Question

PLEASE HELP! 25 PTS!!

Answer 1

Answer:

Solution given:

$f(x)=\frac{x-16}{x^2+6x-40}$

$g(x)=\frac{1}{x+10}$

now

f(x)+g(x)=$\frac{x-16}{x^2+6x-40}+\frac{1}{x+10}$....(1)

now

factoring x²+6x-40

we get

x²+10x-4x-40

x(x+10)-4(x+10)

(x+10)(x-4)

now substituting in equation 1 ,we get

f(x)+g(x)=$\frac{x-16}{(x+10)(x-4)}+\frac{1}{x+10}$

taking l.c.m

=$\frac{(x-16)+(x-4)}{(x-10)(x-4)}$

=now

opening bracket

$\frac{x-16+x-4}{x²-10x-4x+40}$

=$\frac{2x-20}{x²+6x-40}$

So

answer is :

B. $\bold b\frac{2x-20}{x^2+6x-40}$