Use the identity

Since $x+y=12$, the first equation becomes

Now solve for $xy$:

So the correct answer is $\frac{72}{5}$.

Since $f(x)=\frac{1}{x+b}$, use any row of the table to find $b$. From $f(2)=\frac{1}{4}$, we have

so $2+b=4$, which gives $b=2$. This matches the other rows because $f(5)=\frac{1}{5+2}=\frac{1}{7}$ and $f(8)=\frac{1}{8+2}=\frac{1}{10}$. Then

A point in Quadrant II has $x<0$ and $y>0$. So for $y=\frac{x+4}{x-2}$, we need

with $x<0$.

A rational expression is positive when its numerator and denominator have the same sign. Since $x<0$, the denominator $x-2$ is definitely negative. Therefore the numerator must also be negative:

so

That guarantees $x<0$ as well, so any Quadrant II point on the graph must have $x<-4$.

For a rectangle, $\text{area}=\text{length}\times \text{width}$. Since the area is $48$ and the length is $x$, the width is

The perimeter of a rectangle is

So the correct expression is $2x+\frac{96}{x}$.

Set up the condition that the cost per student is at most $36$:

Subtract $12$ from both sides:

Because $x>0$, multiply both sides by $x$ without changing the inequality:

Now divide by $24$:

So the least possible whole number of students is $100$.