## Precalculus: Mathematics for Calculus, 7th Edition

a) $\frac{1}{2}$ b) $\frac{1}{8}$ c) $\frac{9}{4}$
a) $(\frac{5}{3})^{0}$ $\times$ $2^{-1}$= 1 $\times$ $2^{-1}$ = $\frac{1}{2}$. Note that anything to the power 0 is equal to 1 b) $\frac{2^{-3}}{3^{0}}$= $2^{-3}$= $\frac{1}{2^{3}}$ = $\frac{1}{8}$ c) ($\frac{2}{3}$)$^{-2}$ = $\frac{1}{\frac{4}{9}}$ = $\frac{9}{4}$