## Elementary Algebra

$\frac{21}{16}$
Recall the following rule: $(\frac{a}{b})^{-x}$ = $\frac{b^{x}}{a^{x}}$. Thus, we obtain: $4^{-2}$ + $4^{-1}$ + $4^{0}$ = $\frac{1^{2}}{4^{2}}$ + $\frac{1^{1}}{4^{1}}$ + $\frac{1^{0}}{4^{0}}$ = $\frac{1}{16}$ + $\frac{1}{4}$ + 1 = Make all the denominators equal. $\frac{1}{16}$ + $\frac{1}{4}$ $\times$ $\frac{4}{4}$ + 1 $\times$ $\frac{16}{16}$ = $\frac{1}{16}$ + $\frac{4}{16}$ + $\frac{16}{16}$ = $\frac{21}{16}$