Answer:
Option A
Step-by-step explanation:
Given in the question,
[tex]f(x)=\frac{x-16}{x^2+6x-40}[/tex]
[tex]g(x)=\frac{1}{x+10}[/tex]
[tex]f(x)+g(x)=\frac{x-16}{x^2+6x-40}+\frac{1}{x+10}[/tex]
[tex]=\frac{(x+10)(x-16)+(x^2+6x-40)}{(x+10)(x^2+6x-10)}[/tex]
[tex]=\frac{x^2-16x+10x-160+x^2+6x-40}{(x+10)(x^2+6x-40)}[/tex]
[tex]=\frac{2x^2-200}{(x+10)(x^2+6x-40)}[/tex]
[tex]=\frac{2(x-10)(x+10)}{(x+10)(x^2+6x-40)}[/tex]
[tex]=\frac{2(x-10)}{x^2+6x-40}[/tex]
[tex]=\frac{2x-20}{x^2+6x-40}[/tex]
Therefore, Option A will be the correct option.