3 years ago

PLEASE HELP ASAP WILL GIVE BLAINLY PTS

Mathematics

1 answer:

ddd [48]3 years ago

4 0

Answer:

$\large\boxed{C.\ \dfrac{2x-20}{x^2+6x-40}}$

Step-by-step explanation:

$f(x)=\dfrac{x-16}{x^2+6x-40}=\dfrac{x-16}{x+10x-4x-40}=\dfrac{x-16}{x(x+10)-4(x+10)}\\\\f(x)=\dfrac{x-16}{(x-4)(x+10)}\\\\g(x)=\dfrac{1}{x+10}\\\\f(x)+g(x)=\dfrac{x-16}{(x-4)(x+10)}+\dfrac{1}{x+10}=\dfrac{x-16}{(x-4)(x+10)}+\dfrac{1(x-4)}{(x-4)(x+10)}\\\\=\dfrac{x-16+x-4}{(x-4)(x+10)}=\dfrac{2x-20}{(x-4)(x+10)}=\dfrac{2x-20}{x^2+6x-40}$

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Hich is an equation of a circle with center (-5, -7) that passes through the point (0, 0)?

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