\(\displaystyle f (x) = x^{2}\), so, substituting 2 for \(\displaystyle x\), we see that

\(\displaystyle f (2) = 2^{2} =4\).

By definition, 

\(\displaystyle (fg)(2) = f(2) \cdot g(2)\), so

\(\displaystyle (fg)(2) = 64\)

Substituting and solving for \(\displaystyle g(2)\):

\(\displaystyle f(2) \cdot g(2) = 64\)

\(\displaystyle 4\cdot g(2) = 64\)

\(\displaystyle 4\cdot g(2) \div 4 = 64 \div 4\)

\(\displaystyle g(2) = 16\)

Examine all four alternatives by, again, substituting 2 for \(\displaystyle x\) in each and finding for which one \(\displaystyle g(2) \ne 16\):

\(\displaystyle g(x) = x^{2} + 12\)

\(\displaystyle g(2) = 2^{2} + 12 = 4 + 12 = 16\)

\(\displaystyle g(x) = \frac{32}{x}\)

\(\displaystyle g(2) = \frac{32}{2} = 16\)

\(\displaystyle g(x) = 8x\)

\(\displaystyle g(2) = 8 \cdot 2 = 16\)

\(\displaystyle g(x) = x ^{3}\)

\(\displaystyle g(2) = 2 ^{3} = 8\)

Therefore, of the four choices, only \(\displaystyle g(x) = x ^{3}\) is not a valid definition, since it does not match the conditions.

By definition, 

\(\displaystyle \frac{f}{g} (5) = \frac{f(5)}{g(5)}\).

Also, by substitution:

\(\displaystyle f(x) = x^{5}\)

\(\displaystyle f(5) = 5^{5} = 3,125\)

Therefore, the question is equivalent to asking for which definition of \(\displaystyle g(x)\) it holds that

\(\displaystyle \frac{3,125}{g(5)} = 50\)

Each definition in the given choices can be evaluated for \(\displaystyle x = 5\) by substitution, with each value of \(\displaystyle g(5)\) tested in turn:

\(\displaystyle g(x) =\frac{1}{ 2x^{3}}\)

\(\displaystyle g(5) =\frac{1}{ 2 \cdot 5^{3}} = \frac{1}{ 2 \cdot 125} = \frac{1}{250} = 0.004\)

\(\displaystyle \frac{f}{g} (5) = \frac{3,125}{0.008} = 3,125 \div 0.008 = 390,625\)

\(\displaystyle g(x) =\frac{2}{ x^{3}}\)

\(\displaystyle g(x) =\frac{2}{ 5^{3}}= \frac{2}{125} = 0.016\)

\(\displaystyle \frac{f}{g} (5) = \frac{3,125}{0.016} = 3,125 \div 0.016 =195, 312 .5\)

\(\displaystyle g(x) = 2x^{3}\)

\(\displaystyle g(5) = 2 \cdot 5^{3} = 2 \cdot 125 = 250\)

\(\displaystyle \frac{f}{g} (5) = \frac{3,125}{250} = 12.5\)

\(\displaystyle g(x) =\frac{ x^{3}}{2}\)

\(\displaystyle g(x) =\frac{ 5^{3}}{2} = \frac{ 125}{2} = 62.5\)

\(\displaystyle \frac{f}{g} (5) = \frac{3,125}{62.5} = 50\)

This makes \(\displaystyle g(x) =\frac{ x^{3}}{2}\) the correct choice.

\(\displaystyle f (x) = 7x - 23\), so, substituting 8 for \(\displaystyle x\), we see that

\(\displaystyle f (8) = 7 \cdot 8 - 23= 56 - 23 = 33\)

By definition, 

\(\displaystyle (f-g) (8) = f(8) - g (8)\), so

\(\displaystyle (f-g) (8) = 34\)

\(\displaystyle f(8) - g (8) = 34\)

\(\displaystyle 33 - g (8) = 34\)

\(\displaystyle 33 - g (8) - 33 = 34 - 33\)

\(\displaystyle -g(8)= 1\)

\(\displaystyle g(8) = -1\)

Examine all four alternatives by, again, substituting 8 for \(\displaystyle x\), and find the one for which\(\displaystyle g(8) = -1\).

