Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises: 131

Answer

$\dfrac{1}{6y^{13}}$

Work Step by Step

Using the laws of exponents, the given expression, $ \dfrac{(-y^{-4})^2}{6(y^{-5})^{-1}} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{(-1)^2y^{-4(2)}}{6(y^{-5(-1)})} \\\\= \dfrac{1y^{-8}}{6y^{5}} \\\\= \dfrac{y^{-8-5}}{6} \\\\= \dfrac{y^{-13}}{6} \\\\= \dfrac{1}{6y^{13}} .\end{array}
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