Substitute the assigned values into the expression.

$\displaystyle 3b^2-5c^3 = 3(2)^2-5(-1)^3$

Simplify by order of operations.

$\displaystyle = 3(4)-5(-1) = 12+5 = 17$

The answer is:  $\displaystyle 17$

Substitute the values of $\displaystyle a=1, b=2,c=3, d=5$ into the expression.

$\displaystyle \frac{a}{c}-\frac{b}{d} = \frac{1}{3}-\frac{2}{5}$

In order to subtract these fractions, we will need a least common denominator.

Multiply the denominators together for the LCD.  Convert the two fractions.

$\displaystyle \frac{1}{3}-\frac{2}{5} = \frac{1(5)}{3(5)}-\frac{2(3)}{5(3)} = \frac{5}{15}-\frac{6}{15}$

Subtract the numerators now that the denominators are common.

The answer is:  $\displaystyle -\frac{1}{15}$

Given the expression $\displaystyle a^3-b^2-c$ and the assigned values, substitute the values into the expression.

$\displaystyle (-1)^3-(-3)^2-(-5)$

Simplify this expression by order of operations.

$\displaystyle -1-(9)+5 = -10+5 = -5$

The answer is:   $\displaystyle -5$

Substitute the known values into the expression.

$\displaystyle a^2-b^3 = (-6)^2-(-3)^3$

Simplify the expression.

$\displaystyle 36-(-27) = 36+27 = 63$

The answer is:  $\displaystyle 63$

Substitute the assigned values into the expression.

$\displaystyle (-1)^{(-2)}\times (-2)^{(-1)}$

Simplify the negative exponents by rewriting both terms as fractions.

$\displaystyle x^{-a} =\frac{1}{x^a}$

$\displaystyle (-1)^{(-2)}\times (-2)^{(-1)}= \frac{1}{(-1)^2} \times \frac{1}{(-2)^{1}}$

Simplify the fractions.

$\displaystyle 1\times \frac{1}{-2} = -\frac{1}{2}$

The answer is:  $\displaystyle -\frac{1}{2}$

Substitute the known value of $\displaystyle x=2, y=5$ into the equation.

$\displaystyle 5= 3[2](A-7)$

Simplify the equation.

$\displaystyle 5=6(A-7)$

Solve the right side by distribution.

$\displaystyle 5=6A-42$

Add 42 on both sides.

$\displaystyle 5+42=6A-42+42$

$\displaystyle 47=6A$

Divide by six on both sides.

$\displaystyle \frac{47}{6}=\frac{6A}{6}$

The answer is:  $\displaystyle \frac{47}{6}$

The term $\displaystyle f(2) = 4$ means that $\displaystyle f(x)=4$ when $\displaystyle x=2$.

Substitute the terms in the function to solve for $\displaystyle k$.

$\displaystyle 4 = 5(2)-k$

Solve for $\displaystyle k$.

$\displaystyle 4 = 10-k$

Subtract 10 on both sides.

$\displaystyle 4 -10= 10-k-10$

$\displaystyle -6 =-k$

Divide by negative one to eliminate the negative signs.

$\displaystyle \frac{-6 }{-1}=\frac{-k}{-1}$

The answer is:  $\displaystyle 6$

The vertical asymptote(s) of the graph of a rational function such as $\displaystyle f(x)$ can be found by evaluating the zeroes of the denominator after the rational expression is reduced. The expression is in simplest form, so set the denominator equal to 0 and solve for $\displaystyle x$:

$\displaystyle x^{2} - 16 = 0$

Add 16 to both sides:

$\displaystyle x^{2} - 16 + 16 = 0+ 16$

$\displaystyle x^{2} =16$

Take the positive and negative square roots:

$\displaystyle x = -\sqrt{16 } = -4$

or 

$\displaystyle x = \sqrt{16 } = 4$

The graph of $\displaystyle f(x)$ has two vertical asymptotes, the graphs of the lines $\displaystyle x = -4$ and $\displaystyle x= 4$.

$\displaystyle g(f(x))$ is a composite function which means that the inside function is plugged into the outside function. So in this case, $\displaystyle f(x)$ is plugged into $\displaystyle g(x)$. In other words, replace the $\displaystyle f(x)$ expression each time there is an $\displaystyle x$ in the $\displaystyle g(x)$ expression. 

In this case $\displaystyle x-4$ would be plugged into each $\displaystyle x$ in the $\displaystyle g(x)$ expression. See below:

$\displaystyle 5(x-4)+6(x-4)^2+3$

This is then simplified to:

$\displaystyle 5x-20+6x^2-48x+96+3$

And then further simplified to:

$\displaystyle 6x^2-43x+79$