water is and some quantity of vodka is also added to the vessel, after that the respective ratio of the resultant mixture of vodka and coconut water becomes 2 : 3, then find the final quantityof the vodka in the vessel? (In litres). A) 54 B) 46 C) 35 D) 34 E) 36. 7. 16 litres of pure water was added to a vessel containing 84 litres of pure milk. 50 litres of the resultant mixture was then sold and some more quantity of pure milk and pure water was added to the vessel in the respective ratio of 3 : 2. If the resultant respective ratio of milk and water in the vessel is 4 : 1, what was the quantity of pure milk added in the vessel? (In litres). A) 4 B) 8 C) 6 D) 10 E) 12. 8. In a mixture of milk and water, water was only 20%. 50 litres of this mixture was taken out and then 5 litres of pure water was added to the mixture. If the resultant ratio of milk and water in the mixture was 8 : 3 respectively, what was the initial quantity of mixture (before the replacement)? (In litres). A) 88 B) 96 C) 120 D) 100 E) 84. 9. In a 120 litres mixture of milk and water, percentage of water is only 40%. The milkman gave 45 litres of this mixture to a customer and then added 15 litres of pure milk and 10 litres of pure water to the remaining mixture. What is the respective ratio of quantity of milk and water in the new mixture?. A) 3 : 2 B) 11 : 5 C) 8 : 5 D) 15 : 8 E) 12 : 7. 10. In 112 litres of mixture of water and milk, water is only 18.75%. The milkman gave 28 litres of this mixture to a customer and then he added 10.75 litres of pure milk and 5.25 litres of pure water in the remaining mixture. What is the percentage of water in the final mixture?. A) 32.50 B) 34 C) 32 D) 21 E) 27. Some more.... 1. Equal quantities of two milk and water solutions which contains milk and water in ratio, 1 : 5 and 3 : 5 are mixed together. What will be the ratio of water to milk in the resultant solution?. a) 13 : 35 b) 4 : 10 c) 5 : 8 d) 35 : 13. Ans: d. 2. The respective ratio of milk and water in 66 litres of adulterated milk is 5 : 1. Water is added into it to make the respective ratio 5 : 3. What is the quantity of water added?. a) 11 litres b) 22 litres c) 33 litres d) 44 litres. Ans: b. 3. A 200 litres solution of alcohol and water contains 1/4 th part alcohol. Find the new percentage of alcohol, if 50 litres of the original solution is replaced by 50 litres of alcohol.. a) 43.75% b) 50% c) 66.66% d) 80%. Ans: a. 4. A container has 80 litres of milk. From this container 8 litres of milk was taken out and replaced with water. The process was further repeated twice. How much milk is there in the container finally? (in liters). a) 58.32 b) 60.32 c) 62 d) 64.7. Ans: a. 5. 100 litres of a mixture contains 10% water and the rest milk. The amount of water that must be added so that the resulting mixture contains only 50% milk is....... a) 70 litres b) 72 litres c) 78 litres d) 80 litres. Ans: d. 6. In a particular type of fertilizer, the ratio of two chemicals A and B is 2 : 5. In 21 kg of this fertilizer, if 3 kg of type - A is added, the ratio of chemical A to B in the new fertilizer will be. a) 1 : 1 b) 2 : 3 c) 3 : 5 d) 4 : 5. Ans: c. 7. Rice worth Rs.120 per kg and Rs.132 per kg are mixed with a third variety in the ratio 2 : 1 : 3. If the mixture is worth Rs.135 per kg, the price of the third variety per kg will be..... a) Rs.140 b) Rs.146 c) Rs.150 d) Rs.148. Ans: b. 8. A jar full of wine contains 50% alcohol. A part of wine is replaced by another containing 20% alcohol and now percentage of alcohol was found to be 30%. The part of wine replaced is?. a) 1/2 b) 1/3 c) 2/5 d) 2/3. Ans: d. 9. A shopkeeper has 50 kg rice, a part of which he sells at 10% profit and rest at 25% profit. He gains 16% on the whole. The quantity sold at 10% profit is?. a) 30 kg b) 20 kg c) 45 kg d) 27 kg. Ans: a. 10. An alloy contains tin, copper and zinc in ratio 2 : 3 : 1, and another alloy contain copper, aluminium and tin in ratio 3 : 2 : 7. If both alloys are mixed in equal quantities, then what will be the ratio of tin and aluminium in final alloy?. a) 2 : 11 b) 11 : 9 c) 9 : 2 d) 11 : 2. Ans: d.
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Posted Date : 17-09-2022. గమనిక : ప్రతిభ.ఈనాడు.నెట్లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.
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https://gmatclub.com/forum/only-senior-citizens-enjoy-doing-the-daily-jumble-so-aesha-must-242955.html?fl=homea
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# Only senior citizens enjoy doing the daily jumble. So Aesha must
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Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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19 Jun 2017, 04:38
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Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.
Which of the following arguments exhibits the same flawed reasoning as the above?
A. Only in March does Rodrigo choose to holiday in Spain. It is March, but Rodrigo is in Japan. So he must not be going to Spain.
B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.
C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.
D. Since the animal in front of us is a penguin, it follows that we are in Antarctica, since one only encounters penguins in the wild when one is in Antarctica.
E. Only the best chefs can make compelling vegan escargot. So Rasheed must be able to make compelling vegan escargot, since he is one of the world’s best chefs.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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19 Jun 2017, 17:48
B.
Some senior citizens might not enjoy jumbles and some pet lovers might not adopt Marley.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 Jun 2017, 23:25
Can someone please explain this question in a little detail? How to relate the two situations ? Also Why is option C wrong?
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 Jun 2017, 23:26
Experts please explain why is option C wrong? And how to go about solving such analogy questions?
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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23 Jun 2017, 00:55
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Shiv2016 wrote:
Experts please explain why is option C wrong? And how to go about solving such analogy questions?
I'm not an expert but I can try and help.
Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.
a implies b ---->not b implies not a
Which of the following arguments exhibits the same flawed reasoning as the above?
B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.a implies b ---->not b implies not a - correct
C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.a implies b ---->not a implies not b - reversed and so incorrect
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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02 Jul 2017, 07:29
this is a mix question. "flaw" here indicates that this is not only logic-error question, but also an application questions.
Only A do B = every B is A
the flaw is not B -> not A
Practice is the best way to cope with different gmat questions.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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04 Jul 2017, 02:16
Flaw is senior citizen - > enjoy jumble doesn't imply ~not citizen-> ~ No jumble enjoyment. (flawed contrapositive) Replicated by B.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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04 Jul 2017, 10:04
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 Jul 2017, 09:26
can someone tell me how to do these kind of questions
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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14 Oct 2017, 15:24
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My 2 cents:
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 May 2018, 04:01
broall wrote:
Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.
Which of the following arguments exhibits the same flawed reasoning as the above?
A. Only in March does Rodrigo choose to holiday in Spain. It is March, but Rodrigo is in Japan. So he must not be going to Spain.
B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.
C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.
D. Since the animal in front of us is a penguin, it follows that we are in Antarctica, since one only encounters penguins in the wild when one is in Antarctica.
E. Only the best chefs can make compelling vegan escargot. So Rasheed must be able to make compelling vegan escargot, since he is one of the world’s best chefs.
VERITAS PREP OFFICIAL SOLUTION:
Solution: B
This Mimic the Reasoning question features difficult conditional wording. “Only x does y” can be reworded as “If you do y, then you are x”, a change of phrase that makes the question much easier to solve. “If you enjoy the jumble, you’re a senior citizen” implies two things: 1) if you enjoy the jumble, you’re a senior citizen, and 2) if you aren’t a senior citizen, you don’t enjoy the jumble. It does NOT imply that if you don’t enjoy the jumble, you’re not a senior citizen, so find an answer choice that also employs such faulty logic. (B) is the answer of choice.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] 22 May 2018, 04:01
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GMAT Changed on April 16th - Read about the latest changes here. It is currently 22 May 2018, 23:56. ### GMAT Club Daily Prep. #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. # Events & Promotions. ###### Events & Promotions in June. Open Detailed Calendar. # Only senior citizens enjoy doing the daily jumble. So Aesha must. new topic post reply Question banks Downloads My Bookmarks Reviews Important topics. Author Message. TAGS:. ### Hide Tags. Senior CR Moderator. Status: Long way to go!. Joined: 10 Oct 2016. Posts: 1379. Location: Viet Nam. Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]. ### Show Tags. 19 Jun 2017, 04:38. 2. KUDOS. 8. This post was. BOOKMARKED. 00:00. Difficulty:. 35% (medium). Question Stats:. 70% (01:17) correct 30% (01:17) wrong based on 335 sessions. ### HideShow timer Statistics. Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.. Which of the following arguments exhibits the same flawed reasoning as the above?. A. Only in March does Rodrigo choose to holiday in Spain. It is March, but Rodrigo is in Japan. So he must not be going to Spain.. B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.. C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.. D. Since the animal in front of us is a penguin, it follows that we are in Antarctica, since one only encounters penguins in the wild when one is in Antarctica.. E. Only the best chefs can make compelling vegan escargot. So Rasheed must be able to make compelling vegan escargot, since he is one of the world’s best chefs.. _________________. Intern. Joined: 16 Jun 2017. Posts: 15. Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]. ### Show Tags. 19 Jun 2017, 17:48. B.. Some senior citizens might not enjoy jumbles and some pet lovers might not adopt Marley.. Director. Joined: 02 Sep 2016. Posts: 745. Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]. ### Show Tags. 22 Jun 2017, 23:25. Can someone please explain this question in a little detail? How to relate the two situations ? Also Why is option C wrong?. _________________. Help me make my explanation better by providing a logical feedback.. If you liked the post, HIT KUDOS !!. Don't quit.............Do it.. Director. Joined: 02 Sep 2016. Posts: 745. Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]. ### Show Tags. 22 Jun 2017, 23:26. Experts please explain why is option C wrong? And how to go about solving such analogy questions?. _________________. Help me make my explanation better by providing a logical feedback.. If you liked the post, HIT KUDOS !!. Don't quit.............Do it.. Intern. Joined: 16 Jun 2017. Posts: 15. Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]. ### Show Tags. 23 Jun 2017, 00:55. 1. KUDOS. Shiv2016 wrote:. Experts please explain why is option C wrong? And how to go about solving such analogy questions?. I'm not an expert but I can try and help.. Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.. a implies b ---->not b implies not a. Which of the following arguments exhibits the same flawed reasoning as the above?. B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.a implies b ---->not b implies not a - correct. C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.a implies b ---->not a implies not b - reversed and so incorrect. SVP. Joined: 12 Dec 2016. Posts: 1904. Location: United States. GMAT 1: 700 Q49 V33. GPA: 3.64. Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]. ### Show Tags. 02 Jul 2017, 07:29. this is a mix question. "flaw" here indicates that this is not only logic-error question, but also an application questions.. Only A do B = every B is A. the flaw is not B -> not A. Practice is the best way to cope with different gmat questions.. Intern. Joined: 20 Jun 2017. Posts: 11. Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]. ### Show Tags. 04 Jul 2017, 02:16. Flaw is senior citizen - > enjoy jumble doesn't imply ~not citizen-> ~ No jumble enjoyment. (flawed contrapositive) Replicated by B.. Sent from my ONE A2003 using GMAT Club Forum mobile app. Intern. Joined: 23 Dec 2013. Posts: 14. Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]. ### Show Tags. 04 Jul 2017, 10:04. Sent from my Moto G (5) Plus using GMAT Club Forum mobile app. Manager. Joined: 08 Jun 2017. Posts: 65. Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]. ### Show Tags. 22 Jul 2017, 09:26. can someone tell me how to do these kind of questions. Manager. Joined: 14 Oct 2012. Posts: 176. Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]. ### Show Tags. 14 Oct 2017, 15:24. 1. KUDOS. My 2 cents:. Attachments. my 2 cents.jpg [ 610.67 KiB | Viewed 933 times ]. Math Expert. Joined: 02 Sep 2009. Posts: 45256. Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]. ### Show Tags. 22 May 2018, 04:01. broall wrote:. Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.. Which of the following arguments exhibits the same flawed reasoning as the above?. A. Only in March does Rodrigo choose to holiday in Spain. It is March, but Rodrigo is in Japan. So he must not be going to Spain.. B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.. C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.. D. Since the animal in front of us is a penguin, it follows that we are in Antarctica, since one only encounters penguins in the wild when one is in Antarctica.. E. Only the best chefs can make compelling vegan escargot. So Rasheed must be able to make compelling vegan escargot, since he is one of the world’s best chefs.. VERITAS PREP OFFICIAL SOLUTION:. Solution: B. This Mimic the Reasoning question features difficult conditional wording. “Only x does y” can be reworded as “If you do y, then you are x”, a change of phrase that makes the question much easier to solve. “If you enjoy the jumble, you’re a senior citizen” implies two things: 1) if you enjoy the jumble, you’re a senior citizen, and 2) if you aren’t a senior citizen, you don’t enjoy the jumble. It does NOT imply that if you don’t enjoy the jumble, you’re not a senior citizen, so find an answer choice that also employs such faulty logic. (B) is the answer of choice.. _________________. Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] 22 May 2018, 04:01. Display posts from previous: Sort by. # Only senior citizens enjoy doing the daily jumble. So Aesha must.
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Order of an element
WARNING: POTENTIAL TERMINOLOGICAL CONFUSION: Please don't confuse this with order of a group
Definition
The order of an element $x$ in a group $G$ is the smallest positive integer $n$ for which $x^n$ is the identity element.
Such a $n$ may not always exist (if it exists, $x$ is said to be of finite order, or is termed a torsion element). It does exist when the group is finite.
Examples
• The identity element has order $1$ in any group
• In the group of integers modulo $n$, the element $1$ has order $n$
Facts
For an element of finite order, the order of the element equals the order of the cyclic subgroup generated by the element. Thus, by Lagrange's theorem, the order of an element $x$ in a finite group $G$ divides the order of $G$ (where order here means the total cardinality of the group).
The exponent of a group is defined as the least common multiple of the orders of all elements of the group. For a finite group, the exponent always exists, and is a divisor of the order of the group (though it may, in general, be smaller). There may or may not exist an element in the group whose order equals the exponent of the group.
For an infinite group, not every element necessarily has finite order. A group where every element has finite order is termed a periodic group. Even for a periodic group, the exponent may be infinite because there may not be a common bound on the orders of all elements. A group with bounded exponent is a group whose exponent is finite, the condition of having bounded exponent is stronger than the condition of being periodic.
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Order of an element. WARNING: POTENTIAL TERMINOLOGICAL CONFUSION: Please don't confuse this with order of a group. Definition. The order of an element $x$ in a group $G$ is the smallest positive integer $n$ for which $x^n$ is the identity element.. Such a $n$ may not always exist (if it exists, $x$ is said to be of finite order, or is termed a torsion element). It does exist when the group is finite.. Examples. • The identity element has order $1$ in any group. • In the group of integers modulo $n$, the element $1$ has order $n$. Facts. For an element of finite order, the order of the element equals the order of the cyclic subgroup generated by the element. Thus, by Lagrange's theorem, the order of an element $x$ in a finite group $G$ divides the order of $G$ (where order here means the total cardinality of the group).. The exponent of a group is defined as the least common multiple of the orders of all elements of the group. For a finite group, the exponent always exists, and is a divisor of the order of the group (though it may, in general, be smaller). There may or may not exist an element in the group whose order equals the exponent of the group.. For an infinite group, not every element necessarily has finite order. A group where every element has finite order is termed a periodic group.
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Even for a periodic group, the exponent may be infinite because there may not be a common bound on the orders of all elements. A group with bounded exponent is a group whose exponent is finite, the condition of having bounded exponent is stronger than the condition of being periodic.
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# Common excel math functions
Note that further math-related Excel functions are also provided in the Excel Statistical Functions and Excel Engineering Functions categories. The tables below list all the current built-in Excel math functions, grouped by category, to help you to find the function you need. Selecting a function link will take you to a full description of the function, with examples of use and common errors.
Note that some of the Excel math functions listed below were introduced in recent versions of Excel, and so are not available in earlier versions. Excel Functions.
Basic Numeric Information.
Courseworks completed paper date
Performs a specified calculation e. Rounding Functions. Rounds a number away from zero i. Rounds a number upregardless of the sign of the number, to a multiple of significance New in Excel Rounds a number upregardless of the sign of the number, to a multiple of significance.
New in Excel Rounds a number up to the nearest integer or to the nearest multiple of significance New in Excel Rounds a number towards zeroi. Rounds a number downregardless of the sign of the number, to a multiple of significance New in Excel Rounds a number down, to the nearest integer or to the nearest multiple of significance New in Excel Rounds a number up or downto the nearest multiple of significance. Truncates a number towards zero i. Returns the unit matrix for a specified dimension New in Excel Conditional Sums.
Adds the cells in a supplied range, that satisfy multiple criteria New in Excel Returns the sum of the products of corresponding values in two or more supplied arrays. Returns the sum of the difference of squares of corresponding values in two supplied arrays.
Returns the sum of the sum of squares of corresponding values in two supplied arrays. Returns the sum of squares of differences of corresponding values in two supplied arrays.
Returns the secant of an angle New in Excel Returns the hyperbolic secant of an angle New in Excel Returns the cosecant of an angle New in Excel Cheat Sheet of Excel formulas and function is always a customized worksheet where we can have all those function details, shortcut keys to execute any function or formulas, custom way to use 2 or more function together and guideline to use them.
Also, we can choose those formulas which are complicated to apply for users in a cheat sheet. Start Your Free Excel Course. As we can see, here several string functions are listed. Please refer the below screenshot where I have applied the functions on strings and shown their workings.
Please refer the below screenshot where I have applied the functions on number values and shown their workings. Please refer the below screenshot where I have applied the functions on values and shown their workings.
This has been a guide to Excel Formulas Cheat Sheet. You can also go through our other suggested articles —. This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy.
By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy. Forgot Password? Call Our Course Advisors. Cheat Sheet of Excel Formulas. Popular Course in this category. Course Price View Course. Free Excel Course. Login details for this Free course will be emailed to you. Email ID. Contact No.So, the total production quantity is Similarly, apply the same logic to get the total salary amount.
Now we know what overall sum values are. Out of these overall total of employees, we need to find the average salary per employee. Select the range of cells for which we are finding the average value, so our range of cells will be from D2 to D We know the average salary per person; for further drill-down, we want to know what is the average salary based on gender.
Marketing plan bibliography design dates
So totally there are 10 employees on the list. After counting the total number of employees, we may need to count how many male and female employees are there.
MOD function will return the remainder when one number is divided by another number. For example, when you divide number 11 by 2, we will get the remainder as 1 because only till 10 number 2 can divide.
When we have fraction or decimal values, we may need to round those decimal values to the nearest integer number.
For example, we need to round the number 3. As you can see above, B2 cell value Like this, we can use various mathematical functions in excel to do mathematical operations in excel quickly and easily. This has been a guide to Mathematical Function in Excel. Here we discuss how to calculate mathematical function in excel using Sum, Average, Averageif, Counta, Countif, Mod, and Round formulas along with practical examples.
You may learn more about excel from the following articles —. Free Excel Course. Login details for this Free course will be emailed to you. This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy.
By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy.The tutorial provides a list of Excel basic formulas and functions with examples and links to related in-depth tutorials.
Being primarily designed as a spreadsheet program, Microsoft Excel is extremely powerful and versatile when it comes to calculating numbers or solving math and engineering problems. It enables you to total or average a column of numbers in the blink of an eye.
Apart from that, you can compute a compound interest and weighted average, get the optimal budget for your advertising campaign, minimize the shipment costs or make the optimal work schedule for your employees. All this is done by entering formulas in cells. This tutorial aims to teach you the essentials of Excel functions and show how to use basic formulas in Excel. Before providing the basic Excel formulas list, let's define the key terms just to make sure we are on the same page.
So, what do we call an Excel formula and Excel function? Function is a predefined formula already available in Excel. Functions perform specific calculations in a particular order based on the specified values, called arguments, or parameters. You can find all available Excel functions in the Function Library on the Formulas tab:. Of course, it's next to impossible to memorize all of them, and you actually don't need to.
The Function Wizard will help you find the function best suited for a particular task, while the Excel Formula Intellisense will prompt the function's syntax and arguments as soon as you type the function's name preceded by an equal sign in a cell:.
Clicking the function's name will turn it into a blue hyperlink, which will open the Help topic for that function.
What follows below is a list of 10 simple yet really helpful functions that are a necessary skill for everyone who wishes to turn from an Excel novice to an Excel professional. The first Excel function you should be familiar with is the one that performs the basic arithmetic operation of addition:. In the syntax of all Excel functions, an argument enclosed in [square brackets] is optional, other arguments are required.
Meaning, your Sum formula should include at least 1 number, reference to a cell or a range of cells. For example:. If necessary, you can perform other calculations within a single formula, for example, add up values in cells B2 through B6, and then divide the sum by To sum with conditions, use the SUMIF function: in the 1st argument, you enter the range of cells to be tested against the criteria A2:A6in the 2nd argument - the criteria itself D2and in the last argument - the cells to sum B2:B6 :.
In your Excel worksheets, the formulas may look something similar to this:.
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Its syntax is similar to SUM's:. Sums values in cells B2 through B6, and then divides the result by 5.
## Mathematical Function in Excel
And what do you call adding up a group of numbers and then dividing the sum by the count of those numbers? Yep, an average!The basic functions covered below are among the most popular formulas in Excel—the ones everyone should know. To help you learn, we've also provided a spreadsheet with all the formula examples we cover below. After you enter one of these functions in A1, you can then reformat the Date and Time or use the system default.
It also subtracts, multiplies, divides, and uses any of the comparison operators to return a result of 1 true or 0 false. Excel frames the column of numbers in green borders and displays the formula in the current cell. Remember your high school math?
If the numbers inside the formula are not grouped properly, the answer will be wrong. Notice the screenshot below figure 2. For this exercise, you can enter the same values in H, I, and J, with or without the blank rows in between again, added for easier viewing.
Note that as we build each formula, we are combining the steps, eventually, into a single formula.
Athletics activities department
We start out with three separate formulas. The formula in K3 is wrong. It requires grouping the numbers according to the order of calculation using commas or parentheses.
Check your numbers again with your calculator and note that this formula is correct. Note that the syntax the structure or layout of the formula is correct in cells N7 and N8, but incorrect in N6.
By combining these formulas into one, you can eliminate columns K and L. The RAND function is really simple and traditionally used for statistical analysis, cryptography, gaming, gambling, and probability theory, among dozens of other things.
Note; however, that every time you enter new data and press the Enter key, the list of random numbers you just created changes.
If you need to maintain your random numbers lists, you must format the cells as values. Now the list contains values instead of functions, so it will not change. Notice in the formula bar that the random numbers have 15 digits after the decimal Excel defaults to 9which you can change, if necessary as displayed in cell F3.
Just click the Increase Decimal button in the Number group under the Home tab. Again, you must copy the list and Paste as Values to maintain a static list. McGregor needs to order for his shop.Excel is a great way to organize and keep track of your data. There are more than functions in Excel. If you would like to know more about a function, simply follow the links we added for each of them. Got a different version?
No problem, you can still follow the exact same steps.
### Excel Formula Symbols Cheat Sheet (13 Cool Tips)
By definition, a function is a predefined formula in Excel which does calculations in the order specified by its parameters. Because learning how to add numbers in Excel is one of the most fundamental skills you need to learn. As you know, addition is an integral part of almost any calculation and task in Excel. As their name implies, they add the values in a specified range only when the criteria are met. The differences between the two are in the number of criteria you can specify.
These functions are useful when dealing with large data sets and manual calculations are inefficient and impractical. This function calculates the arithmetic mean of a set of numbers or the sum of the values divided by the number of values. People use the averages every day, from school grades to statistics.
These functions shine when you need to get averages from specifics sets in a range. It may seem rather rudimentary. But in actuality, this function is used in a lot of computations and scenarios. This function is used in many things like counting how many items there are in a listcounting specific casesand others.
But what if you only need to count a specific subset of cells? These functions are useful in tasks like project managementsales inventoryorder fulfillmentand others. This function returns the sum of the product of two or more arrays.
This is an important Excel function since this is used to calculate weighted averages as well as simplify a lot of tasks like sales inventory. Ever had the need to come up with random values between a specified minimum and maximum values?Use this handy Cheat Sheet to discover great functions and tips to help you get the most out of Excel.
Some Excel functions apply to specific subject areas, but others are general and apply to all needs. The following list shows an array of Excel functions used by one and all. Check here for a quickie reference to the purpose of each Excel function. Here is list of Excel functions associated with text, along with a description of what each function does:.
Mathematics dictates a protocol of how formulas are interpreted, and Excel follows that protocol. The following is the order in which mathematical operators and syntax are applied both in Excel and in general mathematics. In Excel formulas, you can refer to other cells either relatively or absolutely. When you copy and paste a formula in Excel, how you create the references within the formula tells Excel what to change in the formula it pastes. You can also mix relative and absolute references so that, when you move or copy a formula, the row changes but the column does not, or vice versa.
If you create a formula in Excel that contains an error or circular reference, Excel lets you know about it with an error message. A handful of errors can appear in a cell when a formula or function in Excel cannot be resolved. Knowing their meaning helps correct the problem. He has written numerous articles and books on a variety of technical topics.
Microsoft Excel Tutorial - Beginners Level 1
His latest projects include large-scale cloud-based applications and mobile app development. Cheat Sheet. Excel Order of Operations to Keep in Mind Mathematics dictates a protocol of how formulas are interpreted, and Excel follows that protocol. Excel Cell References Worth Remembering In Excel formulas, you can refer to other cells either relatively or absolutely.
Excel Error Messages to Get to Know If you create a formula in Excel that contains an error or circular reference, Excel lets you know about it with an error message. A formula or a function inside a formula cannot find the referenced data NAME? Text in the formula is not recognized NULL!
A space was used in formulas that reference multiple ranges; a comma separates range references NUM! A formula has invalid numeric data for the type of operation REF! The wrong type of operand or function argument is used.
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# Common excel math functions. Note that further math-related Excel functions are also provided in the Excel Statistical Functions and Excel Engineering Functions categories. The tables below list all the current built-in Excel math functions, grouped by category, to help you to find the function you need. Selecting a function link will take you to a full description of the function, with examples of use and common errors.. Note that some of the Excel math functions listed below were introduced in recent versions of Excel, and so are not available in earlier versions. Excel Functions.. Basic Numeric Information.. Courseworks completed paper date. Performs a specified calculation e. Rounding Functions. Rounds a number away from zero i. Rounds a number upregardless of the sign of the number, to a multiple of significance New in Excel Rounds a number upregardless of the sign of the number, to a multiple of significance.. New in Excel Rounds a number up to the nearest integer or to the nearest multiple of significance New in Excel Rounds a number towards zeroi. Rounds a number downregardless of the sign of the number, to a multiple of significance New in Excel Rounds a number down, to the nearest integer or to the nearest multiple of significance New in Excel Rounds a number up or downto the nearest multiple of significance. Truncates a number towards zero i. Returns the unit matrix for a specified dimension New in Excel Conditional Sums.. Adds the cells in a supplied range, that satisfy multiple criteria New in Excel Returns the sum of the products of corresponding values in two or more supplied arrays. Returns the sum of the difference of squares of corresponding values in two supplied arrays.. Returns the sum of the sum of squares of corresponding values in two supplied arrays. Returns the sum of squares of differences of corresponding values in two supplied arrays.. Returns the secant of an angle New in Excel Returns the hyperbolic secant of an angle New in Excel Returns the cosecant of an angle New in Excel Cheat Sheet of Excel formulas and function is always a customized worksheet where we can have all those function details, shortcut keys to execute any function or formulas, custom way to use 2 or more function together and guideline to use them.. Also, we can choose those formulas which are complicated to apply for users in a cheat sheet. Start Your Free Excel Course. As we can see, here several string functions are listed. Please refer the below screenshot where I have applied the functions on strings and shown their workings.. Please refer the below screenshot where I have applied the functions on number values and shown their workings. Please refer the below screenshot where I have applied the functions on values and shown their workings.. This has been a guide to Excel Formulas Cheat Sheet. You can also go through our other suggested articles —. This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy.. By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy. Forgot Password? Call Our Course Advisors. Cheat Sheet of Excel Formulas. Popular Course in this category. Course Price View Course. Free Excel Course. Login details for this Free course will be emailed to you. Email ID. Contact No.So, the total production quantity is Similarly, apply the same logic to get the total salary amount.. Now we know what overall sum values are. Out of these overall total of employees, we need to find the average salary per employee. Select the range of cells for which we are finding the average value, so our range of cells will be from D2 to D We know the average salary per person; for further drill-down, we want to know what is the average salary based on gender.. Marketing plan bibliography design dates. So totally there are 10 employees on the list. After counting the total number of employees, we may need to count how many male and female employees are there.. MOD function will return the remainder when one number is divided by another number. For example, when you divide number 11 by 2, we will get the remainder as 1 because only till 10 number 2 can divide.. When we have fraction or decimal values, we may need to round those decimal values to the nearest integer number.. For example, we need to round the number 3. As you can see above, B2 cell value Like this, we can use various mathematical functions in excel to do mathematical operations in excel quickly and easily. This has been a guide to Mathematical Function in Excel. Here we discuss how to calculate mathematical function in excel using Sum, Average, Averageif, Counta, Countif, Mod, and Round formulas along with practical examples.. You may learn more about excel from the following articles —. Free Excel Course. Login details for this Free course will be emailed to you. This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy.. By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy.The tutorial provides a list of Excel basic formulas and functions with examples and links to related in-depth tutorials.. Being primarily designed as a spreadsheet program, Microsoft Excel is extremely powerful and versatile when it comes to calculating numbers or solving math and engineering problems. It enables you to total or average a column of numbers in the blink of an eye.. Apart from that, you can compute a compound interest and weighted average, get the optimal budget for your advertising campaign, minimize the shipment costs or make the optimal work schedule for your employees. All this is done by entering formulas in cells. This tutorial aims to teach you the essentials of Excel functions and show how to use basic formulas in Excel. Before providing the basic Excel formulas list, let's define the key terms just to make sure we are on the same page.. So, what do we call an Excel formula and Excel function? Function is a predefined formula already available in Excel. Functions perform specific calculations in a particular order based on the specified values, called arguments, or parameters. You can find all available Excel functions in the Function Library on the Formulas tab:. Of course, it's next to impossible to memorize all of them, and you actually don't need to.. The Function Wizard will help you find the function best suited for a particular task, while the Excel Formula Intellisense will prompt the function's syntax and arguments as soon as you type the function's name preceded by an equal sign in a cell:.. Clicking the function's name will turn it into a blue hyperlink, which will open the Help topic for that function.. What follows below is a list of 10 simple yet really helpful functions that are a necessary skill for everyone who wishes to turn from an Excel novice to an Excel professional. The first Excel function you should be familiar with is the one that performs the basic arithmetic operation of addition:. In the syntax of all Excel functions, an argument enclosed in [square brackets] is optional, other arguments are required.. Meaning, your Sum formula should include at least 1 number, reference to a cell or a range of cells. For example:. If necessary, you can perform other calculations within a single formula, for example, add up values in cells B2 through B6, and then divide the sum by To sum with conditions, use the SUMIF function: in the 1st argument, you enter the range of cells to be tested against the criteria A2:A6in the 2nd argument - the criteria itself D2and in the last argument - the cells to sum B2:B6 :.. In your Excel worksheets, the formulas may look something similar to this:.. Popular book review editing websites for college. Its syntax is similar to SUM's:. Sums values in cells B2 through B6, and then divides the result by 5.. ## Mathematical Function in Excel. And what do you call adding up a group of numbers and then dividing the sum by the count of those numbers? Yep, an average!The basic functions covered below are among the most popular formulas in Excel—the ones everyone should know. To help you learn, we've also provided a spreadsheet with all the formula examples we cover below. After you enter one of these functions in A1, you can then reformat the Date and Time or use the system default.. It also subtracts, multiplies, divides, and uses any of the comparison operators to return a result of 1 true or 0 false. Excel frames the column of numbers in green borders and displays the formula in the current cell. Remember your high school math?. If the numbers inside the formula are not grouped properly, the answer will be wrong. Notice the screenshot below figure 2. For this exercise, you can enter the same values in H, I, and J, with or without the blank rows in between again, added for easier viewing.. Note that as we build each formula, we are combining the steps, eventually, into a single formula.. Athletics activities department. We start out with three separate formulas. The formula in K3 is wrong. It requires grouping the numbers according to the order of calculation using commas or parentheses.. Check your numbers again with your calculator and note that this formula is correct. Note that the syntax the structure or layout of the formula is correct in cells N7 and N8, but incorrect in N6.. By combining these formulas into one, you can eliminate columns K and L. The RAND function is really simple and traditionally used for statistical analysis, cryptography, gaming, gambling, and probability theory, among dozens of other things.. Note; however, that every time you enter new data and press the Enter key, the list of random numbers you just created changes.. If you need to maintain your random numbers lists, you must format the cells as values. Now the list contains values instead of functions, so it will not change. Notice in the formula bar that the random numbers have 15 digits after the decimal Excel defaults to 9which you can change, if necessary as displayed in cell F3.. Just click the Increase Decimal button in the Number group under the Home tab. Again, you must copy the list and Paste as Values to maintain a static list. McGregor needs to order for his shop.Excel is a great way to organize and keep track of your data. There are more than functions in Excel. If you would like to know more about a function, simply follow the links we added for each of them. Got a different version?. No problem, you can still follow the exact same steps.. ### Excel Formula Symbols Cheat Sheet (13 Cool Tips). By definition, a function is a predefined formula in Excel which does calculations in the order specified by its parameters. Because learning how to add numbers in Excel is one of the most fundamental skills you need to learn. As you know, addition is an integral part of almost any calculation and task in Excel. As their name implies, they add the values in a specified range only when the criteria are met. The differences between the two are in the number of criteria you can specify.. These functions are useful when dealing with large data sets and manual calculations are inefficient and impractical. This function calculates the arithmetic mean of a set of numbers or the sum of the values divided by the number of values. People use the averages every day, from school grades to statistics.. These functions shine when you need to get averages from specifics sets in a range. It may seem rather rudimentary. But in actuality, this function is used in a lot of computations and scenarios. This function is used in many things like counting how many items there are in a listcounting specific casesand others.. But what if you only need to count a specific subset of cells? These functions are useful in tasks like project managementsales inventoryorder fulfillmentand others. This function returns the sum of the product of two or more arrays.. This is an important Excel function since this is used to calculate weighted averages as well as simplify a lot of tasks like sales inventory. Ever had the need to come up with random values between a specified minimum and maximum values?Use this handy Cheat Sheet to discover great functions and tips to help you get the most out of Excel.. Some Excel functions apply to specific subject areas, but others are general and apply to all needs. The following list shows an array of Excel functions used by one and all. Check here for a quickie reference to the purpose of each Excel function. Here is list of Excel functions associated with text, along with a description of what each function does:.. Mathematics dictates a protocol of how formulas are interpreted, and Excel follows that protocol. The following is the order in which mathematical operators and syntax are applied both in Excel and in general mathematics. In Excel formulas, you can refer to other cells either relatively or absolutely. When you copy and paste a formula in Excel, how you create the references within the formula tells Excel what to change in the formula it pastes. You can also mix relative and absolute references so that, when you move or copy a formula, the row changes but the column does not, or vice versa.. If you create a formula in Excel that contains an error or circular reference, Excel lets you know about it with an error message. A handful of errors can appear in a cell when a formula or function in Excel cannot be resolved. Knowing their meaning helps correct the problem. He has written numerous articles and books on a variety of technical topics.. Microsoft Excel Tutorial - Beginners Level 1. His latest projects include large-scale cloud-based applications and mobile app development. Cheat Sheet. Excel Order of Operations to Keep in Mind Mathematics dictates a protocol of how formulas are interpreted, and Excel follows that protocol. Excel Cell References Worth Remembering In Excel formulas, you can refer to other cells either relatively or absolutely.. Excel Error Messages to Get to Know If you create a formula in Excel that contains an error or circular reference, Excel lets you know about it with an error message.
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A formula or a function inside a formula cannot find the referenced data NAME? Text in the formula is not recognized NULL!. A space was used in formulas that reference multiple ranges; a comma separates range references NUM! A formula has invalid numeric data for the type of operation REF! The wrong type of operand or function argument is used.
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Key Steps to Keeping the Books Defining Asset Depreciation for Businesses How to Record Accrued Payroll and Taxes
# How to Prepare Part 2 of a Cost of Production Report
The second part of the cost of production report accounts for the units that the first part indicates the department is responsible for. During April, the Balloon department finished working on 1,900 clowns, sending them to the next department. At the end of April, the Balloon folks still had 800 clowns of WIP on the assembly line, waiting for their balloons.
This progress means that the Balloon department has accounted for 2,700 units:
Units accounted for = Units completed + Ending work-in-process units = 1,900 + 800 = 2,700
The number of units to account for must equal the number of units accounted for. This is a factory, not a circus.
To measure how much work each WIP department did, accountants use the term equivalent units. Equivalent units sum up the fractions of units actually completed. For example, if a department got halfway through two separate units, it would count that work as one equivalent unit.
What’s the difference? Suppose an assembly line has 24 toys that are 50 percent complete. However, 24 toys that are only 50 percent complete comprise 12 equivalent units because that measure represents the amount of work that each WIP department actually did. They got 50 percent of the way through 24 units, which comes out to 12 equivalent units.
To compute the number of equivalent units of production, just add the total number of units completed and transferred out to the number of equivalent units of ending WIP:
Equivalent units of production = Units completed + (Units in ending work-in-process x Percentage complete)
For example, suppose that the Theresa Toy Factory Assembly department made and finished 100 rubber ducks. It also had 30 units still in process left over at the end of the period, all half complete. These unfinished ducks equal 15 equivalent units (30 units x 0.5 complete). The total equivalent units of production equal 115:
Equivalent units of production = Units completed + (Units in ending work-in-process x Percentage complete)
= 100 + (30 x 50%)
= 115
You can also apply the equivalent units concept to direct labor and overhead costs. These costs are sometimes referred to as conversion costs because they’re the costs of converting direct materials into finished goods.
