3 years ago

Eep help I'll give brainliest if you answer all

Mathematics

2 answers:

ikadub [295]3 years ago

5 0

Answer: 28π, 35π,7π

Step-by-step explanation:

klemol [59]3 years ago

3 0

for number 2 it's 18.45 I think

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<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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