## Algebra 1: Common Core (15th Edition)

The answer is $m^{-5}$
To solve this $m^{x}$ $\times$ $m^{-4}$ = $m^{-9}$ $m^{x}$ = $\frac{m^{-9}}{m^{-4}}$ $m^{x}$ = $m^{-9-(-4)}$ $m^{x}$ = $m^{(-9+4)}$ $m^{x}$ = $m^{-5}$ --> same base, bring the exponents down x=-5 Therefore the missing part is $m^{-5}$.