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.

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 , 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:

 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 .

 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 .

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 a^ba^c = a^b+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 (a^b)^c = a^bc.

y³ = 2^3•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 (a^b)^c = a^bc.

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 , meaning that for some constant of variation ,

.

Square both sides, and the expression becomes

  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 ; therefore, for some constant of variation ,

Setting  and , the formula becomes 

Setting  as the new constant of variation, we have a new variation equation

,

meaning that  varies directly as .