Every minute Phillip completes another \(\displaystyle y\) 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.

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

If he completes \(\displaystyle y\) square feet per minute, then we can multiply \(\displaystyle y\) by the total minutes to find the final answer.

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

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 = kx²z³, 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) = 8k

24 = 8k

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 = kx²z³

y = 3(3)²(1)³ = 3(9)(1) = 27

y = 27

The answer is 27. 

We know that the initial population is 3, and that every week the population will triple.

The equation to model this growth will be \(\displaystyle i(r)^n\), where \(\displaystyle i\) is the initial size, \(\displaystyle r\) is the rate of growth, and \(\displaystyle n\) is the time.

In this case, the equation will be \(\displaystyle 3(3)^4\).

\(\displaystyle 3(3)^4=3(81)=243\)

Alternatively, you can evaluate for each consecutive week.

Week 1: \(\displaystyle 3(3)=9\)

Week 2: \(\displaystyle 3(9)=27\)

Week 3: \(\displaystyle 3(27)=81\)

Week 4: \(\displaystyle 3(81)=243\)

If \(\displaystyle d\) and \(\displaystyle C\) are the diameter and circumference, respectively, of the same circle, then

\(\displaystyle C = \pi d\).

By substitution,

\(\displaystyle t^{2} = \pi \cdot 12 L^{2}\)

\(\displaystyle t^{2} = 12 \pi L^{2}\)

Taking the square root of both sides:

\(\displaystyle \sqrt{t^{2} }= \sqrt{12 \pi L^{2}}\)

\(\displaystyle t = \sqrt{12 \pi} \cdot \sqrt{L^{2}}\)

\(\displaystyle t = \sqrt{12 \pi} \cdot L\)

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

\(\displaystyle t =K L\),

meaning that \(\displaystyle t\) varies directly as \(\displaystyle L\).

The volume of a cone can be calculated from the radius of its base \(\displaystyle r\), and the height \(\displaystyle h\), using the formula

\(\displaystyle V = \frac{1}{3} \pi r^{2}h\)

\(\displaystyle n = 4r\), so \(\displaystyle r = \frac{n}{4}\).

\(\displaystyle n = h^{2}\), so \(\displaystyle h = \sqrt{n}\).

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

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

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

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

Square both sides:

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

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

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

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

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

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

\(\displaystyle m=K n^{5}}\),

and \(\displaystyle m\) varies directly as the fifth power of \(\displaystyle n\).

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

\(\displaystyle V = \frac{4}{3} \pi r^{3}\)

Therefore, squaring both sides, we get

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

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

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

\(\displaystyle W = \sqrt[3]{V}\)

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

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

Substituting:

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

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

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

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

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

\(\displaystyle Q = K W ^{6 }\),

and \(\displaystyle Q\) varies directly as \(\displaystyle W^{6}\), the sixth power of \(\displaystyle W \:\).

This may seem confusing, but is pretty straightforward.

\(\displaystyle x_1=0\)

\(\displaystyle x_2=-2500\)

\(\displaystyle y_1=25\)

\(\displaystyle y_2=5\)

\(\displaystyle \text{Slope}=\frac{y_2-y_1}{x_2-x_1}\)

\(\displaystyle \text{Slope}=\frac{5-25}{-2500-0}\)

\(\displaystyle \text{Slope}=\frac{-20}{-2500}=\frac{125}{1}\)

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:

\(\displaystyle \frac{100}{125}=\frac{4}{5}\)

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