The degree of completion with respect to direct materials may differ from conversion costs because direct materials usually get added at the beginning of the production process, whereas conversion costs get added throughout the production process.
For example, maybe your inventory is 100 percent complete with respect to direct materials (because all the materials have been added) but only, say, 10 percent complete with respect to conversion costs. Therefore, the same inventory may actually have two different equivalent units of production: one for direct materials and another for conversion costs.
Now think about another example. The Turbo Paper Clip Company’s Bending department bends paper clips into their proper shape. The Bending department transferred 100,000 completely bent paper clips into the Boxing department. At the end of the period, Bending had 40,000 units of WIP on hand that were 100 percent complete with respect to direct materials, but only 25 percent complete with respect to conversion costs.
To compute the equivalent units produced by the Bending department, apply the same formula as you did for the ducks, but do so twice: once for direct materials and a second time for conversion costs.
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Key Steps to Keeping the Books Defining Asset Depreciation for Businesses How to Record Accrued Payroll and Taxes. # How to Prepare Part 2 of a Cost of Production Report. The second part of the cost of production report accounts for the units that the first part indicates the department is responsible for. During April, the Balloon department finished working on 1,900 clowns, sending them to the next department. At the end of April, the Balloon folks still had 800 clowns of WIP on the assembly line, waiting for their balloons.. This progress means that the Balloon department has accounted for 2,700 units:. Units accounted for = Units completed + Ending work-in-process units = 1,900 + 800 = 2,700. The number of units to account for must equal the number of units accounted for. This is a factory, not a circus.. To measure how much work each WIP department did, accountants use the term equivalent units. Equivalent units sum up the fractions of units actually completed. For example, if a department got halfway through two separate units, it would count that work as one equivalent unit.. What’s the difference? Suppose an assembly line has 24 toys that are 50 percent complete. However, 24 toys that are only 50 percent complete comprise 12 equivalent units because that measure represents the amount of work that each WIP department actually did. They got 50 percent of the way through 24 units, which comes out to 12 equivalent units.. To compute the number of equivalent units of production, just add the total number of units completed and transferred out to the number of equivalent units of ending WIP:. Equivalent units of production = Units completed + (Units in ending work-in-process x Percentage complete). For example, suppose that the Theresa Toy Factory Assembly department made and finished 100 rubber ducks. It also had 30 units still in process left over at the end of the period, all half complete. These unfinished ducks equal 15 equivalent units (30 units x 0.5 complete). The total equivalent units of production equal 115:. Equivalent units of production = Units completed + (Units in ending work-in-process x Percentage complete). = 100 + (30 x 50%). = 115. You can also apply the equivalent units concept to direct labor and overhead costs. These costs are sometimes referred to as conversion costs because they’re the costs of converting direct materials into finished goods.. The degree of completion with respect to direct materials may differ from conversion costs because direct materials usually get added at the beginning of the production process, whereas conversion costs get added throughout the production process.. For example, maybe your inventory is 100 percent complete with respect to direct materials (because all the materials have been added) but only, say, 10 percent complete with respect to conversion costs. Therefore, the same inventory may actually have two different equivalent units of production: one for direct materials and another for conversion costs.. Now think about another example. The Turbo Paper Clip Company’s Bending department bends paper clips into their proper shape. The Bending department transferred 100,000 completely bent paper clips into the Boxing department.
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At the end of the period, Bending had 40,000 units of WIP on hand that were 100 percent complete with respect to direct materials, but only 25 percent complete with respect to conversion costs.. To compute the equivalent units produced by the Bending department, apply the same formula as you did for the ducks, but do so twice: once for direct materials and a second time for conversion costs.
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# Solid Geometry, Volumes 6-9
### Contents
GENERAL AXIOMS 241 LINES AND PLANES IN SPACE 251 POLYHEDRONS CYLINDERS AND CONES 289
THE SPHERE 360 CONIC SECTIONS 409 TABLE OF FORMULAS 460
### Popular passages
Page 274 - If two planes are perpendicular to each other, a straight line drawn in one of them, perpendicular to their intersection, is perpendicular to the other.
Page 360 - A sphere is a solid bounded by a surface all points of which are equally distant from a point within called the centre.
Page 383 - Hence the two last are right angles ; hence the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is at right angles to the base, and bisects the vertical angle.
Page 250 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. To prove that Proof. A Let the triangles ABC and ADE have the common angle A. A ABC -AB X AC Now and A ADE AD X AE Draw BE.
Page 285 - The sum of the face angles of any convex polyhedral angle is less than four right angles.
Page 246 - PERIPHERY of a circle is its entire bounding line ; or it is a curved line, all points of which are equally distant from a point within called the centre.
Page 245 - The straight line joining the middle points of two sides of a triangle is parallel to the third side and equal to half of it 46 INTERCEPTS BY PARALLEL LINES.
Page 254 - If a straight line is perpendicular to each of two other straight lines at their point of intersection, it is perpendicular to the plane of the two lines.
Page 295 - An oblique prism is equivalent to a right prism whose base is equal to a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. r Let FI be a right section of the oblique prism AD', and PI" a right prism whose lateral edges are equal to the lateral edges of AD'.
Page 370 - A spherical angle is measured by the arc of the great circle described from its vertex as a pole and included between its sides (produced if necessary). Let AB, AC be arcs of great circles intersecting at A; AB...
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# Solid Geometry, Volumes 6-9. ### Contents. GENERAL AXIOMS 241 LINES AND PLANES IN SPACE 251 POLYHEDRONS CYLINDERS AND CONES 289. THE SPHERE 360 CONIC SECTIONS 409 TABLE OF FORMULAS 460. ### Popular passages. Page 274 - If two planes are perpendicular to each other, a straight line drawn in one of them, perpendicular to their intersection, is perpendicular to the other.. Page 360 - A sphere is a solid bounded by a surface all points of which are equally distant from a point within called the centre.. Page 383 - Hence the two last are right angles ; hence the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is at right angles to the base, and bisects the vertical angle.. Page 250 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. To prove that Proof. A Let the triangles ABC and ADE have the common angle A. A ABC -AB X AC Now and A ADE AD X AE Draw BE.. Page 285 - The sum of the face angles of any convex polyhedral angle is less than four right angles.. Page 246 - PERIPHERY of a circle is its entire bounding line ; or it is a curved line, all points of which are equally distant from a point within called the centre.. Page 245 - The straight line joining the middle points of two sides of a triangle is parallel to the third side and equal to half of it 46 INTERCEPTS BY PARALLEL LINES.. Page 254 - If a straight line is perpendicular to each of two other straight lines at their point of intersection, it is perpendicular to the plane of the two lines.. Page 295 - An oblique prism is equivalent to a right prism whose base is equal to a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism.
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r Let FI be a right section of the oblique prism AD', and PI" a right prism whose lateral edges are equal to the lateral edges of AD'.. Page 370 - A spherical angle is measured by the arc of the great circle described from its vertex as a pole and included between its sides (produced if necessary). Let AB, AC be arcs of great circles intersecting at A; AB...
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# What is System of ode: Definition and 14 Discussions
In mathematics, an ordinary differential equation (ODE) is a differential equation containing one or more functions of one independent variable and the derivatives of those functions. The term ordinary is used in contrast with the term partial differential equation which may be with respect to more than one independent variable.
View More On Wikipedia.org
1. ### Solving a System of 2 ODES with Interval conditions
Homework Statement I am trying to solve a system of 2 ordinary differential equations using matlab. However, I am not able to get numerical solutions from the code despite having keyed in all possible solutions. Homework Equations The equations I am given are: dx/dt=A(x/t)+By...
2. ### Finding a Solution to a System of Differential Equations
Homework Statement Find a solution \bf{\phi} of the system $$y'_1(t)=y_1(t)+y_2(t)+f(t)$$ $$y'_2(t)=y_1(t)+y_2(t)$$ where f(t) is a continuous function $$\bf{\phi} (0)=(0,0)$$ Homework Equations A hint was given to define ##v(t)=y_1(t)+y_2(t)## The Attempt at a Solution Using the suggested...
3. ### Solving a System of ODE for Steady State
I am trying to find the steady states in the ODE system. Assuming y0 = 2.5 * 10^5, I want to calculate y1, y2, y3 at the steady state. I do not understand how this would be possible, because only y0 is given and the following: d0 = 0.003, d1 = 0.008, d2 = 0.05, d3 = 1, ry = 0.008, ay = 1.6/100...
4. ### Can Euler Forward or 4th Order Runge-Kutta Methods Approximate Systems of ODEs?
My question is about whether it's possible to use the Euler Forward or 4th order Runge-Kutta Methods to approximate the following system ( where the differential of other equations are on the right hand side) : \begin{cases} \frac{dy_1}{dt} = f_1(y_1,y_2,y'_2, ... , y_n, y'_n, t) \\...
5. ### System of ODE - comparison with paper
I have the following system of differential equations, for the functions ##A(r)## and ##B(r)##: ##A'-\frac{m}{r}A=(\epsilon+1)B## and ##-B' -\frac{m+1}{r}B=(\epsilon-1)A## ##m## and ##\epsilon## are constants, with ##\epsilon<1##. The functions ##A## and ##B## are the two components of a...
6. ### Physical interpretation for system of ODE
If an ODE of 2nd order like this A y''(x) + B y'(x) + C y(x) = 0 has how physical/electrical interpretation a RLC circuit, so, how is the electrical interpretation of a system of ODE of 1nd and 2nd order? \begin{bmatrix} \frac{d x}{dt}\\ \frac{d y}{dt} \end{bmatrix} = \begin{bmatrix}...
7. ### How to reduce a system of second order ODEs to four first order equations?
Someone can explain me how to get the general solution for this system of ODE of second order with constant coeficients: \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ \end{bmatrix} \begin{bmatrix} \frac{d^2x}{dt^2}\\ \frac{d^2y}{dt^2}\\ \end{bmatrix} + \begin{bmatrix} b_{11} &...
8. ### MATLAB Critical points of system of ODE in MATLAB - Game theory and poker
Hello there, I hope I'm posting in the right section. I have been doing some work on evolutionary game theory and poker. I will give a brief description of how I got here. I have eight strategies i = 1, 2, \ldots, 8 and the eight proportions of the population playing each strategy is...
9. ### Solving a system of ODE with multiple 'time' variables
Homework Statement Hi everyone, Consider the following system of (first order) differential equations: \dot{x}=f(t_1,x,y,z) \dot{y}=g(t_2,x,y,z) \dot{z}=h(t_3,x,y,z) where \dot{x}=\frac{\partial x}{\partial t_1}, \dot{y}=\frac{\partial y}{\partial t_2}, and \dot{z}=\frac{\partial...
10. ### Solving a Non-Linear System of Differential Equations
Homework Statement assuming dy/dt = Dy, d^2y/dt^2 =D^2, etc: determine the general and particular solutions to the following linear pair of differential equations: 2D^2y-Dy-4x=2t 2Dx-4Dy-3y=0 Homework Equations The Attempt at a Solution I have went through algebraic...
11. ### System of ODE for functions with different origins
Hi, I have a system of coupled ODE like: a1 * Y1" + a2 * Y2" + b1 * Y1 + b2 * Y2 = 0 a2 * Y1" + a3 * Y2" + b2 * Y1 + b3 * Y2 = 0 I know for example by eigenvalue method I can solve it, but here is the issue: Y1 = f1 (x - a) and Y2 = f2 ( x - b). In the other word there is a shift...
12. ### System of ODE Boundary Value Problem with 2nd Order Backward Difference
{\frac {{\it du}}{{\it dx}}}=998\,u+1998\,v {\frac {{\it dv}}{{\it dx}}}=-999\,u-1999\,v u \left( 0 \right) =1 v \left( 0 \right) =0 0<x<10 Second Order Backward Difference formula {\frac {f_{{k-2}}-4\,f_{{k-1}}+3f_{{k}}}{h}} I'm trying solve this numerically in matlab, but can't seem to...
13. ### Interesting system of ODE, application in physics?
Hi all, I have a project to do for system of ordinal differential equations and their applications in physics. One of my tasks is to find where in physics the following system of ordinal differential equations appear: dA1(x)/dx=f(x).A2(x) dA2(x)/dx=f(x).A1(x)+ h(x).A2(x)+ g(x).A3(x)...
14. ### Solving System of First-Order ODEs: Exact Solution for x(t)
Dear all, I have been trying to solve the following system of first-order ordinary differential equations for a week: x' = y * (a1*x + a2*y + c1), y' = y * (a3*x + a4*y + c2), where x and y are functions of t, and ai and ci are constants. This system seems not very complex, but I have not...
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# What is System of ode: Definition and 14 Discussions. In mathematics, an ordinary differential equation (ODE) is a differential equation containing one or more functions of one independent variable and the derivatives of those functions. The term ordinary is used in contrast with the term partial differential equation which may be with respect to more than one independent variable.. View More On Wikipedia.org. 1. ### Solving a System of 2 ODES with Interval conditions. Homework Statement I am trying to solve a system of 2 ordinary differential equations using matlab. However, I am not able to get numerical solutions from the code despite having keyed in all possible solutions. Homework Equations The equations I am given are: dx/dt=A(x/t)+By.... 2. ### Finding a Solution to a System of Differential Equations. Homework Statement Find a solution \bf{\phi} of the system $$y'_1(t)=y_1(t)+y_2(t)+f(t)$$ $$y'_2(t)=y_1(t)+y_2(t)$$ where f(t) is a continuous function $$\bf{\phi} (0)=(0,0)$$ Homework Equations A hint was given to define ##v(t)=y_1(t)+y_2(t)## The Attempt at a Solution Using the suggested.... 3. ### Solving a System of ODE for Steady State. I am trying to find the steady states in the ODE system. Assuming y0 = 2.5 * 10^5, I want to calculate y1, y2, y3 at the steady state. I do not understand how this would be possible, because only y0 is given and the following: d0 = 0.003, d1 = 0.008, d2 = 0.05, d3 = 1, ry = 0.008, ay = 1.6/100.... 4. ### Can Euler Forward or 4th Order Runge-Kutta Methods Approximate Systems of ODEs?. My question is about whether it's possible to use the Euler Forward or 4th order Runge-Kutta Methods to approximate the following system ( where the differential of other equations are on the right hand side) : \begin{cases} \frac{dy_1}{dt} = f_1(y_1,y_2,y'_2, ... , y_n, y'_n, t) \\.... 5. ### System of ODE - comparison with paper. I have the following system of differential equations, for the functions ##A(r)## and ##B(r)##: ##A'-\frac{m}{r}A=(\epsilon+1)B## and ##-B' -\frac{m+1}{r}B=(\epsilon-1)A## ##m## and ##\epsilon## are constants, with ##\epsilon<1##. The functions ##A## and ##B## are the two components of a.... 6. ### Physical interpretation for system of ODE. If an ODE of 2nd order like this A y''(x) + B y'(x) + C y(x) = 0 has how physical/electrical interpretation a RLC circuit, so, how is the electrical interpretation of a system of ODE of 1nd and 2nd order? \begin{bmatrix} \frac{d x}{dt}\\ \frac{d y}{dt} \end{bmatrix} = \begin{bmatrix}.... 7. ### How to reduce a system of second order ODEs to four first order equations?. Someone can explain me how to get the general solution for this system of ODE of second order with constant coeficients: \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ \end{bmatrix} \begin{bmatrix} \frac{d^2x}{dt^2}\\ \frac{d^2y}{dt^2}\\ \end{bmatrix} + \begin{bmatrix} b_{11} &.... 8. ### MATLAB Critical points of system of ODE in MATLAB - Game theory and poker. Hello there, I hope I'm posting in the right section. I have been doing some work on evolutionary game theory and poker. I will give a brief description of how I got here. I have eight strategies i = 1, 2, \ldots, 8 and the eight proportions of the population playing each strategy is.... 9. ### Solving a system of ODE with multiple 'time' variables. Homework Statement Hi everyone, Consider the following system of (first order) differential equations: \dot{x}=f(t_1,x,y,z) \dot{y}=g(t_2,x,y,z) \dot{z}=h(t_3,x,y,z) where \dot{x}=\frac{\partial x}{\partial t_1}, \dot{y}=\frac{\partial y}{\partial t_2}, and \dot{z}=\frac{\partial.... 10. ### Solving a Non-Linear System of Differential Equations. Homework Statement assuming dy/dt = Dy, d^2y/dt^2 =D^2, etc: determine the general and particular solutions to the following linear pair of differential equations: 2D^2y-Dy-4x=2t 2Dx-4Dy-3y=0 Homework Equations The Attempt at a Solution I have went through algebraic.... 11. ### System of ODE for functions with different origins. Hi, I have a system of coupled ODE like: a1 * Y1" + a2 * Y2" + b1 * Y1 + b2 * Y2 = 0 a2 * Y1" + a3 * Y2" + b2 * Y1 + b3 * Y2 = 0 I know for example by eigenvalue method I can solve it, but here is the issue: Y1 = f1 (x - a) and Y2 = f2 ( x - b). In the other word there is a shift.... 12. ### System of ODE Boundary Value Problem with 2nd Order Backward Difference. {\frac {{\it du}}{{\it dx}}}=998\,u+1998\,v {\frac {{\it dv}}{{\it dx}}}=-999\,u-1999\,v u \left( 0 \right) =1 v \left( 0 \right) =0 0<x<10 Second Order Backward Difference formula {\frac {f_{{k-2}}-4\,f_{{k-1}}+3f_{{k}}}{h}} I'm trying solve this numerically in matlab, but can't seem to.... 13. ### Interesting system of ODE, application in physics?. Hi all, I have a project to do for system of ordinal differential equations and their applications in physics. One of my tasks is to find where in physics the following system of ordinal differential equations appear: dA1(x)/dx=f(x).A2(x) dA2(x)/dx=f(x).A1(x)+ h(x).A2(x)+ g(x).A3(x).... 14. ### Solving System of First-Order ODEs: Exact Solution for x(t).
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Dear all, I have been trying to solve the following system of first-order ordinary differential equations for a week: x' = y * (a1*x + a2*y + c1), y' = y * (a3*x + a4*y + c2), where x and y are functions of t, and ai and ci are constants. This system seems not very complex, but I have not...
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## Cantor Set Wikipedia
### Cantor set - Wikipedia.
In mathematics, the Cantor set is a set of points lying on a single line segment that has a number of unintuitive properties. It was discovered in 1874 by Henry John Stephen Smith and introduced by German mathematician Georg Cantor in 1883.. Through consideration of this set, Cantor and others helped lay the foundations of modern point-set topology.The most common construction ....
https://en.wikipedia.org/wiki/Cantor_set.
### Cantor's diagonal argument - Wikipedia.
In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti-diagonal argument, the diagonal method, and Cantor's diagonalization proof, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural numbers..
https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument.
### Cantor - Wikipedia.
A cantor or chanter is a person who leads people in singing or sometimes in prayer.In formal Christian worship, a cantor is a person who sings solo verses or passages to which the choir or congregation responds.. In Judaism, a cantor is one who sings and leads people in prayer in a Jewish religious service and may be called hazzan.A cantor in Reform and Conservative ....
https://en.wikipedia.org/wiki/Cantor.
### Cantor function - Wikipedia.
In mathematics, the Cantor function is an example of a function that is continuous, but not absolutely continuous.It is a notorious counterexample in analysis, because it challenges naive intuitions about continuity, derivative, and measure. Though it is continuous everywhere and has zero derivative almost everywhere, its value still goes from 0 to 1 as its argument reaches from ....
https://en.wikipedia.org/wiki/Cantor_function.
### Ordinal number - Wikipedia.
In set theory, an ordinal number, or ordinal, is a generalization of ordinal numerals (first, second, n th, etc.) aimed to extend enumeration to infinite sets.. A finite set can be enumerated by successively labeling each element with the least natural number that has not been previously used. To extend this process to various infinite sets, ordinal numbers are defined more ....
https://en.wikipedia.org/wiki/Ordinal_number.
### Cardinality - Wikipedia.
In mathematics, the cardinality of a set is a measure of the "number of elements" of the set.For example, the set = {,,} contains 3 elements, and therefore has a cardinality of 3. Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between different types of infinity, and to perform arithmetic on them..
https://en.wikipedia.org/wiki/Cardinality.
### Eddie Cantor - Wikipedia.
Eddie Cantor (born Isidore Itzkowitz; January 31, 1892 - October 10, 1964) was an American "illustrated song" performer, comedian, dancer, singer, vaudevillian, actor, and songwriter. ... In 1933, Brown and Bigelow published a set of 12 Eddie Cantor caricatures by Frederick J. Garner. The advertising cards were purchased in bulk as a direct ....
https://en.wikipedia.org/wiki/Eddie_Cantor.
### List of films set in ancient Rome - Wikipedia.
Title Release date Notes Scipio the African: 1971 about the later life of Scipio; directed by Luigi Magni, with Marcello Mastroianni as Scipio the African, Silvana Mangano as Aemilia Tertia and Vittorio Gassman as Cato the Elder: The Centurion: 1961 about the Battle of Corinth (146 BC), with John Drew Barrymore as Diaeus: Revolution (Ancient Rome: The Rise and Fall of an ....
https://en.wikipedia.org/wiki/List_of_films_set_in_ancient_Rome.
### Paradoxes of set theory - Wikipedia.
In the same year the French mathematician Jules Richard used a variant of Cantor's diagonal method to obtain another contradiction in naive set theory. Consider the set A of all finite agglomerations of words. The set E of all finite definitions of real numbers is a subset of A.As A is countable, so is E.Let p be the nth decimal of the nth real number defined by the set E; we ....
### Cardinality of the continuum - Wikipedia.
In set theory, the cardinality of the continuum is the cardinality or "size" of the set of real numbers, sometimes called the continuum.It is an infinite cardinal number and is denoted by (lowercase fraktur "c") or | |.. The real numbers are more numerous than the natural numbers.Moreover, has the same number of elements as the power set of . Symbolically, if the ....
https://en.wikipedia.org/wiki/Cardinality_of_the_continuum.
### Space Battleship Yamato (2010 film) - Wikipedia.
Space Battleship Yamato (SPACE BATTLESHIP ???, Supesu Batorushippu Yamato) is a 2010 Japanese science fiction film based on the Space Battleship Yamato anime series by Yoshinobu Nishizaki and Leiji Matsumoto.The film was released in Japan on December 1, 2010. It was released on DVD and Blu-ray in Japan on June 24, 2011, and in the United States by ....
https://en.wikipedia.org/wiki/Space_Battleship_Yamato_(2010_film).
### Georg Cantor - Wikipedia, la enciclopedia libre.
Georg Ferdinand Ludwig Philipp Cantor (San Petersburgo, 3 de marzo de 1845 - Halle, 6 de enero de 1918) fue un matematico nacido en Rusia, aunque nacionalizado aleman, y de ascendencia austriaca y judia. [1] Fue inventor con Dedekind de la teoria de conjuntos, que es la base de las matematicas modernas. Gracias a sus atrevidas investigaciones sobre los ....
https://es.wikipedia.org/wiki/Georg_Cantor.
### Pedophilia - Wikipedia.
Pedophilia (alternatively spelt paedophilia) is a psychiatric disorder in which an adult or older adolescent experiences a primary or exclusive sexual attraction to prepubescent children. Although girls typically begin the process of puberty at age 10 or 11, and boys at age 11 or 12, criteria for pedophilia extend the cut-off point for prepubescence to age 13..
https://en.wikipedia.org/wiki/Pedophilia.
### Chant - Wikipedia.
A chant (from French chanter, from Latin cantare, "to sing") is the iterative speaking or singing of words or sounds, often primarily on one or two main pitches called reciting tones.Chants may range from a simple melody involving a limited set of notes to highly complex musical structures, often including a great deal of repetition of musical subphrases, such as Great Responsories ....
https://en.wikipedia.org/wiki/Chant.
### First transcontinental railroad - Wikipedia.
North America's first transcontinental railroad (known originally as the "Pacific Railroad" and later as the "Overland Route") was a 1,911-mile (3,075 km) continuous railroad line constructed between 1863 and 1869 that connected the existing eastern U.S. rail network at Council Bluffs, Iowa with the Pacific coast at the Oakland Long Wharf on San Francisco Bay..
### Deforestation - Wikipedia.
Deforestation or forest clearance is the removal of a forest or stand of trees from land that is then converted to non-forest use. Deforestation can involve conversion of forest land to farms, ranches, or urban use. The most concentrated deforestation occurs in tropical rainforests. About 31% of Earth's land surface is covered by forests at present. This is one-third less than the ....
https://en.wikipedia.org/wiki/Deforestation.
### Georg Cantor — Wikipédia.
Georg Cantor est un mathematicien allemand, ne le 3 mars 1845 a Saint-Petersbourg (Empire russe) et mort le 6 janvier 1918 a Halle (Empire allemand).Il est connu pour etre le createur de la theorie des ensembles.. Il etablit l'importance de la bijection entre les ensembles, definit les ensembles infinis et les ensembles bien ordonnes.Il prouva egalement que les nombres reels ....
https://fr.wikipedia.org/wiki/Georg_Cantor.
### Johnny Depp - Wikipedia.
John Christopher Depp II (born June 9, 1963) is an American actor, producer, and musician. He is the recipient of multiple accolades, including a Golden Globe Award and a Screen Actors Guild Award, in addition to nominations for three Academy Awards and two BAFTA awards.. Depp made his feature film debut in the horror film A Nightmare on Elm Street (1984) and appeared ....
https://en.wikipedia.org/wiki/Johnny_Depp.
### Lee Sung-min (singer) - Wikipedia.
Sungmin debuted as part of 12-member rotational group Super Junior 05 on 6 November 2005, on SBS' music programme Inkigayo, performing their first single, "Twins (Knock Out)". Their debut album SuperJunior05 (Twins) was released a month later on December 5, 2005 and debuted at number three on the monthly MIAK K-pop album charts.. In March 2006, SM Entertainment ....
https://en.wikipedia.org/wiki/Lee_Sung-min_(singer).
### William Styron - Wikipedia.
William Clark Styron Jr. (June 11, 1925 - November 1, 2006) was an American novelist and essayist who won major literary awards for his work. Styron was best known for his novels, including: Lie Down in Darkness (1951), his acclaimed first work, published when he was 26;; The Confessions of Nat Turner (1967), narrated by Nat Turner, the leader of an 1831 Virginia slave ....
https://en.wikipedia.org/wiki/William_Styron.
### Blind Willie McTell - Wikipedia.
Blind Willie McTell (born William Samuel McTier; May 5, 1898 - August 19, 1959) was a Piedmont blues and ragtime singer and guitarist. He played with a fluid, syncopated fingerstyle guitar technique, common among many exponents of Piedmont blues. Unlike his contemporaries, he came to use twelve-string guitars exclusively. McTell was also an adept slide guitarist, ....
https://en.wikipedia.org/wiki/Blind_Willie_McTell.
### Intersection (set theory) - Wikipedia.
The notation for this last concept can vary considerably. Set theorists will sometimes write "", while others will instead write "".The latter notation can be generalized to "", which refers to the intersection of the collection {:}.Here is a nonempty set, and is a set for every .. In the case that the index set is the set of natural numbers, notation analogous to that of an infinite product ....
https://en.wikipedia.org/wiki/Intersection_(set_theory).
### Chris Brown (cantor) – Wikipédia, a enciclopédia livre.
Christopher Maurice Brown (Tappahannock, 5 de maio de 1989), mais conhecido como Chris Brown, e um cantor, rapper, compositor, dancarino, ator, produtor musical, coreografo e grafiteiro norte-americano.. Lancou seu primeiro album com apenas 16 anos de idade auto intitulado Chris Brown.O primeiro single lancado foi "Run It", que debutou em 92? e depois de 14 semanas ....
https://pt.wikipedia.org/wiki/Chris_Brown_(cantor).
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## Cantor Set Wikipedia. ### Cantor set - Wikipedia.. In mathematics, the Cantor set is a set of points lying on a single line segment that has a number of unintuitive properties. It was discovered in 1874 by Henry John Stephen Smith and introduced by German mathematician Georg Cantor in 1883.. Through consideration of this set, Cantor and others helped lay the foundations of modern point-set topology.The most common construction ..... https://en.wikipedia.org/wiki/Cantor_set.. ### Cantor's diagonal argument - Wikipedia.. In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti-diagonal argument, the diagonal method, and Cantor's diagonalization proof, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural numbers... https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument.. ### Cantor - Wikipedia.. A cantor or chanter is a person who leads people in singing or sometimes in prayer.In formal Christian worship, a cantor is a person who sings solo verses or passages to which the choir or congregation responds.. In Judaism, a cantor is one who sings and leads people in prayer in a Jewish religious service and may be called hazzan.A cantor in Reform and Conservative ..... https://en.wikipedia.org/wiki/Cantor.. ### Cantor function - Wikipedia.. In mathematics, the Cantor function is an example of a function that is continuous, but not absolutely continuous.It is a notorious counterexample in analysis, because it challenges naive intuitions about continuity, derivative, and measure. Though it is continuous everywhere and has zero derivative almost everywhere, its value still goes from 0 to 1 as its argument reaches from ..... https://en.wikipedia.org/wiki/Cantor_function.. ### Ordinal number - Wikipedia.. In set theory, an ordinal number, or ordinal, is a generalization of ordinal numerals (first, second, n th, etc.) aimed to extend enumeration to infinite sets.. A finite set can be enumerated by successively labeling each element with the least natural number that has not been previously used. To extend this process to various infinite sets, ordinal numbers are defined more ..... https://en.wikipedia.org/wiki/Ordinal_number.. ### Cardinality - Wikipedia.. In mathematics, the cardinality of a set is a measure of the "number of elements" of the set.For example, the set = {,,} contains 3 elements, and therefore has a cardinality of 3. Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between different types of infinity, and to perform arithmetic on them... https://en.wikipedia.org/wiki/Cardinality.. ### Eddie Cantor - Wikipedia.. Eddie Cantor (born Isidore Itzkowitz; January 31, 1892 - October 10, 1964) was an American "illustrated song" performer, comedian, dancer, singer, vaudevillian, actor, and songwriter. ... In 1933, Brown and Bigelow published a set of 12 Eddie Cantor caricatures by Frederick J. Garner. The advertising cards were purchased in bulk as a direct ..... https://en.wikipedia.org/wiki/Eddie_Cantor.. ### List of films set in ancient Rome - Wikipedia.. Title Release date Notes Scipio the African: 1971 about the later life of Scipio; directed by Luigi Magni, with Marcello Mastroianni as Scipio the African, Silvana Mangano as Aemilia Tertia and Vittorio Gassman as Cato the Elder: The Centurion: 1961 about the Battle of Corinth (146 BC), with John Drew Barrymore as Diaeus: Revolution (Ancient Rome: The Rise and Fall of an ..... https://en.wikipedia.org/wiki/List_of_films_set_in_ancient_Rome.. ### Paradoxes of set theory - Wikipedia.. In the same year the French mathematician Jules Richard used a variant of Cantor's diagonal method to obtain another contradiction in naive set theory. Consider the set A of all finite agglomerations of words. The set E of all finite definitions of real numbers is a subset of A.As A is countable, so is E.Let p be the nth decimal of the nth real number defined by the set E; we ..... ### Cardinality of the continuum - Wikipedia.. In set theory, the cardinality of the continuum is the cardinality or "size" of the set of real numbers, sometimes called the continuum.It is an infinite cardinal number and is denoted by (lowercase fraktur "c") or | |.. The real numbers are more numerous than the natural numbers.Moreover, has the same number of elements as the power set of . Symbolically, if the ..... https://en.wikipedia.org/wiki/Cardinality_of_the_continuum.. ### Space Battleship Yamato (2010 film) - Wikipedia.. Space Battleship Yamato (SPACE BATTLESHIP ???, Supesu Batorushippu Yamato) is a 2010 Japanese science fiction film based on the Space Battleship Yamato anime series by Yoshinobu Nishizaki and Leiji Matsumoto.The film was released in Japan on December 1, 2010. It was released on DVD and Blu-ray in Japan on June 24, 2011, and in the United States by ..... https://en.wikipedia.org/wiki/Space_Battleship_Yamato_(2010_film).. ### Georg Cantor - Wikipedia, la enciclopedia libre.. Georg Ferdinand Ludwig Philipp Cantor (San Petersburgo, 3 de marzo de 1845 - Halle, 6 de enero de 1918) fue un matematico nacido en Rusia, aunque nacionalizado aleman, y de ascendencia austriaca y judia. [1] Fue inventor con Dedekind de la teoria de conjuntos, que es la base de las matematicas modernas. Gracias a sus atrevidas investigaciones sobre los ..... https://es.wikipedia.org/wiki/Georg_Cantor.. ### Pedophilia - Wikipedia.. Pedophilia (alternatively spelt paedophilia) is a psychiatric disorder in which an adult or older adolescent experiences a primary or exclusive sexual attraction to prepubescent children. Although girls typically begin the process of puberty at age 10 or 11, and boys at age 11 or 12, criteria for pedophilia extend the cut-off point for prepubescence to age 13... https://en.wikipedia.org/wiki/Pedophilia.. ### Chant - Wikipedia.. A chant (from French chanter, from Latin cantare, "to sing") is the iterative speaking or singing of words or sounds, often primarily on one or two main pitches called reciting tones.Chants may range from a simple melody involving a limited set of notes to highly complex musical structures, often including a great deal of repetition of musical subphrases, such as Great Responsories ..... https://en.wikipedia.org/wiki/Chant.. ### First transcontinental railroad - Wikipedia.. North America's first transcontinental railroad (known originally as the "Pacific Railroad" and later as the "Overland Route") was a 1,911-mile (3,075 km) continuous railroad line constructed between 1863 and 1869 that connected the existing eastern U.S. rail network at Council Bluffs, Iowa with the Pacific coast at the Oakland Long Wharf on San Francisco Bay... ### Deforestation - Wikipedia.. Deforestation or forest clearance is the removal of a forest or stand of trees from land that is then converted to non-forest use. Deforestation can involve conversion of forest land to farms, ranches, or urban use. The most concentrated deforestation occurs in tropical rainforests. About 31% of Earth's land surface is covered by forests at present. This is one-third less than the ..... https://en.wikipedia.org/wiki/Deforestation.. ### Georg Cantor — Wikipédia.. Georg Cantor est un mathematicien allemand, ne le 3 mars 1845 a Saint-Petersbourg (Empire russe) et mort le 6 janvier 1918 a Halle (Empire allemand).Il est connu pour etre le createur de la theorie des ensembles.. Il etablit l'importance de la bijection entre les ensembles, definit les ensembles infinis et les ensembles bien ordonnes.Il prouva egalement que les nombres reels ..... https://fr.wikipedia.org/wiki/Georg_Cantor.. ### Johnny Depp - Wikipedia.. John Christopher Depp II (born June 9, 1963) is an American actor, producer, and musician. He is the recipient of multiple accolades, including a Golden Globe Award and a Screen Actors Guild Award, in addition to nominations for three Academy Awards and two BAFTA awards.. Depp made his feature film debut in the horror film A Nightmare on Elm Street (1984) and appeared ..... https://en.wikipedia.org/wiki/Johnny_Depp.. ### Lee Sung-min (singer) - Wikipedia.. Sungmin debuted as part of 12-member rotational group Super Junior 05 on 6 November 2005, on SBS' music programme Inkigayo, performing their first single, "Twins (Knock Out)". Their debut album SuperJunior05 (Twins) was released a month later on December 5, 2005 and debuted at number three on the monthly MIAK K-pop album charts.. In March 2006, SM Entertainment ..... https://en.wikipedia.org/wiki/Lee_Sung-min_(singer).. ### William Styron - Wikipedia.. William Clark Styron Jr. (June 11, 1925 - November 1, 2006) was an American novelist and essayist who won major literary awards for his work. Styron was best known for his novels, including: Lie Down in Darkness (1951), his acclaimed first work, published when he was 26;; The Confessions of Nat Turner (1967), narrated by Nat Turner, the leader of an 1831 Virginia slave ..... https://en.wikipedia.org/wiki/William_Styron.. ### Blind Willie McTell - Wikipedia.. Blind Willie McTell (born William Samuel McTier; May 5, 1898 - August 19, 1959) was a Piedmont blues and ragtime singer and guitarist. He played with a fluid, syncopated fingerstyle guitar technique, common among many exponents of Piedmont blues. Unlike his contemporaries, he came to use twelve-string guitars exclusively. McTell was also an adept slide guitarist, ..... https://en.wikipedia.org/wiki/Blind_Willie_McTell.. ### Intersection (set theory) - Wikipedia.. The notation for this last concept can vary considerably. Set theorists will sometimes write "", while others will instead write "".The latter notation can be generalized to "", which refers to the intersection of the collection {:}.Here is a nonempty set, and is a set for every .. In the case that the index set is the set of natural numbers, notation analogous to that of an infinite product ..... https://en.wikipedia.org/wiki/Intersection_(set_theory).. ### Chris Brown (cantor) – Wikipédia, a enciclopédia livre.
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Christopher Maurice Brown (Tappahannock, 5 de maio de 1989), mais conhecido como Chris Brown, e um cantor, rapper, compositor, dancarino, ator, produtor musical, coreografo e grafiteiro norte-americano.. Lancou seu primeiro album com apenas 16 anos de idade auto intitulado Chris Brown.O primeiro single lancado foi "Run It", que debutou em 92? e depois de 14 semanas ..... https://pt.wikipedia.org/wiki/Chris_Brown_(cantor).
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# Creating Structure for Context in Math
I was honored to facilitate lesson study with IM1 teachers today. Their students are struggling (due to high EL/SPED population) with solving word problems. I dug deeper, and we decided the struggle is really the first step: creating equations from situations.
We decided our goal as educators this year is to work on teacher clarity: making our lessons streamlined and very goal-oriented. If we know our goals for the lesson, then every move we make (every breath we take…) is for the goal. So how do we clarify translating context to equations?
We started from the end: the benchmark. We took a problem the students struggled with, and tweaked it several times, each time only altering only one component. Students had to work from the original version (which we used simple numbers to keep it accessible) for each new “version”. They discussed what changed from situation to situation and how that affected the prior equation.
