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

The answer is $\frac{4}{x^{5}r^{2}ny^{2}}$
To solve this expression, we need to multiply instead of dividing: = $\frac{2x^{-5}y^{3}}{n^{2}}$ $\div$ $\frac{r^{2}y^{5}}{2n}$ = $\frac{2y^{3}}{x^{5}n^{2}}$ $\times$ $\frac{2n}{r^{2}y^{5}}$ = $\frac{4}{x^{5}r^{2}n^{2-1}y^{5-3}}$ = $\frac{4}{x^{5}r^{2}ny^{2}}$