\(\displaystyle \left ( f + g \right ) (x) = f (x) + g(x) = \sqrt{4-x} + \sqrt{x+3}\)

Substitute \(\displaystyle 4\) and \(\displaystyle -3\) to determine the values of the respective expressions:

\(\displaystyle \left ( f + g \right ) (4) = \sqrt{4-4} + \sqrt{4+3}= \sqrt{0} + \sqrt{7} = \sqrt{7}\)

\(\displaystyle \left ( f + g \right ) (-3) = \sqrt{4-(-3)} + \sqrt{-3+3}= \sqrt{7} + \sqrt{0} = \sqrt{7}\)

The expressions are equal.

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

so

\(\displaystyle N\Leftrightarrow 5 = \frac{N + 5}{ N \cdot 5 +1} = \frac{N + 5}{5 N +1}\)

This expression is undefined if and only if the denominator is equal to 0, so

\(\displaystyle 5N + 1 = 0\)

\(\displaystyle 5N + 1 -1 = 0 -1\)

\(\displaystyle 5N = -1\)

\(\displaystyle 5N \div 5 = -1 \div 5\)

\(\displaystyle N = - \frac{1}{5}\)

Let \(\displaystyle x_{1} = 2, y_{1} = -1.5, x_{2} =6, y_{2} = 3.1\)

We calculate the slope as follows:

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{3.1-(-1.5)}{6-2} = \frac{4.6}{4} = 1.15\)

Apply the point-slope formula setting 

\(\displaystyle x_{1} = 2, y_{1} = -1.5, m = 1.15\):

\(\displaystyle y - y_{1} = m (x - x_{1})\)

\(\displaystyle y - (-1.5)= 1.15 (x - 2)\)

\(\displaystyle y+1.5= 1.15 (x - 2)\)

\(\displaystyle y+1.5= 1.15 x - 2.3\)

Set \(\displaystyle y \, = 0\) to find the \(\displaystyle x\)-coordinate of the \(\displaystyle x\)-intercept:

\(\displaystyle 0+1.5= 1.15 x - 2.3\)

\(\displaystyle 1.5= 1.15 x - 2.3\)

\(\displaystyle 1.5+ 2.3= 1.15 x - 2.3 + 2.3\)

\(\displaystyle 3.8= 1.15 x\)

\(\displaystyle 3.8 \div 1.15 = 1.15 x \div 1.15\)

\(\displaystyle x \approx 3.3\)

The \(\displaystyle x\)-intercept is (approximately at) \(\displaystyle (3.3,0)\)

Let \(\displaystyle x_{1} = 3, y_{1} = -1.7, x_{2} = 5, y_{2} = 3.1\)

We calculate the slope as follows:

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{3.1-(-1.7)}{5-3} = \frac{4.8}{2} = 2.4\)

Apply the point-slope formula setting 

\(\displaystyle x_{1} = 5, y_{1} = 3.1, m= 2.4\)

\(\displaystyle y - y_{1} = m (x - x_{1})\)

\(\displaystyle y - 3.1= 2.4 (x - 5)\)

Set \(\displaystyle x = 0\) to find the \(\displaystyle y\)-coordinate of the \(\displaystyle y\)-intercept:

\(\displaystyle y - 3.1= 2.4 (0- 5)\)

\(\displaystyle y - 3.1= 2.4 (- 5)\)

\(\displaystyle y - 3.1= -12\)

\(\displaystyle y - 3.1+ 3.1 = -12 + 3.1\)

\(\displaystyle y = -8.9\)

The \(\displaystyle y\)-intercept is  \(\displaystyle (0,-8.9)\).

