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

The answer is $\frac{14}{t^{5}m^{2}}$
The answer is $\frac{14}{t^{5}m^{2}}$ for the expression $\frac{7s^{0}t^{-5}}{2^{-1}m^{2}}$. To simplify this expression, we have to change the exponents from negative to positive, and the only way to do it is take the reciprocal of the negative exponent. Also, any number to the power of 0 becomes 1, so $s^{0}$=1. $\frac{7s^{0}t^{-5}}{2^{-1}m^{2}}$. = $\frac{7\times(1)\times2^{1}}{t^{5}m^{2}}$ = $\frac{7\times(1)\times2}{t^{5}m^{2}}$ = $\frac{14}{t^{5}m^{2}}$