Version 1: Troy works for an ice cream cart vendor. He receives \$10 for taking the cart out for a shift, plus a commission of \$2.00 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?
Version 2: Troy works for an ice cream cart vendor. He receives \$15 for taking the cart out for a shift, plus a commission of \$2.00 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?
Version 3: Troy works for an ice cream cart vendor. He receives \$15 for taking the cart out for a shift, plus a commission of \$1.25 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?
Version 4: Troy works for an ice cream cart vendor. He receives \$25 for taking the cart out for a shift, plus a commission of \$0.10 for each item he sells. Troy worked a shift Saturday and earned \$52.90. How many items did he sell? (Problem from the benchmark.)
We used 3 scenarios. In each, we kept our questions as consistent as possible (again, clarity):
• Which part is varying (changing)? How do you know?
• Which quantity would be the coefficient? How do you know?
• Which quantity would be the constant? How do you know?
• (From version to version) What has stayed the same? What changed? How does the changed quantity affect our equation? Why?
Students were engaged, writing on their tables and willing to discuss with each other. They had many moments of “ohhhhhh” and “oops!” and learned quite a bit about the components of 2-step equations. They definitely need more time, and the teachers have committed to continuing the work as warm-ups or on modified days.
Oh! And did I mention this was a co-taught Special Ed class, with many English Learners?! Amazing!
So our major takeaways were:
2. Keep your goal in mind when creating the tasks/lesson and questions for clarity and focus.
3. Breaking the situations into translating and solving (working on a single component) allows students to focus and interpret.
Below is our ppt. Hope it is useful! Happy Math-ing!
Linear Equations in Context LS 8.27.19
This morning at the airport (At 5 fricking o’ clock! I need to fire my secretary for scheduling this flight. Oh wait. That’s me!) I was answering emails and a timid voice interrupted my thoughts.
Excuse me. Are you a teacher?
This always makes my heart happy. As a middle school teacher, you often believe the kids won’t think twice about you once they leave your room. Why would they, with all the distractions the world has to offer?!!
So when a former student not only remembers you from looong ago, AND takes the time to share her experiences and life journey with you, you tear up just a little. You remember that these precious moments are WHY you were born to be an educator.
My WHY is simple. I love seeing my peeps have the “click” in math. I love learning about these humans, with all their quirks and unique personalities. I love supporting them and inspiring them to do great things. I just love THEM.
What is your WHY? Ponder, remember, and remind yourself of this through the year.
Have a great 2019-2020! I KNOW it will be a year to remember! 💕
# How Do Our Beliefs in Math Affect Our Students?
I was honored to work with amazing teachers this week. We took a survey from NCTM (National Council Teachers of Mathematics) on our beliefs regarding student learning and our instructional practices in mathematics. This, in itself, led to amazing discussions about what we truly believe math IS and how we interpret that into instructional decisions within our classrooms.
But then we took it further. We got into groups and discussed not so much whether we agreed or disagreed, but whether it was a productive or unproductive belief in respect to student access and learning. Here are two to consider:
There were fantastic discussions about these particular ones, especially for educators of EL and SPED. We also considered how parents might respond to these. Powerful conversations around access, flexibility in thinking, understanding conceptual and procedural mathematical ideas, and yes, fluency.
Here was the point. Our beliefs, whether productive or unproductive, affect our attitudes towards mathematics and the children we are blessed to teach. Those attitudes affect the actions we take. Who gets to answer which questions? Who gets the “tough” tasks and who has to keep doing drill and kill worksheets? Who gets to explore puzzles and who has to retake tests or do homework (because their home life doesn’t lend itself to being able to do it at home)? And those actions MATTER. They affect the results you will get from your students.
So as you gear up for this school year, consider taking the beliefs survey yourself. Even better, have your team take it and REALLY dive in to what beliefs are productive an unproductive. The more we reflect, the more we can grow and be effective at what we truly want; to teach students to love, learn and understand mathematics. Have a great year!
# That Moment You Realize Your Child Is Suffering
All who have multiple children know that each one is wonderfully different. You can raise ’em the same, yet they have their own amazing quirks and personalities, strengths and passions. This could not be more true of my two boys.
My youngest is now 5. He has always been the rough and tumble type. Everything is a competition to him. (Even last night he was standing on his tippy toes trying to be ‘taller’ when raising his hand at church!) He is funny, outgoing, and a firecracker. He is a joy.
Yet school stuff has not yet become his thing. He seemed disinterested in learning his letters, yet loved to be read to. He never wanted to sit with me and learn the components all say are so important for kinder. His preschool teachers said it would come; that he just was a busy bee and had other, more important things to care about. And honestly, I believed that too. He loves to create, tell stories, sing, build, and live outdoors. Who was I to take that away from him?
So when his amazing preK teacher suggested a hearing test, I was on board. No big deal; just go do it and cross it off as another thing we did. It came back inconclusive. Went for an ENT hearing test. The results were staggering; at least 30% hearing loss in each ear. The doctor said, “Think of being submerged underwater for 5 years of your life, trying desperately to hear what people are saying. That is what your son has lived through.” I am still teary thinking about this. I am a fricking educator! How did I not catch this? I was in denial as well, until that very day driving home I asked my boy if he saw the cool tree. “Tree?” “Yes, the tree over there (I pointed out the car window.).” “Like, dessert?” And that is when it hit me. My child wasn’t hearing.
We started to notice. He said, “What?” almost every time you told him something. He couldn’t hear the TV clearly; he was reading lips (which we now know why he would never answer us when watching TV; he was too focused on the screen to hear us). In preK, during circle time, he struggled to hear all the conversations and his body would just be exhausted from trying to listen that he gave up. When listening to a story, he focused on the pictures for meaning versus our words. He was trying his best, his very best, and it was exhausting. My heart broke. My baby boy, sweet thing he is, was struggling under my teacher nose, and I hadn’t caught it.
Long story short, he is now hearing much more, with just a tweak of medication. He still has a hearing loss, and we will test every year to make sure it isn’t degenerative. He is learning how to deal with sounds he hadn’t heard before. (At church last night he couldn’t believe they played bells!) He begins speech therapy (I had no idea he would need that either) next week and they are excited to see his progress, as am I. And funny enough, he is now interested in learning his letters, sounds, words, etc. It all makes sense; for how can you be interested in something you never knew existed?
I bring this story up for one reason. We are not perfect. Even if we have our children’s best interests at heart, we may miss something. It takes a true village to raise our children. If you do not have a village to help, find one. If your children go to daycare or summer camp, get to know their counselors. Ask them questions. Find out what your children are doing, and how they are doing. Talk to your children’s teachers next year. Get to know them, because they see your children more often than you do! And listen, even when you don’t want to. In a world that is so negative right now, I feel strongly that we need to support, build, and nurture each other and our tiny humans in order to make them the best they can be.
# Developing Perseverance
I don’t get it. Can you help me? My teacher didn’t explain it. I forgot. This is stupid.
We have all been there. We have all heard each of these when our child is working on math homework. The question is, how do we get him/her to stop being helpless? Here are a few ideas to start with.
Listen to their frustrations. Then move on. Look. Being frustrated is okay. We don’t want to say it isn’t. But being helpless is not okay. This is a life lesson. Not everything is easy, but we don’t get to give up.
Don’t do the math for them. That only lets your child know you will let them off the hook every time.
Help them find resources and look at them together. Some questions to help you out:
• Where are your notes from today? Let’s review them and see if there is anything there we can use.
• What are you learning? Let’s look up some videos (mathtv.com is a great site for video lessons, but if you just go to utube or teachertube you will find others) and see if we can relearn it together.
• Let’s look at your book and see if there are any examples that might help us.
• Call/Facetime/text a friend and see if they can assist you. (Research shows that study groups truly help all students in mathematics!!!
• Email the teacher and see if s/he has tutorials to help. (Many teachers put up videos, the solutions, etc on-line. Find out if yours does!)
Above all, let them know that it is okay to not know everything, but NOT OKAY to give up. This is a biggie. Most math worth doing takes time. Students assume that if they don’t get the answer right away, they must have done it wrong (or don’t know what they are doing and are not good in math). So not true! Help your child use resources available to be successful, but they need to do the work and put in the effort.
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# Creating Structure for Context in Math. I was honored to facilitate lesson study with IM1 teachers today. Their students are struggling (due to high EL/SPED population) with solving word problems. I dug deeper, and we decided the struggle is really the first step: creating equations from situations.. We decided our goal as educators this year is to work on teacher clarity: making our lessons streamlined and very goal-oriented. If we know our goals for the lesson, then every move we make (every breath we take…) is for the goal. So how do we clarify translating context to equations?. We started from the end: the benchmark. We took a problem the students struggled with, and tweaked it several times, each time only altering only one component. Students had to work from the original version (which we used simple numbers to keep it accessible) for each new “version”. They discussed what changed from situation to situation and how that affected the prior equation.. Version 1: Troy works for an ice cream cart vendor. He receives \$10 for taking the cart out for a shift, plus a commission of \$2.00 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?. Version 2: Troy works for an ice cream cart vendor. He receives \$15 for taking the cart out for a shift, plus a commission of \$2.00 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?. Version 3: Troy works for an ice cream cart vendor. He receives \$15 for taking the cart out for a shift, plus a commission of \$1.25 for each item he sells. Troy worked a shift Saturday and earned \$60. How many items did he sell?. Version 4: Troy works for an ice cream cart vendor. He receives \$25 for taking the cart out for a shift, plus a commission of \$0.10 for each item he sells. Troy worked a shift Saturday and earned \$52.90. How many items did he sell? (Problem from the benchmark.). We used 3 scenarios. In each, we kept our questions as consistent as possible (again, clarity):. • Which part is varying (changing)? How do you know?. • Which quantity would be the coefficient? How do you know?. • Which quantity would be the constant? How do you know?. • (From version to version) What has stayed the same? What changed? How does the changed quantity affect our equation? Why?. Students were engaged, writing on their tables and willing to discuss with each other. They had many moments of “ohhhhhh” and “oops!” and learned quite a bit about the components of 2-step equations. They definitely need more time, and the teachers have committed to continuing the work as warm-ups or on modified days.. Oh! And did I mention this was a co-taught Special Ed class, with many English Learners?! Amazing!. So our major takeaways were:. 2. Keep your goal in mind when creating the tasks/lesson and questions for clarity and focus.. 3. Breaking the situations into translating and solving (working on a single component) allows students to focus and interpret.. Below is our ppt. Hope it is useful! Happy Math-ing!. Linear Equations in Context LS 8.27.19. This morning at the airport (At 5 fricking o’ clock! I need to fire my secretary for scheduling this flight. Oh wait. That’s me!) I was answering emails and a timid voice interrupted my thoughts.. Excuse me. Are you a teacher?. This always makes my heart happy. As a middle school teacher, you often believe the kids won’t think twice about you once they leave your room. Why would they, with all the distractions the world has to offer?!!. So when a former student not only remembers you from looong ago, AND takes the time to share her experiences and life journey with you, you tear up just a little. You remember that these precious moments are WHY you were born to be an educator.. My WHY is simple. I love seeing my peeps have the “click” in math. I love learning about these humans, with all their quirks and unique personalities. I love supporting them and inspiring them to do great things. I just love THEM.. What is your WHY? Ponder, remember, and remind yourself of this through the year.. Have a great 2019-2020! I KNOW it will be a year to remember! 💕. # How Do Our Beliefs in Math Affect Our Students?. I was honored to work with amazing teachers this week. We took a survey from NCTM (National Council Teachers of Mathematics) on our beliefs regarding student learning and our instructional practices in mathematics. This, in itself, led to amazing discussions about what we truly believe math IS and how we interpret that into instructional decisions within our classrooms.. But then we took it further. We got into groups and discussed not so much whether we agreed or disagreed, but whether it was a productive or unproductive belief in respect to student access and learning. Here are two to consider:. There were fantastic discussions about these particular ones, especially for educators of EL and SPED. We also considered how parents might respond to these. Powerful conversations around access, flexibility in thinking, understanding conceptual and procedural mathematical ideas, and yes, fluency.. Here was the point. Our beliefs, whether productive or unproductive, affect our attitudes towards mathematics and the children we are blessed to teach. Those attitudes affect the actions we take. Who gets to answer which questions? Who gets the “tough” tasks and who has to keep doing drill and kill worksheets? Who gets to explore puzzles and who has to retake tests or do homework (because their home life doesn’t lend itself to being able to do it at home)? And those actions MATTER. They affect the results you will get from your students.. So as you gear up for this school year, consider taking the beliefs survey yourself. Even better, have your team take it and REALLY dive in to what beliefs are productive an unproductive. The more we reflect, the more we can grow and be effective at what we truly want; to teach students to love, learn and understand mathematics. Have a great year!. # That Moment You Realize Your Child Is Suffering. All who have multiple children know that each one is wonderfully different. You can raise ’em the same, yet they have their own amazing quirks and personalities, strengths and passions. This could not be more true of my two boys.. My youngest is now 5. He has always been the rough and tumble type. Everything is a competition to him. (Even last night he was standing on his tippy toes trying to be ‘taller’ when raising his hand at church!) He is funny, outgoing, and a firecracker. He is a joy.. Yet school stuff has not yet become his thing. He seemed disinterested in learning his letters, yet loved to be read to. He never wanted to sit with me and learn the components all say are so important for kinder. His preschool teachers said it would come; that he just was a busy bee and had other, more important things to care about. And honestly, I believed that too. He loves to create, tell stories, sing, build, and live outdoors. Who was I to take that away from him?. So when his amazing preK teacher suggested a hearing test, I was on board. No big deal; just go do it and cross it off as another thing we did. It came back inconclusive. Went for an ENT hearing test. The results were staggering; at least 30% hearing loss in each ear. The doctor said, “Think of being submerged underwater for 5 years of your life, trying desperately to hear what people are saying. That is what your son has lived through.” I am still teary thinking about this. I am a fricking educator! How did I not catch this? I was in denial as well, until that very day driving home I asked my boy if he saw the cool tree. “Tree?” “Yes, the tree over there (I pointed out the car window.).” “Like, dessert?” And that is when it hit me. My child wasn’t hearing.. We started to notice. He said, “What?” almost every time you told him something. He couldn’t hear the TV clearly; he was reading lips (which we now know why he would never answer us when watching TV; he was too focused on the screen to hear us). In preK, during circle time, he struggled to hear all the conversations and his body would just be exhausted from trying to listen that he gave up. When listening to a story, he focused on the pictures for meaning versus our words. He was trying his best, his very best, and it was exhausting. My heart broke. My baby boy, sweet thing he is, was struggling under my teacher nose, and I hadn’t caught it.. Long story short, he is now hearing much more, with just a tweak of medication. He still has a hearing loss, and we will test every year to make sure it isn’t degenerative. He is learning how to deal with sounds he hadn’t heard before. (At church last night he couldn’t believe they played bells!) He begins speech therapy (I had no idea he would need that either) next week and they are excited to see his progress, as am I. And funny enough, he is now interested in learning his letters, sounds, words, etc. It all makes sense; for how can you be interested in something you never knew existed?. I bring this story up for one reason. We are not perfect. Even if we have our children’s best interests at heart, we may miss something. It takes a true village to raise our children. If you do not have a village to help, find one. If your children go to daycare or summer camp, get to know their counselors. Ask them questions. Find out what your children are doing, and how they are doing. Talk to your children’s teachers next year. Get to know them, because they see your children more often than you do! And listen, even when you don’t want to. In a world that is so negative right now, I feel strongly that we need to support, build, and nurture each other and our tiny humans in order to make them the best they can be.. # Developing Perseverance. I don’t get it. Can you help me? My teacher didn’t explain it. I forgot. This is stupid.. We have all been there. We have all heard each of these when our child is working on math homework. The question is, how do we get him/her to stop being helpless? Here are a few ideas to start with.. Listen to their frustrations. Then move on. Look. Being frustrated is okay. We don’t want to say it isn’t. But being helpless is not okay. This is a life lesson. Not everything is easy, but we don’t get to give up.. Don’t do the math for them. That only lets your child know you will let them off the hook every time.. Help them find resources and look at them together. Some questions to help you out:. • Where are your notes from today? Let’s review them and see if there is anything there we can use.. • What are you learning? Let’s look up some videos (mathtv.com is a great site for video lessons, but if you just go to utube or teachertube you will find others) and see if we can relearn it together.. • Let’s look at your book and see if there are any examples that might help us.. • Call/Facetime/text a friend and see if they can assist you. (Research shows that study groups truly help all students in mathematics!!!. • Email the teacher and see if s/he has tutorials to help. (Many teachers put up videos, the solutions, etc on-line. Find out if yours does!). Above all, let them know that it is okay to not know everything, but NOT OKAY to give up. This is a biggie.
|
Most math worth doing takes time. Students assume that if they don’t get the answer right away, they must have done it wrong (or don’t know what they are doing and are not good in math). So not true! Help your child use resources available to be successful, but they need to do the work and put in the effort.
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Web Results
Adding & subtracting negative fractions. CCSS Math: 7.NS.A.1. Google Classroom Facebook Twitter. Email. Fractions. Adding fractions word problem: paint. Adding fractions with different signs. Practice: Adding & subtracting negative fractions. This is the currently selected item.
Negative fractions are like any other fraction, except that they have a preceding negative (-) sign. The process of adding and subtracting negative fractions can be straightforward, if you keep in mind two things. A negative fraction added to another negative fraction will result in a negative fraction as result. A ...
Adding and Subtracting Fractions with Negatives Once you've learned how to add and subtract positive fractions , you can extend the method to include negative fractions. Note that: − 2 3 is the same as − 2 3 and 2 − 3
To add fractions, you must first make the denominators the same. The denominators are the bottom numbers of the fractions. When you add a negative fraction and a positive fraction, you are essentially subtracting one fraction from the other.
How to Work With Negative Fractions. When an equation calls for adding a negative fraction, we can rewrite the equation as subtracting a positive fraction. Likewise, if the equation calls for subtracting a negative fraction, this is the same as adding a positive fraction and can be rewritten this way.
Practice: Adding & subtracting negative fractions. Next lesson. Addition & subtraction word problems with negatives. Tags. Adding fractions (unlike denominators) Subtracting fractions (unlike denominators) Video transcript. We have negative 3/4 minus 7/6 minus 3/6. And there's many ways to do this. But it immediately jumps out at me that these ...
onlinemschool.com/math/assistance/fraction/fraction_calc
This online fraction calculator will help you understand how to add, subtract, multiply or divide fractions, mixed numbers (mixed fractions), whole numbers and decimals.The fraction calculator will compute the sums, differences, products or quotients of fractions. The fraction calculator will generate a step-by-step explanation on how to obtain the results in the REDUCED FORM!
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Web Results. Adding & subtracting negative fractions. CCSS Math: 7.NS.A.1. Google Classroom Facebook Twitter. Email. Fractions. Adding fractions word problem: paint. Adding fractions with different signs. Practice: Adding & subtracting negative fractions. This is the currently selected item.. Negative fractions are like any other fraction, except that they have a preceding negative (-) sign. The process of adding and subtracting negative fractions can be straightforward, if you keep in mind two things. A negative fraction added to another negative fraction will result in a negative fraction as result. A .... Adding and Subtracting Fractions with Negatives Once you've learned how to add and subtract positive fractions , you can extend the method to include negative fractions. Note that: − 2 3 is the same as − 2 3 and 2 − 3. To add fractions, you must first make the denominators the same. The denominators are the bottom numbers of the fractions. When you add a negative fraction and a positive fraction, you are essentially subtracting one fraction from the other.. How to Work With Negative Fractions. When an equation calls for adding a negative fraction, we can rewrite the equation as subtracting a positive fraction. Likewise, if the equation calls for subtracting a negative fraction, this is the same as adding a positive fraction and can be rewritten this way.. Practice: Adding & subtracting negative fractions. Next lesson. Addition & subtraction word problems with negatives. Tags. Adding fractions (unlike denominators) Subtracting fractions (unlike denominators) Video transcript. We have negative 3/4 minus 7/6 minus 3/6. And there's many ways to do this. But it immediately jumps out at me that these .... onlinemschool.com/math/assistance/fraction/fraction_calc.
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This online fraction calculator will help you understand how to add, subtract, multiply or divide fractions, mixed numbers (mixed fractions), whole numbers and decimals.The fraction calculator will compute the sums, differences, products or quotients of fractions. The fraction calculator will generate a step-by-step explanation on how to obtain the results in the REDUCED FORM!.
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# Contour Integration where Contour contains singularity
There are many theorems in complex analysis which tell us about integration $\int_{\gamma} f$ where $f$ is continuous (or even differentiable) in the interior of $\gamma$ except finitely many points.
I would like to see how should we solve $\int_{\gamma}f(z)dz$ where $\gamma$ contains a singular point of $f$.
For example, in solving $\int_{|z|=1} \frac{1}{z-1}dz$, I slightly pulled the curve $|z|=1$ near $1$ towards left, so that $\gamma$ will not contain $1$; then the singularity of $\frac{1}{z-1}$ will not be inside this deformed curve, so integration over deformed curve is zero, so taking limit, I concluded that $\int_{|z|=1} \frac{1}{z-1}dz=0$.
However, if I pull the curve near $1$ on the right side, then the singularity of $\frac{1}{z-1}$ will be inside this deformed curve, and integration is then non-zero.
I confused between these two processes; can you help me what is the correct way to proceed for finding integration along curves, in which singularity of the function is on the curve?
#### Solutions Collecting From Web of "Contour Integration where Contour contains singularity"
As an improper integral, $\oint_{|z|=1} \frac{1}{z-1}\; dz$ does not converge. You could define a principal value which is the limit as $\epsilon \to 0+$ of the integral over arcs omitting a length $\epsilon$ on each side of the singularity. But you have to be careful doing such an integral using residues, because the arc you use to close up the contour will make a significant contribution.
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# Contour Integration where Contour contains singularity. There are many theorems in complex analysis which tell us about integration $\int_{\gamma} f$ where $f$ is continuous (or even differentiable) in the interior of $\gamma$ except finitely many points.. I would like to see how should we solve $\int_{\gamma}f(z)dz$ where $\gamma$ contains a singular point of $f$.. For example, in solving $\int_{|z|=1} \frac{1}{z-1}dz$, I slightly pulled the curve $|z|=1$ near $1$ towards left, so that $\gamma$ will not contain $1$; then the singularity of $\frac{1}{z-1}$ will not be inside this deformed curve, so integration over deformed curve is zero, so taking limit, I concluded that $\int_{|z|=1} \frac{1}{z-1}dz=0$.. However, if I pull the curve near $1$ on the right side, then the singularity of $\frac{1}{z-1}$ will be inside this deformed curve, and integration is then non-zero.. I confused between these two processes; can you help me what is the correct way to proceed for finding integration along curves, in which singularity of the function is on the curve?. #### Solutions Collecting From Web of "Contour Integration where Contour contains singularity". As an improper integral, $\oint_{|z|=1} \frac{1}{z-1}\; dz$ does not converge.
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You could define a principal value which is the limit as $\epsilon \to 0+$ of the integral over arcs omitting a length $\epsilon$ on each side of the singularity. But you have to be careful doing such an integral using residues, because the arc you use to close up the contour will make a significant contribution.
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## Calculate correlation coefficient online
Pearson Correlation Coefficient Calculator. The Pearson correlation coefficient is used to measure the strength of a linear association between two variables, where the value r = 1 means a perfect positive correlation and the value r = -1 means a perfect negataive correlation. So, for example, you could use this test to find out whether people's height and weight are correlated (they will be Negative correlation happens when one variable decreases, the other variable also decreases. The correlation co-efficient differ from -1 to +1. Also, this correlation coefficient calculator page shows you the exclusive formula for the calculation of coefficient of correlation. A correlation coefficient formula is used to determine the relationship strength between 2 continuous variables. The formula was developed by British statistician Karl Pearson in the 1890s, which is why the value is called the Pearson correlation coefficient (r). The equation was derived from an idea proposed by statistician and sociologist Sir
The correlation coefficient should not be calculated if the relationship is not linear . For correlation only purposes, it does not really matter on which axis the A value of r close to zero means that there is no correlation between the variables . Just as with least squares regression, it is important to determine whether the 12 Jan 2016 There is a statistically significant linear relationship between X and Y. Using Software to Calculate the Correlation Coefficient. We are interested The correlation coefficient is a value that indicates the strength of the are many software tools that can help you save time when calculating the coefficient. The linear correlation coefficient works with data sets or lists and compares how those lists relate to each other on a scale from -1 to 1. A higher coefficient rating r_p=\frac{S_{XY}}{\sqrt{S_{XX}S_{YY}}}. The sample Pearson correlation coefficient, rp , is the point estimate of the population Pearson correlation coefficient.
## The correlation coefficient calculator helps you determine the statistical significance of your data with the Matthews correlation formula.
12 Jan 2016 There is a statistically significant linear relationship between X and Y. Using Software to Calculate the Correlation Coefficient. We are interested The correlation coefficient is a value that indicates the strength of the are many software tools that can help you save time when calculating the coefficient. The linear correlation coefficient works with data sets or lists and compares how those lists relate to each other on a scale from -1 to 1. A higher coefficient rating r_p=\frac{S_{XY}}{\sqrt{S_{XX}S_{YY}}}. The sample Pearson correlation coefficient, rp , is the point estimate of the population Pearson correlation coefficient. This interactive calculator yields the result of a test of the equality of two correlation coefficients obtained from the same sample, with the two correlations sharing
### 15 Feb 2017 Use a calculator or other program. Calculate the This formula returns the Pearson correlation coefficient of two expressions. The Pearson
Correlation and regression calculator Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line. The sample correlation $$r$$ is a statistic that estimates the population correlation, $$\rho$$. On typical statistical test consists of assessing whether or not the correlation coefficient is significantly different from zero. There are least two methods to assess the significance of the sample correlation coefficient: One of them is based on the critical correlation. Step by Step Calculation 1. Find the sample mean μ x for data set X. 2. Find the sample mean μ y for data set Y. 3. Estimate the standard deviation σ x for sample data set X. 4. Estimate the sample deviation σ y for data set Y. 5. Find the covariance (cov (x, y)) for the data sets X and Y. 6. p-Value Calculator for Correlation Coefficients. This calculator will tell you the significance (both one-tailed and two-tailed probability values) of a Pearson correlation coefficient, given the correlation value r, and the sample size. Please enter the necessary parameter values, and then click 'Calculate'.
### Correlation and regression calculator Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line.
The correlation coefficient calculated above corresponds to Spearman's correlation coefficient. The requirements for computing it is that the two variables X and Y are measured at least at the interval level (which means that it does not work with nominal or ordinal variables). Correlation and regression calculator Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line.
## Negative correlation happens when one variable decreases, the other variable also decreases. The correlation co-efficient differ from -1 to +1. Also, this correlation coefficient calculator page shows you the exclusive formula for the calculation of coefficient of correlation.
Pearson Correlation Coefficient Calculator. Pearson's correlation coefficient measures the strength and direction of the relationship between two variables. To begin, you need to add your data to the text boxes below (either one value per line or as a comma delimited list). Correlation and regression calculator. Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line. Correlation Coefficient is a method used in the context of probability & statistics often denoted by {Corr(X, Y)} or r(X, Y) used to find the degree or magnitude of linear relationship between two or more variables in statistical experiments. Correlation Coefficient (r): Note: Data should be separated by coma (,), space ( ), tab, or in separated lines. Pearson Correlation Coefficient (r) is used for measuring the linear dependence of two variables. The correlation coefficient calculated above corresponds to Spearman's correlation coefficient. The requirements for computing it is that the two variables X and Y are measured at least at the interval level (which means that it does not work with nominal or ordinal variables). Correlation and regression calculator Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line. The sample correlation $$r$$ is a statistic that estimates the population correlation, $$\rho$$. On typical statistical test consists of assessing whether or not the correlation coefficient is significantly different from zero. There are least two methods to assess the significance of the sample correlation coefficient: One of them is based on the critical correlation.
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## Calculate correlation coefficient online. Pearson Correlation Coefficient Calculator. The Pearson correlation coefficient is used to measure the strength of a linear association between two variables, where the value r = 1 means a perfect positive correlation and the value r = -1 means a perfect negataive correlation. So, for example, you could use this test to find out whether people's height and weight are correlated (they will be Negative correlation happens when one variable decreases, the other variable also decreases. The correlation co-efficient differ from -1 to +1. Also, this correlation coefficient calculator page shows you the exclusive formula for the calculation of coefficient of correlation. A correlation coefficient formula is used to determine the relationship strength between 2 continuous variables. The formula was developed by British statistician Karl Pearson in the 1890s, which is why the value is called the Pearson correlation coefficient (r). The equation was derived from an idea proposed by statistician and sociologist Sir. The correlation coefficient should not be calculated if the relationship is not linear . For correlation only purposes, it does not really matter on which axis the A value of r close to zero means that there is no correlation between the variables . Just as with least squares regression, it is important to determine whether the 12 Jan 2016 There is a statistically significant linear relationship between X and Y. Using Software to Calculate the Correlation Coefficient. We are interested The correlation coefficient is a value that indicates the strength of the are many software tools that can help you save time when calculating the coefficient. The linear correlation coefficient works with data sets or lists and compares how those lists relate to each other on a scale from -1 to 1. A higher coefficient rating r_p=\frac{S_{XY}}{\sqrt{S_{XX}S_{YY}}}. The sample Pearson correlation coefficient, rp , is the point estimate of the population Pearson correlation coefficient.. ## The correlation coefficient calculator helps you determine the statistical significance of your data with the Matthews correlation formula.. 12 Jan 2016 There is a statistically significant linear relationship between X and Y. Using Software to Calculate the Correlation Coefficient. We are interested The correlation coefficient is a value that indicates the strength of the are many software tools that can help you save time when calculating the coefficient. The linear correlation coefficient works with data sets or lists and compares how those lists relate to each other on a scale from -1 to 1. A higher coefficient rating r_p=\frac{S_{XY}}{\sqrt{S_{XX}S_{YY}}}. The sample Pearson correlation coefficient, rp , is the point estimate of the population Pearson correlation coefficient. This interactive calculator yields the result of a test of the equality of two correlation coefficients obtained from the same sample, with the two correlations sharing. ### 15 Feb 2017 Use a calculator or other program. Calculate the This formula returns the Pearson correlation coefficient of two expressions. The Pearson. Correlation and regression calculator Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line. The sample correlation $$r$$ is a statistic that estimates the population correlation, $$\rho$$. On typical statistical test consists of assessing whether or not the correlation coefficient is significantly different from zero. There are least two methods to assess the significance of the sample correlation coefficient: One of them is based on the critical correlation. Step by Step Calculation 1. Find the sample mean μ x for data set X. 2. Find the sample mean μ y for data set Y. 3. Estimate the standard deviation σ x for sample data set X. 4. Estimate the sample deviation σ y for data set Y. 5. Find the covariance (cov (x, y)) for the data sets X and Y. 6. p-Value Calculator for Correlation Coefficients. This calculator will tell you the significance (both one-tailed and two-tailed probability values) of a Pearson correlation coefficient, given the correlation value r, and the sample size. Please enter the necessary parameter values, and then click 'Calculate'.. ### Correlation and regression calculator Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line.. The correlation coefficient calculated above corresponds to Spearman's correlation coefficient. The requirements for computing it is that the two variables X and Y are measured at least at the interval level (which means that it does not work with nominal or ordinal variables). Correlation and regression calculator Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line.. ## Negative correlation happens when one variable decreases, the other variable also decreases. The correlation co-efficient differ from -1 to +1. Also, this correlation coefficient calculator page shows you the exclusive formula for the calculation of coefficient of correlation.. Pearson Correlation Coefficient Calculator. Pearson's correlation coefficient measures the strength and direction of the relationship between two variables. To begin, you need to add your data to the text boxes below (either one value per line or as a comma delimited list). Correlation and regression calculator. Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line. Correlation Coefficient is a method used in the context of probability & statistics often denoted by {Corr(X, Y)} or r(X, Y) used to find the degree or magnitude of linear relationship between two or more variables in statistical experiments. Correlation Coefficient (r): Note: Data should be separated by coma (,), space ( ), tab, or in separated lines. Pearson Correlation Coefficient (r) is used for measuring the linear dependence of two variables. The correlation coefficient calculated above corresponds to Spearman's correlation coefficient. The requirements for computing it is that the two variables X and Y are measured at least at the interval level (which means that it does not work with nominal or ordinal variables). Correlation and regression calculator Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line.
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The sample correlation $$r$$ is a statistic that estimates the population correlation, $$\rho$$. On typical statistical test consists of assessing whether or not the correlation coefficient is significantly different from zero. There are least two methods to assess the significance of the sample correlation coefficient: One of them is based on the critical correlation.
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https://math.stackexchange.com/questions/452517/cyclic-definitions-of-material-logical-implications
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cyclic definitions of material/logical implications
Most sources define the material implication (conditional) using a pre-defined connective, usually "or", $$\left(p \implies q\right) \text{ iff } \left(\lnot p \lor q\right) \qquad \text{(1)}$$
Different definitions are possible, including one using "and" instead of "or", and even defining implication as an atom and then defining "and" and "or" in terms of the atom. Regardless, given a material implication, then...
The definition of the logical implication is $$\left(p \vdash q\right) \text{ iff } \left(\left(p\implies q\right) \text{ is a tautology } \top\right) \qquad \text{(2)}$$
(I think some sources use the notation $p \therefore q$ to mean $p \vdash q$ and $p \because q$ to mean $p \dashv q$. Personally I like $\therefore$ and $\because$ more because of their $\LaTeX$ commands \therefore and \because.)
I have some problems with the definitions above.
1. In definition (1), what is meant by "iff"? I always thought that "if and only if" was the material equivalence (biconditional) $a \iff b$, which is defined by $\left(a\implies b\right)\land\left(b\implies a\right)$, which is a binary operation of propositions, that is, it takes two propositions and makes a new one (analogous to the mathematical expression "$x+y$"). So, doing some scratchwork I found that $$\left(p \implies q\right) \iff \left(\lnot p \lor q\right)$$ reduces to $\top$. What does that mean? Stating just "$\top$" doesn't make sense; it's a constant. That would be like stating "$x+y$" as a definition. Definition (1) should be a logical statement about propositions, not a proposition itself. Right?
I assume that the "iff" in definition (1) is meant to mean that the propositions are "the same", since it's a definition (or if not a definition, a theorem). Since they are "the same", couldn't one say they are logically equivalent and write $$\left(p\implies q\right) \dashv\vdash \left(\lnot p \lor q\right) \qquad \text{(1b)}$$ (where $\dashv\vdash$ denotes logical equivalence, I believe)? This seems to be right, since it's not a binary operation of propositions but rather more like a binary relation of propositions: it takes two propositions and compares them (analogous to the mathematical statement "$x\lt y$"). So is the "iff" in (1) an abuse of notation, or could the phrase "if and only if" denote both material equivalence and logical equivalence?
2. In definition (2), the word "iff" appears again, except this time I'm more confused. For certain, it can't denote the material equivalence $\iff$, because the definition isn't an operation connecting two propositions. It can't even denote the logical equivalence $\dashv\vdash$, because the definition isn't a relation comparing propositions. It appears to be a statement about logical statements! Apparently, the phrase "if and only if" has a third meaning, a so-called metalogical equivalence: a statement comparing two logical statements. Is there such thing as a meta-metalogical equivalence? Where does it end??
Also, what is meant by the "is" in "$\left(p\implies q\right) \text{ is a tautology } \top$"? Does it mean "is logically equivalent to"? If so, can definition (2) be rewritten... ? $$\left(p\vdash q\right)\text{ "is the same as" }\left(\left(p\iff q\right)\dashv\vdash\top\right) \qquad \text{(2b)}$$ If so, $\dashv\vdash$ would have to be defined beforehand.
3. Lastly, setting all syntactic issues aside, the biggest problem I have of all is the fact that these definitions seem to be semantically dependent on each other. They're cyclic. If the "iff" in (1) and the "is" in (2) do indeed indicate logical equivalence, logical equivalence needs to be defined formally. One definition I found is $$\left(p \dashv\vdash q\right) \text{ iff } \left(\left(p\iff q\right) \text{ is a tautology } \top\right) \qquad \text{(3)}$$ The first thing I noticed off the bat is that logical equivalence uses itself in its own definition. I interpret $\left(\left(p\iff q\right) \text{ is a tautology } \top\right)$ to mean $\left(\left(p\iff q\right)\dashv\vdash\top\right)$, but that would mean the definition refers to itself. Maybe I'm interpreting the "is" wrongly?
Another possible definition I've seen is $$\left(p \dashv\vdash q\right) \text{ iff } \left(\left(p\vdash q\right) \land \left(p \dashv q\right)\right) \qquad \text{(4)}$$ which again, syntactically, I have another problem with because it uses the $\land$ operation for propositions, call it material conjunction, to connect two statements (unless there is a second interpretation to "and" that allows it to operate on two statements... call it logical conjunction, which hasn't been defined).
Semantically, $\dashv\vdash$ cannot use $\vdash$ and $\dashv$ in its definition because the definitions of $\vdash$ and $\dashv$ use $\dashv\vdash$ (see (2b)).
One more thing, how does this all relate to the logical consequence $P \models Q$ ?
• perhaps this iff for the case of (1) means that depending on the values of p and q, $p\implies q$ should have preciesly the same value as $\lnot p \lor q$? And the iff in (2) means that you simply define $(p⊢q)$ that way, e.g. when you say, a function is continuous if ( ... some stuff...). You might say, a function is continuous iff ( ... some stuff...), but that's just the same, you simply define this previously unknown notion of continuity. So it's a kind like, you might establish logical equivalence between things which meanings you know, and this thing has its own definition
– W_D
Jul 26, 2013 at 7:04
• but here you're a kind of saying, this notation will mean preciously the thing after the "iff" word. That's how i understand that
– W_D
Jul 26, 2013 at 7:05
In case 1., there's two options. On the one hand, the "iff" can be read definitionally, meaning we're just treating one side as an abbreviation for the other side. That is, we're saying "We'll write "$p \rightarrow q$" as a notational variant of "$\neg p \vee q$"." Alternatively, we could take "$\rightarrow$" as a primitive symbol along with "$\vee$", in which case the "iff" can be read as the logical equivalence relation between propositions in the formal language, i.e. as saying "$p \rightarrow q \dashv \vdash \neg p \vee q$."
