How to use the direct variation formula - PSAT Math
Card 0 of 35
Phillip can paint 
square feet of wall per minute. What area of the wall can he paint in 2.5 hours?
Phillip can paint
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Every minute Phillip completes another _
square feet of painting. To solve for the total area that he completes, we need to find the number of minutes that he works.
There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.
If he completes _
square feet per minute, then we can multiply _
by the total minutes to find the final answer.
Every minute Phillip completes another _
There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.
If he completes _
The value of 
varies directly with the square of _
_and the cube of 
. If 
when 
and 
, then what is the value of _
when 
and 
?
The value of
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Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:
y = kx, where k is a constant.
Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:
y = _kx_2_z_3, where k is a constant.
The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.
24 = k(1)2(2)3 = k(1)(8) = 8_k_
24 = 8_k_
Divide both sides by 8.
3 = k
k = 3
Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.
y = _kx_2_z_3
y = 3(3)2(1)3 = 3(9)(1) = 27
y = 27
The answer is 27.
Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:
y = kx, where k is a constant.
Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:
y = _kx_2_z_3, where k is a constant.
The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.
24 = k(1)2(2)3 = k(1)(8) = 8_k_
24 = 8_k_
Divide both sides by 8.
3 = k
k = 3
Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.
y = _kx_2_z_3
y = 3(3)2(1)3 = 3(9)(1) = 27
y = 27
The answer is 27.
In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?
In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?
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We know that the initial population is 3, and that every week the population will triple.
The equation to model this growth will be 
, where 
is the initial size, 
is the rate of growth, and 
is the time.
In this case, the equation will be 
.
Alternatively, you can evaluate for each consecutive week.
Week 1: 
Week 2: 
Week 3: 
Week 4: 
We know that the initial population is 3, and that every week the population will triple.
The equation to model this growth will be
In this case, the equation will be
Alternatively, you can evaluate for each consecutive week.
Week 1:
Week 2:
Week 3:
Week 4:
varies directly as both 
and the cube of 
. Which statement is true of 
concerning its relationship to 
?
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varies directly as both 
and the cube of 
, meaning that for some constant of variation 
,
.
Take the reciprocal of both sides, and the equation becomes
Take the cube root of both sides, and the equation becomes
takes the role of the new constant of variation here, and we now have
so 
varies inversely as the cube root of 
.
Take the reciprocal of both sides, and the equation becomes
Take the cube root of both sides, and the equation becomes
so
varies directly as 
and inversely as 
; 
and 
. Assuming that no other variables affect 
, which statement is true of 
concerning its relationship to 
?
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varies directly as 
and inversely as 
, meaning that for some constant of variation 
,
.
Setting 
and 
, the formula becomes
Setting 
as the new constant of variation, the variation equation becomes
,
so 
varies inversely as 
.
Setting
Setting
so
Phillip can paint square feet of wall per minute. What area of the wall can he paint in 2.5 hours?
Phillip can paint
Tap to see back →
Every minute Phillip completes another _ square feet of painting. To solve for the total area that he completes, we need to find the number of minutes that he works.
There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.
If he completes _ square feet per minute, then we can multiply _ by the total minutes to find the final answer.
Every minute Phillip completes another _
There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.
If he completes _
The value of varies directly with the square of __and the cube of . If when and , then what is the value of _ when and ?
The value of
Tap to see back →
Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:
y = kx, where k is a constant.
Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:
y = _kx_2_z_3, where k is a constant.
The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.
24 = k(1)2(2)3 = k(1)(8) = 8_k_
24 = 8_k_
Divide both sides by 8.
3 = k
k = 3
Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.
y = _kx_2_z_3
y = 3(3)2(1)3 = 3(9)(1) = 27
y = 27
The answer is 27.
Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:
y = kx, where k is a constant.
Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:
y = _kx_2_z_3, where k is a constant.
The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.
24 = k(1)2(2)3 = k(1)(8) = 8_k_
24 = 8_k_
Divide both sides by 8.
3 = k
k = 3
Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.
y = _kx_2_z_3
y = 3(3)2(1)3 = 3(9)(1) = 27
y = 27
The answer is 27.
In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?
In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?
Tap to see back →
We know that the initial population is 3, and that every week the population will triple.
The equation to model this growth will be , where is the initial size, is the rate of growth, and is the time.
In this case, the equation will be .
Alternatively, you can evaluate for each consecutive week.
Week 1:
Week 2:
Week 3:
Week 4:
We know that the initial population is 3, and that every week the population will triple.
The equation to model this growth will be
In this case, the equation will be
Alternatively, you can evaluate for each consecutive week.
Week 1:
Week 2:
Week 3:
Week 4:
varies directly as both and the cube of . Which statement is true of concerning its relationship to ?
Tap to see back →
varies directly as both and the cube of , meaning that for some constant of variation ,
.
Take the reciprocal of both sides, and the equation becomes
Take the cube root of both sides, and the equation becomes
takes the role of the new constant of variation here, and we now have
so varies inversely as the cube root of .
Take the reciprocal of both sides, and the equation becomes
Take the cube root of both sides, and the equation becomes
so
varies directly as and inversely as ; and . Assuming that no other variables affect , which statement is true of concerning its relationship to ?
Tap to see back →
varies directly as and inversely as , meaning that for some constant of variation ,
.
