Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 11 - Additional Topics - 11.3 - Fractional Exponents - Problem Set 11.3: 60

Answer

$4y^{\frac{1}{15}}$

Work Step by Step

Recall the following rule: $a^{n}\cdot a^{m}=a^{n+m}$. Therefore, we obtain: $(y^{-\frac{1}{3}})(4y^{\frac{2}{5}})$ $=(1\times4)(y^{-\frac{1}{3}+\frac{2}{5}})$ $=(4)(y^{\frac{-1(5)+2(3)}{15}})$ $=4y^{\frac{1}{15}}$
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