# How do you simplify  (5^0)^-2 + 2^-3?

Feb 29, 2016

$\frac{9}{8}$

#### Explanation:

A reminder of how to deal with negative exponents:

${a}^{-} b = \frac{1}{a} ^ b$

Thus, the expression equals

$= \frac{1}{{5}^{0}} ^ 2 + \frac{1}{2} ^ 3$

To deal with ${\left({5}^{0}\right)}^{2}$, recall that anything to the $0$ power is $1$.

$= \frac{1}{1} ^ 2 + \frac{1}{8}$

$= 1 + \frac{1}{8}$

$= \frac{9}{8}$