How to use the inverse variation formula - PSAT Math

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The square of varies inversely with the cube of . If when , then what is the value of when ?

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When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x2 and the cube of y as y3. Now, we can write the equation for inverse variation.

x2y3 = k

We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.

82(83) = k

k = 82(83)

Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that abac = ab+c.

k = 82(83) = 82+3 = 85.

Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 85 in for k.

x2y3 = 85

(1)2y3 = 85

y3 = 85

In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 23.

y3 = (23)5

We can now apply the property of exponents that states that (ab)c = abc.

y3 = 23•5 = 215

In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.

(y3)(1/3) = (215)(1/3)

Once again, let's apply the property (ab)c = abc.

y(3 • 1/3) = 2(15 • 1/3)

y = 25 = 32

The answer is 32.

varies inversely as both the square of and the square root of . Assuming that does not depend on any other variable, which statement is true of concerning its relationship to ?

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varies inversely as both the square of and the square root of , meaning that for some constant of variation ,

.

Square both sides, and the expression becomes

$K'= $K^{2}$$ takes the role of the new constant of variation here, and we now have

,

meaning that varies inversely as the fourth power of .

varies directly as the square of and inversely as ; $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ . Assuming that does not depend on any other variables, which of the following gives the variation relationship of to ?

Tap to see back →

varies directly as the square of and inversely as ; therefore, for some constant of variation ,

$$\frac{QV $}{A^{2}$$} = K$

Setting $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ , the formula becomes

$$\frac{Q \left ( \frac{4}{3}$\pi $r^{3}$ \right) }{(4\pi $r^{2}$$)^{2}$} = K$

$$\frac{Q \left( \frac{4}{3}$\pi $r^{3}$ \right ) $}{(16\pi^{2}$ $r^{4}$)} = K$

$$\frac{Q }{12\pi r }$ = K$

$$\frac{Q }{r }$ = 12 \pi K$

Setting $K' = 12 \pi K$ as the new constant of variation, we have a new variation equation

$$\frac{Q }{r }$ =K'$ ,

meaning that varies directly as .

The square of varies inversely with the cube of . If when , then what is the value of when ?

Tap to see back →

When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x2 and the cube of y as y3. Now, we can write the equation for inverse variation.

x2y3 = k

We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.

82(83) = k

k = 82(83)

Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that abac = ab+c.

k = 82(83) = 82+3 = 85.

Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 85 in for k.

x2y3 = 85

(1)2y3 = 85

y3 = 85

In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 23.

y3 = (23)5

We can now apply the property of exponents that states that (ab)c = abc.

y3 = 23•5 = 215

In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.

(y3)(1/3) = (215)(1/3)

Once again, let's apply the property (ab)c = abc.

y(3 • 1/3) = 2(15 • 1/3)

y = 25 = 32

The answer is 32.

varies inversely as both the square of and the square root of . Assuming that does not depend on any other variable, which statement is true of concerning its relationship to ?

Tap to see back →

varies inversely as both the square of and the square root of , meaning that for some constant of variation ,

.

Square both sides, and the expression becomes

$K'= $K^{2}$$ takes the role of the new constant of variation here, and we now have

,

meaning that varies inversely as the fourth power of .

varies directly as the square of and inversely as ; $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ . Assuming that does not depend on any other variables, which of the following gives the variation relationship of to ?

Tap to see back →

varies directly as the square of and inversely as ; therefore, for some constant of variation ,

$$\frac{QV $}{A^{2}$$} = K$

Setting $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ , the formula becomes

$$\frac{Q \left ( \frac{4}{3}$\pi $r^{3}$ \right) }{(4\pi $r^{2}$$)^{2}$} = K$

$$\frac{Q \left( \frac{4}{3}$\pi $r^{3}$ \right ) $}{(16\pi^{2}$ $r^{4}$)} = K$

$$\frac{Q }{12\pi r }$ = K$

$$\frac{Q }{r }$ = 12 \pi K$

Setting $K' = 12 \pi K$ as the new constant of variation, we have a new variation equation

$$\frac{Q }{r }$ =K'$ ,

meaning that varies directly as .

The square of varies inversely with the cube of . If when , then what is the value of when ?

Tap to see back →

When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x2 and the cube of y as y3. Now, we can write the equation for inverse variation.

x2y3 = k

We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.

82(83) = k

k = 82(83)

Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that abac = ab+c.

k = 82(83) = 82+3 = 85.

Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 85 in for k.

x2y3 = 85

(1)2y3 = 85

y3 = 85

In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 23.

y3 = (23)5

We can now apply the property of exponents that states that (ab)c = abc.

y3 = 23•5 = 215

In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.

(y3)(1/3) = (215)(1/3)

Once again, let's apply the property (ab)c = abc.

y(3 • 1/3) = 2(15 • 1/3)

y = 25 = 32

The answer is 32.

varies inversely as both the square of and the square root of . Assuming that does not depend on any other variable, which statement is true of concerning its relationship to ?

Tap to see back →

varies inversely as both the square of and the square root of , meaning that for some constant of variation ,

.

Square both sides, and the expression becomes

$K'= $K^{2}$$ takes the role of the new constant of variation here, and we now have

,

meaning that varies inversely as the fourth power of .

varies directly as the square of and inversely as ; $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ . Assuming that does not depend on any other variables, which of the following gives the variation relationship of to ?

Tap to see back →

varies directly as the square of and inversely as ; therefore, for some constant of variation ,

$$\frac{QV $}{A^{2}$$} = K$

Setting $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ , the formula becomes

$$\frac{Q \left ( \frac{4}{3}$\pi $r^{3}$ \right) }{(4\pi $r^{2}$$)^{2}$} = K$

$$\frac{Q \left( \frac{4}{3}$\pi $r^{3}$ \right ) $}{(16\pi^{2}$ $r^{4}$)} = K$

$$\frac{Q }{12\pi r }$ = K$

$$\frac{Q }{r }$ = 12 \pi K$

Setting $K' = 12 \pi K$ as the new constant of variation, we have a new variation equation

$$\frac{Q }{r }$ =K'$ ,

meaning that varies directly as .

The square of varies inversely with the cube of . If when , then what is the value of when ?

Tap to see back →

When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x2 and the cube of y as y3. Now, we can write the equation for inverse variation.

x2y3 = k

We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.

82(83) = k

k = 82(83)

Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that abac = ab+c.

k = 82(83) = 82+3 = 85.

Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 85 in for k.

x2y3 = 85

(1)2y3 = 85

y3 = 85

In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 23.

y3 = (23)5

We can now apply the property of exponents that states that (ab)c = abc.

y3 = 23•5 = 215

In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.

(y3)(1/3) = (215)(1/3)

Once again, let's apply the property (ab)c = abc.

y(3 • 1/3) = 2(15 • 1/3)

y = 25 = 32

The answer is 32.

varies inversely as both the square of and the square root of . Assuming that does not depend on any other variable, which statement is true of concerning its relationship to ?

Tap to see back →

varies inversely as both the square of and the square root of , meaning that for some constant of variation ,

.

Square both sides, and the expression becomes

$K'= $K^{2}$$ takes the role of the new constant of variation here, and we now have

,

meaning that varies inversely as the fourth power of .

varies directly as the square of and inversely as ; $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ . Assuming that does not depend on any other variables, which of the following gives the variation relationship of to ?

Tap to see back →

varies directly as the square of and inversely as ; therefore, for some constant of variation ,

$$\frac{QV $}{A^{2}$$} = K$

Setting $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ , the formula becomes

$$\frac{Q \left ( \frac{4}{3}$\pi $r^{3}$ \right) }{(4\pi $r^{2}$$)^{2}$} = K$

$$\frac{Q \left( \frac{4}{3}$\pi $r^{3}$ \right ) $}{(16\pi^{2}$ $r^{4}$)} = K$

$$\frac{Q }{12\pi r }$ = K$

$$\frac{Q }{r }$ = 12 \pi K$

Setting $K' = 12 \pi K$ as the new constant of variation, we have a new variation equation

$$\frac{Q }{r }$ =K'$ ,

meaning that varies directly as .

The square of varies inversely with the cube of . If when , then what is the value of when ?

Tap to see back →

When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x2 and the cube of y as y3. Now, we can write the equation for inverse variation.

x2y3 = k

We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.

82(83) = k

k = 82(83)

Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that abac = ab+c.

k = 82(83) = 82+3 = 85.

Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 85 in for k.

x2y3 = 85

(1)2y3 = 85

y3 = 85

In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 23.

y3 = (23)5

We can now apply the property of exponents that states that (ab)c = abc.

y3 = 23•5 = 215

In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.

(y3)(1/3) = (215)(1/3)

Once again, let's apply the property (ab)c = abc.

y(3 • 1/3) = 2(15 • 1/3)

y = 25 = 32

The answer is 32.

varies inversely as both the square of and the square root of . Assuming that does not depend on any other variable, which statement is true of concerning its relationship to ?

