\(\displaystyle a \blacktriangledown b = a - b^{2}\), so

\(\displaystyle x \blacktriangledown 5 = x - 5^{2} = x - 25\)

Now replace:

\(\displaystyle x \blacktriangledown 5 = 96\)

\(\displaystyle x - 25 = 96\)

\(\displaystyle x - 25+ 25 = 96+ 25\)

\(\displaystyle x = 121\)

Substitute \(\displaystyle A = 100\):

\(\displaystyle \blacktriangle A = A^{2} - \sqrt{A}\)

\(\displaystyle \blacktriangle 100 = 100^{2} - \sqrt{100} = 10,000 - 10 = 9,990\)

The total cost of beans is the price per pound times the number of pounds. The total cost of the mixture is equal to that of the Strawberry Dream beans and that of the Vanilla Madness beans. Each can be found by multiplying the cost per pound by the number of pounds:

Strawberry Dream: \(\displaystyle \$ 9 \cdot S\)

Vanilla Madness: \(\displaystyle \$14 \cdot (80-S)\), since the total number of pounds is eighty.

Total mixture: \(\displaystyle \$12 \cdot 80 = \$960\)

Add to get the equation:

\(\displaystyle 9S + 14 (80-S)= 960\)

Loretta earned $3,000 per month for twelve months, and $3,500 per month for the following twelve months. This is a total of 

\(\displaystyle \$ 3,000 \times 12 + \$3,500 \times 12 = \$36,000 + \$42,000 =\$ 78,000\)

earned over two years. 7.5% of this is 

\(\displaystyle \$ 78,000 \times 0.075 = \$5,850\),

the amount of Loretta's bonus.

Use the slope formula, substituting \(\displaystyle x_1 = 5, y_1 = a , x_2 = 9, y_2 = a + 3\):

\(\displaystyle m = \frac{y_2-y_1}{x_2-x_1} = \frac{\left ( a+ 3 \right )- a}{9-5} = \frac{3}{4}\)

\(\displaystyle \left ( f\circ g\right ) (-4) = f(g(-4))\)

First, evaluate \(\displaystyle g (-4)\):

\(\displaystyle g(x) = x^{2}\)

\(\displaystyle g(-4) = (-4)^{2}= 16\)

Therefore, 

\(\displaystyle \left ( f\circ g\right ) (-4) = f(g(-4 )) = f(16)\)

which can be evaluated using the definition of \(\displaystyle f\) for nonnegative values of \(\displaystyle x\):

\(\displaystyle f (x) = x-2\)

\(\displaystyle f (16) = 16-2 = 14\)

For \(\displaystyle x ^{2} + 9x + C = 0\) to have exactly one solution, the discriminant of the expression \(\displaystyle x ^{2} + 9x + C\) must be 0.

The discriminant of \(\displaystyle Ax ^{2} + Bx + C\) is \(\displaystyle B ^{2} - 4AC\), so we substitute \(\displaystyle A = 1, B = 9\), set the discriminant equal to 0, and solve for \(\displaystyle C\):

\(\displaystyle B ^{2} - 4AC = 0\)

\(\displaystyle 9 ^{2} - 4 \cdot 1 \cdot C = 0\)

\(\displaystyle 81 - 4C = 0\)

\(\displaystyle 81 - 4C+4C = 0+4C\)

\(\displaystyle 4C = 81\)

\(\displaystyle 4C \div 4= 81 \div 4\)

\(\displaystyle C = 20 \frac{1}{4}\)

For \(\displaystyle x ^{2} + Bx + 36 = 0\) to have exactly one solution, the discriminant of the expression \(\displaystyle x ^{2} + Bx + 36\) must be 0.

The discriminant of \(\displaystyle Ax ^{2} + Bx + C\) is \(\displaystyle B ^{2} - 4AC\), so we substitute \(\displaystyle A = 1, C= 36\), set the discriminant equal to 0, and solve for \(\displaystyle B\):

\(\displaystyle B ^{2} - 4AC = 0\)

\(\displaystyle B ^{2} - 4 \cdot 1 \cdot 36 = 0\)

\(\displaystyle B ^{2} - 144 = 0\)

\(\displaystyle B ^{2} =144\)

\(\displaystyle B = -12 \textrm{ or } B = 12\)

The \(\displaystyle y\)-intercept will be point \(\displaystyle (0,b)\) for some real \(\displaystyle b\).

We can substitute \(\displaystyle m = N - 10, x_1 = 3, y_1 = 5, x_2 = 0,y_2 =b\) in the slope formula and solve for \(\displaystyle b\):

\(\displaystyle \frac{y_2-y_1}{x_2-x_1} = m\)

\(\displaystyle \frac{b-5}{0-3} = N - 10\)

\(\displaystyle \frac{b-5}{-3} = N - 10\)

\(\displaystyle \frac{b-5}{-3} \cdot (-3) =\left ( N - 10 \right ) \cdot (-3)\)

\(\displaystyle b-5 =-3 N +30\)

\(\displaystyle b-5+5 =-3 N +30 +5\)

\(\displaystyle b =-3 N +35\)

The \(\displaystyle y\)-intercept is \(\displaystyle (0,-3 N +35)\)

\(\displaystyle x^2-4=5\)

\(\displaystyle x^2-4+4=5+4\)

\(\displaystyle x^2=9\)

\(\displaystyle \sqrt{x^2}=\sqrt9\)

\(\displaystyle x=\pm3\)