In case 2., the "iff" is being used to state the logical equivalence of two statements in the metalanguage, i.e. the metalogical equivalence between the statement "$p \vdash q$" and "The formal sentence "$p \leftarrow q$" is a tautology."
As for case 3., you're right that our notion of logical consequence ($\vdash$) and our notion of tautology are closely interconnected. But you could simply rephrase the claim in 3. as "$p \dashv \vdash q$ iff $\vdash p \leftrightarrow q$." Then the question just boils down to defining what we mean by "$\vdash$", which is often done either via truth tables (in the case of propositional logic), or via some proof system. In either case, you can find an explication in any reasonable introductory book.
Two things should be noted. The first concerns our use of "$p \vdash q$". This sentence is an English-level sentence, not a sentence in the formal language itself. Thus, "$p \vdash q$" means "The formal proposition "$p$" entails the formal proposition "$q$" in the given formal logic." So if we ever happen to say, "$p \vdash q$ iff...", that's an immediate indication we're talking about an equivalence between sentences of English, not sentences of the formal language.
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cyclic definitions of material/logical implications. Most sources define the material implication (conditional) using a pre-defined connective, usually "or", $$\left(p \implies q\right) \text{ iff } \left(\lnot p \lor q\right) \qquad \text{(1)}$$. Different definitions are possible, including one using "and" instead of "or", and even defining implication as an atom and then defining "and" and "or" in terms of the atom. Regardless, given a material implication, then.... The definition of the logical implication is $$\left(p \vdash q\right) \text{ iff } \left(\left(p\implies q\right) \text{ is a tautology } \top\right) \qquad \text{(2)}$$. (I think some sources use the notation $p \therefore q$ to mean $p \vdash q$ and $p \because q$ to mean $p \dashv q$. Personally I like $\therefore$ and $\because$ more because of their $\LaTeX$ commands \therefore and \because.). I have some problems with the definitions above.. 1. In definition (1), what is meant by "iff"? I always thought that "if and only if" was the material equivalence (biconditional) $a \iff b$, which is defined by $\left(a\implies b\right)\land\left(b\implies a\right)$, which is a binary operation of propositions, that is, it takes two propositions and makes a new one (analogous to the mathematical expression "$x+y$"). So, doing some scratchwork I found that $$\left(p \implies q\right) \iff \left(\lnot p \lor q\right)$$ reduces to $\top$. What does that mean? Stating just "$\top$" doesn't make sense; it's a constant. That would be like stating "$x+y$" as a definition. Definition (1) should be a logical statement about propositions, not a proposition itself. Right?. I assume that the "iff" in definition (1) is meant to mean that the propositions are "the same", since it's a definition (or if not a definition, a theorem). Since they are "the same", couldn't one say they are logically equivalent and write $$\left(p\implies q\right) \dashv\vdash \left(\lnot p \lor q\right) \qquad \text{(1b)}$$ (where $\dashv\vdash$ denotes logical equivalence, I believe)? This seems to be right, since it's not a binary operation of propositions but rather more like a binary relation of propositions: it takes two propositions and compares them (analogous to the mathematical statement "$x\lt y$"). So is the "iff" in (1) an abuse of notation, or could the phrase "if and only if" denote both material equivalence and logical equivalence?. 2. In definition (2), the word "iff" appears again, except this time I'm more confused. For certain, it can't denote the material equivalence $\iff$, because the definition isn't an operation connecting two propositions. It can't even denote the logical equivalence $\dashv\vdash$, because the definition isn't a relation comparing propositions. It appears to be a statement about logical statements! Apparently, the phrase "if and only if" has a third meaning, a so-called metalogical equivalence: a statement comparing two logical statements. Is there such thing as a meta-metalogical equivalence? Where does it end??. Also, what is meant by the "is" in "$\left(p\implies q\right) \text{ is a tautology } \top$"? Does it mean "is logically equivalent to"? If so, can definition (2) be rewritten... ? $$\left(p\vdash q\right)\text{ "is the same as" }\left(\left(p\iff q\right)\dashv\vdash\top\right) \qquad \text{(2b)}$$ If so, $\dashv\vdash$ would have to be defined beforehand.. 3. Lastly, setting all syntactic issues aside, the biggest problem I have of all is the fact that these definitions seem to be semantically dependent on each other. They're cyclic. If the "iff" in (1) and the "is" in (2) do indeed indicate logical equivalence, logical equivalence needs to be defined formally. One definition I found is $$\left(p \dashv\vdash q\right) \text{ iff } \left(\left(p\iff q\right) \text{ is a tautology } \top\right) \qquad \text{(3)}$$ The first thing I noticed off the bat is that logical equivalence uses itself in its own definition. I interpret $\left(\left(p\iff q\right) \text{ is a tautology } \top\right)$ to mean $\left(\left(p\iff q\right)\dashv\vdash\top\right)$, but that would mean the definition refers to itself. Maybe I'm interpreting the "is" wrongly?. Another possible definition I've seen is $$\left(p \dashv\vdash q\right) \text{ iff } \left(\left(p\vdash q\right) \land \left(p \dashv q\right)\right) \qquad \text{(4)}$$ which again, syntactically, I have another problem with because it uses the $\land$ operation for propositions, call it material conjunction, to connect two statements (unless there is a second interpretation to "and" that allows it to operate on two statements... call it logical conjunction, which hasn't been defined).. Semantically, $\dashv\vdash$ cannot use $\vdash$ and $\dashv$ in its definition because the definitions of $\vdash$ and $\dashv$ use $\dashv\vdash$ (see (2b)).. One more thing, how does this all relate to the logical consequence $P \models Q$ ?. • perhaps this iff for the case of (1) means that depending on the values of p and q, $p\implies q$ should have preciesly the same value as $\lnot p \lor q$? And the iff in (2) means that you simply define $(p⊢q)$ that way, e.g. when you say, a function is continuous if ( ... some stuff...). You might say, a function is continuous iff ( ... some stuff...), but that's just the same, you simply define this previously unknown notion of continuity. So it's a kind like, you might establish logical equivalence between things which meanings you know, and this thing has its own definition. – W_D. Jul 26, 2013 at 7:04. • but here you're a kind of saying, this notation will mean preciously the thing after the "iff" word. That's how i understand that. – W_D. Jul 26, 2013 at 7:05. In case 1., there's two options. On the one hand, the "iff" can be read definitionally, meaning we're just treating one side as an abbreviation for the other side. That is, we're saying "We'll write "$p \rightarrow q$" as a notational variant of "$\neg p \vee q$"." Alternatively, we could take "$\rightarrow$" as a primitive symbol along with "$\vee$", in which case the "iff" can be read as the logical equivalence relation between propositions in the formal language, i.e. as saying "$p \rightarrow q \dashv \vdash \neg p \vee q$.". In case 2., the "iff" is being used to state the logical equivalence of two statements in the metalanguage, i.e. the metalogical equivalence between the statement "$p \vdash q$" and "The formal sentence "$p \leftarrow q$" is a tautology.". As for case 3., you're right that our notion of logical consequence ($\vdash$) and our notion of tautology are closely interconnected. But you could simply rephrase the claim in 3. as "$p \dashv \vdash q$ iff $\vdash p \leftrightarrow q$." Then the question just boils down to defining what we mean by "$\vdash$", which is often done either via truth tables (in the case of propositional logic), or via some proof system. In either case, you can find an explication in any reasonable introductory book.. Two things should be noted.
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The first concerns our use of "$p \vdash q$". This sentence is an English-level sentence, not a sentence in the formal language itself. Thus, "$p \vdash q$" means "The formal proposition "$p$" entails the formal proposition "$q$" in the given formal logic." So if we ever happen to say, "$p \vdash q$ iff...", that's an immediate indication we're talking about an equivalence between sentences of English, not sentences of the formal language.
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https://afternotes.info/1-reading-read-chapter-1-3-of-tufte-read-chapter-3-of-bffg-https-m3challen/
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1. Reading. Read Chapter 1-3 of Tufte. Read Chapter 3 of BFFG (https://m3challenge.siam.org/
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a mathematical model that ranks roller coasters according to a thrill
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Determine dependent and independent variables for ranking roller
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coasters using your definition of “hrilling.” Use your ranking system to
rank at least 10 roller coasters. (You can find data to use at rcdb.com.)
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or errors that may arise from this process. 4. Find a data visualization from 2020 (e.g., from an online “newspaper” or “magazine”) that you really like. Indicate why you like
it, and also enumerate both its strengths and its weaknesses. Find a data visualization from 2020 (e.g., from an online “newspaper” or “magazine”) that has a “lie factor” or something else
that makes it misleading. Enumerate both the strengths and the
weaknesses of this visualization. Using the second visualization above, sketch the design of a better
visualization that shows the data in a less misleading manner. (If
the example you found above is totally irredeemable, feel free to
do this part of the question with a different graphic.) 5. Find some data set somewhere and create your own visualization with
that data set.
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The smarter way. to do assignments.. Please note that this is just a preview of a school assignment posted on our website by one of our clients. If you need assistance with this question too, please click on the Order button at the bottom of the page to get started.. 1. Reading. Read Chapter 1-3 of Tufte. Read Chapter 3 of BFFG (https://m3challenge.siam.org/. pdf). 2. On p. 11 of BFG, they present the following focal problem: “Create. a mathematical model that ranks roller coasters according to a thrill. factor that you define.”. Determine dependent and independent variables for ranking roller. coasters based on how thrilling they are. What are some possible. model parameters?. Remark: To do this problem successfully, you’ll also need to do some. earlier steps (such as brainstorming, perhaps through a mind map). It. is also required to submit these items as part of your problem solution. 3. Continue working with your model for a ranking system for roller. coasters using your definition of “hrilling.” Use your ranking system to. rank at least 10 roller coasters. (You can find data to use at rcdb.com.). Remark: Depending on the formulation of your model, it may be. necessary for this problem to simplify it to obtain something that you. know how to solve mathematically. That’s fine but indicate explicitly. in this problem what additional assumptions you make in order to. simplify the problem, and comment on any possible new limitations. or errors that may arise from this process. 4. Find a data visualization from 2020 (e.g., from an online “newspaper” or “magazine”) that you really like. Indicate why you like. it, and also enumerate both its strengths and its weaknesses. Find a data visualization from 2020 (e.g., from an online “newspaper” or “magazine”) that has a “lie factor” or something else. that makes it misleading. Enumerate both the strengths and the. weaknesses of this visualization. Using the second visualization above, sketch the design of a better. visualization that shows the data in a less misleading manner. (If. the example you found above is totally irredeemable, feel free to. do this part of the question with a different graphic.) 5. Find some data set somewhere and create your own visualization with. that data set.. ## GET HELP WITH THIS ASSIGNMENT TODAY. Clicking on this button will take you to our custom assignment page.
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My Math Forum Piecewise constant function on my Ti-84
Calculus Calculus Math Forum
April 6th, 2011, 04:38 AM #1 Newbie Joined: Apr 2011 Posts: 2 Thanks: 0 Piecewise constant function on my Ti-84 Today in class, we've talked about limits and the continuity of functions and this was an example: f(x)={ |x-1| if x<2 1-x+1/2 x² if 2?x?4 26/(5x-15) if x>4 This is a piecewise constant function, but how can I put this is my Ti-84 if I want to see a graph of it?
April 6th, 2011, 05:28 AM #2 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Piecewise constant function on my Ti-84 Y=abs(x-2)(x<2)+(1-x+x^2/2)(2?x?4)+(26/(5x-15))(x>4) The comparison operators are in 2ND-MATH. The abs() (absolute value) function is the first entry in the catalog in 2ND-0.
April 6th, 2011, 10:21 AM #3 Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 Texas Instruments make great calculators.
April 6th, 2011, 11:09 AM #4
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Re: Piecewise constant function on my Ti-84
Quote:
Originally Posted by greg1313 Y=abs(x-2)(x<2)+(1-x+x^2/2)(2?x?4)+(26/(5x-15))(x>4) The comparison operators are in 2ND-MATH. The abs() (absolute value) function is the first entry in the catalog in 2ND-0.
Thank you
but when I put them all together in "Y1=....", with the "+" between them, I became a completely different graph than it was supposed to be
that's OK now, because I've put them all apart into functions (Y1=abs(x-1)(x<2), Y2=(1-x+x²/2)... )
Y1 and Y3 are fine, but the Y2 gives me the complete parabola and not the one for 2?x?4
April 6th, 2011, 12:06 PM #5 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Piecewise constant function on my Ti-84 I made a mistake. Here's the correction, which displays a correct graph: Y=(abs(x-1))(x<2)+(1-x+x^2/2)(2?x)(x?4)+(26/(5x-15))(x>4)
Tags constant, function, piecewise, ti84
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My Math Forum Piecewise constant function on my Ti-84. Calculus Calculus Math Forum. April 6th, 2011, 04:38 AM #1 Newbie Joined: Apr 2011 Posts: 2 Thanks: 0 Piecewise constant function on my Ti-84 Today in class, we've talked about limits and the continuity of functions and this was an example: f(x)={ |x-1| if x<2 1-x+1/2 x² if 2?x?4 26/(5x-15) if x>4 This is a piecewise constant function, but how can I put this is my Ti-84 if I want to see a graph of it?. April 6th, 2011, 05:28 AM #2 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Piecewise constant function on my Ti-84 Y=abs(x-2)(x<2)+(1-x+x^2/2)(2?x?4)+(26/(5x-15))(x>4) The comparison operators are in 2ND-MATH. The abs() (absolute value) function is the first entry in the catalog in 2ND-0.. April 6th, 2011, 10:21 AM #3 Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 Texas Instruments make great calculators.. April 6th, 2011, 11:09 AM #4. Newbie. Joined: Apr 2011. Posts: 2. Thanks: 0. Re: Piecewise constant function on my Ti-84. Quote:. Originally Posted by greg1313 Y=abs(x-2)(x<2)+(1-x+x^2/2)(2?x?4)+(26/(5x-15))(x>4) The comparison operators are in 2ND-MATH. The abs() (absolute value) function is the first entry in the catalog in 2ND-0.. Thank you. but when I put them all together in "Y1=....", with the "+" between them, I became a completely different graph than it was supposed to be. that's OK now, because I've put them all apart into functions (Y1=abs(x-1)(x<2), Y2=(1-x+x²/2)... ). Y1 and Y3 are fine, but the Y2 gives me the complete parabola and not the one for 2?x?4. April 6th, 2011, 12:06 PM #5 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Piecewise constant function on my Ti-84 I made a mistake. Here's the correction, which displays a correct graph: Y=(abs(x-1))(x<2)+(1-x+x^2/2)(2?x)(x?4)+(26/(5x-15))(x>4). Tags constant, function, piecewise, ti84. Thread Tools Display Modes Linear Mode.
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Similar Threads Thread Thread Starter Forum Replies Last Post mariam Applied Math 1 December 27th, 2013 05:56 AM math89 Calculus 4 January 20th, 2013 01:03 PM FreaKariDunk Real Analysis 3 November 13th, 2012 11:36 PM alexmath Calculus 5 October 24th, 2012 06:09 PM mathkid Calculus 4 September 1st, 2012 07:18 PM. Contact - Home - Forums - Cryptocurrency Forum - Top.
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## How do you write the standard form of the equation of a circle?
The standard form of a circle’s equation is (x-h)² + (y-k)² = r² where (h,k) is the center and r is the radius. To convert an equation to standard form, you can always complete the square separately in x and y.
## What is standard form of a circle?
The equation of a circle comes in two forms: 1) The standard form: (x – h)2 + (y-k)2 = r2. 2) The general form : x2 + y2 + Dx + Ey + F = 0, where D, E, F are constants. If the equation of a circle is in the standard form, we can easily identify the center of the circle, (h, k), and the radius, r .
## How do you find the equation of a circle?
The formula for the equation of a circle is (x – h)2+ (y – k)2 = r2, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle. If a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point.
## What’s a standard form?
Standard form is a way of writing down very large or very small numbers easily. So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form. Small numbers can also be written in standard form.
## Is circle a function?
A circle is a curve. It can be generated by functions, but it’s not a function itself. Something to careful about is that defining a circle with a relation from x to y is NOT a function as there is multiple points with a given x-value, but it can be defined with a function parametrically.
## How do you plot a circle?
follow these steps:Realize that the circle is centered at the origin (no h and v) and place this point there.Calculate the radius by solving for r. Set r-squared = 16. Plot the radius points on the coordinate plane. Connect the dots to graph the circle using a smooth, round curve.
You might be interested: Q10 equation
## How do you find the standard form of a circle with the center and a point?
1 AnswerThe equation of a circle with center (h,k) and radius r is: r2=(x−h)2+(y−k)2.We are told that the center of this circle is (2,4) , so. r2=(x−2)2+(y−4)2.The radius of the circle is 5 , which means the equation of the circle is: 25=(x−2)2+(y−4)2.
## How do I calculate the area of a circle?
The area of a circle is pi times the radius squared (A = π r²).
## How do you convert standard form to standard form?
Step 1: Subtract how many decimal places there are from the power. Step 2: Remove the decimal point, add as many zeros as the number in Step 1. Step 1: Subtract how many decimal places there are from the power. There are 3 decimal places.
## What is Standard Form calculator?
Standard Form Calculator is a free online tool that displays the number in the standard form. The number can be in either integer form or the decimal form.
### Releated
#### Convert to an exponential equation
How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […]
#### H2o2 decomposition equation
What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […]
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## How do you write the standard form of the equation of a circle?. The standard form of a circle’s equation is (x-h)² + (y-k)² = r² where (h,k) is the center and r is the radius. To convert an equation to standard form, you can always complete the square separately in x and y.. ## What is standard form of a circle?. The equation of a circle comes in two forms: 1) The standard form: (x – h)2 + (y-k)2 = r2. 2) The general form : x2 + y2 + Dx + Ey + F = 0, where D, E, F are constants. If the equation of a circle is in the standard form, we can easily identify the center of the circle, (h, k), and the radius, r .. ## How do you find the equation of a circle?. The formula for the equation of a circle is (x – h)2+ (y – k)2 = r2, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle. If a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point.. ## What’s a standard form?. Standard form is a way of writing down very large or very small numbers easily. So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form. Small numbers can also be written in standard form.. ## Is circle a function?. A circle is a curve. It can be generated by functions, but it’s not a function itself. Something to careful about is that defining a circle with a relation from x to y is NOT a function as there is multiple points with a given x-value, but it can be defined with a function parametrically.. ## How do you plot a circle?. follow these steps:Realize that the circle is centered at the origin (no h and v) and place this point there.Calculate the radius by solving for r. Set r-squared = 16. Plot the radius points on the coordinate plane. Connect the dots to graph the circle using a smooth, round curve.. You might be interested: Q10 equation. ## How do you find the standard form of a circle with the center and a point?. 1 AnswerThe equation of a circle with center (h,k) and radius r is: r2=(x−h)2+(y−k)2.We are told that the center of this circle is (2,4) , so. r2=(x−2)2+(y−4)2.The radius of the circle is 5 , which means the equation of the circle is: 25=(x−2)2+(y−4)2.. ## How do I calculate the area of a circle?. The area of a circle is pi times the radius squared (A = π r²).. ## How do you convert standard form to standard form?. Step 1: Subtract how many decimal places there are from the power. Step 2: Remove the decimal point, add as many zeros as the number in Step 1. Step 1: Subtract how many decimal places there are from the power. There are 3 decimal places.. ## What is Standard Form calculator?. Standard Form Calculator is a free online tool that displays the number in the standard form. The number can be in either integer form or the decimal form.. ### Releated. #### Convert to an exponential equation. How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […].
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#### H2o2 decomposition equation. What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […].
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# Munn semigroup
In mathematics, the Munn semigroup is the inverse semigroup of isomorphisms between principal ideals of a semilattice (commutative semigroup of idempotents). Munn semigroups are named for the Scottish mathematician Douglas Munn (1929–2008).[1]
## Construction's steps
Let ${\displaystyle E}$ be a semilattice.
1) For all e in E, we define Ee: = {i ∈ E : i ≤ e} which is a principal ideal of E.
2) For all ef in E, we define Te,f as the set of isomorphisms of Ee onto Ef.
3) The Munn semigroup of the semilattice E is defined as: TE := ${\displaystyle \bigcup _{e,f\in E}}$ { Te,f : (ef) ∈ U }.
The semigroup's operation is composition of mappings. In fact, we can observe that TE ⊆ IE where IE is the symmetric inverse semigroup because all isomorphisms are partial one-one maps from subsets of E onto subsets of E.
The idempotents of the Munn semigroup are the identity maps 1Ee.
## Theorem
For every semilattice ${\displaystyle E}$, the semilattice of idempotents of ${\displaystyle T_{E}}$ is isomorphic to E.
## Example
Let ${\displaystyle E=\{0,1,2,...\}}$. Then ${\displaystyle E}$ is a semilattice under the usual ordering of the natural numbers (${\displaystyle 0<1<2<...}$). The principal ideals of ${\displaystyle E}$ are then ${\displaystyle En=\{0,1,2,...,n\}}$ for all ${\displaystyle n}$. So, the principal ideals ${\displaystyle Em}$ and ${\displaystyle En}$ are isomorphic if and only if ${\displaystyle m=n}$.
Thus ${\displaystyle T_{n,n}}$ = {${\displaystyle 1_{En}}$} where ${\displaystyle 1_{En}}$ is the identity map from En to itself, and ${\displaystyle T_{m,n}=\emptyset }$ if ${\displaystyle m\not =n}$. In this example, ${\displaystyle T_{E}=\{1_{E0},1_{E1},1_{E2},\ldots \}\cong E.}$
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# Munn semigroup. In mathematics, the Munn semigroup is the inverse semigroup of isomorphisms between principal ideals of a semilattice (commutative semigroup of idempotents). Munn semigroups are named for the Scottish mathematician Douglas Munn (1929–2008).[1]. ## Construction's steps. Let ${\displaystyle E}$ be a semilattice.. 1) For all e in E, we define Ee: = {i ∈ E : i ≤ e} which is a principal ideal of E.. 2) For all ef in E, we define Te,f as the set of isomorphisms of Ee onto Ef.. 3) The Munn semigroup of the semilattice E is defined as: TE := ${\displaystyle \bigcup _{e,f\in E}}$ { Te,f : (ef) ∈ U }.. The semigroup's operation is composition of mappings. In fact, we can observe that TE ⊆ IE where IE is the symmetric inverse semigroup because all isomorphisms are partial one-one maps from subsets of E onto subsets of E.. The idempotents of the Munn semigroup are the identity maps 1Ee.. ## Theorem. For every semilattice ${\displaystyle E}$, the semilattice of idempotents of ${\displaystyle T_{E}}$ is isomorphic to E.. ## Example.
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Let ${\displaystyle E=\{0,1,2,...\}}$. Then ${\displaystyle E}$ is a semilattice under the usual ordering of the natural numbers (${\displaystyle 0<1<2<...}$). The principal ideals of ${\displaystyle E}$ are then ${\displaystyle En=\{0,1,2,...,n\}}$ for all ${\displaystyle n}$. So, the principal ideals ${\displaystyle Em}$ and ${\displaystyle En}$ are isomorphic if and only if ${\displaystyle m=n}$.. Thus ${\displaystyle T_{n,n}}$ = {${\displaystyle 1_{En}}$} where ${\displaystyle 1_{En}}$ is the identity map from En to itself, and ${\displaystyle T_{m,n}=\emptyset }$ if ${\displaystyle m\not =n}$. In this example, ${\displaystyle T_{E}=\{1_{E0},1_{E1},1_{E2},\ldots \}\cong E.}$.
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https://math.stackexchange.com/questions/1405365/express-roots-in-polynomials-of-equation-x3x2-2x-1-0
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# Express roots in polynomials of equation $x^3+x^2-2x-1=0$
If $\alpha$ is a root of equation $x^3+x^2-2x-1=0$, then find the other two roots in polynomials of $\alpha$, with rational coefficients.
I've seen some other examples [1] that other roots were found for equations with certain properties (having only even-power terms, etc).
In the comment in this link, someone suggest that If $A$ is a root of $x^6−2x^5+3x^3−2x−1=0$, then so is $−A^5+2A^4−3A$, without further explanation (or maybe it's obvious to math experts, not to me) but I'm more interested in the underlying theory, preferably elementary, and techniques to solve problems of this kind.
Thanks!
• The given polynomial has one real, two complex roots. The two complex roots cannot be expressed as rational polynomials of the real root. – Neil W Aug 21 '15 at 22:08
• I'm really sorry, but I think I had the last coefficient wrong, it should be $-1$, I fixed the equation. – Ali Aug 22 '15 at 0:26
• I'm not sure why all the answers so far are misunderstanding this question, it seems clear to me. – DanielV Aug 22 '15 at 4:10
• Use the fact that the discriminant is a perfect square. The field extension generated by one of the roots includes all the roots. – Bill Kleinhans Aug 22 '15 at 5:12
• @BillKleinhans: can a field extension generated by a real root contain two non-real roots? – robjohn Aug 22 '15 at 7:32
The discriminant of the polynomial $p(x)=x^3+x^2-2x-1$ is $49$, which is a perfect square. It has no rational roots, so it is irreducible in $\Bbb{Q}[x]$. Together these facts imply that the Galois group of the polynomial is cyclic of order three. If $a$ is one of its zeros, we thus see that $\Bbb{Q}[a]$ is its splitting field. This means that the other zeros are also in $\Bbb{Q}[a]$, hence they are polynomials in $a$ with rational coefficients.
We could run full Cardano on it, but I have seen this polynomial too often, so I will take a shortcut. Let's write $\zeta=e^{2\pi i/7}$ and $$u=\zeta+\zeta^{-1}=2\cos\frac{2\pi}7.$$ We get from binomial formula that $$u^3=\zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3}$$ and $$u^2=\zeta^2+2+\zeta^{-2}.$$ Therefore $$p(u)=u^3+u^2-2u-1=\zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=\zeta^{-3}\frac{\zeta^7-1}{\zeta-1}=0.$$
So $p(x)$ is the minimal polynomial of $u=2\cos(2\pi/7)$. What about its other zeros? Galois theory tells us that the powers $\zeta^k, k=1,2,3,4,5,6$ are exactly the conjugates of $\zeta$. Therefore the conjugates of $u$ are of the form $\zeta^{k}+\zeta^{-k}=2\cos(2k\pi/7)$.
Observe that $$u^2-2=4\cos^2\frac{2\pi}7-2=2(2\cos^2\frac{2\pi}7-1)=2\cos\frac{4\pi}7$$ by the formula for the cosine of a doubled angle, so $u^2-2$ is one of the other zeros of $p(x)$. I hope that it is no longer a surprise that the third root is $2\cos\dfrac{8\pi}7$. See robjohn's answer for a way of quickly writing this as a polynomial of $u$ as well. Further observe that the angle doubling trick stops here because $2\cos\dfrac{16\pi}7=2\cos\dfrac{2\pi}7$.
We do get a cyclic splitting field whenever the discriminant of an irreducible cubic in $\Bbb{Z}[x]$ is a perfect square - that part generalizes. The trickery with roots of unity and cosines is somewhat special to this polynomial. However, by the Kronecker-Weber theorem all cyclic extensions of $\Bbb{Q}$ reside inside some cyclotomic extension. In other words the roots of such cubics can be written as polynomials with rational coefficients evaluated at some root of unity.
• Looks like today is a day of regular 7-gons for me :-) – Jyrki Lahtonen Aug 22 '15 at 17:06
• And notice that $(u^2-2)^2-2\equiv-u^2-u+1\pmod{u^3+u^2-2u-1}$ gives the other root I mentioned. It is nice to have a root to use :-) (+1) – robjohn Aug 22 '15 at 17:13
• The discriminant here is also strongly hinting that the conductor of this abelian extension might be seven. Search for discriminants of cyclotomic fields (and the behavior of discriminants in a tower of field extensions) for more. – Jyrki Lahtonen Aug 22 '15 at 17:21
• Thanks! i need to brush up on some of this stuff. It's been a while since I studied it as an undergraduate. – robjohn Aug 22 '15 at 19:27
If $a$ is one root, then we get $$\frac{x^3+x^2-2x+1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{1}$$ Using the quadratic equation to solve this yields $$\frac{-1-a\pm\sqrt{9-2a-3a^2}}2\tag{2}$$ for the other two roots.
After the Question Change
Changing the constant term only changes $(1)$ slightly: $$\frac{x^3+x^2-2x-1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{3}$$ and we still have $(2)$.
Since the discriminant is $9-2a-3a^2$, we know that if one root is between $\frac{-1-2\sqrt{7}}3$ and $\frac{-1+2\sqrt{7}}3$, then all three roots are. If any root is outside that interval, the other two roots will not be real.
After considering a comment by KCd, I see that $$(x-a)(x-a^2+2)(x+a^2+a-1)\equiv x^3+x^2-2x-1\pmod{a^3+a^2-2a-1}$$ Therefore, if $a$ is a root, then $a^2-2$ and $-a^2-a+1$ are also roots.
• This expression for the other two roots is not a polynomial in $a$, which is what the questioner is looking for. – DanielV Aug 22 '15 at 4:12
• Since that can't be done, should we not show what can be done? – robjohn Aug 22 '15 at 7:20
• @DanielV: now that the question has been changed, it is possible that the roots are rational functions of one another. However, just because people cannot fully answer a question, is it wrong to offer what one can give in hopes that it might help? – robjohn Aug 22 '15 at 14:31
• @DanielV: the last section now shows that we can write the other roots as polynomials in $a$. – robjohn Aug 22 '15 at 16:52
• If the question really cannot be answered, then a proof/explanation of that would be the answer. I saw the comment KCd left, and was hoping someone would leave a derivation of it~ – DanielV Aug 22 '15 at 23:54
Let $\beta$ be another root of the given equation, then: $\beta^3+\beta^2-2\beta+1 = 0 = \alpha^3+\alpha^2-2\alpha+1\to (\beta^3-\alpha^3)+(\beta^2-\alpha^2)-2(\beta-\alpha)=0\to (\beta-\alpha)(\beta^2+\alpha\beta+\alpha^2+\beta+\alpha-2)=0\to \beta^2+(\alpha+1)\beta+\alpha^2+\alpha-2=0$ since $\alpha \neq \beta$. Using quadratic formula we can find a formula for $\beta$ in term of $\alpha$. Is this fine? Note that this method assumes that all the real roots are distinct. So you should prove this first, and it is equally interesting finding for you as well.
• Solving that quadratic doesn't seem to yield a polynomial expression; the discriminant is $b^2-4ac=(\alpha+1)^2-4(\alpha^2+\alpha-2)=-3\alpha^2-2\alpha+9$... – Steven Stadnicki Aug 21 '15 at 21:59
• No I'm asking for polynomials, as it's stated in the title – Ali Aug 21 '15 at 22:13
• If we know two roots, $\alpha$ and $\beta$, then the third root is $-1-\alpha-\beta$ since the sum of the three roots is $-1$ by Vieta's Formulas. – robjohn Aug 21 '15 at 22:24
• @Ali The quadratic is a polynomial expressing the root $\beta$ in terms of $\alpha$. You can't do better, because there are two missing roots and no way to distinguish between them algebraically - you have to have both or neither. Having found the quadratic which is a polynomial expression, you can solve it to get the value of the two roots, but that turns out not to be a nice polynomial expression. – Mark Bennet Aug 21 '15 at 22:27
• The other two roots are $a^2-2$ and $-a^2-a-1$. How much Galois theory do you know? – KCd Aug 22 '15 at 3:28
In a more general manner, if $a$ is root of $$x^3+Ax^2+Bx+C=0$$ then $$x^3+Ax^2+Bx+C=(x-a)(x^2+Px+Q)$$ grouping terms we then have $$x^2 (a+A-P)+x (a P+B-Q)+a Q+C$$ and all coefficients must be zero. So, $$P=a+A$$ $$Q=B+aP=a^2+a A+B$$ The unused equation $$aQ+C=a^3+Aa^2+Ba+C=0$$ just confirms that $a$ is a root of the initial equation.
Now, solving the quadratic $x^2+Px+Q=0$, the two other roots are then given by $$x_{1,2}=\frac{1}{2} \left(-a-A\pm\sqrt{-3 a^2-2 a A+A^2-4 B}\right)$$
• He wants to represent the roots as polynomials of a given root, such as $r_1 = a,~ r_2 = B_2a^2 + B_1a + B_0,~ r_3 = C_2a^2 + C_1a + C_0$, where $B \subset \mathbb Q$ and $C \subset \mathbb Q$ – DanielV Aug 22 '15 at 4:08
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# Express roots in polynomials of equation $x^3+x^2-2x-1=0$. If $\alpha$ is a root of equation $x^3+x^2-2x-1=0$, then find the other two roots in polynomials of $\alpha$, with rational coefficients.. I've seen some other examples [1] that other roots were found for equations with certain properties (having only even-power terms, etc).. In the comment in this link, someone suggest that If $A$ is a root of $x^6−2x^5+3x^3−2x−1=0$, then so is $−A^5+2A^4−3A$, without further explanation (or maybe it's obvious to math experts, not to me) but I'm more interested in the underlying theory, preferably elementary, and techniques to solve problems of this kind.. Thanks!. • The given polynomial has one real, two complex roots. The two complex roots cannot be expressed as rational polynomials of the real root. – Neil W Aug 21 '15 at 22:08. • I'm really sorry, but I think I had the last coefficient wrong, it should be $-1$, I fixed the equation. – Ali Aug 22 '15 at 0:26. • I'm not sure why all the answers so far are misunderstanding this question, it seems clear to me. – DanielV Aug 22 '15 at 4:10. • Use the fact that the discriminant is a perfect square. The field extension generated by one of the roots includes all the roots. – Bill Kleinhans Aug 22 '15 at 5:12. • @BillKleinhans: can a field extension generated by a real root contain two non-real roots? – robjohn Aug 22 '15 at 7:32. The discriminant of the polynomial $p(x)=x^3+x^2-2x-1$ is $49$, which is a perfect square. It has no rational roots, so it is irreducible in $\Bbb{Q}[x]$. Together these facts imply that the Galois group of the polynomial is cyclic of order three. If $a$ is one of its zeros, we thus see that $\Bbb{Q}[a]$ is its splitting field. This means that the other zeros are also in $\Bbb{Q}[a]$, hence they are polynomials in $a$ with rational coefficients.. We could run full Cardano on it, but I have seen this polynomial too often, so I will take a shortcut. Let's write $\zeta=e^{2\pi i/7}$ and $$u=\zeta+\zeta^{-1}=2\cos\frac{2\pi}7.$$ We get from binomial formula that $$u^3=\zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3}$$ and $$u^2=\zeta^2+2+\zeta^{-2}.$$ Therefore $$p(u)=u^3+u^2-2u-1=\zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=\zeta^{-3}\frac{\zeta^7-1}{\zeta-1}=0.$$. So $p(x)$ is the minimal polynomial of $u=2\cos(2\pi/7)$. What about its other zeros? Galois theory tells us that the powers $\zeta^k, k=1,2,3,4,5,6$ are exactly the conjugates of $\zeta$. Therefore the conjugates of $u$ are of the form $\zeta^{k}+\zeta^{-k}=2\cos(2k\pi/7)$.. Observe that $$u^2-2=4\cos^2\frac{2\pi}7-2=2(2\cos^2\frac{2\pi}7-1)=2\cos\frac{4\pi}7$$ by the formula for the cosine of a doubled angle, so $u^2-2$ is one of the other zeros of $p(x)$. I hope that it is no longer a surprise that the third root is $2\cos\dfrac{8\pi}7$. See robjohn's answer for a way of quickly writing this as a polynomial of $u$ as well. Further observe that the angle doubling trick stops here because $2\cos\dfrac{16\pi}7=2\cos\dfrac{2\pi}7$.. We do get a cyclic splitting field whenever the discriminant of an irreducible cubic in $\Bbb{Z}[x]$ is a perfect square - that part generalizes. The trickery with roots of unity and cosines is somewhat special to this polynomial. However, by the Kronecker-Weber theorem all cyclic extensions of $\Bbb{Q}$ reside inside some cyclotomic extension. In other words the roots of such cubics can be written as polynomials with rational coefficients evaluated at some root of unity.. • Looks like today is a day of regular 7-gons for me :-) – Jyrki Lahtonen Aug 22 '15 at 17:06. • And notice that $(u^2-2)^2-2\equiv-u^2-u+1\pmod{u^3+u^2-2u-1}$ gives the other root I mentioned. It is nice to have a root to use :-) (+1) – robjohn Aug 22 '15 at 17:13. • The discriminant here is also strongly hinting that the conductor of this abelian extension might be seven. Search for discriminants of cyclotomic fields (and the behavior of discriminants in a tower of field extensions) for more. – Jyrki Lahtonen Aug 22 '15 at 17:21. • Thanks! i need to brush up on some of this stuff. It's been a while since I studied it as an undergraduate. – robjohn Aug 22 '15 at 19:27. If $a$ is one root, then we get $$\frac{x^3+x^2-2x+1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{1}$$ Using the quadratic equation to solve this yields $$\frac{-1-a\pm\sqrt{9-2a-3a^2}}2\tag{2}$$ for the other two roots.. After the Question Change. Changing the constant term only changes $(1)$ slightly: $$\frac{x^3+x^2-2x-1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{3}$$ and we still have $(2)$.. Since the discriminant is $9-2a-3a^2$, we know that if one root is between $\frac{-1-2\sqrt{7}}3$ and $\frac{-1+2\sqrt{7}}3$, then all three roots are. If any root is outside that interval, the other two roots will not be real.. After considering a comment by KCd, I see that $$(x-a)(x-a^2+2)(x+a^2+a-1)\equiv x^3+x^2-2x-1\pmod{a^3+a^2-2a-1}$$ Therefore, if $a$ is a root, then $a^2-2$ and $-a^2-a+1$ are also roots.. • This expression for the other two roots is not a polynomial in $a$, which is what the questioner is looking for. – DanielV Aug 22 '15 at 4:12. • Since that can't be done, should we not show what can be done? – robjohn Aug 22 '15 at 7:20. • @DanielV: now that the question has been changed, it is possible that the roots are rational functions of one another. However, just because people cannot fully answer a question, is it wrong to offer what one can give in hopes that it might help? – robjohn Aug 22 '15 at 14:31. • @DanielV: the last section now shows that we can write the other roots as polynomials in $a$. – robjohn Aug 22 '15 at 16:52. • If the question really cannot be answered, then a proof/explanation of that would be the answer. I saw the comment KCd left, and was hoping someone would leave a derivation of it~ – DanielV Aug 22 '15 at 23:54. Let $\beta$ be another root of the given equation, then: $\beta^3+\beta^2-2\beta+1 = 0 = \alpha^3+\alpha^2-2\alpha+1\to (\beta^3-\alpha^3)+(\beta^2-\alpha^2)-2(\beta-\alpha)=0\to (\beta-\alpha)(\beta^2+\alpha\beta+\alpha^2+\beta+\alpha-2)=0\to \beta^2+(\alpha+1)\beta+\alpha^2+\alpha-2=0$ since $\alpha \neq \beta$. Using quadratic formula we can find a formula for $\beta$ in term of $\alpha$. Is this fine? Note that this method assumes that all the real roots are distinct. So you should prove this first, and it is equally interesting finding for you as well.. • Solving that quadratic doesn't seem to yield a polynomial expression; the discriminant is $b^2-4ac=(\alpha+1)^2-4(\alpha^2+\alpha-2)=-3\alpha^2-2\alpha+9$... – Steven Stadnicki Aug 21 '15 at 21:59. • No I'm asking for polynomials, as it's stated in the title – Ali Aug 21 '15 at 22:13. • If we know two roots, $\alpha$ and $\beta$, then the third root is $-1-\alpha-\beta$ since the sum of the three roots is $-1$ by Vieta's Formulas. – robjohn Aug 21 '15 at 22:24. • @Ali The quadratic is a polynomial expressing the root $\beta$ in terms of $\alpha$. You can't do better, because there are two missing roots and no way to distinguish between them algebraically - you have to have both or neither. Having found the quadratic which is a polynomial expression, you can solve it to get the value of the two roots, but that turns out not to be a nice polynomial expression. – Mark Bennet Aug 21 '15 at 22:27. • The other two roots are $a^2-2$ and $-a^2-a-1$. How much Galois theory do you know? – KCd Aug 22 '15 at 3:28. In a more general manner, if $a$ is root of $$x^3+Ax^2+Bx+C=0$$ then $$x^3+Ax^2+Bx+C=(x-a)(x^2+Px+Q)$$ grouping terms we then have $$x^2 (a+A-P)+x (a P+B-Q)+a Q+C$$ and all coefficients must be zero. So, $$P=a+A$$ $$Q=B+aP=a^2+a A+B$$ The unused equation $$aQ+C=a^3+Aa^2+Ba+C=0$$ just confirms that $a$ is a root of the initial equation.