Let \(\displaystyle x_{1} = 3, y_{1} = -1.8 , x_{2} = 5, y_{2} = -5.2\)

We calculate the slope as follows:

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{-5.2-(-1.8)}{5-3} = \frac{-3.4}{2} = -1.7 = -\frac{17}{10}\)

Substitute each pair of expressions:

\(\displaystyle a \ast b = \frac{a + 2b }{ a + 3b}\)

\(\displaystyle t \ast 4t = \frac{t + 2 \cdot 4t }{ t + 3 \cdot 4t } = \frac{t + 8t }{ t + 12t } = \frac{9t }{ 13t } = \frac{9 }{ 13 }\)

\(\displaystyle 4t \ast t = \frac{4t + 2 \cdot t }{4t + 3 \cdot t } = \frac{4t + 2t }{ 4t + 3t } = \frac{6t }{ 7t } = \frac{6 }{ 7 }\)

We can compare these fractions by writing them with a common denominator:

\(\displaystyle t \ast 4t = \frac{9 }{ 13 } = \frac{9 \times 7 }{ 13 \times 7 } = \frac{63}{91}\)

\(\displaystyle 4t \ast t =\frac{6 }{ 7 }=\frac{6 \times 13 }{ 7 \times 13 } = \frac{78}{91}\)

\(\displaystyle 4t \ast t > t \ast 4t\) regardless of the value of \(\displaystyle t\),  making (B) greater.

\(\displaystyle \left (f+g \right ) (x) = f(x) + g(x) = \frac{1}{x+6}+ \frac{1}{x+8}\)

\(\displaystyle \left (f+g \right ) (1) = f(1) + g(1) = \frac{1}{1+6}+ \frac{1}{1+8} = \frac{1}{7}+ \frac{1}{9}\)

\(\displaystyle \left (f+g \right ) (-1) = f(-1) + g(-1) = \frac{1}{-1+6}+ \frac{1}{-1+8} = \frac{1}{5}+ \frac{1}{7}\)

Since \(\displaystyle \frac{1}{9} < \frac{1}{5}\),

\(\displaystyle \frac{1}{7}+ \frac{1}{9} < \frac{1}{5} + \frac{1}{7}\)

and 

\(\displaystyle \left (f+g \right ) (1) < \left (f+g \right ) (-1)\)

making (B) greater.

\(\displaystyle \left ( f \circ g\right ) (t) = f (g(t)) = f (3t-8) = 4(3t-8) - 7 = 12t -32 - 7 = 12t -39\)

\(\displaystyle \left ( g \circ f\right ) (t) = g (f(t)) =g (4t-7) = 3 (4t-7) -8 = 12 t - 21 - 8 = 12t - 29\)

Regardless of the value of \(\displaystyle t\),

\(\displaystyle \left ( g \circ f\right ) (t) = \left ( f \circ g\right ) (t) + 10\), 

which means that 

\(\displaystyle \left ( g \circ f\right ) (t) > \left ( f \circ g\right ) (t)\).

That is, (B) is greater.

To get the slope of each line, use the slope formula 

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1 }}\)

For Line A, \(\displaystyle x_{1} =-5, y_{1} =0, x_{2} = 0, y_{2} = 9\). Substitute in the slope formula.

The slope is 

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{9-0}{0-(-5)} = \frac{9}{5}\)

For Line B, \(\displaystyle x_{1} =-7, y_{1} =0, x_{2} = 0, y_{2} = 9\). Substitute in the slope formula.

The slope is 

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{9-0}{0-(-7)} = \frac{9}{7}\)

Since 

\(\displaystyle \frac{9}{5} = \frac{63}{35}> \frac{45}{35} = \frac{9}{7}\),

Line A has the greater slope, and (A) is greater.

To get the slope of Line A and Line C, use the slope formula 

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1 }}\)

For Line A, \(\displaystyle x_{1} =5, y_{1} =0, x_{2} = 0, y_{2} = 7\). Substitute in the slope formula.

The slope is 

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1 }}= \frac{7-0}{0-5} = \frac{7}{-5} = - \frac{7}{5}\)

For Line C, \(\displaystyle x_{1} =14, y_{1} =0, x_{2} = 0, y_{2} = -10\). Substitute in the slope formula.

The slope is 

\(\displaystyle m = \frac{y_{2}-y_{1}}{x_{2}-x_{1 }}= \frac{-10-0}{0-14} = \frac{-10}{-14} = \frac{5}{7}\)

Since Line B is perpendicular to Line C, its slope is the opposite of the reciprocal of that of Line C; this is \(\displaystyle - \frac{5}{7}\), which is equal to the slope of Line A.

The two quantities are equal.