\(\displaystyle g(x) = x - 6\)

\(\displaystyle g(x) = 8 - 6 = 2\)

\(\displaystyle g(x) = x - 7\)

\(\displaystyle g(x) = 8 - 7 = 1\)

\(\displaystyle g(x) = x- 8\)

\(\displaystyle g(x) = 8- 8 = 0\)

\(\displaystyle g(x) = x - 9\)

\(\displaystyle g(x) = 8 - 9 = -1\)

Of the four choices, \(\displaystyle g(x) = x - 9\) is the definition such that \(\displaystyle g(8) = -1\).

Solve the following equation for t, when d is 6.

\(\displaystyle 5t^2=6d+34\)

Let's begin by plugging 6 in for d and then, using algebra, we will find t.

\(\displaystyle 5t^2=6(6)+34\)

\(\displaystyle 5t^2=36+34\)

\(\displaystyle 5t^2=70\)

Divide by 5:

\(\displaystyle t^2=\frac{70}{5}=14\)

Square root both sides to finish up

\(\displaystyle t=\sqrt{14}\)

Almost there, but because we are square rooting, we need plus or minus the square root of 14

\(\displaystyle t=\pm\sqrt{14}\)

The reason for this is that positive or negative square root of 14 will get us positive 14 when we square it. Therefore, we technically have two answers.

To solve for x, we want x to stand alone.  So, we get

\(\displaystyle 5x-8=12\)

\(\displaystyle 5x-8+8=12+8\)

\(\displaystyle 5x-0=20\)

\(\displaystyle 5x=20\)

\(\displaystyle \frac{5x}{5} = \frac{20}{5}\)

\(\displaystyle x=4\)

Solve the following equation when y is equal to 12.

\(\displaystyle \sqrt{y^2}-4=c^3\)

To begin, let's realize that 

\(\displaystyle \sqrt{y^2}=y\)

So, let's update our original equation.

\(\displaystyle y-4=c^3\)

Now, let's plug in 12 for y

\(\displaystyle 12-4=c^3\)

\(\displaystyle 8=c^3\)

So, what number cubed equals 8?

\(\displaystyle c=\sqrt[3]{8}=2\)

So our answer is 2

Solve the following equation for q.

\(\displaystyle 3q^3-147=12\)

First, let's divide everything by three. This will work because 147 and 12 are both divisible by 3

\(\displaystyle (3q^3-147=12)\div3\rightarrow q^3-49=4\)

Now, add 49 to both sides to get the variable by itself

\(\displaystyle q^3=53\)

\(\displaystyle q=\sqrt[3]{53}\)

Now, this doesn't come out to a neat decimal, so we will just leave it as is.

To answer the question, we will need to solve for y.  We get

\(\displaystyle 7y-13=29\)

\(\displaystyle 7y-13+13=29+13\)

\(\displaystyle 7y-0=42\)

\(\displaystyle 7y=42\)

\(\displaystyle \frac{7y}{7} = \frac{42}{7}\)

\(\displaystyle y=6\)

Solve the following equation for j.

\(\displaystyle 14j-12=58\)

We can solve this with basic algebra.

First, add 12 to both sides:

\(\displaystyle 14j-12(+12)=58(+12)\)

\(\displaystyle 14j=70\)

Next, divide both sides by 14

\(\displaystyle 14j\div14=70 \div 14\)

\(\displaystyle j=\frac{70}{14}=5\)

And we end up with...

\(\displaystyle j=5\)

To find the answer, we will solve for x. So, we get

\(\displaystyle \frac{x}{5} +9 = 12\)

\(\displaystyle \frac{x}{5} +9 -9 = 12-9\)

\(\displaystyle \frac{x}{5} + 0 = 3\)

\(\displaystyle \frac{x}{5} = 3\)

\(\displaystyle \frac{x}{5} \cdot 5 = 3 \cdot 5\)

\(\displaystyle x=15\)