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

\(\displaystyle L\) varies directly as \(\displaystyle A\) and inversely as \(\displaystyle V\), meaning that for some constant of variation \(\displaystyle K\), 

\(\displaystyle \frac{LV}{A}= K\).

Setting \(\displaystyle A = 6s^{2}\) and \(\displaystyle V= s^{3}\), the formula becomes

\(\displaystyle \frac{Ls^{3}}{6s^{2}}= K\)

\(\displaystyle \frac{Ls }{6 }= K\)

\(\displaystyle Ls= 6K\)

Setting \(\displaystyle K' = 6K\) as the new constant of variation, the variation equation becomes

\(\displaystyle Ls= K'\),

so \(\displaystyle L\) varies inversely as \(\displaystyle s\).

\(\displaystyle x\) varies directly as both \(\displaystyle y\) and the cube of \(\displaystyle z\), meaning that for some constant of variation \(\displaystyle K\),

\(\displaystyle \frac{x}{yz^{3}} = K\).

Take the reciprocal of both sides, and the equation becomes

\(\displaystyle \frac{yz^{3}}{x} =\frac{1}{ K}\)

Take the cube root of both sides, and the equation becomes

\(\displaystyle \frac{z\sqrt[3]{y}}{\sqrt[3]{x}} = \sqrt[3]{\frac{1}{ K}}\)

\(\displaystyle K' = \sqrt[3]{\frac{1}{ K}}\) takes the role of the new constant of variation here, and we now have

\(\displaystyle \frac{z\sqrt[3]{y}}{\sqrt[3]{x}} =K'\)

so \(\displaystyle z\) varies inversely as the cube root of \(\displaystyle y\).

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. 

\(\displaystyle x\) varies inversely as both the square of \(\displaystyle y\) and the square root of \(\displaystyle z\), meaning that for some constant of variation \(\displaystyle K\),

\(\displaystyle xy^{2}\sqrt{z}= K\).

Square both sides, and the expression becomes

\(\displaystyle x^{2}y^{4}z= K^{2}\)

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

\(\displaystyle zx^{2}y^{4} = K'\),

meaning that \(\displaystyle z\) varies inversely as the fourth power of \(\displaystyle y\).

\(\displaystyle Q\) varies directly as the square of \(\displaystyle A\) and inversely as \(\displaystyle V\); therefore, for some constant of variation \(\displaystyle K\),

\(\displaystyle \frac{QV }{A^{2}} = K\)

Setting \(\displaystyle A= 4 \pi r^{2}\) and \(\displaystyle V = \frac{4}{3} \pi r^{3}\), the formula becomes 

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

\(\displaystyle \frac{Q \left ( \frac{4}{3}\pi r^{3} \right ) }{(16\pi^{2} r^{4})} = K\)

\(\displaystyle \frac{Q }{12\pi r } = K\)

\(\displaystyle \frac{Q }{r } = 12 \pi K\)

Setting \(\displaystyle K' = 12 \pi K\) as the new constant of variation, we have a new variation equation

\(\displaystyle \frac{Q }{r } =K'\),

meaning that \(\displaystyle Q\) varies directly as \(\displaystyle r\).

To find the area of a rectangle, multiply the length times the width.  Therefore, you must multiply \(\displaystyle 3n\) times \(\displaystyle 7n-3y +2\).  To do that, you must multiply the monomial times each part of the trinomial, like so:

\(\displaystyle 7n(3n)-3y(3n) +2(3n)\)

\(\displaystyle 21n^2-9ny+6n\)

Use the distributive property:

\(\displaystyle 6x^3y^5(3x^2z^4+x^3y+7y)\)

\(\displaystyle (6x^3y^5)(3x^2z^4)+(6x^3y^5)(x^3y)+(6x^3y^5)(7y)\)

Simplify: don't forget to use the rules of multiplying exponents (add them)

\(\displaystyle 18x^5y^5z^4 + 6x^6y^6+42x^3y^6\)

Find the product:

\(\displaystyle 4a^2(ab^2+a^2b+ac^3)\)

Use the distributive property:

\(\displaystyle (4a^2\cdot ab^2)+(4a^2\cdot a^2b)+(4a^2\cdot ac^3)\)

When multiplying variables with exponents, add the exponents:

\(\displaystyle 4a^3b^2+4a^4b+4a^3c^3\)