How to use the direct variation formula - SAT Math

Card 0 of 7

Question

and are the radius and volume, respectively, of a given sphere.

$r ^{6} = Q$

$W = \sqrt[3]{V}$ .

Which of the following is a true statement?

Answer

The volume of a sphere can be calculated from its radius as follows:

$V = \frac{4}{3} \pi r^{3}$

Therefore, squaring both sides, we get

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

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

$V ^{2}=\left ( \frac{4}{3} \pi \right )^{2}r^{6}$

$W = \sqrt[3]{V}$

$W ^{6 }= \left (\sqrt[3]{V} \right ) ^{6}$

$W ^{6 }= \left ( V ^{\frac{1}{3}}\right ) ^{6} = V ^{\frac{1}{3} \cdot 6} = V^{2}$

Substituting:

$V ^{2}=\left ( \frac{4}{3} \pi \right )^{2}r^{6}$

$W ^{6 }=\left ( \frac{4}{3} \pi \right )^{2}Q$

$\frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}W ^{6 }= \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}} \cdot \left ( \frac{4}{3} \pi \right )^{2}Q$

$Q = \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}W ^{6 }$

If we let the constant of variation be $K = \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}$ , we see that

$Q = K W ^{6 }$ ,

and varies directly as $W^{6}$ , the sixth power of .

Compare your answer with the correct one above

Question

Phillip can paint square feet of wall per minute. What area of the wall can he paint in 2.5 hours?

Answer

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.

$(60\frac{min}{hr})(2.5hr)=150min$

If he completes _ square feet per minute, then we can multiply _ by the total minutes to find the final answer.

$(y\frac{ft^2}{min})(150min)=150y\ ft^2$

Compare your answer with the correct one above

Question

The value of varies directly with the square of __and the cube of . If when and , then what is the value of _ when and ?

Answer

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.

Compare your answer with the correct one above

Question

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?

Answer

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:

Compare your answer with the correct one above

Question

and are the diameter and circumference, respectively, of the same circle.

$d = 12L^{2}$

$C = t^{2}$

Which of the following is a true statement? (Assume all quantities are positive)

Answer

If and are the diameter and circumference, respectively, of the same circle, then

$C = \pi d$ .

By substitution,

$t^{2} = \pi \cdot 12 L^{2}$

$t^{2} = 12 \pi L^{2}$

Taking the square root of both sides:

$\sqrt{t^{2} }= \sqrt{12 \pi L^{2}}$

$t = \sqrt{12 \pi} \cdot \sqrt{L^{2}}$

$t = \sqrt{12 \pi} \cdot L$

Taking $K= \sqrt{12 \pi}$ as the constant of variation, we get

,

meaning that varies directly as .

Compare your answer with the correct one above

Question

is the radius of the base of a cone; is its height; is its volume.

$n = 4r = h^{2}$ ; $m = V ^{2}$ .

Which of the following is a true statement?

Answer

The volume of a cone can be calculated from the radius of its base , and the height , using the formula

$V = \frac{1}{3} \pi r^{2}h$

, so $r = \frac{n}{4}$ .

$n = h^{2}$ , so $h = \sqrt{n}$ .

$V = \frac{1}{3} \pi r^{2}h$ , so by substitution,

$V = \frac{1}{3} \pi\left ( \frac{n}{4} \right ) ^{2} \cdot \sqrt{n}$

$V = \frac{1}{3} \pi\left ( \frac{n^{2}}{16} \right ) \cdot \sqrt{n}$

$V = \frac{1}{48} \pi \cdot n^{2}} \sqrt{n}$

Square both sides:

$V ^{2 }=\left ( \frac{1}{48} \pi \cdot n^{2}} \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot \left ( n^{2}} \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot \left ( n^{2}} \right )^{2} \cdot \left ( \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot n^{4}} \cdot n$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot n^{5}}$

If we take $K = \left ( \frac{1}{48} \pi \right )^{2}$ as the constant of variation, then

$m=K n^{5}}$ ,

and varies directly as the fifth power of .

Compare your answer with the correct one above

Question

The temperature at the surface of the ocean is $25^{\circ}C$ . At meters below the surface, the ocean temperature is $5^{\circ}C$ . By how much does the temperature decrease for every meters below the ocean's surface?

Answer

This may seem confusing, but is pretty straightforward.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{5-25}{-2500-0}$

$\text{Slope}=\frac{-20}{-2500}=\frac{125}{1}$

Thus, for every 125 meters below the surface, the temperature decreases by one degree.

To find how much it decreases with every 100 meters, we need to do the following:

$\frac{100}{125}=\frac{4}{5}$

Thus, the temperature decreases by $\frac{4}{5}^{\circ}$ every 100 meters.

Compare your answer with the correct one above

Tap the card to reveal the answer