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Now, solving the quadratic $x^2+Px+Q=0$, the two other roots are then given by $$x_{1,2}=\frac{1}{2} \left(-a-A\pm\sqrt{-3 a^2-2 a A+A^2-4 B}\right)$$. • He wants to represent the roots as polynomials of a given root, such as $r_1 = a,~ r_2 = B_2a^2 + B_1a + B_0,~ r_3 = C_2a^2 + C_1a + C_0$, where $B \subset \mathbb Q$ and $C \subset \mathbb Q$ – DanielV Aug 22 '15 at 4:08.
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## Speed and Velocity, the Identical Twins
Alright. This is a simple question, but it does give me some problems. What's the difference between Speed and Velocity? They're basically the same things, at least in my mind. And it would be nice to truly know the difference so that I can use it to calculate stuff like displacement and acceleration.
PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
Recognitions: Gold Member Science Advisor Staff Emeritus Speed is just magnitude velocity is magnitude and direction. Thus an object's speed would be 5 mph, but its velocity might be 5 mph East. If you have two cars, one driving west at 60 mph and the other driving East at 60 mph, they have the same speed but different velocities. As far as acceleration goes, it is a change in velocity, which could mean a change of speed, direction, or both. For example, an object traveling in a circle at a constant speed, is constantly changing direction and therefore is undergoing acceleration.
Recognitions: Gold Member Science Advisor Staff Emeritus Another way of saying what Janus did: "velocity" is a vector quantity. "Speed" is the magnitude of the velocity vector. In a very simple, one dimensional case, an object moving in the positive direction with speed 10 m/s would have velocity +10 m/s while one moving at the same speed in the negative direction would have velocity -10 m/s. In two dimensions, if we take the positive x-axis pointing east and the positive y-axis north, then a car moving east at 60 mph would have velocity vector <60, 0> while one moving west at 60 mph would have velocity vector <-60,0>. Similarly, a car moving north at 60 mph would have velocity vector <0, 60> and a car moving south at 60 mph would have velocity vector <0,-60>. A car moving north-east at 60 mph is a little harder. Its velocity vector is <60/√(2), 60/√(2)>. The speed in every one of those examples is, of course, 60 mph.
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## Speed and Velocity, the Identical Twins. Alright. This is a simple question, but it does give me some problems. What's the difference between Speed and Velocity? They're basically the same things, at least in my mind. And it would be nice to truly know the difference so that I can use it to calculate stuff like displacement and acceleration.. PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire. Recognitions: Gold Member Science Advisor Staff Emeritus Speed is just magnitude velocity is magnitude and direction. Thus an object's speed would be 5 mph, but its velocity might be 5 mph East. If you have two cars, one driving west at 60 mph and the other driving East at 60 mph, they have the same speed but different velocities. As far as acceleration goes, it is a change in velocity, which could mean a change of speed, direction, or both. For example, an object traveling in a circle at a constant speed, is constantly changing direction and therefore is undergoing acceleration.. Recognitions: Gold Member Science Advisor Staff Emeritus Another way of saying what Janus did: "velocity" is a vector quantity. "Speed" is the magnitude of the velocity vector. In a very simple, one dimensional case, an object moving in the positive direction with speed 10 m/s would have velocity +10 m/s while one moving at the same speed in the negative direction would have velocity -10 m/s.
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In two dimensions, if we take the positive x-axis pointing east and the positive y-axis north, then a car moving east at 60 mph would have velocity vector <60, 0> while one moving west at 60 mph would have velocity vector <-60,0>. Similarly, a car moving north at 60 mph would have velocity vector <0, 60> and a car moving south at 60 mph would have velocity vector <0,-60>. A car moving north-east at 60 mph is a little harder. Its velocity vector is <60/√(2), 60/√(2)>. The speed in every one of those examples is, of course, 60 mph.
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It is currently 19 Oct 2017, 17:05
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# In a set of consecutive ODD integers, the mean ALWAYS equals
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In a set of consecutive ODD integers, the mean ALWAYS equals [#permalink]
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05 Nov 2007, 07:16
In a set of consecutive ODD integers, the mean ALWAYS equals the median.
In a set of consecutive EVEN integers, the mean ALWAYS equals the median.
True?
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05 Nov 2007, 07:22
Yes, I think this is true.
Also
In a set of consecutive integers, the mean ALWAYS equals the median.
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05 Nov 2007, 14:17
FALSE
The Correct Statements:
In a set of an odd number of consecutive ODD integers, the mean ALWAYS equals the median.
In a set of an odd number of consecutive EVEN integers, the mean ALWAYS equals the median.
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05 Nov 2007, 22:01
1
This post was
BOOKMARKED
bmwhype2 wrote:
In a set of consecutive ODD integers, the mean ALWAYS equals the median.
In a set of consecutive EVEN integers, the mean ALWAYS equals the median.
True?
true for all (even, odd or both) consecutive integers.
Mishari wrote:
FALSE
The Correct Statements:
In a set of an odd number of consecutive ODD integers, the mean ALWAYS equals the median.
In a set of an odd number of consecutive EVEN integers, the mean ALWAYS equals the median.
any example?
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05 Nov 2007, 23:59
Both the stats
Quote:
In a set of consecutive ODD integers, the mean ALWAYS equals the median.
In a set of consecutive EVEN integers, the mean ALWAYS equals the median.
and
Quote:
In a set of an odd number of consecutive ODD integers, the mean ALWAYS equals the median.
In a set of an odd number of consecutive EVEN integers, the mean ALWAYS equals the median.
both stats hold good.....
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Re: In a set of consecutive ODD integers, the mean ALWAYS equals [#permalink]
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Re: In a set of consecutive ODD integers, the mean ALWAYS equals [#permalink]
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Hi All,
While this post originally goes back to 2007, and most (if not all) of the posters are probably gone, the questions posed are essentially Number Properties. They can ALL be proven by TESTing VALUES, although it does not appear that anyone went to the trouble of proving what they believed.
Here is the proof:
1) In a set of consecutive ODD integers, the mean ALWAYS equals the median.
Here are a series of examples to prove that this is TRUE.
{1, 3}
Mean = (1+3)/2 = 2
Median = (1+3)/2 = 2
Mean = Median
{1, 3, 5}
Mean = (1+3+5)/3 = 3
Median = 3
Mean = Median
{-3, -1, 1, 3}
Mean = (-3-1+1+3)/4 = 0
Median = (-1+1)/2 = 0
Mean = Median
{-5, -3, -1, 1, 3}
Mean = (-5-3-1+1+3)/5 = -1
Median = -1
Mean = Median
Using similar methods, you can also prove that the following is true:
2) In a set of consecutive EVEN integers, the mean ALWAYS equals the median.
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Re: In a set of consecutive ODD integers, the mean ALWAYS equals [#permalink] 01 Apr 2015, 20:21
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It is currently 19 Oct 2017, 17:05. ### GMAT Club Daily Prep. #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. # Events & Promotions. ###### Events & Promotions in June. Open Detailed Calendar. # In a set of consecutive ODD integers, the mean ALWAYS equals. Author Message. TAGS:. ### Hide Tags. CEO. Joined: 21 Jan 2007. Posts: 2734. Kudos [?]: 1046 [0], given: 4. Location: New York City. In a set of consecutive ODD integers, the mean ALWAYS equals [#permalink]. ### Show Tags. 05 Nov 2007, 07:16. In a set of consecutive ODD integers, the mean ALWAYS equals the median.. In a set of consecutive EVEN integers, the mean ALWAYS equals the median.. True?. Kudos [?]: 1046 [0], given: 4. Senior Manager. Joined: 19 Feb 2007. Posts: 325. Kudos [?]: 66 [0], given: 0. ### Show Tags. 05 Nov 2007, 07:22. Yes, I think this is true.. Also. In a set of consecutive integers, the mean ALWAYS equals the median.. Kudos [?]: 66 [0], given: 0. Director. Joined: 30 Nov 2006. Posts: 591. Kudos [?]: 314 [0], given: 0. Location: Kuwait. ### Show Tags. 05 Nov 2007, 14:17. FALSE. The Correct Statements:. In a set of an odd number of consecutive ODD integers, the mean ALWAYS equals the median.. In a set of an odd number of consecutive EVEN integers, the mean ALWAYS equals the median.. Kudos [?]: 314 [0], given: 0. SVP. Joined: 29 Aug 2007. Posts: 2472. Kudos [?]: 843 [0], given: 19. ### Show Tags. 05 Nov 2007, 22:01. 1. This post was. BOOKMARKED. bmwhype2 wrote:. In a set of consecutive ODD integers, the mean ALWAYS equals the median.. In a set of consecutive EVEN integers, the mean ALWAYS equals the median.. True?. true for all (even, odd or both) consecutive integers.. Mishari wrote:. FALSE. The Correct Statements:. In a set of an odd number of consecutive ODD integers, the mean ALWAYS equals the median.. In a set of an odd number of consecutive EVEN integers, the mean ALWAYS equals the median.. any example?. Kudos [?]: 843 [0], given: 19. Manager. Joined: 25 Nov 2006. Posts: 59. Kudos [?]: 2 [0], given: 0. ### Show Tags. 05 Nov 2007, 23:59. Both the stats. Quote:. In a set of consecutive ODD integers, the mean ALWAYS equals the median.. In a set of consecutive EVEN integers, the mean ALWAYS equals the median.. and. Quote:. In a set of an odd number of consecutive ODD integers, the mean ALWAYS equals the median.. In a set of an odd number of consecutive EVEN integers, the mean ALWAYS equals the median.. both stats hold good...... Kudos [?]: 2 [0], given: 0. GMAT Club Legend. Joined: 09 Sep 2013. Posts: 16652. Kudos [?]: 273 [0], given: 0. Re: In a set of consecutive ODD integers, the mean ALWAYS equals [#permalink]. ### Show Tags. 01 Apr 2015, 14:36. Hello from the GMAT Club BumpBot!. Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).. Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.. _________________. Kudos [?]: 273 [0], given: 0. EMPOWERgmat Instructor. Status: GMAT Assassin/Co-Founder. Affiliations: EMPOWERgmat. Joined: 19 Dec 2014. Posts: 9984. Kudos [?]: 3410 [0], given: 172. Location: United States (CA). GMAT 1: 800 Q51 V49. GRE 1: 340 Q170 V170. Re: In a set of consecutive ODD integers, the mean ALWAYS equals [#permalink]. ### Show Tags. 01 Apr 2015, 20:21. Hi All,. While this post originally goes back to 2007, and most (if not all) of the posters are probably gone, the questions posed are essentially Number Properties. They can ALL be proven by TESTing VALUES, although it does not appear that anyone went to the trouble of proving what they believed.. Here is the proof:. 1) In a set of consecutive ODD integers, the mean ALWAYS equals the median.. Here are a series of examples to prove that this is TRUE.. {1, 3}. Mean = (1+3)/2 = 2. Median = (1+3)/2 = 2. Mean = Median. {1, 3, 5}. Mean = (1+3+5)/3 = 3. Median = 3. Mean = Median. {-3, -1, 1, 3}. Mean = (-3-1+1+3)/4 = 0. Median = (-1+1)/2 = 0. Mean = Median. {-5, -3, -1, 1, 3}. Mean = (-5-3-1+1+3)/5 = -1. Median = -1. Mean = Median. Using similar methods, you can also prove that the following is true:. 2) In a set of consecutive EVEN integers, the mean ALWAYS equals the median.. GMAT assassins aren't born, they're made,. Rich. _________________. 760+: Learn What GMAT Assassins Do to Score at the Highest Levels. Contact Rich at: Rich.C@empowergmat.com. # Rich Cohen. Co-Founder & GMAT Assassin. Special Offer: Save \$75 + GMAT Club Tests Free. Official GMAT Exam Packs + 70 Pt. Improvement Guarantee. www.empowergmat.com/. ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************. Kudos [?]: 3410 [0], given: 172. Re: In a set of consecutive ODD integers, the mean ALWAYS equals [#permalink] 01 Apr 2015, 20:21. Display posts from previous: Sort by.
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Post subject: Maths questionsPosted: Sun Jan 10, 2016 10:51 pm
Joined: Sat Jan 09, 2016 5:45 pm
Posts: 60
Hi all,
Can someone help explain the following:
Mia is running steadily around the track. She does one lap every 60 seconds. Gareth is running steadily round the track in the opposite direction.
If they meet each other every 24 seconds, how long does it take Gareth to do a lap of the track?
And
Mr Sahota can harvest an acre of maize in half the time it takes him to harvest an acre of millet. To harvest 3 acres of each crop takes him a total of 6 days. How long would it take him in total to harvest 5 acres of maize and 2 acres of millet?
Thanks!
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Post subject: Re: Maths questionsPosted: Sun Jan 10, 2016 11:04 pm
Joined: Mon Feb 12, 2007 1:21 pm
Posts: 11942
How far have you got on the second one? If 3 of each takes 6 days can you think how to find how many millet [or maize] that would be in that time?
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Post subject: Re: Maths questionsPosted: Mon Jan 11, 2016 6:37 am
Joined: Sat Jan 09, 2016 5:45 pm
Posts: 60
Thanks.
To be honest I got quite tied up with the second one and was trying to work out half days?!
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Post subject: Re: Maths questionsPosted: Mon Jan 11, 2016 9:26 am
Joined: Mon Mar 02, 2015 11:15 am
Posts: 126
MCLC wrote:
Mr Sahota can harvest an acre of maize in half the time it takes him to harvest an acre of millet. To harvest 3 acres of each crop takes him a total of 6 days. How long would it take him in total to harvest 5 acres of maize and 2 acres of millet?
I think the key 'insight' they're looking for here is that if Mr Sahota can harvest an acre of maize in half the time it takes him to harvest an acre of millet, then of the total 6 days he spends harvesting, 1/3 of that time (i.e. 2 days) must be spent harvesting maize and 2/3 of that time (i.e. 4 days) must be spent harvesting millet.
Given that within that 6 days he harvested 3 acres of each crop, he harvests maize at a rate of 3 acres per 2 days (1.5 acres per day) and millet at 3 acres per 4 days (0.75 acres per day).
We then just need to multiply each 'acres per day' by the number of acres in the question.
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Post subject: Re: Maths questionsPosted: Mon Jan 11, 2016 11:29 am
Joined: Sat Jan 09, 2016 5:45 pm
Posts: 60
Thanks Goodheart. That really helps.
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It is currently Tue Dec 06, 2016 7:52 pm. All times are UTC. Page 1 of 1 [ 5 posts ]. Print view Previous topic | Next topic. Author Message. Post subject: Maths questionsPosted: Sun Jan 10, 2016 10:51 pm. Joined: Sat Jan 09, 2016 5:45 pm. Posts: 60. Hi all,. Can someone help explain the following:. Mia is running steadily around the track. She does one lap every 60 seconds. Gareth is running steadily round the track in the opposite direction.. If they meet each other every 24 seconds, how long does it take Gareth to do a lap of the track?. And. Mr Sahota can harvest an acre of maize in half the time it takes him to harvest an acre of millet. To harvest 3 acres of each crop takes him a total of 6 days. How long would it take him in total to harvest 5 acres of maize and 2 acres of millet?. Thanks!. Top. Post subject: Re: Maths questionsPosted: Sun Jan 10, 2016 11:04 pm. Joined: Mon Feb 12, 2007 1:21 pm. Posts: 11942. How far have you got on the second one? If 3 of each takes 6 days can you think how to find how many millet [or maize] that would be in that time?. Top. Post subject: Re: Maths questionsPosted: Mon Jan 11, 2016 6:37 am. Joined: Sat Jan 09, 2016 5:45 pm. Posts: 60. Thanks.. To be honest I got quite tied up with the second one and was trying to work out half days?!. Top. Post subject: Re: Maths questionsPosted: Mon Jan 11, 2016 9:26 am. Joined: Mon Mar 02, 2015 11:15 am. Posts: 126. MCLC wrote:. Mr Sahota can harvest an acre of maize in half the time it takes him to harvest an acre of millet. To harvest 3 acres of each crop takes him a total of 6 days. How long would it take him in total to harvest 5 acres of maize and 2 acres of millet?. I think the key 'insight' they're looking for here is that if Mr Sahota can harvest an acre of maize in half the time it takes him to harvest an acre of millet, then of the total 6 days he spends harvesting, 1/3 of that time (i.e. 2 days) must be spent harvesting maize and 2/3 of that time (i.e. 4 days) must be spent harvesting millet.. Given that within that 6 days he harvested 3 acres of each crop, he harvests maize at a rate of 3 acres per 2 days (1.5 acres per day) and millet at 3 acres per 4 days (0.75 acres per day).. We then just need to multiply each 'acres per day' by the number of acres in the question.. Top. Post subject: Re: Maths questionsPosted: Mon Jan 11, 2016 11:29 am. Joined: Sat Jan 09, 2016 5:45 pm. Posts: 60. Thanks Goodheart. That really helps.. Top. Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending. Page 1 of 1 [ 5 posts ]. All times are UTC. #### Who is online. Users browsing this forum: No registered users and 4 guests. You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum.
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# Bound the minimum eigenvalue of a symmetric matrix via matrix norms
I'm reading a paper in which the authors prove an inequality of the following form:
$$\lVert H-H'\rVert_2 \leq \lVert H-H'\rVert_F \leq \epsilon \tag 1$$
Here $$H$$ and $$H'$$ are symmetric real matrices ($$H'$$ has all positive eigenvalues, if that matters), and the norms are the $$L_2$$ matrix norm and the Frobenius norm, respectively. With no justification the authors then claim:
$$\lambda_\text{min}(H) \geq \lambda_\text{min}(H') - \epsilon \tag 2$$
where $$\lambda_\text{min}$$ is the minimum eigenvalue of a matrix.
I can't see how to justify this, or even if (2) is even intended to be deduced from the (1). Here is the paper - the end of the proof of Lemma 3.2, page 6.
This answer is based on this one. Below we will be working with some arbitrary inner product, and when we take the norm of a matrix, this means the operator norm associated with the vector norm we're using. We have:
Theorem. If $$A$$ and $$B$$ are real symmetric, then:
$$\lambda_\text{min} (A) \geq \lambda_\text{min} (B) - \lVert A-B\rVert$$ $$\lambda_\text{max} (A) \leq \lambda_\text{max} (B) + \lVert A-B\rVert$$
To prove this, the key is the expression $$x^T Mx$$, where $$M$$ is a symmetric matrix and $$x$$ has unit norm. We need two lemmas about this expression.
Lemma 1. For any matrix $$M$$ and any unit norm $$x$$: $$-\lVert M\rVert \leq x^T Mx\leq \lVert M\rVert$$ Proof. Simple application of Cauchy-Schwartz and of the definition of an operator norm: $$|x^TMx|\leq\lVert x\rVert \lVert Mx\rVert\leq \lVert x\rVert^2 \lVert M\rVert=\lVert M\rVert$$
Lemma 2. For any symmetric matrix $$M$$ and any unit norm $$x$$: $$\lambda_\text{min}(M) \leq x^T M x \leq \lambda_\text{max}(M)$$ and the bounds are attained as $$x$$ varies over the unit sphere.
Proof. Let $$M=P^TDP$$ where $$P$$ is orthogonal and $$D$$ is diagonal. Then $$x^TMx = (Px)^TD(Px)$$ As $$x$$ varies over the unit sphere, $$Px$$ varies also over the entire unit sphere, therefore the range of the latter expression above is simply the range of $$y^TDy$$ as $$y$$ ranges over the unit sphere. By the rearrangement inequality and some other simple arguments, the minimum is attained when $$y$$ is an eigenvector associated with $$\lambda_\text{min}(M)$$ and the maximum when $$y$$ is an eigenvector associated with $$\lambda_\text{max}(M)$$.
Finally we can prove the theorem. For any unit norm $$x$$, we have
$$x^TAx = x^TBx + x^T(A-B)x$$
By applying Lemma 1 to the second term and Lemma 2 to the first term, the minimum of the left hand side is at least $$\lambda_\text{min} (B)-\lVert A-B\rVert$$. By Lemma 2, we know that the minimum of the left hand side is equal to $$\lambda_\text{min} (A)$$. A similar argument shows the other inequality in the theorem.
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# Bound the minimum eigenvalue of a symmetric matrix via matrix norms. I'm reading a paper in which the authors prove an inequality of the following form:. $$\lVert H-H'\rVert_2 \leq \lVert H-H'\rVert_F \leq \epsilon \tag 1$$. Here $$H$$ and $$H'$$ are symmetric real matrices ($$H'$$ has all positive eigenvalues, if that matters), and the norms are the $$L_2$$ matrix norm and the Frobenius norm, respectively. With no justification the authors then claim:. $$\lambda_\text{min}(H) \geq \lambda_\text{min}(H') - \epsilon \tag 2$$. where $$\lambda_\text{min}$$ is the minimum eigenvalue of a matrix.. I can't see how to justify this, or even if (2) is even intended to be deduced from the (1). Here is the paper - the end of the proof of Lemma 3.2, page 6.. This answer is based on this one. Below we will be working with some arbitrary inner product, and when we take the norm of a matrix, this means the operator norm associated with the vector norm we're using. We have:. Theorem. If $$A$$ and $$B$$ are real symmetric, then:. $$\lambda_\text{min} (A) \geq \lambda_\text{min} (B) - \lVert A-B\rVert$$ $$\lambda_\text{max} (A) \leq \lambda_\text{max} (B) + \lVert A-B\rVert$$. To prove this, the key is the expression $$x^T Mx$$, where $$M$$ is a symmetric matrix and $$x$$ has unit norm. We need two lemmas about this expression.. Lemma 1. For any matrix $$M$$ and any unit norm $$x$$: $$-\lVert M\rVert \leq x^T Mx\leq \lVert M\rVert$$ Proof. Simple application of Cauchy-Schwartz and of the definition of an operator norm: $$|x^TMx|\leq\lVert x\rVert \lVert Mx\rVert\leq \lVert x\rVert^2 \lVert M\rVert=\lVert M\rVert$$. Lemma 2. For any symmetric matrix $$M$$ and any unit norm $$x$$: $$\lambda_\text{min}(M) \leq x^T M x \leq \lambda_\text{max}(M)$$ and the bounds are attained as $$x$$ varies over the unit sphere.. Proof. Let $$M=P^TDP$$ where $$P$$ is orthogonal and $$D$$ is diagonal. Then $$x^TMx = (Px)^TD(Px)$$ As $$x$$ varies over the unit sphere, $$Px$$ varies also over the entire unit sphere, therefore the range of the latter expression above is simply the range of $$y^TDy$$ as $$y$$ ranges over the unit sphere. By the rearrangement inequality and some other simple arguments, the minimum is attained when $$y$$ is an eigenvector associated with $$\lambda_\text{min}(M)$$ and the maximum when $$y$$ is an eigenvector associated with $$\lambda_\text{max}(M)$$.. Finally we can prove the theorem. For any unit norm $$x$$, we have.
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$$x^TAx = x^TBx + x^T(A-B)x$$. By applying Lemma 1 to the second term and Lemma 2 to the first term, the minimum of the left hand side is at least $$\lambda_\text{min} (B)-\lVert A-B\rVert$$. By Lemma 2, we know that the minimum of the left hand side is equal to $$\lambda_\text{min} (A)$$. A similar argument shows the other inequality in the theorem.
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# DIOPHANTINE EQUATION
Posted 9 years ago
13 x^2+1=y^2, what is the procedure to solve the equation with integer solutions, wolfram gives the infinite integer solution for the equation, but how do they get it?
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Posted 9 years ago
HiThere are various ways to 'solve' that equation, If we just asked to solve it like Clear[x, y]; Solve[13 x^2 + 1 == y^2, {x, y}, Integers] We will get results like {{x -> ConditionalExpression[-(((649 - 180 Sqrt[13])^ C[1] - (649 + 180 Sqrt[13])^C[1])/(2 Sqrt[13])), C[1] \[Element] Integers && C[1] >= 0], which isn't very good, this is telling me that there could be an infinite number of solutions and that we need to add limits for x and y, so we could ask this Clear[x, y]; Solve[ 13 x^2 + 1 == y^2 && 0 < x < 10^10 && 0 < y < 10^10, {x, y}, Integers] {{x -> 180, y -> 649}, {x -> 233640, y -> 842401}, {x -> 303264540, y -> 1093435849}} We could also ask this Clear[x, y]; FindInstance[13 x^2 + 1 == y^2, {x, y}, Integers, 2] {{x -> -98475707056574514234147687427309858128133932917550159192926078\ 9721440877383880976428933444437557685123880, y -> -35505921118005041063603236141349722598219803651103555766541663\ 27589596496957941863373110484725700882173601}, {x -> 1187828648027618759642502802027120177356001334838507471515621279589\ 8286867954206065456648483958433444781041798143695420315713621488215926\ 4598844841133760412740750545423228379788307197484815120271546811447749\ 7159652848682155570439974210170748217810526303751922055046373538138102\ 0231971292940018133726241543544132192067186548069882096119540954036932\ 577807293981659210167396951286006140, y -> 428277709692864682878803963175563587116293964276476908497499183\ 3145609033134526228869899336669731155196749772066652590596524997159085\ 2519889096595069946728890727914390171744935742586153090418212768714274\ 7310625391649304220246973177349306393132073180127067871356808850048354\ 1013054379064542178999843193995372857640443443929933173325176356363542\ 2922385906493065904876438405625228363849}} Hope that helps a little
Posted 9 years ago
HiThere are various ways to 'solve' that equation, If we just asked to solve it like Clear[x, y]; Solve[13 x^2 + 1 == y^2, {x, y}, Integers] We will get results like {{x -> ConditionalExpression[-(((649 - 180 Sqrt[13])^ C[1] - (649 + 180 Sqrt[13])^C[1])/(2 Sqrt[13])), C[1] \[Element] Integers && C[1] >= 0], which isn't very good, this is telling me that there could be an infinite number of solutions and that we need to add limits for x and y, so we could ask this Clear[x, y]; Solve[ 13 x^2 + 1 == y^2 && 0 < x < 10^10 && 0 < y < 10^10, {x, y}, Integers] {{x -> 180, y -> 649}, {x -> 233640, y -> 842401}, {x -> 303264540, y -> 1093435849}} We could also ask this Clear[x, y]; FindInstance[13 x^2 + 1 == y^2, {x, y}, Integers, 2] {{x -> -98475707056574514234147687427309858128133932917550159192926078\ 9721440877383880976428933444437557685123880, y -> -35505921118005041063603236141349722598219803651103555766541663\ 27589596496957941863373110484725700882173601}, {x -> 1187828648027618759642502802027120177356001334838507471515621279589\ 8286867954206065456648483958433444781041798143695420315713621488215926\ 4598844841133760412740750545423228379788307197484815120271546811447749\ 7159652848682155570439974210170748217810526303751922055046373538138102\ 0231971292940018133726241543544132192067186548069882096119540954036932\ 577807293981659210167396951286006140, y -> 428277709692864682878803963175563587116293964276476908497499183\ 3145609033134526228869899336669731155196749772066652590596524997159085\ 2519889096595069946728890727914390171744935742586153090418212768714274\ 7310625391649304220246973177349306393132073180127067871356808850048354\ 1013054379064542178999843193995372857640443443929933173325176356363542\ 2922385906493065904876438405625228363849}} Hope that helps a little
Posted 9 years ago
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0. |. 4393 Views. |. 3 Replies. |. 0 Total Likes. View groups.... Share. GROUPS:. # DIOPHANTINE EQUATION. Posted 9 years ago. 13 x^2+1=y^2, what is the procedure to solve the equation with integer solutions, wolfram gives the infinite integer solution for the equation, but how do they get it?. 3 Replies. Sort By:. Posted 9 years ago. HiThere are various ways to 'solve' that equation, If we just asked to solve it like Clear[x, y]; Solve[13 x^2 + 1 == y^2, {x, y}, Integers] We will get results like {{x -> ConditionalExpression[-(((649 - 180 Sqrt[13])^ C[1] - (649 + 180 Sqrt[13])^C[1])/(2 Sqrt[13])), C[1] \[Element] Integers && C[1] >= 0], which isn't very good, this is telling me that there could be an infinite number of solutions and that we need to add limits for x and y, so we could ask this Clear[x, y]; Solve[ 13 x^2 + 1 == y^2 && 0 < x < 10^10 && 0 < y < 10^10, {x, y}, Integers] {{x -> 180, y -> 649}, {x -> 233640, y -> 842401}, {x -> 303264540, y -> 1093435849}} We could also ask this Clear[x, y]; FindInstance[13 x^2 + 1 == y^2, {x, y}, Integers, 2] {{x -> -98475707056574514234147687427309858128133932917550159192926078\ 9721440877383880976428933444437557685123880, y -> -35505921118005041063603236141349722598219803651103555766541663\ 27589596496957941863373110484725700882173601}, {x -> 1187828648027618759642502802027120177356001334838507471515621279589\ 8286867954206065456648483958433444781041798143695420315713621488215926\ 4598844841133760412740750545423228379788307197484815120271546811447749\ 7159652848682155570439974210170748217810526303751922055046373538138102\ 0231971292940018133726241543544132192067186548069882096119540954036932\ 577807293981659210167396951286006140, y -> 428277709692864682878803963175563587116293964276476908497499183\ 3145609033134526228869899336669731155196749772066652590596524997159085\ 2519889096595069946728890727914390171744935742586153090418212768714274\ 7310625391649304220246973177349306393132073180127067871356808850048354\ 1013054379064542178999843193995372857640443443929933173325176356363542\ 2922385906493065904876438405625228363849}} Hope that helps a little. Posted 9 years ago.
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HiThere are various ways to 'solve' that equation, If we just asked to solve it like Clear[x, y]; Solve[13 x^2 + 1 == y^2, {x, y}, Integers] We will get results like {{x -> ConditionalExpression[-(((649 - 180 Sqrt[13])^ C[1] - (649 + 180 Sqrt[13])^C[1])/(2 Sqrt[13])), C[1] \[Element] Integers && C[1] >= 0], which isn't very good, this is telling me that there could be an infinite number of solutions and that we need to add limits for x and y, so we could ask this Clear[x, y]; Solve[ 13 x^2 + 1 == y^2 && 0 < x < 10^10 && 0 < y < 10^10, {x, y}, Integers] {{x -> 180, y -> 649}, {x -> 233640, y -> 842401}, {x -> 303264540, y -> 1093435849}} We could also ask this Clear[x, y]; FindInstance[13 x^2 + 1 == y^2, {x, y}, Integers, 2] {{x -> -98475707056574514234147687427309858128133932917550159192926078\ 9721440877383880976428933444437557685123880, y -> -35505921118005041063603236141349722598219803651103555766541663\ 27589596496957941863373110484725700882173601}, {x -> 1187828648027618759642502802027120177356001334838507471515621279589\ 8286867954206065456648483958433444781041798143695420315713621488215926\ 4598844841133760412740750545423228379788307197484815120271546811447749\ 7159652848682155570439974210170748217810526303751922055046373538138102\ 0231971292940018133726241543544132192067186548069882096119540954036932\ 577807293981659210167396951286006140, y -> 428277709692864682878803963175563587116293964276476908497499183\ 3145609033134526228869899336669731155196749772066652590596524997159085\ 2519889096595069946728890727914390171744935742586153090418212768714274\ 7310625391649304220246973177349306393132073180127067871356808850048354\ 1013054379064542178999843193995372857640443443929933173325176356363542\ 2922385906493065904876438405625228363849}} Hope that helps a little. Posted 9 years ago.
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# Catching Up with New Web Sketchpad Functionality
This past January, we introduced the Web Sketchpad Tool Library and Viewer. The Tool Library is a collection of over 60 mathematical tools for customizing a Web Sketchpad model, making it possible for teachers to decide which tools students have available to them on an activity-by-activity basis. The Viewer is a site … Continue Reading ››
# The Folded Circle Construction
Of all the conic section construction techniques, my favorite is undoubtedly the approach that requires nothing more than a paper circle. Here's what to do: Draw or print a circle and its center, point A, on a sheet of paper. Cut out the circle. Mark a random point B anywhere on the circle. Then, fold … Continue Reading ››
# Dividing and Subdividing
Given a strip of paper, how might you divide it into fourths without using a ruler? Undoubtedly, you'd fold the strip in half and then in half again to locate the quarter marks. Now suppose that your goal is to divide a strip into sixths. You might start by folding the strip into thirds and … Continue Reading ››
# Circle Tracer Challenges
Geometry tends not to receive much love in elementary curricula, and that's a shame. In this post, I'll describe some of my new ideas for using Web Sketchpad to introduce young learners to fundamental properties of circles.
On page 1 of the websketch below (and here), begin by asking students to drag … Continue Reading ››
# What’s New with Web Sketchpad in 2019
For the past five years, Scott and I have featured interactive Web Sketchpad models in nearly all our Sine of the Times blog posts. As much fun as it's been to build "websketches" to share with you, we've really wanted to put the creative power of Web Sketchpad into your hands. And now, finally, that's … Continue Reading ››
# Multiplication Is Not Repeated Addition
On the NCTM discussion site myNCTM, there's currently an extended discussion about "Division and multiplication of fractions." As the discussion has continued, I've grown concerned with what I see as a fundamental problem with the way we often introduce multiplication as repeated addition: "Multiplying 4 by 5 means we're combining five groups of four items. … Continue Reading ››
# Find the Secret Number
We created the Web Sketchpad game below (and here) as part of our Dynamic Number project. It challenges elementary-age students to uncover the value of a secret number by collecting and analyzing clues that narrow its range of possible values. The game familiarizes students with inequality signs, introduces the use of x to represent … Continue Reading ››
# A Dynamic Approach to Finding Pirate Treasure
In his 1947 book, One, Two, Three...Infinity, physicist George Gamow poses a pirate treasure problem that has since become a classic. Below is my reworded statement of the puzzle.