Tap to see back →

varies inversely as both the square of and the square root of , meaning that for some constant of variation ,

.

Square both sides, and the expression becomes

$K'= $K^{2}$$ takes the role of the new constant of variation here, and we now have

,

meaning that varies inversely as the fourth power of .

varies directly as the square of and inversely as ; $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ . Assuming that does not depend on any other variables, which of the following gives the variation relationship of to ?

Tap to see back →

varies directly as the square of and inversely as ; therefore, for some constant of variation ,

$$\frac{QV $}{A^{2}$$} = K$

Setting $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ , the formula becomes

$$\frac{Q \left ( \frac{4}{3}$\pi $r^{3}$ \right) }{(4\pi $r^{2}$$)^{2}$} = K$

$$\frac{Q \left( \frac{4}{3}$\pi $r^{3}$ \right ) $}{(16\pi^{2}$ $r^{4}$)} = K$

$$\frac{Q }{12\pi r }$ = K$

$$\frac{Q }{r }$ = 12 \pi K$

Setting $K' = 12 \pi K$ as the new constant of variation, we have a new variation equation

$$\frac{Q }{r }$ =K'$ ,

meaning that varies directly as .

The square of varies inversely with the cube of . If when , then what is the value of when ?

Tap to see back →

When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x2 and the cube of y as y3. Now, we can write the equation for inverse variation.

x2y3 = k

We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.

82(83) = k

k = 82(83)

Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that abac = ab+c.

k = 82(83) = 82+3 = 85.

Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 85 in for k.

x2y3 = 85

(1)2y3 = 85

y3 = 85

In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 23.

y3 = (23)5

We can now apply the property of exponents that states that (ab)c = abc.

y3 = 23•5 = 215

In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.

(y3)(1/3) = (215)(1/3)

Once again, let's apply the property (ab)c = abc.

y(3 • 1/3) = 2(15 • 1/3)

y = 25 = 32

The answer is 32.

varies inversely as both the square of and the square root of . Assuming that does not depend on any other variable, which statement is true of concerning its relationship to ?

Tap to see back →

varies inversely as both the square of and the square root of , meaning that for some constant of variation ,

.

Square both sides, and the expression becomes

$K'= $K^{2}$$ takes the role of the new constant of variation here, and we now have

,

meaning that varies inversely as the fourth power of .

varies directly as the square of and inversely as ; $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ . Assuming that does not depend on any other variables, which of the following gives the variation relationship of to ?

Tap to see back →

varies directly as the square of and inversely as ; therefore, for some constant of variation ,

$$\frac{QV $}{A^{2}$$} = K$

Setting $A= 4 \pi $r^{2}$$ and $V = $\frac{4}{3}$ \pi $r^{3}$$ , the formula becomes

$$\frac{Q \left ( \frac{4}{3}$\pi $r^{3}$ \right) }{(4\pi $r^{2}$$)^{2}$} = K$

$$\frac{Q \left( \frac{4}{3}$\pi $r^{3}$ \right ) $}{(16\pi^{2}$ $r^{4}$)} = K$

$$\frac{Q }{12\pi r }$ = K$

$$\frac{Q }{r }$ = 12 \pi K$

Setting $K' = 12 \pi K$ as the new constant of variation, we have a new variation equation

$$\frac{Q }{r }$ =K'$ ,

meaning that varies directly as .

The square of varies inversely with the cube of . If when , then what is the value of when ?

Tap to see back →

When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x2 and the cube of y as y3. Now, we can write the equation for inverse variation.

x2y3 = k

We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.

82(83) = k

k = 82(83)

Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that abac = ab+c.

k = 82(83) = 82+3 = 85.

Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 85 in for k.

x2y3 = 85

(1)2y3 = 85

y3 = 85

In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 23.

y3 = (23)5

We can now apply the property of exponents that states that (ab)c = abc.

y3 = 23•5 = 215

In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.

(y3)(1/3) = (215)(1/3)

Once again, let's apply the property (ab)c = abc.

y(3 • 1/3) = 2(15 • 1/3)

y = 25 = 32

The answer is 32.

varies inversely as both the square of and the square root of . Assuming that does not depend on any other variable, which statement is true of concerning its relationship to ?

Tap to see back →

varies inversely as both the square of and the square root of , meaning that for some constant of variation ,

.

Square both sides, and the expression becomes

$K'= $K^{2}$$ takes the role of the new constant of variation here, and we now have

,

meaning that varies inversely as the fourth power of .