<h3>
Answer: Choice D) </h3>
Work Shown:
![(f+g)(x) = f(x)+g(x)\\\\(f+g)(x) = \frac{x-23}{x^2+9x-36}+\frac{1}{x+12}\\\\(f+g)(x) = \frac{x-23}{(x+12)(x-3)}+\frac{1(x-3)}{(x+12)(x-3)}\\\\(f+g)(x) = \frac{x-23+1(x-3)}{(x+12)(x-3)}\\\\(f+g)(x) = \frac{x-23+x-3}{x^2+9x-36}\\\\(f+g)(x) = \frac{2x-26}{x^2+9x-36}\\\\](https://tex.z-dn.net/?f=%28f%2Bg%29%28x%29%20%3D%20f%28x%29%2Bg%28x%29%5C%5C%5C%5C%28f%2Bg%29%28x%29%20%3D%20%5Cfrac%7Bx-23%7D%7Bx%5E2%2B9x-36%7D%2B%5Cfrac%7B1%7D%7Bx%2B12%7D%5C%5C%5C%5C%28f%2Bg%29%28x%29%20%3D%20%5Cfrac%7Bx-23%7D%7B%28x%2B12%29%28x-3%29%7D%2B%5Cfrac%7B1%28x-3%29%7D%7B%28x%2B12%29%28x-3%29%7D%5C%5C%5C%5C%28f%2Bg%29%28x%29%20%3D%20%5Cfrac%7Bx-23%2B1%28x-3%29%7D%7B%28x%2B12%29%28x-3%29%7D%5C%5C%5C%5C%28f%2Bg%29%28x%29%20%3D%20%5Cfrac%7Bx-23%2Bx-3%7D%7Bx%5E2%2B9x-36%7D%5C%5C%5C%5C%28f%2Bg%29%28x%29%20%3D%20%5Cfrac%7B2x-26%7D%7Bx%5E2%2B9x-36%7D%5C%5C%5C%5C)
We must require that
and
to avoid having 0 in the denominator. This is why choice D is the answer.
This translates to "a number" is greater than 45. All you have to do now is translate these words into an algebraic statement. Basically, you replace "a number" with the variable which it is defined for, and you use the "greater than" symbol to show that the variable is greater than the value of 45.
$20.25 because 8.50 and 8.50 equal 17 and then subtract 2.00 makes 15, and then add 5.25 and you end up with 20.25