How to use the inverse variation formula - PSAT Math

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Question

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

Answer

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.

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Question

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 ?

Answer

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

$xy^{2}\sqrt{z}= K$ .

Square both sides, and the expression becomes

$x^{2}y^{4}z= K^{2}$

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

$zx^{2}y^{4} = K'$ ,

meaning that varies inversely as the fourth power of .

Compare your answer with the correct one above

Question

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 ?

Answer

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 .

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

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 ?

Answer

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

$xy^{2}\sqrt{z}= K$ .

Square both sides, and the expression becomes

$x^{2}y^{4}z= K^{2}$

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

$zx^{2}y^{4} = K'$ ,

meaning that varies inversely as the fourth power of .

Compare your answer with the correct one above

Question

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 ?

Answer

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 .

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

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 ?

Answer

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

$xy^{2}\sqrt{z}= K$ .

Square both sides, and the expression becomes

$x^{2}y^{4}z= K^{2}$

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

$zx^{2}y^{4} = K'$ ,

meaning that varies inversely as the fourth power of .

Compare your answer with the correct one above

Question

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 ?

Answer

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 .

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

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 ?

Answer

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

$xy^{2}\sqrt{z}= K$ .

Square both sides, and the expression becomes

$x^{2}y^{4}z= K^{2}$

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

$zx^{2}y^{4} = K'$ ,

meaning that varies inversely as the fourth power of .

Compare your answer with the correct one above

Question

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 ?

Answer

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 .

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

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 ?

Answer

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

$xy^{2}\sqrt{z}= K$ .

Square both sides, and the expression becomes

$x^{2}y^{4}z= K^{2}$

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

$zx^{2}y^{4} = K'$ ,

meaning that varies inversely as the fourth power of .

Compare your answer with the correct one above

Question

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 ?

Answer

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 .

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

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 ?

Answer

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

$xy^{2}\sqrt{z}= K$ .

Square both sides, and the expression becomes

$x^{2}y^{4}z= K^{2}$

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

$zx^{2}y^{4} = K'$ ,

meaning that varies inversely as the fourth power of .

Compare your answer with the correct one above

Question

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 ?

Answer

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 .

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

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 ?

Answer

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

$xy^{2}\sqrt{z}= K$ .

Square both sides, and the expression becomes

$x^{2}y^{4}z= K^{2}$

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

$zx^{2}y^{4} = K'$ ,

meaning that varies inversely as the fourth power of .

Compare your answer with the correct one above

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