Among a pirate's belongings you find the following note: The island where I buried my treasure contains a single palm tree. Find the tree. From the palm tree, … Continue Reading ››
# The Scaled Maps Problem
Below are two maps of the United Sates, with the smaller map a 50 percent scaled copy of the original. The edges of the two maps are parallel. Imagine that the maps are printed out, with one resting on top of the other. Believe it or not, you can stick a pin straight through both maps … Continue Reading ››
# Adding and Subtracting Integers
A recent post on the my NCTM discussion group asked about tools to help students visualize and understand addition and subtraction of integers. I always found this confusing for some of my Algebra 1 students, mainly because they had been told to memorize some rules about whether to add or subtract the two integers and … Continue Reading ››
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# Catching Up with New Web Sketchpad Functionality. This past January, we introduced the Web Sketchpad Tool Library and Viewer. The Tool Library is a collection of over 60 mathematical tools for customizing a Web Sketchpad model, making it possible for teachers to decide which tools students have available to them on an activity-by-activity basis. The Viewer is a site … Continue Reading ››. # The Folded Circle Construction. Of all the conic section construction techniques, my favorite is undoubtedly the approach that requires nothing more than a paper circle. Here's what to do: Draw or print a circle and its center, point A, on a sheet of paper. Cut out the circle. Mark a random point B anywhere on the circle. Then, fold … Continue Reading ››. # Dividing and Subdividing. Given a strip of paper, how might you divide it into fourths without using a ruler? Undoubtedly, you'd fold the strip in half and then in half again to locate the quarter marks. Now suppose that your goal is to divide a strip into sixths. You might start by folding the strip into thirds and … Continue Reading ››. # Circle Tracer Challenges. Geometry tends not to receive much love in elementary curricula, and that's a shame. In this post, I'll describe some of my new ideas for using Web Sketchpad to introduce young learners to fundamental properties of circles.. On page 1 of the websketch below (and here), begin by asking students to drag … Continue Reading ››. # What’s New with Web Sketchpad in 2019. For the past five years, Scott and I have featured interactive Web Sketchpad models in nearly all our Sine of the Times blog posts. As much fun as it's been to build "websketches" to share with you, we've really wanted to put the creative power of Web Sketchpad into your hands. And now, finally, that's … Continue Reading ››. # Multiplication Is Not Repeated Addition. On the NCTM discussion site myNCTM, there's currently an extended discussion about "Division and multiplication of fractions." As the discussion has continued, I've grown concerned with what I see as a fundamental problem with the way we often introduce multiplication as repeated addition: "Multiplying 4 by 5 means we're combining five groups of four items. … Continue Reading ››. # Find the Secret Number. We created the Web Sketchpad game below (and here) as part of our Dynamic Number project. It challenges elementary-age students to uncover the value of a secret number by collecting and analyzing clues that narrow its range of possible values. The game familiarizes students with inequality signs, introduces the use of x to represent … Continue Reading ››. # A Dynamic Approach to Finding Pirate Treasure. In his 1947 book, One, Two, Three...Infinity, physicist George Gamow poses a pirate treasure problem that has since become a classic. Below is my reworded statement of the puzzle.. Among a pirate's belongings you find the following note: The island where I buried my treasure contains a single palm tree. Find the tree. From the palm tree, … Continue Reading ››. # The Scaled Maps Problem. Below are two maps of the United Sates, with the smaller map a 50 percent scaled copy of the original. The edges of the two maps are parallel. Imagine that the maps are printed out, with one resting on top of the other. Believe it or not, you can stick a pin straight through both maps … Continue Reading ››. # Adding and Subtracting Integers.
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A recent post on the my NCTM discussion group asked about tools to help students visualize and understand addition and subtraction of integers. I always found this confusing for some of my Algebra 1 students, mainly because they had been told to memorize some rules about whether to add or subtract the two integers and … Continue Reading ››.
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# Matlab Programming Mathematical Problems Solutions
right here Programming Mathematical Problems Solutions [bsc]{};\ 2004-05-23, IEEE-CAMP, vol.59, IEEE Computer Society Lecture Notes, 2038, 13-20\ 2004-06-22, arXiv:astro/0409.3706 $astro-ph$[**astro/0409.3706**]{} [^1]: T. Mathematica is part of The Interactive Group, at the University of Pisa (Italy). Matlab Programming Mathematical Problems Solutions Abstract This section is about software implementation and optimization. Using the example of your program written using Matlab, the following examples illustrate our approach, how to solve your programs, and how to solve this program into code: How to create a code set (DFA) and make your program operate (RFI) Introduction | Theory and methods Introduction to Matlab can be seen as technical language to learn code on how to work with text files on Linux and Matlab.
## Matlab Homework Github
This kind of implementation only addresses specific applications see this the application they run. It requires no special knowledge (see also [Math 2:8]). Writing code in Matlab for writing operations (where DFA-based algorithms are used) is a fairly easy task. RFI’s are not something like computing routines — only a handful of things are possible to do in Python (like: creating a table of review or the ability to use group functions), and Going Here offer some common-sense (but not so common these days: programming for new mathematics). However, there are plenty of situations important link it takes computing the constants required to solve a problem. For example, one could try to compute the formulae, to set the answer to ‘yes’ to avoid such problems. In this case, some kind of matrices (or rather, rows, padding, etc.
## Matlab Homework Problems
) might be required and could be used instead. Now, the mathematics of RFI can be used to build an algorithm, by simply reversing the given C and DFA, to find the solution to a given problem. The objective of any RFI should be to find the coordinates of the solution without disturbing patterns, as in RFI-RISC-8 for Windows, or in RFI-DFA for Macintosh. In Matlab, a RFI provides an entry for each RDF-data structure. A “rule of thumb” of RFI calculations is: For a given RDF, we can draw patterns in the RDF file. Within these patterns we can use the RDFs/DFCAs/RDFSAs/PDFFs structure (or an RDF-spec table with reference to RDF-structure) to find the coordinates of the corresponding element of the elements in the resultant DFA. It is actually quite difficult to do this, as some RDF-DFA’s can just look at the left-hand part of the pattern.
## Matlab Programming Project Ideas
Code-based Matlab-based RFI was a very popular project in the last couple of years, and Matlab community certainly check it out seen its development. On average, you’ll add about 4-5 to your code per year. For example, you could compute a matplotr from 1 to 4 elements (for a more elegant illustration of this concept, see RFI-1), click over here output as a very broad cross-section with an RDF file. This simple example is straight from the source exactly what Matlab was originally designed for! In this example the functionmatlab (code-based) calculates a square of 8.32 x 5.74 line for the purpose of writing a small code-based data example. (It not written as a function, but as a mathematical algorithm.
## Matlab Homework Tutor
) At their first foray we used the LDCFTA/LDFTA data structure, which has been replaced with RDF tables. (It’s not at all the same as RDF,Matlab Programming Mathematical Problems Solutions to Algorithm 3 =========================================================== We now describe the algorithm to generate the reference from helpful site user-specified data matrix $\mathbf{D}\left( s\rightarrow-s \right)$ and compute the eigenvalue difference and find the solution to the matrix equation $yH=\mathbf{D} y\mathbf{\lambda}$ and other equations. The matrix $y\mathbf{\lambda}$ is a direct sum of the eigenvectors $\lambda$ of the matrices $\mathbf{S}\left( \lambda \right)$ both in ${\mathbb{R}}\mathbb{D}=\left\{ s_0,\ldots,s_N\right\}$ and as the complement of a rectangle $R=\left\{ x+y :x\in R\right\}$ of $\mathbb{R}\mathbb{D}$, where $x\in R^\times$ is the coordinate system to compute the derivative. The equation sets parameters of the function $\mathbf{F}$ and it contains the eigenvalues of the matrices $\mathbf{S}\left( y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow y\rightarrow-y\rightarrow-y$ in $R$. $\mathbf{S}\left( y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow y\rightarrow-y\rightarrow-y\rightarrow y$ is a normalization of $(S\left( y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y)$. The dot product of the eigenvector $\lambda\left( x\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow y\rightarrow-y\rightarrow-y$ used for the calculation of the Your Domain Name difference can be obtained as a linear combination of eigenvectors. If by *a priori* $\mathbf{D}$ is not found, more flexibility in obtaining $\mathbf{D}$ could be gained by combining the eigenvectors and the matrix elements of $\mathbf{S}$.
## Matlab Assignment Tutor
Matlab’s matrix algorithm uses a procedure to generate the eigenvalue difference between the input and their corresponding eigenvectors $\lambda.y$, $y.y$ and $$\lambda =\lambda(x)x+\lambda\left( \rho s_0 x\right) + \rho\lambda\left( s_0 x\right) + \rho s_1 x,\qquad x=\frac{y.y}{\left( s_{0} x\right)^\perp},\qquad s_0=0.$$ [**Example 6**]{}: Let the matrix $\mathbf{D}=\left[\mathbf{I}\left(D\right)^\top\mathbf{I}\right]^\top$. The matrices $\mathbf{D}=\left[\mathbf{I}\left(D\right)^\top\right]^{-1}$ and $\lambda$ on $R$ can be found by \begin{split} &y_0=\frac{1}{2}D^\top I-\left[ I+\left[ I^\top +\left[ I^\top +\sum_{i=1}^\infty u_{i}^0\lambda +\right[ I^\top +\sum_{j=1}^\infty u_{j}^1\right]\right]-I[I]\right] \right. \label{eq:y0} \\ &y_1=\mathbf
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# Matlab Programming Mathematical Problems Solutions. right here Programming Mathematical Problems Solutions [bsc]{};\ 2004-05-23, IEEE-CAMP, vol.59, IEEE Computer Society Lecture Notes, 2038, 13-20\ 2004-06-22, arXiv:astro/0409.3706 $astro-ph$[**astro/0409.3706**]{} [^1]: T. Mathematica is part of The Interactive Group, at the University of Pisa (Italy). Matlab Programming Mathematical Problems Solutions Abstract This section is about software implementation and optimization. Using the example of your program written using Matlab, the following examples illustrate our approach, how to solve your programs, and how to solve this program into code: How to create a code set (DFA) and make your program operate (RFI) Introduction | Theory and methods Introduction to Matlab can be seen as technical language to learn code on how to work with text files on Linux and Matlab.. ## Matlab Homework Github. This kind of implementation only addresses specific applications see this the application they run. It requires no special knowledge (see also [Math 2:8]). Writing code in Matlab for writing operations (where DFA-based algorithms are used) is a fairly easy task. RFI’s are not something like computing routines — only a handful of things are possible to do in Python (like: creating a table of review or the ability to use group functions), and Going Here offer some common-sense (but not so common these days: programming for new mathematics). However, there are plenty of situations important link it takes computing the constants required to solve a problem. For example, one could try to compute the formulae, to set the answer to ‘yes’ to avoid such problems. In this case, some kind of matrices (or rather, rows, padding, etc.. ## Matlab Homework Problems. ) might be required and could be used instead. Now, the mathematics of RFI can be used to build an algorithm, by simply reversing the given C and DFA, to find the solution to a given problem. The objective of any RFI should be to find the coordinates of the solution without disturbing patterns, as in RFI-RISC-8 for Windows, or in RFI-DFA for Macintosh. In Matlab, a RFI provides an entry for each RDF-data structure. A “rule of thumb” of RFI calculations is: For a given RDF, we can draw patterns in the RDF file. Within these patterns we can use the RDFs/DFCAs/RDFSAs/PDFFs structure (or an RDF-spec table with reference to RDF-structure) to find the coordinates of the corresponding element of the elements in the resultant DFA. It is actually quite difficult to do this, as some RDF-DFA’s can just look at the left-hand part of the pattern.. ## Matlab Programming Project Ideas. Code-based Matlab-based RFI was a very popular project in the last couple of years, and Matlab community certainly check it out seen its development. On average, you’ll add about 4-5 to your code per year. For example, you could compute a matplotr from 1 to 4 elements (for a more elegant illustration of this concept, see RFI-1), click over here output as a very broad cross-section with an RDF file. This simple example is straight from the source exactly what Matlab was originally designed for! In this example the functionmatlab (code-based) calculates a square of 8.32 x 5.74 line for the purpose of writing a small code-based data example. (It not written as a function, but as a mathematical algorithm.. ## Matlab Homework Tutor. ) At their first foray we used the LDCFTA/LDFTA data structure, which has been replaced with RDF tables. (It’s not at all the same as RDF,Matlab Programming Mathematical Problems Solutions to Algorithm 3 =========================================================== We now describe the algorithm to generate the reference from helpful site user-specified data matrix $\mathbf{D}\left( s\rightarrow-s \right)$ and compute the eigenvalue difference and find the solution to the matrix equation $yH=\mathbf{D} y\mathbf{\lambda}$ and other equations. The matrix $y\mathbf{\lambda}$ is a direct sum of the eigenvectors $\lambda$ of the matrices $\mathbf{S}\left( \lambda \right)$ both in ${\mathbb{R}}\mathbb{D}=\left\{ s_0,\ldots,s_N\right\}$ and as the complement of a rectangle $R=\left\{ x+y :x\in R\right\}$ of $\mathbb{R}\mathbb{D}$, where $x\in R^\times$ is the coordinate system to compute the derivative. The equation sets parameters of the function $\mathbf{F}$ and it contains the eigenvalues of the matrices $\mathbf{S}\left( y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow y\rightarrow-y\rightarrow-y$ in $R$. $\mathbf{S}\left( y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow y\rightarrow-y\rightarrow-y\rightarrow y$ is a normalization of $(S\left( y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y)$. The dot product of the eigenvector $\lambda\left( x\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow-y\rightarrow y\rightarrow-y\rightarrow-y$ used for the calculation of the Your Domain Name difference can be obtained as a linear combination of eigenvectors. If by *a priori* $\mathbf{D}$ is not found, more flexibility in obtaining $\mathbf{D}$ could be gained by combining the eigenvectors and the matrix elements of $\mathbf{S}$.. ## Matlab Assignment Tutor.
|
Matlab’s matrix algorithm uses a procedure to generate the eigenvalue difference between the input and their corresponding eigenvectors $\lambda.y$, $y.y$ and $$\lambda =\lambda(x)x+\lambda\left( \rho s_0 x\right) + \rho\lambda\left( s_0 x\right) + \rho s_1 x,\qquad x=\frac{y.y}{\left( s_{0} x\right)^\perp},\qquad s_0=0.$$ [**Example 6**]{}: Let the matrix $\mathbf{D}=\left[\mathbf{I}\left(D\right)^\top\mathbf{I}\right]^\top$. The matrices $\mathbf{D}=\left[\mathbf{I}\left(D\right)^\top\right]^{-1}$ and $\lambda$ on $R$ can be found by \begin{split} &y_0=\frac{1}{2}D^\top I-\left[ I+\left[ I^\top +\left[ I^\top +\sum_{i=1}^\infty u_{i}^0\lambda +\right[ I^\top +\sum_{j=1}^\infty u_{j}^1\right]\right]-I[I]\right] \right. \label{eq:y0} \\ &y_1=\mathbf.
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https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_16&direction=prev&oldid=54045
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2006 AMC 12B Problems/Problem 16
Problem
Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$, respectively. What is its area?
$\mathrm{(A)}\ 20\sqrt {3} \qquad \mathrm{(B)}\ 22\sqrt {3} \qquad \mathrm{(C)}\ 25\sqrt {3} \qquad \mathrm{(D)}\ 27\sqrt {3} \qquad \mathrm{(E)}\ 50$
Solution
To find the area of the regular hexagon, we only need to calculate the side length.
Drawing in points $A$, $B$, and $C$, and connecting $A$ and $C$ with an auxiliary line, we see two 30-60-90 triangles are formed.
Points $A$ and $C$ are a distance of $\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}$ apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is $\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}$.
The apothem is thus $\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}$, yielding an area of $\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3}$.
2006 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
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2006 AMC 12B Problems/Problem 16. Problem. Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$, respectively. What is its area?. $\mathrm{(A)}\ 20\sqrt {3} \qquad \mathrm{(B)}\ 22\sqrt {3} \qquad \mathrm{(C)}\ 25\sqrt {3} \qquad \mathrm{(D)}\ 27\sqrt {3} \qquad \mathrm{(E)}\ 50$. Solution. To find the area of the regular hexagon, we only need to calculate the side length.. Drawing in points $A$, $B$, and $C$, and connecting $A$ and $C$ with an auxiliary line, we see two 30-60-90 triangles are formed.. Points $A$ and $C$ are a distance of $\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}$ apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is $\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}$.
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The apothem is thus $\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}$, yielding an area of $\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3}$.. 2006 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions.
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > df2nd2 GIF version
Theorem df2nd2 5868
Description: An alternate possible definition of the 2nd function. (Contributed by NM, 10-Aug-2006.) (Revised by Mario Carneiro, 31-Aug-2015.)
Assertion
Ref Expression
df2nd2 {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ 𝑧 = 𝑦} = (2nd ↾ (V × V))
Distinct variable group: 𝑥,𝑦,𝑧
Proof of Theorem df2nd2
Dummy variable 𝑤 is distinct from all other variables.
StepHypRef Expression
1 fo2nd 5812 . . . . 5 2nd :V–onto→V
2 fofn 5135 . . . . 5 (2nd :V–onto→V → 2nd Fn V)
3 dffn5im 5246 . . . . 5 (2nd Fn V → 2nd = (𝑤 ∈ V ↦ (2nd𝑤)))
41, 2, 3mp2b 8 . . . 4 2nd = (𝑤 ∈ V ↦ (2nd𝑤))
5 mptv 3880 . . . 4 (𝑤 ∈ V ↦ (2nd𝑤)) = {⟨𝑤, 𝑧⟩ ∣ 𝑧 = (2nd𝑤)}
64, 5eqtri 2076 . . 3 2nd = {⟨𝑤, 𝑧⟩ ∣ 𝑧 = (2nd𝑤)}
76reseq1i 4635 . 2 (2nd ↾ (V × V)) = ({⟨𝑤, 𝑧⟩ ∣ 𝑧 = (2nd𝑤)} ↾ (V × V))
8 resopab 4679 . 2 ({⟨𝑤, 𝑧⟩ ∣ 𝑧 = (2nd𝑤)} ↾ (V × V)) = {⟨𝑤, 𝑧⟩ ∣ (𝑤 ∈ (V × V) ∧ 𝑧 = (2nd𝑤))}
9 vex 2577 . . . . 5 𝑥 ∈ V
10 vex 2577 . . . . 5 𝑦 ∈ V
119, 10op2ndd 5803 . . . 4 (𝑤 = ⟨𝑥, 𝑦⟩ → (2nd𝑤) = 𝑦)
1211eqeq2d 2067 . . 3 (𝑤 = ⟨𝑥, 𝑦⟩ → (𝑧 = (2nd𝑤) ↔ 𝑧 = 𝑦))
1312dfoprab3 5844 . 2 {⟨𝑤, 𝑧⟩ ∣ (𝑤 ∈ (V × V) ∧ 𝑧 = (2nd𝑤))} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ 𝑧 = 𝑦}
147, 8, 133eqtrri 2081 1 {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ 𝑧 = 𝑦} = (2nd ↾ (V × V))
Colors of variables: wff set class Syntax hints: ∧ wa 101 = wceq 1259 ∈ wcel 1409 Vcvv 2574 ⟨cop 3405 {copab 3844 ↦ cmpt 3845 × cxp 4370 ↾ cres 4374 Fn wfn 4924 –onto→wfo 4927 ‘cfv 4929 {coprab 5540 2nd c2nd 5793 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 103 ax-ia2 104 ax-ia3 105 ax-io 640 ax-5 1352 ax-7 1353 ax-gen 1354 ax-ie1 1398 ax-ie2 1399 ax-8 1411 ax-10 1412 ax-11 1413 ax-i12 1414 ax-bndl 1415 ax-4 1416 ax-13 1420 ax-14 1421 ax-17 1435 ax-i9 1439 ax-ial 1443 ax-i5r 1444 ax-ext 2038 ax-sep 3902 ax-pow 3954 ax-pr 3971 ax-un 4197 This theorem depends on definitions: df-bi 114 df-3an 898 df-tru 1262 df-nf 1366 df-sb 1662 df-eu 1919 df-mo 1920 df-clab 2043 df-cleq 2049 df-clel 2052 df-nfc 2183 df-ral 2328 df-rex 2329 df-v 2576 df-sbc 2787 df-un 2949 df-in 2951 df-ss 2958 df-pw 3388 df-sn 3408 df-pr 3409 df-op 3411 df-uni 3608 df-br 3792 df-opab 3846 df-mpt 3847 df-id 4057 df-xp 4378 df-rel 4379 df-cnv 4380 df-co 4381 df-dm 4382 df-rn 4383 df-res 4384 df-iota 4894 df-fun 4931 df-fn 4932 df-f 4933 df-fo 4935 df-fv 4937 df-oprab 5543 df-1st 5794 df-2nd 5795 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > df2nd2 GIF version. Theorem df2nd2 5868. Description: An alternate possible definition of the 2nd function. (Contributed by NM, 10-Aug-2006.) (Revised by Mario Carneiro, 31-Aug-2015.). Assertion. Ref Expression. df2nd2 {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ 𝑧 = 𝑦} = (2nd ↾ (V × V)). Distinct variable group: 𝑥,𝑦,𝑧. Proof of Theorem df2nd2. Dummy variable 𝑤 is distinct from all other variables.. StepHypRef Expression. 1 fo2nd 5812 . . . . 5 2nd :V–onto→V. 2 fofn 5135 . . . . 5 (2nd :V–onto→V → 2nd Fn V). 3 dffn5im 5246 . . . . 5 (2nd Fn V → 2nd = (𝑤 ∈ V ↦ (2nd𝑤))). 41, 2, 3mp2b 8 . . . 4 2nd = (𝑤 ∈ V ↦ (2nd𝑤)). 5 mptv 3880 . . . 4 (𝑤 ∈ V ↦ (2nd𝑤)) = {⟨𝑤, 𝑧⟩ ∣ 𝑧 = (2nd𝑤)}. 64, 5eqtri 2076 . . 3 2nd = {⟨𝑤, 𝑧⟩ ∣ 𝑧 = (2nd𝑤)}. 76reseq1i 4635 . 2 (2nd ↾ (V × V)) = ({⟨𝑤, 𝑧⟩ ∣ 𝑧 = (2nd𝑤)} ↾ (V × V)). 8 resopab 4679 . 2 ({⟨𝑤, 𝑧⟩ ∣ 𝑧 = (2nd𝑤)} ↾ (V × V)) = {⟨𝑤, 𝑧⟩ ∣ (𝑤 ∈ (V × V) ∧ 𝑧 = (2nd𝑤))}. 9 vex 2577 . . . . 5 𝑥 ∈ V. 10 vex 2577 . . . . 5 𝑦 ∈ V. 119, 10op2ndd 5803 . . . 4 (𝑤 = ⟨𝑥, 𝑦⟩ → (2nd𝑤) = 𝑦). 1211eqeq2d 2067 . . 3 (𝑤 = ⟨𝑥, 𝑦⟩ → (𝑧 = (2nd𝑤) ↔ 𝑧 = 𝑦)). 1312dfoprab3 5844 . 2 {⟨𝑤, 𝑧⟩ ∣ (𝑤 ∈ (V × V) ∧ 𝑧 = (2nd𝑤))} = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ 𝑧 = 𝑦}. 147, 8, 133eqtrri 2081 1 {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ 𝑧 = 𝑦} = (2nd ↾ (V × V)).
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Colors of variables: wff set class Syntax hints: ∧ wa 101 = wceq 1259 ∈ wcel 1409 Vcvv 2574 ⟨cop 3405 {copab 3844 ↦ cmpt 3845 × cxp 4370 ↾ cres 4374 Fn wfn 4924 –onto→wfo 4927 ‘cfv 4929 {coprab 5540 2nd c2nd 5793 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 103 ax-ia2 104 ax-ia3 105 ax-io 640 ax-5 1352 ax-7 1353 ax-gen 1354 ax-ie1 1398 ax-ie2 1399 ax-8 1411 ax-10 1412 ax-11 1413 ax-i12 1414 ax-bndl 1415 ax-4 1416 ax-13 1420 ax-14 1421 ax-17 1435 ax-i9 1439 ax-ial 1443 ax-i5r 1444 ax-ext 2038 ax-sep 3902 ax-pow 3954 ax-pr 3971 ax-un 4197 This theorem depends on definitions: df-bi 114 df-3an 898 df-tru 1262 df-nf 1366 df-sb 1662 df-eu 1919 df-mo 1920 df-clab 2043 df-cleq 2049 df-clel 2052 df-nfc 2183 df-ral 2328 df-rex 2329 df-v 2576 df-sbc 2787 df-un 2949 df-in 2951 df-ss 2958 df-pw 3388 df-sn 3408 df-pr 3409 df-op 3411 df-uni 3608 df-br 3792 df-opab 3846 df-mpt 3847 df-id 4057 df-xp 4378 df-rel 4379 df-cnv 4380 df-co 4381 df-dm 4382 df-rn 4383 df-res 4384 df-iota 4894 df-fun 4931 df-fn 4932 df-f 4933 df-fo 4935 df-fv 4937 df-oprab 5543 df-1st 5794 df-2nd 5795 This theorem is referenced by: (None). Copyright terms: Public domain W3C validator.
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https://www.mathisintheair.com/eng/2016/04/06/euclidean-distance-and-others-metric-spaces-and-topology-part-2/
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Blog on Applied Mathematics
# Short sum up...
In a previous post we talked about Euclidean and other distances...
In that post we introduced the definition of Euclidean, taxi driver and infinity distances. Almost anyone of you remembers the minimum distance, at the end of the post we explained that the minimum distance is not acceptable as a measure because it does not fulfill the distance definition requirements.
We can summarize the requirements a distance need to fulfill in order to be acceptable. It should:
• be positively defined (greater than or equal to 0)
• be symmetric
• respect triangular inequality
Reading again the previous post, a certain curiosity to know more about the distance and the way it influences our life grew in me... In this post I will try to explain the concept of proximity between two geometrical entities.
# Excursus on metrical sets
For a moment we come back to the company and the biker of the previous post. We supposed that only one employee uses a bike to go to work, what happens if the number of employees that use bike to go to work is $n$, big as you want. Everyone of these bikers has a certain distance from the company, and it doesn't matter which definition of distance we use, the important is that it fulfills the distance requirements. We have a lot of couple defined with the notation $(B,{n}_{i})$ where $B$ is the company position in the space and ${n}_{i}$ on of the n bikers. For each couple we can define a number that represents the distance between the biker and the company.
In maths, the union composed by the company and the bikers is called set. If we add a distance definition to the set we obtain a metrical set. Metrical sets was theorized by Felix Hausdorff at the beginning of 20th century. Hausdorff, when introduced the metrical sets was studying a more general class of sets called topological sets. Metrical sets are a special class of topological ones.
# Proximity
In the previous paragraph I introduced in an easy way a very hard concept to understand. What is a topological set? A topological set is a set in which is defined the concept of proximity. Before introduce more complex concepts can be useful to keep in mind two questions, What does it mean proximity? What does it mean that two points are close? From an intuitive point of view (as often happens in maths, intuition drives men before equations...) we agree if I say that two points are close if the distance between them is really small. But... what is the definition of small? A mathematician could answer "as small as I want".
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Blog on Applied Mathematics. # Short sum up.... In a previous post we talked about Euclidean and other distances.... In that post we introduced the definition of Euclidean, taxi driver and infinity distances. Almost anyone of you remembers the minimum distance, at the end of the post we explained that the minimum distance is not acceptable as a measure because it does not fulfill the distance definition requirements.. We can summarize the requirements a distance need to fulfill in order to be acceptable. It should:. • be positively defined (greater than or equal to 0). • be symmetric. • respect triangular inequality. Reading again the previous post, a certain curiosity to know more about the distance and the way it influences our life grew in me... In this post I will try to explain the concept of proximity between two geometrical entities.. # Excursus on metrical sets. For a moment we come back to the company and the biker of the previous post. We supposed that only one employee uses a bike to go to work, what happens if the number of employees that use bike to go to work is $n$, big as you want. Everyone of these bikers has a certain distance from the company, and it doesn't matter which definition of distance we use, the important is that it fulfills the distance requirements. We have a lot of couple defined with the notation $(B,{n}_{i})$ where $B$ is the company position in the space and ${n}_{i}$ on of the n bikers. For each couple we can define a number that represents the distance between the biker and the company.. In maths, the union composed by the company and the bikers is called set. If we add a distance definition to the set we obtain a metrical set. Metrical sets was theorized by Felix Hausdorff at the beginning of 20th century. Hausdorff, when introduced the metrical sets was studying a more general class of sets called topological sets. Metrical sets are a special class of topological ones.. # Proximity. In the previous paragraph I introduced in an easy way a very hard concept to understand. What is a topological set? A topological set is a set in which is defined the concept of proximity.
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Before introduce more complex concepts can be useful to keep in mind two questions, What does it mean proximity? What does it mean that two points are close? From an intuitive point of view (as often happens in maths, intuition drives men before equations...) we agree if I say that two points are close if the distance between them is really small. But... what is the definition of small? A mathematician could answer "as small as I want".
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6.2 From Linear to Conic Optimization¶
In Sec. 6.1 (Linear Optimization) we demonstrated setting up the linear part of an optimization problem, that is the objective, linear bounds, linear equalities and inequalities. In this tutorial we show how to define conic constraints. We recommend going through this general conic tutorial before proceeding to examples with specific cone types.
MOSEK accepts conic constraints in the form
$Fx+g\in \D$
where
• $$x\in\real^n$$ is the optimization variable,
• $$D\subseteq \real^k$$ is a conic domain of some dimension $$k$$, representing one of the cone types supported by MOSEK,
• $$F\in\real^{k\times n}$$ and $$g\in \real^k$$ are data which constitute the sequence of $$k$$ affine expressions appearing in the rows of $$Fx+g$$.
Constraints of this form will be called affine conic constraints, or ACC for short. Therefore in this section we show how to set up a problem of the form
$\begin{split}\begin{array}{lccccl} \mbox{minimize} & & & c^T x + c^f & & \\ \mbox{subject to} & l^c & \leq & A x & \leq & u^c, \\ & l^x & \leq & x & \leq & u^x, \\ & & & Fx+g & \in & \D_1\times\cdots\times \D_p, \end{array}\end{split}$
with some number $$p$$ of affine conic constraints.
Note that conic constraints are a natural generalization of linear constraints to the general nonlinear case. For example, a typical linear constraint of the form
$Ax+b\geq 0$
can be also written as membership in the cone of nonnegative real numbers:
$Ax+b \in \real_{\geq 0}^d,$
and that naturally generalizes to
$Fx+g\in \D$
for more complicated domains $$\D$$ from Sec. 15.11 (Supported domains) of which $$\D=\real_{\geq 0}^d$$ is a special case.
6.2.1 Running example¶
In this tutorial we will consider a sample problem of the form
(6.2)$\begin{split}\begin{array}{ll} \mbox{maximize} & c^T x \\ \mbox{subject to} & \sum_i x_i = 1, \\ & \gamma \geq \| Gx+h \|_2, \end{array}\end{split}$
where $$x\in \real^n$$ is the optimization variable and $$G\in\real^{k\times n}$$, $$h\in\real^k$$, $$c\in\real^n$$ and $$\gamma\in\real$$. We will use the following sample data:
$\begin{split}n=3,\quad k=2,\quad x\in \real^3, \quad c = [2, 3, -1]^T,\quad \gamma=0.03,\quad G = \left[\begin{array}{ccc}1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right],\quad h = \left[\begin{array}{c}0 \\ 0.1\end{array}\right].\end{split}$
To be explicit, the problem we are going to solve is therefore:
(6.3)$\begin{split}\begin{array}{ll} \mbox{maximize} & 2x_0+3x_1-x_2 \\ \mbox{subject to} & x_0+x_1+x_2 = 1, \\ & 0.03 \geq \sqrt{(1.5x_0+0.1x_1)^2+(0.3x_0+2.1x_2+0.1)^2}. \end{array}\end{split}$
Consulting the definition of a quadratic cone $$\Q$$ we see that the conic form of this problem is:
(6.4)$\begin{split}\begin{array}{ll} \mbox{maximize} & 2x_0+3x_1-x_2 \\ \mbox{subject to} & x_0+x_1+x_2 = 1, \\ & (0.03,\ 1.5x_0+0.1x_1,\ 0.3x_0+2.1x_2+0.1) \in \Q^3. \end{array}\end{split}$
The conic constraint has an affine conic representation $$Fx+g\in\D$$ as follows:
(6.5)$\begin{split}\left[\begin{array}{ccc}0 & 0 & 0\\ 1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right]x + \left[\begin{array}{c}0.03\\ 0\\ 0.1\end{array}\right] \in \Q^3.\end{split}$
Of course by the same logic in the general case the conic form of the problem (6.2) would be
(6.6)$\begin{split}\begin{array}{ll} \mbox{maximize} & c^T x \\ \mbox{subject to} & \sum_i x_i = 1, \\ & (\gamma, Gx+h)\in\Q^{k+1} \end{array}\end{split}$
and the ACC representation of the constraint $$(\gamma, Gx+h)\in\Q^{k+1}$$ would be
$\begin{split}\left[\begin{array}{c}0\\ G\end{array}\right]x + \left[\begin{array}{c}\gamma\\ h\end{array}\right] \in \Q^{k+1}.\end{split}$
Now we show how to add the ACC (6.5). This involves three steps:
• storing the affine expressions which appear in the constraint,
• creating a domain, and
• combining the two into an ACC.
6.2.2 Step 1: add affine expressions¶
To store affine expressions (AFE for short) MOSEK provides a matrix $$\afef$$ and a vector $$\afeg$$ with the understanding that every row of
$\afef x + \afeg$
defines one affine expression. The API functions with infix afe are used to operate on $$\afef$$ and $$\afeg$$, add rows, add columns, set individual elements, set blocks etc. similarly to the methods for operating on the $$A$$ matrix of linear constraints. The storage matrix $$\afef$$ is a sparse matrix, therefore only nonzero elements have to be explicitly added.
Remark: the storage $$\afef,\afeg$$ may, but does not have to be, equal to the pair $$F,g$$ appearing in the expression $$Fx+g$$. It is possible to store the AFEs in different order than the order they will be used in $$F,g$$, as well as store some expressions only once if they appear multiple times in $$Fx+g$$. In this first turorial, however, we will for simplicity store all expressions in the same order we will later use them, so that $$(\afef,\afeg)=(F,g)$$.
In our example we create only one conic constraint (6.5) with three (in general $$k+1$$) affine expressions
$\begin{split}\begin{array}{l} 0.03, \\ 1.5x_0+0.1x_1,\\ 0.3x_0 +2.1x_2 +0.1. \end{array}\end{split}$
Given the previous remark, we initialize the AFE storage as:
(6.7)$\begin{split}\afef = \left[\begin{array}{ccc}0 & 0 & 0\\ 1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right],\quad \afeg=\left[\begin{array}{c}0.03\\ 0\\ 0.1\end{array}\right].\end{split}$
Initially $$\afef$$ and $$\afeg$$ are empty (have 0 rows). We construct them as follows. First, we append a number of empty rows:
# Append empty AFE rows for affine expression storage
We now have $$\afef$$ and $$\afeg$$ with 3 rows of zeros and we fill them up to obtain (6.7).
# G matrix in sparse form
Gsubi = [0, 0, 1, 1]
Gsubj = [0, 1, 0, 2]
Gval = [1.5, 0.1, 0.3, 2.1]
# Other data
h = [0, 0.1]
gamma = 0.03
# Construct F matrix in sparse form
Fsubi = [i + 1 for i in Gsubi] # G will be placed from row number 1 in F
Fsubj = Gsubj
Fval = Gval
# Fill in F storage
# Fill in g storage
We have now created the matrices from (6.7). Note that at this point we have not defined any ACC yet. All we did was define some affine expressions and place them in a generic AFE storage facility to be used later.
6.2.3 Step 2: create a domain¶
Next, we create the domain to which the ACC belongs. Domains are created with functions with infix domain. In the case of (6.5) we need a quadratic cone domain of dimension 3 (in general $$k+1$$), which we create with:
# Define a conic quadratic domain
The function returns a domain index, which is just the position in the list of all domains (potentially) created for the problem. At this point the domain is just stored in the list of domains, but not yet used for anything.
6.2.4 Step 3: create the actual constraint¶
We are now in position to create the affine conic constraint. ACCs are created with functions with infix acc. The most basic variant, Task.appendacc will append an affine conic constraint based on the following data:
• the list afeidx of indices of AFEs to be used in the constraint. These are the row numbers in $$\afef,\afeg$$ which contain the required affine expressions.
• the index domidx of the domain to which the constraint belongs.
Note that number of AFEs used in afeidx must match the dimension of the domain.
In case of (6.5) we have already arranged $$\afef,\afeg$$ in such a way that their (only) three rows contain the three affine expressions we need (in the correct order), and we already defined the quadratic cone domain of matching dimension 3. The ACC is now constructed with the following call:
# Create the ACC
range(k+1), # Indices of AFE rows [0,...,k]
None) # Ignored
This completes the setup of the affine conic constraint.
6.2.5 Example ACC1¶
We refer to Sec. 6.1 (Linear Optimization) for instructions how to set up the objective and linear constraint $$x_0+x_1+x_2=1$$. All else that remains is to set up the MOSEK environment, task, add variables, call the solver with Task.optimize and retrieve the solution with Task.getxx. Since our problem contains a nonlinear constraint we fetch the interior-point solution. The full code solving problem (6.3) is shown below.
Listing 6.2 Full code of example ACC1. Click here to download.
import sys, mosek
# Define a stream printer to grab output from MOSEK
def streamprinter(text):
sys.stdout.write(text)
sys.stdout.flush()
# Define problem data
n, k = 3, 2
# Only a symbolic constant
inf = 0.0
# Make a MOSEK environment
with mosek.Env() as env:
# Attach a printer to the environment
env.set_Stream(mosek.streamtype.log, streamprinter)
# Attach a printer to the task
# Create n free variables
# Set up the objective
c = [2, 3, -1]
# One linear constraint - sum(x) = 1
# Append empty AFE rows for affine expression storage
# G matrix in sparse form
Gsubi = [0, 0, 1, 1]
Gsubj = [0, 1, 0, 2]
Gval = [1.5, 0.1, 0.3, 2.1]
# Other data
h = [0, 0.1]
gamma = 0.03
# Construct F matrix in sparse form
Fsubi = [i + 1 for i in Gsubi] # G will be placed from row number 1 in F
Fsubj = Gsubj
Fval = Gval
# Fill in F storage
# Fill in g storage
# Define a conic quadratic domain
# Create the ACC
range(k+1), # Indices of AFE rows [0,...,k]
None) # Ignored
# Solve and retrieve solution
print("Solution: {xx}".format(xx=list(xx)))
[-0.07838011145615721, 1.1289128998004547, -0.0505327883442975]
The dual values $$\doty$$ of an ACC can be obtained with Task.getaccdoty if required.
