If we let $\displaystyle x$ be the number, "the sum of a number and sixty" can be written as 

$\displaystyle x+60$

"One-third of the sum of a number and sixty" can be written as 

$\displaystyle \frac{1}{3} (x+60)$

Set this equal to ninety-three and solve:

$\displaystyle \frac{1}{3} (x+60) = 93$

$\displaystyle 3 \cdot \frac{1}{3} (x+60) =3 \cdot 93$

$\displaystyle x+60 = 279$

$\displaystyle x+60 - 60= 279 - 60$

$\displaystyle x = 219$

If we let $\displaystyle x$ be the number, "two-fifths of a number" can be written as

$\displaystyle \frac{2}{5} x$.

"Twelve added to two-fifths of a number" can be written as

$\displaystyle \frac{2}{5} x + 12$.

Then "Twelve added to two-fifths of a number is equal to sixty" can be written and solved for $\displaystyle x$ as follows:

$\displaystyle \frac{2}{5} x + 12 = 60$

$\displaystyle \frac{2}{5} x + 12 - 12 = 60 - 12$

$\displaystyle \frac{2}{5} x =48$

$\displaystyle \frac{5} {2}\cdot \frac{2}{5}x = \frac{5} {2}\cdot 48$

$\displaystyle x = 120$

If we let $\displaystyle x$ be the number, "two-fifths of a number" can be written as 

$\displaystyle \frac{2}{5}x$.

"Five less than two-fifths of a number can be written as 

$\displaystyle \frac{2}{5}x -5$

"Ninety-seven is five less than two-fifths of a number" can be written and solved as follows:

$\displaystyle 97 = \frac{2}{5}x -5$

$\displaystyle 97 + 5 = \frac{2}{5}x -5 + 5$

$\displaystyle 102 = \frac{2}{5}x$

$\displaystyle \frac{5}{2} \cdot 102 = \frac{5}{2} \cdot \frac{2}{5}x$

$\displaystyle 255 = x$

$\displaystyle f (x) = \sqrt{3x+8 }$ has as its domain the set of values of $\displaystyle x$ for which its radicand is nonnegative; that is,

$\displaystyle 3x+8 \geq 0$

$\displaystyle 3x+8-8 \geq 0 -8$

$\displaystyle 3x \geq -8$

$\displaystyle 3x\div 3 \geq -8 \div 3$

$\displaystyle x \geq -2 \frac{2}{3}$ or $\displaystyle \left [-2 \frac{2}{3}, \infty \right )$

Similarly, $\displaystyle g(x) = \sqrt{2x+7}$ has as its domain the set of values of $\displaystyle x$ for which its radicand is nonnegative; that is,

$\displaystyle 2x+7 \geq 0$

$\displaystyle 2x+7-7 \geq 0 -7$

$\displaystyle 2x \geq -7$

$\displaystyle 2x \div 2 \geq -7 \div 2$

$\displaystyle x \geq -3\frac{1}{2}$ or $\displaystyle \left [-3 \frac{1} {2}, \infty \right )$

The domain of the sum of two functions is the intersection of the domains of the two individual functions. This intersection is 

$\displaystyle \left [-2 \frac{2}{3}, \infty \right ) \cap \left [-3 \frac{1} {2}, \infty \right ) = \left [-2 \frac{2}{3}, \infty \right )$

$\displaystyle a \ast b = \frac{a b + 1}{a+ b + 1 }$ , so

$\displaystyle N \ast \left (-7 \right ) = \frac{N (-7) + 1}{N+ (-7) + 1 } = \frac{-7N + 1}{N-6 }$

This is is undefined if and only if the denominator is equal to zero, which happens when

$\displaystyle N - 6 = 0$, or $\displaystyle N = 6$

$\displaystyle a \Lambda b = \frac{a^{2} b - 16}{16a^{2} + b}$, so

$\displaystyle N \Lambda 4 = \frac{N^{2} \cdot 4 - 16}{16 \cdot N^{2} +4} = \frac{4N^{2} - 16}{16 N^{2} +4}$

This expression is undefined if and only if the denominator is equal to 0. However, for all values of $\displaystyle N$, 

$\displaystyle 16 N^{2} \geq 0$, so 

$\displaystyle 16 N^{2} +4 > 0$

It is impossible for $\displaystyle N \Lambda 4$ to be undefined, so none of the four values of $\displaystyle N$ given gives a correct response.

$\displaystyle \left ( g \circ f \right )(3) = g (f(3))$

$\displaystyle f (x)= 2x+ 5$, so

$\displaystyle f (3)= 2 \cdot 3 + 5 = 6 + 5 = 11$

$\displaystyle g (x) = 2x+3$, so

$\displaystyle g (f(3)) = g (11) = 2 \cdot 11 + 3 = 22 + 3 = 25$,

which is the correct response.

$\displaystyle \left ( f \circ g\right ) (x) = f(g(x)) = f(2x+3) = 2 (2x+3) +5 = 4x + 6 + 5 = 4x + 11$

Therefore, if 

$\displaystyle \left ( f \circ g\right ) (N) = 100$,

we solve for $\displaystyle N$ in the equation

$\displaystyle 4N + 11 = 100$

$\displaystyle 4N + 11- 11 = 100 - 11$

$\displaystyle 4N = 89$

$\displaystyle 4N \div 4 = 89 \div 4$

$\displaystyle N = 22 \frac{1}{4}$

Substitute for both in the expression:

$\displaystyle a\Leftrightarrow b = \frac{a + b}{ ab+1}$

$\displaystyle t\Leftrightarrow 4t = \frac{t + 4t}{ t \cdot 4t +1} = \frac{5t}{4t^{2}+1}$

$\displaystyle 2t\Leftrightarrow 2t = \frac{2t + 2t}{ 2t \cdot 2t +1} = \frac{4t}{4t^{2}+1}$

The quantities share a denominator that is positve, being one greater than for times a square of a positive. Therefore, we need only compare numerators. For any positive $\displaystyle t$, $\displaystyle 5t > 4t$. Therefore,

$\displaystyle \frac{5t}{4t^{2}+1} > \frac{4t}{4t^{2}+1}$

and 

$\displaystyle t\Leftrightarrow 4t > 2t\Leftrightarrow 2t$

making (A) greater.

The statement is equivalent to 

$\displaystyle y = \frac{4}{3}x + 20$

If $\displaystyle x$ is positive, then 

$\displaystyle \frac{4}{3} x > x$

and 

$\displaystyle y = \frac{4}{3} x + 20 > \frac{4}{3} x > x$

and $\displaystyle y > x$,

making (B) greater.