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Ksju [112]
3 years ago
14

Someone help me please and thank you!! ASAP :(

Mathematics
1 answer:
pantera1 [17]3 years ago
3 0
<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}\\\\

We must require that x \ne -12 and x \ne 3 to avoid having 0 in the denominator. This is why choice D is the answer.

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