In this case, we need to solve for the volume of the water tank, so we set the full volume of the water tank as $\displaystyle x$. According to the question, $\displaystyle \frac{4}{7}$-full  can be replaced as $\displaystyle \frac{4}{7}x$. $\displaystyle \frac{13}{14}$-full  would be $\displaystyle \frac{13}{14}x$. Therefore, we can write out the equation as: 

$\displaystyle \frac{4}{7}x+5=\frac{13}{14}x$.

Then we can solve the equation and find the answer is 14 gallons.

Let's look at $\displaystyle f,g,and\ h$ and see if any of them are functions.

1. $\displaystyle f$ = {(2, 3), (1, 4), (2, 1), (3, 2), (4, 4)}: This cannot be a function of $\displaystyle X$ because two of the ordered pairs, (2, 3) and (2, 1) have the same number (2) as the first coordinate.

2. $\displaystyle g$ = {(3, 1), (4, 2), (1, 1)}: This cannot be a function of $\displaystyle X$ because it contains no ordered pair with first coordinate 2.  Because the set $\displaystyle X$ = {1, 2, 3, 4}, we need an ordered pair of the form (2,  ) .

3. $\displaystyle h$ = {(2, 1), (3, 4), (1, 4), (2, 1), (4, 4)}: This is a function.  Even though two of the ordered pairs have the same number (2) as the first coordinate, $\displaystyle h$ is still a function of $\displaystyle X$ because (2, 1) is simply repeated twice, so the two ordered pairs with first coordinate 2 are equal.

This is a definition question.  The only choice that does not equal the others is $\displaystyle f(y)=x^{2}$.  This describes a function that assigns $\displaystyle x^{2}$ to some number $\displaystyle y$, instead of assigning $\displaystyle x^{2}$ to its own square root, $\displaystyle x$.

We are given f(x) and h, so the only missing piece is f(x + h).

$\displaystyle f(x+h)=(x+h)^{2}=x^{2}+2xh+h^{2}$

Then $\displaystyle \frac{f(x+h)-f(x)}{h}= \frac{x^{2}+2xh+h^{2}-x^{2}}{h} = \frac{2xh+h^{2}}{h}=2x+h$

We look at the range of the function on each of the three parts of the domain. The overall range is the union of these three intervals.

On $\displaystyle (-\infty , -2)$, $\displaystyle f (x)$ takes the values:

$\displaystyle x < -2$

$\displaystyle 2x - 1 < 2 (-2) - 1$

$\displaystyle 2x - 1 < -5$

$\displaystyle f\left (x \right )< -5$

or $\displaystyle (-\infty , -5)$

On $\displaystyle [-2, 2]$, $\displaystyle f (x)$ takes the values:

$\displaystyle -2 \leq x\leq 2$

$\displaystyle -3(-2) \geq -3x\geq -3 (2)$

$\displaystyle 6 \geq -3x\geq -6$

$\displaystyle -6 \leq f(x)\leq 6$,

or $\displaystyle [-6, 6]$

On $\displaystyle (2, \infty)$, $\displaystyle f (x)$ takes only value 5.

The range of $\displaystyle f (x)$ is therefore $\displaystyle (-\infty , -5) \cup [-6, 6] \cup \left \{ 5\right \}$ , which simplifies to $\displaystyle (-\infty , 6]$.

Each term of the Fibonacci sequence is formed by adding the previous two terms. Therefore, do the same to form this sequence:

$\displaystyle 8 + 12= 20$

$\displaystyle 12+20=32$

The easiest way to find the inverse of $\displaystyle f (x)$ is to replace $\displaystyle f (x)$ in the definition with $\displaystyle y$ , switch $\displaystyle y$ with $\displaystyle x$, and solve for $\displaystyle y$ in the new equation.

$\displaystyle y = 5x - 7$

$\displaystyle x = 5y - 7$

$\displaystyle x + 7 = 5y - 7 + 7$

$\displaystyle x + 7 = 5y$

$\displaystyle \frac{1}{5} \cdot\left ( x + 7 \right ) = \frac{1}{5} \cdot 5y$

$\displaystyle \frac{1}{5} x + \frac{7}{5} = y$

$\displaystyle f^{-1}(x)= \frac{1}{5}x + \frac{7}{5}$

The easiest way to find the inverse of $\displaystyle f (x)$ is to replace $\displaystyle f (x)$ in the definition with $\displaystyle y$ , switch $\displaystyle y$ with $\displaystyle x$, and solve for $\displaystyle y$ in the new equation.

$\displaystyle y = x^{3} - 8$

$\displaystyle x= y^{3} - 8$

$\displaystyle x +8= y^{3} - 8 +8$

$\displaystyle x +8= y^{3}$

$\displaystyle \sqrt[3]{x +8}= \sqrt[3]{y^{3}}$

$\displaystyle y = \sqrt[3]{x +8}$

$\displaystyle (f \circ g)(x) = f(g(x))$

$\displaystyle = f(x^{2}+4)$

$\displaystyle = (x^{2}+4)^{2}-4$

$\displaystyle =x^{4} + 8x^{2}+16-4$

$\displaystyle =x^{4} + 8x^{2}+12$

Solve for $\displaystyle A$ in this equation:

$\displaystyle g (A) = 5A - 2 = 11$

$\displaystyle 5A - 2 + 2 = 11+ 2$

$\displaystyle 5A = 13$

$\displaystyle 5A \div 5 = 13\div 5$

$\displaystyle A = 2.6$