Intermediate Algebra (12th Edition)

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

Chapter 4 - Mixed Review Exercises - Page 319: 3

Answer

$\dfrac{y^{4}}{36}$

Work Step by Step

Using the laws of exponents, the given expression, $ \dfrac{6^{-1}y^3(y^2)^{-2}}{6y^{-4}(y^{-1})} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{y^3(y^{2(-2)})}{6(6^{1})y^{-4}(y^{-1})} \\\\= \dfrac{y^3(y^{-4})}{36y^{-4}(y^{-1})} \\\\= \dfrac{y^{3+(-4)-(-4)-(-1)}}{36} \\\\= \dfrac{y^{3-4+4+1}}{36} \\\\= \dfrac{y^{4}}{36} .\end{array}
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