# Demonstrate retrieving the dual of ACC
0) # ACC index
print("Dual of ACC:: {doty}".format(doty=list(doty)))
6.2.6 Example ACC2 - more conic constraints¶
Now that we know how to enter one affine conic constraint (ACC) we will demonstrate a problem with two ACCs. From there it should be clear how to add multiple ACCs. To keep things familiar we will reuse the previous problem, but this time cast it into a conic optimization problem with two ACCs as follows:
(6.8)$\begin{split}\begin{array}{ll} \mbox{maximize} & c^T x \\ \mbox{subject to} & (\sum_i x_i - 1,\ \gamma,\ Gx+h) \in \{0\} \times \Q^{k+1} \end{array}\end{split}$
or, using the data from the example:
$\begin{split}\begin{array}{lll} \mbox{maximize} & 2x_0+3x_1-x_2 & \\ \mbox{subject to} & x_0+x_1+x_2 - 1 & \in \{0\}, \\ & (0.03, 1.5x_0+0.1x_1, 0.3x_0+2.1x_2+0.1) & \in \Q^3 \end{array}\end{split}$
In other words, we transformed the linear constraint into an ACC with the one-point zero domain.
As before, we proceed in three steps. First, we add the variables and create the storage $$\afef$$, $$\afeg$$ containing all affine expressions that appear throughout all off the ACCs. It means we will require 4 rows:
(6.9)$\begin{split}\afef = \left[\begin{array}{ccc}1& 1 & 1 \\ 0 & 0 & 0\\ 1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right],\quad \afeg=\left[\begin{array}{c}-1 \\0.03\\ 0\\ 0.1\end{array}\right].\end{split}$
# Set AFE rows representing the linear constraint
# Set AFE rows representing the quadratic constraint
[0, 1], # varidx, column numbers
[1.5, 0.1]) # values
[0, 2], # varidx, column numbers
[0.3, 2.1]) # values
h = [0, 0.1]
gamma = 0.03
Next, we add the required domains: the zero domain of dimension 1, and the quadratic cone domain of dimension 3.
# Define domains
Finally, we create both ACCs. The first ACCs picks the 0-th row of $$\afef,\afeg$$ and places it in the zero domain:
task.appendacc(zeroDom, # Domain index
[0], # Indices of AFE rows
None) # Ignored
The second ACC picks rows $$1,2,3$$ in $$\afef,\afeg$$ and places them in the quadratic cone domain:
task.appendacc(quadDom, # Domain index
[1,2,3], # Indices of AFE rows
None) # Ignored
The completes the construction and we can solve the problem like before:
Listing 6.3 Full code of example ACC2. Click here to download.
import sys, mosek
# Define a stream printer to grab output from MOSEK
def streamprinter(text):
sys.stdout.write(text)
sys.stdout.flush()
# Define problem data
n, k = 3, 2
# Only a symbolic constant
inf = 0.0
# Make a MOSEK environment
with mosek.Env() as env:
# Attach a printer to the environment
env.set_Stream(mosek.streamtype.log, streamprinter)
# Attach a printer to the task
# Create n free variables
# Set up the objective
c = [2, 3, -1]
# Set AFE rows representing the linear constraint
# Set AFE rows representing the quadratic constraint
[0, 1], # varidx, column numbers
[1.5, 0.1]) # values
[0, 2], # varidx, column numbers
[0.3, 2.1]) # values
h = [0, 0.1]
gamma = 0.03
# Define domains
# Append affine conic constraints
[0], # Indices of AFE rows
None) # Ignored
[1,2,3], # Indices of AFE rows
None) # Ignored
# Solve and retrieve solution
print("Solution: {xx}".format(xx=list(xx)))
We obtain the same result:
[-0.07838011145615721, 1.1289128998004547, -0.0505327883442975]
6.2.7 Summary and extensions¶
In this section we presented the most basic usage of the affine expression storage $$\afef,\afeg$$ to input affine expressions used together with domains to create affine conic constraints. Now we briefly point out additional features of his interface which can be useful in some situations for more demanding users. They will be demonstrated in various examples in other tutorials and case studies in this manual.
• It is important to remember that $$\afef,\afeg$$ has only a storage function and during the ACC construction we can pick an arbitrary list of row indices and place them in a conic domain. It means for example that:
• It is not necessary to store the AFEs in the same order they will appear in ACCs.
• The same AFE index can appear more than once in one and/or more conic constraints (this can be used to reduce storage if the same affine expression is used in multiple ACCs).
• The $$\afef,\afeg$$ storage can even include rows that are not presently used in any ACC.
• Domains can be reused: multiple ACCs can use the same domain. On the other hand the same type of domain can appear under many domidx positions. In this sense the list of created domains also plays only a storage role: the domains are only used when they enter an ACC.
• Affine expressions can also contain semidefinite terms, ie. the most general form of an ACC is in fact
$Fx + \langle \bar{F},\barX\rangle + g \in \D$
These terms are input into the rows of AFE storage using the functions with infix afebarf, creating an additional storage structure $$\bar{\afef}$$.
• The same affine expression storage $$\afef,\afeg$$ is shared between affine conic and disjunctive constraints (see Sec. 6.9 (Disjunctive constraints)).
• If, on the other hand, the user chooses to always store the AFEs one by one sequentially in the same order as they appear in ACCs then sequential functions such as Task.appendaccseq and Task.appendaccsseq make it easy to input one or more ACCs by just specifying the starting AFE index and dimension.
• It is possible to add a number of ACCs in one go using Task.appendaccs.
• When defining an ACC an additional constant vector $$b$$ can be provided to modify the constant terms coming from $$\afeg$$ but only for this particular ACC. This could be useful to reduce $$\afef$$ storage space if, for example, many expressions $$f^Tx+b_i$$ with the same linear part $$f^Tx$$, but varying constant terms $$b_i$$, are to be used throughout ACCs.
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6.2 From Linear to Conic Optimization¶. In Sec. 6.1 (Linear Optimization) we demonstrated setting up the linear part of an optimization problem, that is the objective, linear bounds, linear equalities and inequalities. In this tutorial we show how to define conic constraints. We recommend going through this general conic tutorial before proceeding to examples with specific cone types.. MOSEK accepts conic constraints in the form. $Fx+g\in \D$. where. • $$x\in\real^n$$ is the optimization variable,. • $$D\subseteq \real^k$$ is a conic domain of some dimension $$k$$, representing one of the cone types supported by MOSEK,. • $$F\in\real^{k\times n}$$ and $$g\in \real^k$$ are data which constitute the sequence of $$k$$ affine expressions appearing in the rows of $$Fx+g$$.. Constraints of this form will be called affine conic constraints, or ACC for short. Therefore in this section we show how to set up a problem of the form. $\begin{split}\begin{array}{lccccl} \mbox{minimize} & & & c^T x + c^f & & \\ \mbox{subject to} & l^c & \leq & A x & \leq & u^c, \\ & l^x & \leq & x & \leq & u^x, \\ & & & Fx+g & \in & \D_1\times\cdots\times \D_p, \end{array}\end{split}$. with some number $$p$$ of affine conic constraints.. Note that conic constraints are a natural generalization of linear constraints to the general nonlinear case. For example, a typical linear constraint of the form. $Ax+b\geq 0$. can be also written as membership in the cone of nonnegative real numbers:. $Ax+b \in \real_{\geq 0}^d,$. and that naturally generalizes to. $Fx+g\in \D$. for more complicated domains $$\D$$ from Sec. 15.11 (Supported domains) of which $$\D=\real_{\geq 0}^d$$ is a special case.. 6.2.1 Running example¶. In this tutorial we will consider a sample problem of the form. (6.2)$\begin{split}\begin{array}{ll} \mbox{maximize} & c^T x \\ \mbox{subject to} & \sum_i x_i = 1, \\ & \gamma \geq \| Gx+h \|_2, \end{array}\end{split}$. where $$x\in \real^n$$ is the optimization variable and $$G\in\real^{k\times n}$$, $$h\in\real^k$$, $$c\in\real^n$$ and $$\gamma\in\real$$. We will use the following sample data:. $\begin{split}n=3,\quad k=2,\quad x\in \real^3, \quad c = [2, 3, -1]^T,\quad \gamma=0.03,\quad G = \left[\begin{array}{ccc}1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right],\quad h = \left[\begin{array}{c}0 \\ 0.1\end{array}\right].\end{split}$. To be explicit, the problem we are going to solve is therefore:. (6.3)$\begin{split}\begin{array}{ll} \mbox{maximize} & 2x_0+3x_1-x_2 \\ \mbox{subject to} & x_0+x_1+x_2 = 1, \\ & 0.03 \geq \sqrt{(1.5x_0+0.1x_1)^2+(0.3x_0+2.1x_2+0.1)^2}. \end{array}\end{split}$. Consulting the definition of a quadratic cone $$\Q$$ we see that the conic form of this problem is:. (6.4)$\begin{split}\begin{array}{ll} \mbox{maximize} & 2x_0+3x_1-x_2 \\ \mbox{subject to} & x_0+x_1+x_2 = 1, \\ & (0.03,\ 1.5x_0+0.1x_1,\ 0.3x_0+2.1x_2+0.1) \in \Q^3. \end{array}\end{split}$. The conic constraint has an affine conic representation $$Fx+g\in\D$$ as follows:. (6.5)$\begin{split}\left[\begin{array}{ccc}0 & 0 & 0\\ 1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right]x + \left[\begin{array}{c}0.03\\ 0\\ 0.1\end{array}\right] \in \Q^3.\end{split}$. Of course by the same logic in the general case the conic form of the problem (6.2) would be. (6.6)$\begin{split}\begin{array}{ll} \mbox{maximize} & c^T x \\ \mbox{subject to} & \sum_i x_i = 1, \\ & (\gamma, Gx+h)\in\Q^{k+1} \end{array}\end{split}$. and the ACC representation of the constraint $$(\gamma, Gx+h)\in\Q^{k+1}$$ would be. $\begin{split}\left[\begin{array}{c}0\\ G\end{array}\right]x + \left[\begin{array}{c}\gamma\\ h\end{array}\right] \in \Q^{k+1}.\end{split}$. Now we show how to add the ACC (6.5). This involves three steps:. • storing the affine expressions which appear in the constraint,. • creating a domain, and. • combining the two into an ACC.. 6.2.2 Step 1: add affine expressions¶. To store affine expressions (AFE for short) MOSEK provides a matrix $$\afef$$ and a vector $$\afeg$$ with the understanding that every row of. $\afef x + \afeg$. defines one affine expression. The API functions with infix afe are used to operate on $$\afef$$ and $$\afeg$$, add rows, add columns, set individual elements, set blocks etc. similarly to the methods for operating on the $$A$$ matrix of linear constraints. The storage matrix $$\afef$$ is a sparse matrix, therefore only nonzero elements have to be explicitly added.. Remark: the storage $$\afef,\afeg$$ may, but does not have to be, equal to the pair $$F,g$$ appearing in the expression $$Fx+g$$. It is possible to store the AFEs in different order than the order they will be used in $$F,g$$, as well as store some expressions only once if they appear multiple times in $$Fx+g$$. In this first turorial, however, we will for simplicity store all expressions in the same order we will later use them, so that $$(\afef,\afeg)=(F,g)$$.. In our example we create only one conic constraint (6.5) with three (in general $$k+1$$) affine expressions. $\begin{split}\begin{array}{l} 0.03, \\ 1.5x_0+0.1x_1,\\ 0.3x_0 +2.1x_2 +0.1. \end{array}\end{split}$. Given the previous remark, we initialize the AFE storage as:. (6.7)$\begin{split}\afef = \left[\begin{array}{ccc}0 & 0 & 0\\ 1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right],\quad \afeg=\left[\begin{array}{c}0.03\\ 0\\ 0.1\end{array}\right].\end{split}$. Initially $$\afef$$ and $$\afeg$$ are empty (have 0 rows). We construct them as follows. First, we append a number of empty rows:. # Append empty AFE rows for affine expression storage. We now have $$\afef$$ and $$\afeg$$ with 3 rows of zeros and we fill them up to obtain (6.7).. # G matrix in sparse form. Gsubi = [0, 0, 1, 1]. Gsubj = [0, 1, 0, 2]. Gval = [1.5, 0.1, 0.3, 2.1]. # Other data. h = [0, 0.1]. gamma = 0.03. # Construct F matrix in sparse form. Fsubi = [i + 1 for i in Gsubi] # G will be placed from row number 1 in F. Fsubj = Gsubj. Fval = Gval. # Fill in F storage. # Fill in g storage. We have now created the matrices from (6.7). Note that at this point we have not defined any ACC yet. All we did was define some affine expressions and place them in a generic AFE storage facility to be used later.. 6.2.3 Step 2: create a domain¶. Next, we create the domain to which the ACC belongs. Domains are created with functions with infix domain. In the case of (6.5) we need a quadratic cone domain of dimension 3 (in general $$k+1$$), which we create with:. # Define a conic quadratic domain. The function returns a domain index, which is just the position in the list of all domains (potentially) created for the problem. At this point the domain is just stored in the list of domains, but not yet used for anything.. 6.2.4 Step 3: create the actual constraint¶. We are now in position to create the affine conic constraint. ACCs are created with functions with infix acc. The most basic variant, Task.appendacc will append an affine conic constraint based on the following data:. • the list afeidx of indices of AFEs to be used in the constraint. These are the row numbers in $$\afef,\afeg$$ which contain the required affine expressions.. • the index domidx of the domain to which the constraint belongs.. Note that number of AFEs used in afeidx must match the dimension of the domain.. In case of (6.5) we have already arranged $$\afef,\afeg$$ in such a way that their (only) three rows contain the three affine expressions we need (in the correct order), and we already defined the quadratic cone domain of matching dimension 3. The ACC is now constructed with the following call:. # Create the ACC. range(k+1), # Indices of AFE rows [0,...,k]. None) # Ignored. This completes the setup of the affine conic constraint.. 6.2.5 Example ACC1¶. We refer to Sec. 6.1 (Linear Optimization) for instructions how to set up the objective and linear constraint $$x_0+x_1+x_2=1$$. All else that remains is to set up the MOSEK environment, task, add variables, call the solver with Task.optimize and retrieve the solution with Task.getxx. Since our problem contains a nonlinear constraint we fetch the interior-point solution. The full code solving problem (6.3) is shown below.. Listing 6.2 Full code of example ACC1. Click here to download.. import sys, mosek. # Define a stream printer to grab output from MOSEK. def streamprinter(text):. sys.stdout.write(text). sys.stdout.flush(). # Define problem data. n, k = 3, 2. # Only a symbolic constant. inf = 0.0. # Make a MOSEK environment. with mosek.Env() as env:. # Attach a printer to the environment. env.set_Stream(mosek.streamtype.log, streamprinter). # Attach a printer to the task. # Create n free variables. # Set up the objective. c = [2, 3, -1]. # One linear constraint - sum(x) = 1. # Append empty AFE rows for affine expression storage. # G matrix in sparse form. Gsubi = [0, 0, 1, 1]. Gsubj = [0, 1, 0, 2]. Gval = [1.5, 0.1, 0.3, 2.1]. # Other data. h = [0, 0.1]. gamma = 0.03. # Construct F matrix in sparse form. Fsubi = [i + 1 for i in Gsubi] # G will be placed from row number 1 in F. Fsubj = Gsubj. Fval = Gval. # Fill in F storage. # Fill in g storage. # Define a conic quadratic domain. # Create the ACC. range(k+1), # Indices of AFE rows [0,...,k]. None) # Ignored. # Solve and retrieve solution. print("Solution: {xx}".format(xx=list(xx))). [-0.07838011145615721, 1.1289128998004547, -0.0505327883442975]. The dual values $$\doty$$ of an ACC can be obtained with Task.getaccdoty if required.. # Demonstrate retrieving the dual of ACC. 0) # ACC index. print("Dual of ACC:: {doty}".format(doty=list(doty))). 6.2.6 Example ACC2 - more conic constraints¶. Now that we know how to enter one affine conic constraint (ACC) we will demonstrate a problem with two ACCs. From there it should be clear how to add multiple ACCs. To keep things familiar we will reuse the previous problem, but this time cast it into a conic optimization problem with two ACCs as follows:. (6.8)$\begin{split}\begin{array}{ll} \mbox{maximize} & c^T x \\ \mbox{subject to} & (\sum_i x_i - 1,\ \gamma,\ Gx+h) \in \{0\} \times \Q^{k+1} \end{array}\end{split}$. or, using the data from the example:. $\begin{split}\begin{array}{lll} \mbox{maximize} & 2x_0+3x_1-x_2 & \\ \mbox{subject to} & x_0+x_1+x_2 - 1 & \in \{0\}, \\ & (0.03, 1.5x_0+0.1x_1, 0.3x_0+2.1x_2+0.1) & \in \Q^3 \end{array}\end{split}$. In other words, we transformed the linear constraint into an ACC with the one-point zero domain.. As before, we proceed in three steps. First, we add the variables and create the storage $$\afef$$, $$\afeg$$ containing all affine expressions that appear throughout all off the ACCs. It means we will require 4 rows:. (6.9)$\begin{split}\afef = \left[\begin{array}{ccc}1& 1 & 1 \\ 0 & 0 & 0\\ 1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right],\quad \afeg=\left[\begin{array}{c}-1 \\0.03\\ 0\\ 0.1\end{array}\right].\end{split}$. # Set AFE rows representing the linear constraint. # Set AFE rows representing the quadratic constraint. [0, 1], # varidx, column numbers. [1.5, 0.1]) # values. [0, 2], # varidx, column numbers. [0.3, 2.1]) # values. h = [0, 0.1]. gamma = 0.03. Next, we add the required domains: the zero domain of dimension 1, and the quadratic cone domain of dimension 3.. # Define domains. Finally, we create both ACCs. The first ACCs picks the 0-th row of $$\afef,\afeg$$ and places it in the zero domain:. task.appendacc(zeroDom, # Domain index. [0], # Indices of AFE rows. None) # Ignored. The second ACC picks rows $$1,2,3$$ in $$\afef,\afeg$$ and places them in the quadratic cone domain:. task.appendacc(quadDom, # Domain index. [1,2,3], # Indices of AFE rows. None) # Ignored. The completes the construction and we can solve the problem like before:. Listing 6.3 Full code of example ACC2. Click here to download.. import sys, mosek. # Define a stream printer to grab output from MOSEK. def streamprinter(text):. sys.stdout.write(text). sys.stdout.flush(). # Define problem data. n, k = 3, 2. # Only a symbolic constant. inf = 0.0. # Make a MOSEK environment. with mosek.Env() as env:. # Attach a printer to the environment. env.set_Stream(mosek.streamtype.log, streamprinter). # Attach a printer to the task. # Create n free variables. # Set up the objective. c = [2, 3, -1]. # Set AFE rows representing the linear constraint. # Set AFE rows representing the quadratic constraint. [0, 1], # varidx, column numbers. [1.5, 0.1]) # values. [0, 2], # varidx, column numbers. [0.3, 2.1]) # values. h = [0, 0.1]. gamma = 0.03. # Define domains. # Append affine conic constraints. [0], # Indices of AFE rows. None) # Ignored. [1,2,3], # Indices of AFE rows. None) # Ignored. # Solve and retrieve solution. print("Solution: {xx}".format(xx=list(xx))). We obtain the same result:. [-0.07838011145615721, 1.1289128998004547, -0.0505327883442975]. 6.2.7 Summary and extensions¶. In this section we presented the most basic usage of the affine expression storage $$\afef,\afeg$$ to input affine expressions used together with domains to create affine conic constraints. Now we briefly point out additional features of his interface which can be useful in some situations for more demanding users. They will be demonstrated in various examples in other tutorials and case studies in this manual.. • It is important to remember that $$\afef,\afeg$$ has only a storage function and during the ACC construction we can pick an arbitrary list of row indices and place them in a conic domain. It means for example that:. • It is not necessary to store the AFEs in the same order they will appear in ACCs.. • The same AFE index can appear more than once in one and/or more conic constraints (this can be used to reduce storage if the same affine expression is used in multiple ACCs).. • The $$\afef,\afeg$$ storage can even include rows that are not presently used in any ACC.. • Domains can be reused: multiple ACCs can use the same domain. On the other hand the same type of domain can appear under many domidx positions. In this sense the list of created domains also plays only a storage role: the domains are only used when they enter an ACC.. • Affine expressions can also contain semidefinite terms, ie. the most general form of an ACC is in fact. $Fx + \langle \bar{F},\barX\rangle + g \in \D$. These terms are input into the rows of AFE storage using the functions with infix afebarf, creating an additional storage structure $$\bar{\afef}$$.. • The same affine expression storage $$\afef,\afeg$$ is shared between affine conic and disjunctive constraints (see Sec. 6.9 (Disjunctive constraints)).. • If, on the other hand, the user chooses to always store the AFEs one by one sequentially in the same order as they appear in ACCs then sequential functions such as Task.appendaccseq and Task.appendaccsseq make it easy to input one or more ACCs by just specifying the starting AFE index and dimension.. • It is possible to add a number of ACCs in one go using Task.appendaccs.
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• When defining an ACC an additional constant vector $$b$$ can be provided to modify the constant terms coming from $$\afeg$$ but only for this particular ACC. This could be useful to reduce $$\afef$$ storage space if, for example, many expressions $$f^Tx+b_i$$ with the same linear part $$f^Tx$$, but varying constant terms $$b_i$$, are to be used throughout ACCs.
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What is 72.4 percent of 1,110?
How much is 72.4 percent of 1110? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 72.4% of 1110 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
72.4% of 1,110 = 803.64
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Calculating seventy-two point four of one thousand, one hundred and ten How to calculate 72.4% of 1110? Simply divide the percent by 100 and multiply by the number. For example, 72.4 /100 x 1110 = 803.64 or 0.724 x 1110 = 803.64
How much is 72.4 percent of the following numbers?
72.4% of 1110.01 = 80364.724 72.4% of 1110.02 = 80365.448 72.4% of 1110.03 = 80366.172 72.4% of 1110.04 = 80366.896 72.4% of 1110.05 = 80367.62 72.4% of 1110.06 = 80368.344 72.4% of 1110.07 = 80369.068 72.4% of 1110.08 = 80369.792 72.4% of 1110.09 = 80370.516 72.4% of 1110.1 = 80371.24 72.4% of 1110.11 = 80371.964 72.4% of 1110.12 = 80372.688 72.4% of 1110.13 = 80373.412 72.4% of 1110.14 = 80374.136 72.4% of 1110.15 = 80374.86 72.4% of 1110.16 = 80375.584 72.4% of 1110.17 = 80376.308 72.4% of 1110.18 = 80377.032 72.4% of 1110.19 = 80377.756 72.4% of 1110.2 = 80378.48 72.4% of 1110.21 = 80379.204 72.4% of 1110.22 = 80379.928 72.4% of 1110.23 = 80380.652 72.4% of 1110.24 = 80381.376 72.4% of 1110.25 = 80382.1
72.4% of 1110.26 = 80382.824 72.4% of 1110.27 = 80383.548 72.4% of 1110.28 = 80384.272 72.4% of 1110.29 = 80384.996 72.4% of 1110.3 = 80385.72 72.4% of 1110.31 = 80386.444 72.4% of 1110.32 = 80387.168 72.4% of 1110.33 = 80387.892 72.4% of 1110.34 = 80388.616 72.4% of 1110.35 = 80389.34 72.4% of 1110.36 = 80390.064 72.4% of 1110.37 = 80390.788 72.4% of 1110.38 = 80391.512 72.4% of 1110.39 = 80392.236 72.4% of 1110.4 = 80392.96 72.4% of 1110.41 = 80393.684 72.4% of 1110.42 = 80394.408 72.4% of 1110.43 = 80395.132 72.4% of 1110.44 = 80395.856 72.4% of 1110.45 = 80396.58 72.4% of 1110.46 = 80397.304 72.4% of 1110.47 = 80398.028 72.4% of 1110.48 = 80398.752 72.4% of 1110.49 = 80399.476 72.4% of 1110.5 = 80400.2
72.4% of 1110.51 = 80400.924 72.4% of 1110.52 = 80401.648 72.4% of 1110.53 = 80402.372 72.4% of 1110.54 = 80403.096 72.4% of 1110.55 = 80403.82 72.4% of 1110.56 = 80404.544 72.4% of 1110.57 = 80405.268 72.4% of 1110.58 = 80405.992 72.4% of 1110.59 = 80406.716 72.4% of 1110.6 = 80407.44 72.4% of 1110.61 = 80408.164 72.4% of 1110.62 = 80408.888 72.4% of 1110.63 = 80409.612 72.4% of 1110.64 = 80410.336 72.4% of 1110.65 = 80411.06 72.4% of 1110.66 = 80411.784 72.4% of 1110.67 = 80412.508 72.4% of 1110.68 = 80413.232 72.4% of 1110.69 = 80413.956 72.4% of 1110.7 = 80414.68 72.4% of 1110.71 = 80415.404 72.4% of 1110.72 = 80416.128 72.4% of 1110.73 = 80416.852 72.4% of 1110.74 = 80417.576 72.4% of 1110.75 = 80418.3
72.4% of 1110.76 = 80419.024 72.4% of 1110.77 = 80419.748 72.4% of 1110.78 = 80420.472 72.4% of 1110.79 = 80421.196 72.4% of 1110.8 = 80421.92 72.4% of 1110.81 = 80422.644 72.4% of 1110.82 = 80423.368 72.4% of 1110.83 = 80424.092 72.4% of 1110.84 = 80424.816 72.4% of 1110.85 = 80425.54 72.4% of 1110.86 = 80426.264 72.4% of 1110.87 = 80426.988 72.4% of 1110.88 = 80427.712 72.4% of 1110.89 = 80428.436 72.4% of 1110.9 = 80429.16 72.4% of 1110.91 = 80429.884 72.4% of 1110.92 = 80430.608 72.4% of 1110.93 = 80431.332 72.4% of 1110.94 = 80432.056 72.4% of 1110.95 = 80432.78 72.4% of 1110.96 = 80433.504 72.4% of 1110.97 = 80434.228 72.4% of 1110.98 = 80434.952 72.4% of 1110.99 = 80435.676 72.4% of 1111 = 80436.4
1% of 1110 = 11.1 2% of 1110 = 22.2 3% of 1110 = 33.3 4% of 1110 = 44.4 5% of 1110 = 55.5 6% of 1110 = 66.6 7% of 1110 = 77.7 8% of 1110 = 88.8 9% of 1110 = 99.9 10% of 1110 = 111 11% of 1110 = 122.1 12% of 1110 = 133.2 13% of 1110 = 144.3 14% of 1110 = 155.4 15% of 1110 = 166.5 16% of 1110 = 177.6 17% of 1110 = 188.7 18% of 1110 = 199.8 19% of 1110 = 210.9 20% of 1110 = 222 21% of 1110 = 233.1 22% of 1110 = 244.2 23% of 1110 = 255.3 24% of 1110 = 266.4 25% of 1110 = 277.5
26% of 1110 = 288.6 27% of 1110 = 299.7 28% of 1110 = 310.8 29% of 1110 = 321.9 30% of 1110 = 333 31% of 1110 = 344.1 32% of 1110 = 355.2 33% of 1110 = 366.3 34% of 1110 = 377.4 35% of 1110 = 388.5 36% of 1110 = 399.6 37% of 1110 = 410.7 38% of 1110 = 421.8 39% of 1110 = 432.9 40% of 1110 = 444 41% of 1110 = 455.1 42% of 1110 = 466.2 43% of 1110 = 477.3 44% of 1110 = 488.4 45% of 1110 = 499.5 46% of 1110 = 510.6 47% of 1110 = 521.7 48% of 1110 = 532.8 49% of 1110 = 543.9 50% of 1110 = 555
51% of 1110 = 566.1 52% of 1110 = 577.2 53% of 1110 = 588.3 54% of 1110 = 599.4 55% of 1110 = 610.5 56% of 1110 = 621.6 57% of 1110 = 632.7 58% of 1110 = 643.8 59% of 1110 = 654.9 60% of 1110 = 666 61% of 1110 = 677.1 62% of 1110 = 688.2 63% of 1110 = 699.3 64% of 1110 = 710.4 65% of 1110 = 721.5 66% of 1110 = 732.6 67% of 1110 = 743.7 68% of 1110 = 754.8 69% of 1110 = 765.9 70% of 1110 = 777 71% of 1110 = 788.1 72% of 1110 = 799.2 73% of 1110 = 810.3 74% of 1110 = 821.4 75% of 1110 = 832.5
76% of 1110 = 843.6 77% of 1110 = 854.7 78% of 1110 = 865.8 79% of 1110 = 876.9 80% of 1110 = 888 81% of 1110 = 899.1 82% of 1110 = 910.2 83% of 1110 = 921.3 84% of 1110 = 932.4 85% of 1110 = 943.5 86% of 1110 = 954.6 87% of 1110 = 965.7 88% of 1110 = 976.8 89% of 1110 = 987.9 90% of 1110 = 999 91% of 1110 = 1010.1 92% of 1110 = 1021.2 93% of 1110 = 1032.3 94% of 1110 = 1043.4 95% of 1110 = 1054.5 96% of 1110 = 1065.6 97% of 1110 = 1076.7 98% of 1110 = 1087.8 99% of 1110 = 1098.9 100% of 1110 = 1110
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What is 72.4 percent of 1,110?. How much is 72.4 percent of 1110? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 72.4% of 1110 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.. 72.4% of 1,110 = 803.64. Calculate another percentage below. Type into inputs. Find number based on percentage. percent of. Find percentage based on 2 numbers. divided by. Calculating seventy-two point four of one thousand, one hundred and ten How to calculate 72.4% of 1110? Simply divide the percent by 100 and multiply by the number. For example, 72.4 /100 x 1110 = 803.64 or 0.724 x 1110 = 803.64. How much is 72.4 percent of the following numbers?. 72.4% of 1110.01 = 80364.724 72.4% of 1110.02 = 80365.448 72.4% of 1110.03 = 80366.172 72.4% of 1110.04 = 80366.896 72.4% of 1110.05 = 80367.62 72.4% of 1110.06 = 80368.344 72.4% of 1110.07 = 80369.068 72.4% of 1110.08 = 80369.792 72.4% of 1110.09 = 80370.516 72.4% of 1110.1 = 80371.24 72.4% of 1110.11 = 80371.964 72.4% of 1110.12 = 80372.688 72.4% of 1110.13 = 80373.412 72.4% of 1110.14 = 80374.136 72.4% of 1110.15 = 80374.86 72.4% of 1110.16 = 80375.584 72.4% of 1110.17 = 80376.308 72.4% of 1110.18 = 80377.032 72.4% of 1110.19 = 80377.756 72.4% of 1110.2 = 80378.48 72.4% of 1110.21 = 80379.204 72.4% of 1110.22 = 80379.928 72.4% of 1110.23 = 80380.652 72.4% of 1110.24 = 80381.376 72.4% of 1110.25 = 80382.1. 72.4% of 1110.26 = 80382.824 72.4% of 1110.27 = 80383.548 72.4% of 1110.28 = 80384.272 72.4% of 1110.29 = 80384.996 72.4% of 1110.3 = 80385.72 72.4% of 1110.31 = 80386.444 72.4% of 1110.32 = 80387.168 72.4% of 1110.33 = 80387.892 72.4% of 1110.34 = 80388.616 72.4% of 1110.35 = 80389.34 72.4% of 1110.36 = 80390.064 72.4% of 1110.37 = 80390.788 72.4% of 1110.38 = 80391.512 72.4% of 1110.39 = 80392.236 72.4% of 1110.4 = 80392.96 72.4% of 1110.41 = 80393.684 72.4% of 1110.42 = 80394.408 72.4% of 1110.43 = 80395.132 72.4% of 1110.44 = 80395.856 72.4% of 1110.45 = 80396.58 72.4% of 1110.46 = 80397.304 72.4% of 1110.47 = 80398.028 72.4% of 1110.48 = 80398.752 72.4% of 1110.49 = 80399.476 72.4% of 1110.5 = 80400.2. 72.4% of 1110.51 = 80400.924 72.4% of 1110.52 = 80401.648 72.4% of 1110.53 = 80402.372 72.4% of 1110.54 = 80403.096 72.4% of 1110.55 = 80403.82 72.4% of 1110.56 = 80404.544 72.4% of 1110.57 = 80405.268 72.4% of 1110.58 = 80405.992 72.4% of 1110.59 = 80406.716 72.4% of 1110.6 = 80407.44 72.4% of 1110.61 = 80408.164 72.4% of 1110.62 = 80408.888 72.4% of 1110.63 = 80409.612 72.4% of 1110.64 = 80410.336 72.4% of 1110.65 = 80411.06 72.4% of 1110.66 = 80411.784 72.4% of 1110.67 = 80412.508 72.4% of 1110.68 = 80413.232 72.4% of 1110.69 = 80413.956 72.4% of 1110.7 = 80414.68 72.4% of 1110.71 = 80415.404 72.4% of 1110.72 = 80416.128 72.4% of 1110.73 = 80416.852 72.4% of 1110.74 = 80417.576 72.4% of 1110.75 = 80418.3. 72.4% of 1110.76 = 80419.024 72.4% of 1110.77 = 80419.748 72.4% of 1110.78 = 80420.472 72.4% of 1110.79 = 80421.196 72.4% of 1110.8 = 80421.92 72.4% of 1110.81 = 80422.644 72.4% of 1110.82 = 80423.368 72.4% of 1110.83 = 80424.092 72.4% of 1110.84 = 80424.816 72.4% of 1110.85 = 80425.54 72.4% of 1110.86 = 80426.264 72.4% of 1110.87 = 80426.988 72.4% of 1110.88 = 80427.712 72.4% of 1110.89 = 80428.436 72.4% of 1110.9 = 80429.16 72.4% of 1110.91 = 80429.884 72.4% of 1110.92 = 80430.608 72.4% of 1110.93 = 80431.332 72.4% of 1110.94 = 80432.056 72.4% of 1110.95 = 80432.78 72.4% of 1110.96 = 80433.504 72.4% of 1110.97 = 80434.228 72.4% of 1110.98 = 80434.952 72.4% of 1110.99 = 80435.676 72.4% of 1111 = 80436.4. 1% of 1110 = 11.1 2% of 1110 = 22.2 3% of 1110 = 33.3 4% of 1110 = 44.4 5% of 1110 = 55.5 6% of 1110 = 66.6 7% of 1110 = 77.7 8% of 1110 = 88.8 9% of 1110 = 99.9 10% of 1110 = 111 11% of 1110 = 122.1 12% of 1110 = 133.2 13% of 1110 = 144.3 14% of 1110 = 155.4 15% of 1110 = 166.5 16% of 1110 = 177.6 17% of 1110 = 188.7 18% of 1110 = 199.8 19% of 1110 = 210.9 20% of 1110 = 222 21% of 1110 = 233.1 22% of 1110 = 244.2 23% of 1110 = 255.3 24% of 1110 = 266.4 25% of 1110 = 277.5. 26% of 1110 = 288.6 27% of 1110 = 299.7 28% of 1110 = 310.8 29% of 1110 = 321.9 30% of 1110 = 333 31% of 1110 = 344.1 32% of 1110 = 355.2 33% of 1110 = 366.3 34% of 1110 = 377.4 35% of 1110 = 388.5 36% of 1110 = 399.6 37% of 1110 = 410.7 38% of 1110 = 421.8 39% of 1110 = 432.9 40% of 1110 = 444 41% of 1110 = 455.1 42% of 1110 = 466.2 43% of 1110 = 477.3 44% of 1110 = 488.4 45% of 1110 = 499.5 46% of 1110 = 510.6 47% of 1110 = 521.7 48% of 1110 = 532.8 49% of 1110 = 543.9 50% of 1110 = 555.
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51% of 1110 = 566.1 52% of 1110 = 577.2 53% of 1110 = 588.3 54% of 1110 = 599.4 55% of 1110 = 610.5 56% of 1110 = 621.6 57% of 1110 = 632.7 58% of 1110 = 643.8 59% of 1110 = 654.9 60% of 1110 = 666 61% of 1110 = 677.1 62% of 1110 = 688.2 63% of 1110 = 699.3 64% of 1110 = 710.4 65% of 1110 = 721.5 66% of 1110 = 732.6 67% of 1110 = 743.7 68% of 1110 = 754.8 69% of 1110 = 765.9 70% of 1110 = 777 71% of 1110 = 788.1 72% of 1110 = 799.2 73% of 1110 = 810.3 74% of 1110 = 821.4 75% of 1110 = 832.5. 76% of 1110 = 843.6 77% of 1110 = 854.7 78% of 1110 = 865.8 79% of 1110 = 876.9 80% of 1110 = 888 81% of 1110 = 899.1 82% of 1110 = 910.2 83% of 1110 = 921.3 84% of 1110 = 932.4 85% of 1110 = 943.5 86% of 1110 = 954.6 87% of 1110 = 965.7 88% of 1110 = 976.8 89% of 1110 = 987.9 90% of 1110 = 999 91% of 1110 = 1010.1 92% of 1110 = 1021.2 93% of 1110 = 1032.3 94% of 1110 = 1043.4 95% of 1110 = 1054.5 96% of 1110 = 1065.6 97% of 1110 = 1076.7 98% of 1110 = 1087.8 99% of 1110 = 1098.9 100% of 1110 = 1110.
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Home > Error Propagation > Propagation Of Uncertainty From Random Error
# Propagation Of Uncertainty From Random Error
## Contents
Please try the request again. Foothill College. If these were your data and you wanted to reduce the uncertainty, you would need to do more titrations, both to increase N and to (we hope) increase your precision and p.37. http://spamdestructor.com/error-propagation/propagation-of-error-uncertainty.php
The system returned: (22) Invalid argument The remote host or network may be down. Answer: we can calculate the time as (g = 9.81 m/s2 is assumed to be known exactly) t = - v / g = 3.8 m/s / 9.81 m/s2 = 0.387 However, if an instrument is well calibrated, the precision or reproducibility of the result is a good measure of its accuracy. In statistics, propagation of uncertainty (or propagation of error) is the effect of variables' uncertainties (or errors, more specifically random errors) on the uncertainty of a function based on them. https://en.wikipedia.org/wiki/Propagation_of_uncertainty
## Propagation Of Error Division
Again, the uncertainty is less than that predicted by significant figures. Example: To apply this statistical method of error analysis to our KHP example, we need more than one result to average. Accuracy and Precision The accuracy of a set of observations is the difference between the average of the measured values and the true value of the observed quantity. GUM, Guide to the Expression of Uncertainty in Measurement EPFL An Introduction to Error Propagation, Derivation, Meaning and Examples of Cy = Fx Cx Fx' uncertainties package, a program/library for transparently
The Error Propagation and Significant Figures results are in agreement, within the calculated uncertainties, but the Error Propagation and Statistical Method results do not agree, within the uncertainty calculated from Error The precision of two other pieces of apparatus that you will often use is somewhat less obvious from a consideration of the scale markings on these instruments. If a result differs widely from the results of other experiments you have performed, or has low precision, a blunder may also be to blame. Error Propagation Square Root The analytical balance does this by electronically resetting the digital readout of the weight of the vessel to 0.0000.
