Find the reciprocal.
-$\frac{x^{2}+8}{2-x}$
A) $\frac{8-x}{x^{2}+2}$
B) $\frac{x^{2}+8}{x-2}$
C) $\frac{x^{2}+2}{8-x}$
D) $\frac{2-x}{x^{2}+8}$

Name: Find the reciprocal. -$\frac{x^{2}+8}{2-x}$ A) $\frac{8-x}{x^{2}+2}$ B) $\frac{x^{2}+8}{x-2}$ C) $\frac{x^{2}+2}{8-x}$ D) $\frac{2-x}{x^{2}+8}$
Brand: Quizplus
Rating: 4.8 (38 reviews)

Question

Quizplus · Accepted Answer

The reciprocal of a fraction is obtained by swapping the numerator and the denominator. Therefore, the reciprocal of $\frac{x^{2}+8}{2-x}$ is $\frac{2-x}{x^{2}+8}$.

Find the Reciprocal x2+82−x\frac{x^{2}+8}{2-x}2−xx2+8​ A) 8−xx2+2\frac{8-x}{x^{2}+2}x2+28−x​ B) x2+8x−2\frac{x^{2}+8}{x-2}x−2x2+8​ C) x2+28−x\frac{x^{2}+2}{8-x}8−xx2+2​ D) 2−xx2+8\frac{2-x}{x^{2}+8}x2+82−x​

Find the Reciprocal $\frac{x^{2}+8}{2-x}$ A) $\frac{8-x}{x^{2}+2}$ B) $\frac{x^{2}+8}{x-2}$ C) $\frac{x^{2}+2}{8-x}$ D) $\frac{2-x}{x^{2}+8}$