Setting and , the formula becomes
Setting as the new constant of variation, the variation equation becomes
,
so varies inversely as .
Setting
Setting
so
Phillip can paint square feet of wall per minute. What area of the wall can he paint in 2.5 hours?
Phillip can paint
Tap to see back →
Every minute Phillip completes another _ square feet of painting. To solve for the total area that he completes, we need to find the number of minutes that he works.
There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.
If he completes _ square feet per minute, then we can multiply _ by the total minutes to find the final answer.
Every minute Phillip completes another _
There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.
If he completes _
The value of varies directly with the square of __and the cube of . If when and , then what is the value of _ when and ?
The value of
Tap to see back →
Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:
y = kx, where k is a constant.
Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:
y = _kx_2_z_3, where k is a constant.
The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.
24 = k(1)2(2)3 = k(1)(8) = 8_k_
24 = 8_k_
Divide both sides by 8.
3 = k
k = 3
Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.
y = _kx_2_z_3
y = 3(3)2(1)3 = 3(9)(1) = 27
y = 27
The answer is 27.
Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:
y = kx, where k is a constant.
Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:
y = _kx_2_z_3, where k is a constant.
The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.
24 = k(1)2(2)3 = k(1)(8) = 8_k_
24 = 8_k_
Divide both sides by 8.
3 = k
k = 3
Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.
y = _kx_2_z_3
y = 3(3)2(1)3 = 3(9)(1) = 27
y = 27
The answer is 27.
In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?
In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?
Tap to see back →
We know that the initial population is 3, and that every week the population will triple.
The equation to model this growth will be , where is the initial size, is the rate of growth, and is the time.
In this case, the equation will be .
Alternatively, you can evaluate for each consecutive week.
Week 1:
Week 2:
Week 3:
Week 4:
We know that the initial population is 3, and that every week the population will triple.
The equation to model this growth will be
In this case, the equation will be
Alternatively, you can evaluate for each consecutive week.
Week 1:
Week 2:
Week 3:
Week 4:
varies directly as both and the cube of . Which statement is true of concerning its relationship to ?
Tap to see back →
varies directly as both and the cube of , meaning that for some constant of variation ,
.
Take the reciprocal of both sides, and the equation becomes
Take the cube root of both sides, and the equation becomes
takes the role of the new constant of variation here, and we now have
so varies inversely as the cube root of .
Take the reciprocal of both sides, and the equation becomes
Take the cube root of both sides, and the equation becomes
so
varies directly as and inversely as ; and . Assuming that no other variables affect , which statement is true of concerning its relationship to ?
Tap to see back →
varies directly as and inversely as , meaning that for some constant of variation ,
.
Setting and , the formula becomes
Setting as the new constant of variation, the variation equation becomes
,
so varies inversely as .
Setting
Setting
so
Phillip can paint square feet of wall per minute. What area of the wall can he paint in 2.5 hours?
Phillip can paint
Tap to see back →
Every minute Phillip completes another _ square feet of painting. To solve for the total area that he completes, we need to find the number of minutes that he works.
There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.
If he completes _ square feet per minute, then we can multiply _ by the total minutes to find the final answer.
Every minute Phillip completes another _
There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.
If he completes _
The value of varies directly with the square of __and the cube of . If when and , then what is the value of _ when and ?
The value of
Tap to see back →
Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:
y = kx, where k is a constant.
Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:
y = _kx_2_z_3, where k is a constant.
The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.
24 = k(1)2(2)3 = k(1)(8) = 8_k_
24 = 8_k_
Divide both sides by 8.
3 = k
k = 3
Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.
y = _kx_2_z_3
y = 3(3)2(1)3 = 3(9)(1) = 27
y = 27
The answer is 27.
Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:
y = kx, where k is a constant.
Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:
y = _kx_2_z_3, where k is a constant.
The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.
24 = k(1)2(2)3 = k(1)(8) = 8_k_
24 = 8_k_
Divide both sides by 8.
3 = k
k = 3
Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.
y = _kx_2_z_3
y = 3(3)2(1)3 = 3(9)(1) = 27
y = 27
The answer is 27.
In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?
In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?
Tap to see back →
We know that the initial population is 3, and that every week the population will triple.
The equation to model this growth will be , where is the initial size, is the rate of growth, and is the time.
In this case, the equation will be .
Alternatively, you can evaluate for each consecutive week.
Week 1:
Week 2:
Week 3:
Week 4:
We know that the initial population is 3, and that every week the population will triple.
The equation to model this growth will be
In this case, the equation will be
Alternatively, you can evaluate for each consecutive week.
Week 1:
Week 2:
Week 3:
Week 4:
varies directly as both and the cube of . Which statement is true of concerning its relationship to ?
Tap to see back →
varies directly as both and the cube of , meaning that for some constant of variation ,
.
Take the reciprocal of both sides, and the equation becomes
Take the cube root of both sides, and the equation becomes
takes the role of the new constant of variation here, and we now have
so varies inversely as the cube root of .
Take the reciprocal of both sides, and the equation becomes
Take the cube root of both sides, and the equation becomes
so
varies directly as and inversely as ; and . Assuming that no other variables affect , which statement is true of concerning its relationship to ?
Tap to see back →
varies directly as and inversely as , meaning that for some constant of variation ,
.
Setting and , the formula becomes
Setting as the new constant of variation, the variation equation becomes
,
so varies inversely as .
Setting
Setting
so