An instrument might produce a blunder if a poor electrical connection causes the display to read an occasional incorrect value. Error Propagation Formula Physics In a similar vein, an experimenter may consistently overshoot the endpoint of a titration because she is wearing tinted glasses and cannot see the first color change of the indicator. Taring involves subtraction of the weight of the vessel from the weight of the sample and vessel to determine the weight of the sample. These are tabulated values that relate the standard error of a mean to a confidence interval.
The number of significant figures, used in the significant figure rules for multiplication and division, is related to the relative uncertainty. Error Propagation Inverse If the statistical probability distribution of the variable is known or can be assumed, it is possible to derive confidence limits to describe the region within which the true value of Joint Committee for Guides in Metrology (2011). Trustees of Dartmouth College, Copyright 1997-2010 ERROR PROPAGATION 1. Measurement of Physical Properties The value of a physical property often depends on one or more measured quantities Example: Volume
## Error Propagation Formula Physics
Berkeley Seismology Laboratory. When the variables are the values of experimental measurements they have uncertainties due to measurement limitations (e.g., instrument precision) which propagate to the combination of variables in the function. Propagation Of Error Division Journal of Sound and Vibrations. 332 (11): 2750–2776. Error Propagation Calculator Reciprocal In the special case of the inverse or reciprocal 1 / B {\displaystyle 1/B} , where B = N ( 0 , 1 ) {\displaystyle B=N(0,1)} , the distribution is
Please note that the rule is the same for addition and subtraction of quantities. my review here In this example that would be written 0.118 ± 0.002 (95%, N = 4). This can be rearranged and the calculated molarity substituted to give σM = (3 x 10–3) (0.11892 M) = 4 × 10–4 M The final result would be reported as 0.1189 To reduce the uncertainty, you would need to measure the volume more accurately, not the mass. Error Propagation Chemistry
For example, a balance may always read 0.001 g too light because it was zeroed incorrectly. Trial [NaOH] 1 0.1180 M 2 0.1176 3 0.1159 4 0.1192 The first step is to calculate the mean value of the molarity, using Equation 3. In this case, expressions for more complicated functions can be derived by combining simpler functions. click site Precision of Instrument Readings and Other Raw Data The first step in determining the uncertainty in calculated results is to estimate the precision of the raw data used in the calculation.
You fill the buret to the top mark and record 0.00 mL as your starting volume. Error Propagation Average Please try the request again. The correct procedures are these: A.
## For the example of the three weighings, with an average of 6.3302 ± 0.0001 g, the absolute uncertainty is 0.0001 g.
Addition and subtraction: The result will have a last significant digit in the same place as the left-most of the last significant digits of all the numbers used in the calculation. The values in parentheses indicate the confidence interval and the number of measurements. This means that the true value of the volume is determined by the experiment to be in the range between 8.95 and 9.01 mL Multiplication and division: Uncertainty in results depends Error Propagation Definition Journal of Research of the National Bureau of Standards.
For example, the 68% confidence limits for a one-dimensional variable belonging to a normal distribution are ± one standard deviation from the value, that is, there is approximately a 68% probability Simplification Neglecting correlations or assuming independent variables yields a common formula among engineers and experimental scientists to calculate error propagation, the variance formula:[4] s f = ( ∂ f ∂ x Journal of Sound and Vibrations. 332 (11). navigate to this website First, here are some fundamental things you should realize about uncertainty: • Every measurement has an uncertainty associated with it, unless it is an exact, counted integer, such as the number
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Home > Error Propagation > Propagation Of Uncertainty From Random Error. # Propagation Of Uncertainty From Random Error. ## Contents. Please try the request again. Foothill College. If these were your data and you wanted to reduce the uncertainty, you would need to do more titrations, both to increase N and to (we hope) increase your precision and p.37. http://spamdestructor.com/error-propagation/propagation-of-error-uncertainty.php. The system returned: (22) Invalid argument The remote host or network may be down. Answer: we can calculate the time as (g = 9.81 m/s2 is assumed to be known exactly) t = - v / g = 3.8 m/s / 9.81 m/s2 = 0.387 However, if an instrument is well calibrated, the precision or reproducibility of the result is a good measure of its accuracy. In statistics, propagation of uncertainty (or propagation of error) is the effect of variables' uncertainties (or errors, more specifically random errors) on the uncertainty of a function based on them. https://en.wikipedia.org/wiki/Propagation_of_uncertainty. ## Propagation Of Error Division. Again, the uncertainty is less than that predicted by significant figures. Example: To apply this statistical method of error analysis to our KHP example, we need more than one result to average. Accuracy and Precision The accuracy of a set of observations is the difference between the average of the measured values and the true value of the observed quantity. GUM, Guide to the Expression of Uncertainty in Measurement EPFL An Introduction to Error Propagation, Derivation, Meaning and Examples of Cy = Fx Cx Fx' uncertainties package, a program/library for transparently. The Error Propagation and Significant Figures results are in agreement, within the calculated uncertainties, but the Error Propagation and Statistical Method results do not agree, within the uncertainty calculated from Error The precision of two other pieces of apparatus that you will often use is somewhat less obvious from a consideration of the scale markings on these instruments. If a result differs widely from the results of other experiments you have performed, or has low precision, a blunder may also be to blame. Error Propagation Square Root The analytical balance does this by electronically resetting the digital readout of the weight of the vessel to 0.0000.. An instrument might produce a blunder if a poor electrical connection causes the display to read an occasional incorrect value. Error Propagation Formula Physics In a similar vein, an experimenter may consistently overshoot the endpoint of a titration because she is wearing tinted glasses and cannot see the first color change of the indicator. Taring involves subtraction of the weight of the vessel from the weight of the sample and vessel to determine the weight of the sample. These are tabulated values that relate the standard error of a mean to a confidence interval.. The number of significant figures, used in the significant figure rules for multiplication and division, is related to the relative uncertainty. Error Propagation Inverse If the statistical probability distribution of the variable is known or can be assumed, it is possible to derive confidence limits to describe the region within which the true value of Joint Committee for Guides in Metrology (2011). Trustees of Dartmouth College, Copyright 1997-2010 ERROR PROPAGATION 1. Measurement of Physical Properties The value of a physical property often depends on one or more measured quantities Example: Volume. ## Error Propagation Formula Physics. Berkeley Seismology Laboratory. When the variables are the values of experimental measurements they have uncertainties due to measurement limitations (e.g., instrument precision) which propagate to the combination of variables in the function. Propagation Of Error Division Journal of Sound and Vibrations. 332 (11): 2750–2776. Error Propagation Calculator Reciprocal In the special case of the inverse or reciprocal 1 / B {\displaystyle 1/B} , where B = N ( 0 , 1 ) {\displaystyle B=N(0,1)} , the distribution is. Please note that the rule is the same for addition and subtraction of quantities. my review here In this example that would be written 0.118 ± 0.002 (95%, N = 4). This can be rearranged and the calculated molarity substituted to give σM = (3 x 10–3) (0.11892 M) = 4 × 10–4 M The final result would be reported as 0.1189 To reduce the uncertainty, you would need to measure the volume more accurately, not the mass. Error Propagation Chemistry. For example, a balance may always read 0.001 g too light because it was zeroed incorrectly. Trial [NaOH] 1 0.1180 M 2 0.1176 3 0.1159 4 0.1192 The first step is to calculate the mean value of the molarity, using Equation 3. In this case, expressions for more complicated functions can be derived by combining simpler functions. click site Precision of Instrument Readings and Other Raw Data The first step in determining the uncertainty in calculated results is to estimate the precision of the raw data used in the calculation.. You fill the buret to the top mark and record 0.00 mL as your starting volume. Error Propagation Average Please try the request again. The correct procedures are these: A.. ## For the example of the three weighings, with an average of 6.3302 ± 0.0001 g, the absolute uncertainty is 0.0001 g.. Addition and subtraction: The result will have a last significant digit in the same place as the left-most of the last significant digits of all the numbers used in the calculation. The values in parentheses indicate the confidence interval and the number of measurements. This means that the true value of the volume is determined by the experiment to be in the range between 8.95 and 9.01 mL Multiplication and division: Uncertainty in results depends Error Propagation Definition Journal of Research of the National Bureau of Standards.
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For example, the 68% confidence limits for a one-dimensional variable belonging to a normal distribution are ± one standard deviation from the value, that is, there is approximately a 68% probability Simplification Neglecting correlations or assuming independent variables yields a common formula among engineers and experimental scientists to calculate error propagation, the variance formula:[4] s f = ( ∂ f ∂ x Journal of Sound and Vibrations. 332 (11). navigate to this website First, here are some fundamental things you should realize about uncertainty: • Every measurement has an uncertainty associated with it, unless it is an exact, counted integer, such as the number.
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What's my computational basis if I want to define a unitary operator that implements a function such as $f(i) = 2^{i+1} \text{mod 21}$?
I know I must define $$U_f$$, the unitary operator, on the computational basis. But what's my computational basis here?
Presumably you want to work with qubits? So the usual computation basis applies: $$|0\rangle$$ and $$|1\rangle$$ for a single qubit, and a composite basis of $$|x\rangle$$ for $$x\in\{0,1\}^n$$ when composing $$n$$ qubits.
What I guess you're asking is how you translate your problem onto qubits. For that, you need to make the decimal values correspond to particular bit values. The conventional way of doing this is using the binary string $$x$$ like a binary number, which corresponds to a decimal value. There are a number of different numbering conventions you can pick from but, for example, you might have $$x=x_0x_1x_2\ldots x_{n-1},$$ meaning that the corresponding decimal value is $$x_{n-1}+2x_{n-2}+4x_{n-3}+\ldots+2^{n-2}x_1+2^{n-1}x_0.$$ $$n$$ bits lets you represent any decimal number 0 to $$2^n-1$$.
It depends on the $$i$$ you want to apply this function, which is to be represented as a bitstring (in this case, I guess an unsigned integer representation) that will be one of the computational basis. It can be a particular one or just a superposition. You need to define the registers (the one containing $$i$$, garbage register containing intermediary results and the one containing the result). Mathematically, your operator will have the following effect if you uncompute intermediary results : $$U_f | i \rangle | 0 \rangle_g | 0 \rangle_f = | i \rangle | 0 \rangle_g | f(i) \rangle_f$$
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What's my computational basis if I want to define a unitary operator that implements a function such as $f(i) = 2^{i+1} \text{mod 21}$?. I know I must define $$U_f$$, the unitary operator, on the computational basis. But what's my computational basis here?. Presumably you want to work with qubits? So the usual computation basis applies: $$|0\rangle$$ and $$|1\rangle$$ for a single qubit, and a composite basis of $$|x\rangle$$ for $$x\in\{0,1\}^n$$ when composing $$n$$ qubits.. What I guess you're asking is how you translate your problem onto qubits. For that, you need to make the decimal values correspond to particular bit values. The conventional way of doing this is using the binary string $$x$$ like a binary number, which corresponds to a decimal value. There are a number of different numbering conventions you can pick from but, for example, you might have $$x=x_0x_1x_2\ldots x_{n-1},$$ meaning that the corresponding decimal value is $$x_{n-1}+2x_{n-2}+4x_{n-3}+\ldots+2^{n-2}x_1+2^{n-1}x_0.$$ $$n$$ bits lets you represent any decimal number 0 to $$2^n-1$$.. It depends on the $$i$$ you want to apply this function, which is to be represented as a bitstring (in this case, I guess an unsigned integer representation) that will be one of the computational basis.
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It can be a particular one or just a superposition. You need to define the registers (the one containing $$i$$, garbage register containing intermediary results and the one containing the result). Mathematically, your operator will have the following effect if you uncompute intermediary results : $$U_f | i \rangle | 0 \rangle_g | 0 \rangle_f = | i \rangle | 0 \rangle_g | f(i) \rangle_f$$.
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# When to use L'Hospital's Rule...
• Oct 31st 2007, 06:45 AM
Esiuol
When to use L'Hospital's Rule...
I'm having some issues determining when I can and cannot use L'Hospital's Rule. Tips? Suggested approaches?
• Oct 31st 2007, 07:37 AM
Krizalid
You can use it when you have the following typical cases:
#1 Indetermination of the form $\displaystyle \frac00.$
#2 Indetermination of the form $\displaystyle \frac\infty\infty.$
Of course, if you're solving a problem and this one says that you cannot apply the Rule, don't do it :D:D
Many limits can be computed without the Rule, that makes them more interesting.
• Oct 31st 2007, 09:17 AM
Linnus
Just to add to his comment, there are sometime that you manipulate the problem to get to 0/0 or ∞/∞
Such cases are when the limits become one of those forms : 0^0 1^∞ ∞ - ∞, 0⋅∞ ∞^0
• Oct 31st 2007, 09:37 AM
Esiuol
That's what I'm having issues with -- problems where it isn't apparent that l'hospital's rule applies. My professor just told me to practice, but I'm not getting anywhere...I feel like I'm missing something, (probably extremely simple), but we glossed over the material so fast, I'm not surprised I'm having a little difficulty.
• Oct 31st 2007, 09:44 AM
Linnus
when you plug x in and you get any of the form I listed above, usually there is a way to manipluated it to 0/0 or $\displaystyle \frac\infty\infty$
then you can use the L'Hospital's rule
• Oct 31st 2007, 10:18 AM
Esiuol
For example, the limit as t approaches 0 of (e^t)-1/t^3
It looks like "0/0" so I use l'hopital's rule and get (e^t)/3t^2
Looking at my calculator, I know it's infinity, but I don't understand how we get that from using the rule. I get 1/0 then...
• Oct 31st 2007, 10:24 AM
Krizalid
People sometimes apply the Rule thousand of times, but always you need to check if you got indetermination.
So, when you evaluate directly $\displaystyle \frac{e^t}{3t^2},$ there's no indetermination, and the conclusion follows.
• Oct 31st 2007, 10:31 AM
Linnus
Quote:
Originally Posted by Esiuol
For example, the limit as t approaches 0 of (e^t)-1/t^3
It looks like "0/0" so I use l'hopital's rule and get (e^t)/3t^2
Looking at my calculator, I know it's infinity, but I don't understand how we get that from using the rule. I get 1/0 then...
hmm you probably forgot this...
Lim as t-->0 for (e^t)/3t^2 = infinity because at as the denominator goes to zero, the number gets bigger and bigger...
So yea, the L'hospital rule did work for that problem
• Dec 7th 2007, 07:17 PM
angel.white
Can I also use L'Hospital's rule in:
a) $\displaystyle \frac{-\infty}{\infty}$
b) $\displaystyle \frac{\infty}{-\infty}$
c) $\displaystyle \frac{-\infty}{-\infty}$
d) $\displaystyle \frac{-0}{0}$
e) $\displaystyle \frac{0}{-0}$
f) $\displaystyle \frac{-0}{-0}$
(added the zeros, figure they all take the form 0/0, but may as well make sure :))
• Dec 8th 2007, 12:45 AM
Yes because:
Lim (-f(x)) = -Lim(f(x))
and$\displaystyle \frac {d}{dx} (-f(x))$ = $\displaystyle -\frac{d}{dx}(f(x))$
Quote:
Such cases are when the limits become one of those forms : 0^0 1^∞ ∞ - ∞, 0⋅∞ ∞^0
each of these cases have general methods which you would do well to memorise:
$\displaystyle 0^0$:
$\displaystyle lim 0^0 = e^{lim log(0^0)}$
=$\displaystyle e^{lim 0log0}$
then use the method for 0⋅∞
$\displaystyle 1^\infty$:
lim $\displaystyle 1^\infty$ =$\displaystyle e^{lim log(1^{\infty})}$
= $\displaystyle e^{lim \infty log1}$
then use the method for 0⋅∞
$\displaystyle \infty^0$:
$\displaystyle lim \infty^0$ = $\displaystyle e^{lim log (\infty^0)}$
=$\displaystyle e^{lim 0log(\infty)}$
then use method for 0⋅∞
0⋅∞:
Do either of the following:
0⋅∞ = $\displaystyle \frac{0}{1/\infty}$
=0/0
or
0⋅∞ = $\displaystyle \frac {\infty}{1/0}$
=$\displaystyle \infty/\infty$
$\displaystyle \infty - \infty$:
$\displaystyle lim \infty-\infty = log(lim e^{\infty-\infty})$$\displaystyle =log (lim \frac{e^\infty}{e^\infty}) \displaystyle =log (lim \infty/\infty) • Dec 8th 2007, 08:35 AM angel.white Quote: Originally Posted by badgerigar Yes because: Lim (-f(x)) = -Lim(f(x)) and\displaystyle \frac {d}{dx} (-f(x)) = \displaystyle -\frac{d}{dx}(f(x)) each of these cases have general methods which you would do well to memorise: \displaystyle 0^0: \displaystyle lim 0^0 = e^{lim log(0^0)} =\displaystyle e^{lim 0log0} then use the method for 0⋅∞ \displaystyle 1^\infty: lim \displaystyle 1^\infty =\displaystyle e^{lim log(1^{\infty})} = \displaystyle e^{lim \infty log1} then use the method for 0⋅∞ \displaystyle \infty^0: \displaystyle lim \infty^0 = \displaystyle e^{lim log (\infty^0)} =\displaystyle e^{lim 0log(\infty)} then use method for 0⋅∞ 0⋅∞: Do either of the following: 0⋅∞ = \displaystyle \frac{0}{1/\infty} =0/0 or 0⋅∞ = \displaystyle \frac {\infty}{1/0} =\displaystyle \infty/\infty \displaystyle \infty - \infty: \displaystyle lim \infty-\infty = log(lim e^{\infty-\infty})$$\displaystyle =log (lim \frac{e^\infty}{e^\infty})$
$\displaystyle =log (lim \infty/\infty)$
I see, my book was so vague about what cases I could use it, I remember several times that I got it into usable form, but I didn't think it was usable and kept manipulating it until I got something crazy that would work. And when I finished I would look at it bewildered and go "I really hope that's not on the test"
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# When to use L'Hospital's Rule.... • Oct 31st 2007, 06:45 AM. Esiuol. When to use L'Hospital's Rule.... I'm having some issues determining when I can and cannot use L'Hospital's Rule. Tips? Suggested approaches?. • Oct 31st 2007, 07:37 AM. Krizalid. You can use it when you have the following typical cases:. #1 Indetermination of the form $\displaystyle \frac00.$. #2 Indetermination of the form $\displaystyle \frac\infty\infty.$. Of course, if you're solving a problem and this one says that you cannot apply the Rule, don't do it :D:D. Many limits can be computed without the Rule, that makes them more interesting.. • Oct 31st 2007, 09:17 AM. Linnus. Just to add to his comment, there are sometime that you manipulate the problem to get to 0/0 or ∞/∞. Such cases are when the limits become one of those forms : 0^0 1^∞ ∞ - ∞, 0⋅∞ ∞^0. • Oct 31st 2007, 09:37 AM. Esiuol. That's what I'm having issues with -- problems where it isn't apparent that l'hospital's rule applies. My professor just told me to practice, but I'm not getting anywhere...I feel like I'm missing something, (probably extremely simple), but we glossed over the material so fast, I'm not surprised I'm having a little difficulty.. • Oct 31st 2007, 09:44 AM. Linnus. when you plug x in and you get any of the form I listed above, usually there is a way to manipluated it to 0/0 or $\displaystyle \frac\infty\infty$. then you can use the L'Hospital's rule. • Oct 31st 2007, 10:18 AM. Esiuol. For example, the limit as t approaches 0 of (e^t)-1/t^3. It looks like "0/0" so I use l'hopital's rule and get (e^t)/3t^2. Looking at my calculator, I know it's infinity, but I don't understand how we get that from using the rule. I get 1/0 then.... • Oct 31st 2007, 10:24 AM. Krizalid. People sometimes apply the Rule thousand of times, but always you need to check if you got indetermination.. So, when you evaluate directly $\displaystyle \frac{e^t}{3t^2},$ there's no indetermination, and the conclusion follows.. • Oct 31st 2007, 10:31 AM. Linnus. Quote:. Originally Posted by Esiuol. For example, the limit as t approaches 0 of (e^t)-1/t^3. It looks like "0/0" so I use l'hopital's rule and get (e^t)/3t^2. Looking at my calculator, I know it's infinity, but I don't understand how we get that from using the rule. I get 1/0 then.... hmm you probably forgot this.... Lim as t-->0 for (e^t)/3t^2 = infinity because at as the denominator goes to zero, the number gets bigger and bigger.... So yea, the L'hospital rule did work for that problem. • Dec 7th 2007, 07:17 PM. angel.white. Can I also use L'Hospital's rule in:. a) $\displaystyle \frac{-\infty}{\infty}$. b) $\displaystyle \frac{\infty}{-\infty}$. c) $\displaystyle \frac{-\infty}{-\infty}$. d) $\displaystyle \frac{-0}{0}$. e) $\displaystyle \frac{0}{-0}$. f) $\displaystyle \frac{-0}{-0}$. (added the zeros, figure they all take the form 0/0, but may as well make sure :)). • Dec 8th 2007, 12:45 AM. Yes because:. Lim (-f(x)) = -Lim(f(x)). and$\displaystyle \frac {d}{dx} (-f(x))$ = $\displaystyle -\frac{d}{dx}(f(x))$. Quote:. Such cases are when the limits become one of those forms : 0^0 1^∞ ∞ - ∞, 0⋅∞ ∞^0. each of these cases have general methods which you would do well to memorise:. $\displaystyle 0^0$:. $\displaystyle lim 0^0 = e^{lim log(0^0)}$. =$\displaystyle e^{lim 0log0}$. then use the method for 0⋅∞. $\displaystyle 1^\infty$:. lim $\displaystyle 1^\infty$ =$\displaystyle e^{lim log(1^{\infty})}$. = $\displaystyle e^{lim \infty log1}$. then use the method for 0⋅∞. $\displaystyle \infty^0$:. $\displaystyle lim \infty^0$ = $\displaystyle e^{lim log (\infty^0)}$. =$\displaystyle e^{lim 0log(\infty)}$. then use method for 0⋅∞. 0⋅∞:. Do either of the following:. 0⋅∞ = $\displaystyle \frac{0}{1/\infty}$. =0/0. or. 0⋅∞ = $\displaystyle \frac {\infty}{1/0}$. =$\displaystyle \infty/\infty$. $\displaystyle \infty - \infty$:. $\displaystyle lim \infty-\infty = log(lim e^{\infty-\infty})$$\displaystyle =log (lim \frac{e^\infty}{e^\infty}) \displaystyle =log (lim \infty/\infty) • Dec 8th 2007, 08:35 AM angel.white Quote: Originally Posted by badgerigar Yes because: Lim (-f(x)) = -Lim(f(x)) and\displaystyle \frac {d}{dx} (-f(x)) = \displaystyle -\frac{d}{dx}(f(x)) each of these cases have general methods which you would do well to memorise: \displaystyle 0^0: \displaystyle lim 0^0 = e^{lim log(0^0)} =\displaystyle e^{lim 0log0} then use the method for 0⋅∞ \displaystyle 1^\infty: lim \displaystyle 1^\infty =\displaystyle e^{lim log(1^{\infty})} = \displaystyle e^{lim \infty log1} then use the method for 0⋅∞ \displaystyle \infty^0: \displaystyle lim \infty^0 = \displaystyle e^{lim log (\infty^0)} =\displaystyle e^{lim 0log(\infty)} then use method for 0⋅∞ 0⋅∞: Do either of the following: 0⋅∞ = \displaystyle \frac{0}{1/\infty} =0/0 or 0⋅∞ = \displaystyle \frac {\infty}{1/0} =\displaystyle \infty/\infty \displaystyle \infty - \infty: \displaystyle lim \infty-\infty = log(lim e^{\infty-\infty})$$\displaystyle =log (lim \frac{e^\infty}{e^\infty})$. $\displaystyle =log (lim \infty/\infty)$.
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I see, my book was so vague about what cases I could use it, I remember several times that I got it into usable form, but I didn't think it was usable and kept manipulating it until I got something crazy that would work. And when I finished I would look at it bewildered and go "I really hope that's not on the test".
|
https://de.maplesoft.com/support/help/maplesim/view.aspx?path=Student/ODEs/ODESteps/FirstOrderIVPs
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First Order IVPs - Maple Help
ODE Steps for First Order IVPs
Overview
• This help page gives a few examples of using the command ODESteps to solve first order initial value problems.
• See Student[ODEs][ODESteps] for a general description of the command ODESteps and its calling sequence.
Examples
> $\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right):$
> $\mathrm{ivp1}≔\left\{{t}^{2}\left(z\left(t\right)+1\right)+{z\left(t\right)}^{2}\left(t-1\right)\mathrm{diff}\left(z\left(t\right),t\right)=0,z\left(3\right)=1\right\}$
${\mathrm{ivp1}}{≔}\left\{{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}{,}{z}{}\left({3}\right){=}{1}\right\}$ (1)
> $\mathrm{ODESteps}\left(\mathrm{ivp1}\right)$
$\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}{,}{z}{}\left({3}\right){=}{1}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\\ \text{•}& {}& \text{Separate variables}\\ {}& {}& \frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}{=}{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\\ \text{•}& {}& \text{Integrate both sides with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}t\\ {}& {}& {\int }\frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{=}{\int }{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{\mathrm{C1}}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}z{}\left(3\right)=1\\ {}& {}& {-}\frac{{1}}{{2}}{+}{\mathrm{ln}}{}\left({2}\right){=}{-}\frac{{15}}{{2}}{-}{\mathrm{ln}}{}\left({2}\right){+}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=7+2{}\mathrm{ln}{}\left(2\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\end{array}$ (2)
> $\mathrm{ivp2}≔\left\{2xy\left(x\right)-9{x}^{2}+\left(2y\left(x\right)+{x}^{2}+1\right)\mathrm{diff}\left(y\left(x\right),x\right)=0,y\left(0\right)=1\right\}$
${\mathrm{ivp2}}{≔}\left\{{2}{}{x}{}{y}{}\left({x}\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}$ (3)
> $\mathrm{ODESteps}\left(\mathrm{ivp2}\right)$
$\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{{2}{}{x}{}{y}{}\left({x}\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{▫}& {}& \text{Check if ODE is exact}\\ {}& \text{◦}& \text{ODE is exact if the lhs is the total derivative of a}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{C}^{2}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{function}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}{}\left({x}\right)\right){=}{0}\\ {}& \text{◦}& \text{Compute derivative of lhs}\\ {}& {}& \frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right){+}\left(\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}\\ {}& \text{◦}& \text{Evaluate derivatives}\\ {}& {}& {2}{}{x}{=}{2}{}{x}\\ {}& \text{◦}& \text{Condition met, ODE is exact}\\ \text{•}& {}& \text{Exact ODE implies solution will be of this form}\\ {}& {}& \left[{F}{}\left({x}{,}{y}\right){=}{\mathrm{C1}}{,}{M}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right){,}{N}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\right]\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{by integrating}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}M{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{\int }\left({-}{9}{}{{x}}^{{2}}{+}{2}{}{x}{}{y}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Take derivative of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y\\ {}& {}& {N}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\\ \text{•}& {}& \text{Compute derivative}\\ {}& {}& {{x}}^{{2}}{+}{2}{}{y}{+}{1}{=}{{x}}^{{2}}{+}\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Isolate for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆy}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{_F1}{}\left(y\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({y}\right){=}{2}{}{y}{+}{1}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(y\right)\\ {}& {}& {\mathrm{_F1}}{}\left({y}\right){=}{{y}}^{{2}}{+}{y}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into equation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{{y}}^{{2}}{+}{y}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into the solution of the ODE}\\ {}& {}& {-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{{y}}^{{2}}{+}{y}{=}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& \left\{{y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}{,}{y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right\}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(0\right)=1\\ {}& {}& {1}{=}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Solution does not satisfy initial condition}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(0\right)=1\\ {}& {}& {1}{=}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{2}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& {y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{9}}}{{2}}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{9}}}{{2}}\end{array}$ (4)
> $\mathrm{ivp3}≔\left\{\mathrm{diff}\left(y\left(x\right),x\right)-y\left(x\right)-x\mathrm{exp}\left(x\right)=0,y\left(a\right)=b\right\}$
${\mathrm{ivp3}}{≔}\left\{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}{y}{}\left({a}\right){=}{b}\right\}$ (5)
> $\mathrm{ODESteps}\left(\mathrm{ivp3}\right)$
$\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}{y}{}\left({a}\right){=}{b}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Isolate the derivative}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{y}{}\left({x}\right){+}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{Group terms with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{on the lhs of the ODE and the rest on the rhs of the ODE}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){=}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{The ODE is linear; multiply by an integrating factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{\mu }{}\left(x\right)\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right)\right){=}{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{Assume the lhs of the ODE is the total derivative}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left(\mathrm{\mu }{}\left(x\right){}y{}\left(x\right)\right)\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right)\right){=}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\mu }}{}\left({x}\right)\right){}{y}{}\left({x}\right){+}{\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)\\ \text{•}& {}& \text{Isolate}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{\mu }{}\left(x\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\mu }}{}\left({x}\right){=}{-}{\mathrm{\mu }}{}\left({x}\right)\\ \text{•}& {}& \text{Solve to find the integrating factor}\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){=}{{ⅇ}}^{{-}{x}}\\ \text{•}& {}& \text{Integrate both sides with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {\int }\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({\mathrm{\mu }}{}\left({x}\right){}{y}{}\left({x}\right)\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Evaluate the integral on the lhs}\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}{y}{}\left({x}\right){=}{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {y}{}\left({x}\right){=}\frac{{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}}{{\mathrm{\mu }}{}\left({x}\right)}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{\mu }{}\left(x\right)={ⅇ}^{-x}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{\int }{{ⅇ}}^{{-}{x}}{}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}}{{{ⅇ}}^{{-}{x}}}\\ \text{•}& {}& \text{Evaluate the integrals on the rhs}\\ {}& {}& {y}{}\left({x}\right){=}\frac{\frac{{{x}}^{{2}}}{{2}}{+}{\mathrm{C1}}}{{{ⅇ}}^{{-}{x}}}\\ \text{•}& {}& \text{Simplify}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({{x}}^{{2}}{+}{2}{}{\mathrm{C1}}\right)}{{2}}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(a\right)=b\\ {}& {}& {b}{=}\frac{{{ⅇ}}^{{a}}{}\left({{a}}^{{2}}{+}{2}{}{\mathrm{C1}}\right)}{{2}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{-}\frac{{{ⅇ}}^{{a}}{}{{a}}^{{2}}{-}{2}{}{b}}{{2}{}{{ⅇ}}^{{a}}}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=-\frac{{ⅇ}^{a}{}{a}^{2}-2{}b}{2{}{ⅇ}^{a}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({-}{{a}}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{a}}{}{b}{+}{{x}}^{{2}}\right)}{{2}}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({-}{{a}}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{a}}{}{b}{+}{{x}}^{{2}}\right)}{{2}}\end{array}$ (6)
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First Order IVPs - Maple Help. ODE Steps for First Order IVPs. Overview. • This help page gives a few examples of using the command ODESteps to solve first order initial value problems.. • See Student[ODEs][ODESteps] for a general description of the command ODESteps and its calling sequence.. Examples. > $\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right):$. > $\mathrm{ivp1}≔\left\{{t}^{2}\left(z\left(t\right)+1\right)+{z\left(t\right)}^{2}\left(t-1\right)\mathrm{diff}\left(z\left(t\right),t\right)=0,z\left(3\right)=1\right\}$. ${\mathrm{ivp1}}{≔}\left\{{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}{,}{z}{}\left({3}\right){=}{1}\right\}$ (1). > $\mathrm{ODESteps}\left(\mathrm{ivp1}\right)$. $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}{,}{z}{}\left({3}\right){=}{1}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\\ \text{•}& {}& \text{Separate variables}\\ {}& {}& \frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}{=}{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\\ \text{•}& {}& \text{Integrate both sides with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}t\\ {}& {}& {\int }\frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{=}{\int }{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{\mathrm{C1}}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}z{}\left(3\right)=1\\ {}& {}& {-}\frac{{1}}{{2}}{+}{\mathrm{ln}}{}\left({2}\right){=}{-}\frac{{15}}{{2}}{-}{\mathrm{ln}}{}\left({2}\right){+}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=7+2{}\mathrm{ln}{}\left(2\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\end{array}$ (2). > $\mathrm{ivp2}≔\left\{2xy\left(x\right)-9{x}^{2}+\left(2y\left(x\right)+{x}^{2}+1\right)\mathrm{diff}\left(y\left(x\right),x\right)=0,y\left(0\right)=1\right\}$. ${\mathrm{ivp2}}{≔}\left\{{2}{}{x}{}{y}{}\left({x}\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}$ (3). > $\mathrm{ODESteps}\left(\mathrm{ivp2}\right)$. $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{{2}{}{x}{}{y}{}\left({x}\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{▫}& {}& \text{Check if ODE is exact}\\ {}& \text{◦}& \text{ODE is exact if the lhs is the total derivative of a}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{C}^{2}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{function}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}{}\left({x}\right)\right){=}{0}\\ {}& \text{◦}& \text{Compute derivative of lhs}\\ {}& {}& \frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right){+}\left(\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}\\ {}& \text{◦}& \text{Evaluate derivatives}\\ {}& {}& {2}{}{x}{=}{2}{}{x}\\ {}& \text{◦}& \text{Condition met, ODE is exact}\\ \text{•}& {}& \text{Exact ODE implies solution will be of this form}\\ {}& {}& \left[{F}{}\left({x}{,}{y}\right){=}{\mathrm{C1}}{,}{M}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right){,}{N}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\right]\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{by integrating}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}M{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{\int }\left({-}{9}{}{{x}}^{{2}}{+}{2}{}{x}{}{y}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Take derivative of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y\\ {}& {}& {N}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\\ \text{•}& {}& \text{Compute derivative}\\ {}& {}& {{x}}^{{2}}{+}{2}{}{y}{+}{1}{=}{{x}}^{{2}}{+}\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Isolate for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆy}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{_F1}{}\left(y\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({y}\right){=}{2}{}{y}{+}{1}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(y\right)\\ {}& {}& {\mathrm{_F1}}{}\left({y}\right){=}{{y}}^{{2}}{+}{y}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into equation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{{y}}^{{2}}{+}{y}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into the solution of the ODE}\\ {}& {}& {-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{{y}}^{{2}}{+}{y}{=}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& \left\{{y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}{,}{y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right\}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(0\right)=1\\ {}& {}& {1}{=}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Solution does not satisfy initial condition}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(0\right)=1\\ {}& {}& {1}{=}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{2}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& {y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{9}}}{{2}}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{9}}}{{2}}\end{array}$ (4). > $\mathrm{ivp3}≔\left\{\mathrm{diff}\left(y\left(x\right),x\right)-y\left(x\right)-x\mathrm{exp}\left(x\right)=0,y\left(a\right)=b\right\}$. ${\mathrm{ivp3}}{≔}\left\{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}{y}{}\left({a}\right){=}{b}\right\}$ (5).
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> $\mathrm{ODESteps}\left(\mathrm{ivp3}\right)$. $\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}{y}{}\left({a}\right){=}{b}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Isolate the derivative}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{y}{}\left({x}\right){+}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{Group terms with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{on the lhs of the ODE and the rest on the rhs of the ODE}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){=}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{The ODE is linear; multiply by an integrating factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{\mu }{}\left(x\right)\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right)\right){=}{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{Assume the lhs of the ODE is the total derivative}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left(\mathrm{\mu }{}\left(x\right){}y{}\left(x\right)\right)\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right)\right){=}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\mu }}{}\left({x}\right)\right){}{y}{}\left({x}\right){+}{\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)\\ \text{•}& {}& \text{Isolate}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{\mu }{}\left(x\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\mu }}{}\left({x}\right){=}{-}{\mathrm{\mu }}{}\left({x}\right)\\ \text{•}& {}& \text{Solve to find the integrating factor}\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){=}{{ⅇ}}^{{-}{x}}\\ \text{•}& {}& \text{Integrate both sides with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {\int }\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({\mathrm{\mu }}{}\left({x}\right){}{y}{}\left({x}\right)\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Evaluate the integral on the lhs}\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}{y}{}\left({x}\right){=}{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {y}{}\left({x}\right){=}\frac{{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}}{{\mathrm{\mu }}{}\left({x}\right)}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{\mu }{}\left(x\right)={ⅇ}^{-x}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{\int }{{ⅇ}}^{{-}{x}}{}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}}{{{ⅇ}}^{{-}{x}}}\\ \text{•}& {}& \text{Evaluate the integrals on the rhs}\\ {}& {}& {y}{}\left({x}\right){=}\frac{\frac{{{x}}^{{2}}}{{2}}{+}{\mathrm{C1}}}{{{ⅇ}}^{{-}{x}}}\\ \text{•}& {}& \text{Simplify}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({{x}}^{{2}}{+}{2}{}{\mathrm{C1}}\right)}{{2}}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(a\right)=b\\ {}& {}& {b}{=}\frac{{{ⅇ}}^{{a}}{}\left({{a}}^{{2}}{+}{2}{}{\mathrm{C1}}\right)}{{2}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{-}\frac{{{ⅇ}}^{{a}}{}{{a}}^{{2}}{-}{2}{}{b}}{{2}{}{{ⅇ}}^{{a}}}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=-\frac{{ⅇ}^{a}{}{a}^{2}-2{}b}{2{}{ⅇ}^{a}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({-}{{a}}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{a}}{}{b}{+}{{x}}^{{2}}\right)}{{2}}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({-}{{a}}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{a}}{}{b}{+}{{x}}^{{2}}\right)}{{2}}\end{array}$ (6).
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