Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 7 - Exponents and Exponential Functions - 7-1 Zero and Negative Exponents - Practice and Problem-Solving Exercises - Page 423: 70


The answer is $\frac{4}{x^{5}r^{2}ny^{2}}$

Work Step by Step

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