Intermediate Algebra (12th Edition)

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

Chapter 4 - Mixed Review Exercises: 6

Answer

$\dfrac{1}{5z^{9}}$

Work Step by Step

Using the laws of exponents, the given expression, $ \dfrac{(-z^{-2})^3}{5(z^{-3})^{-1}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{(-z)^{-2(3)}}{5z^{-3(-1)}} \\\\= \dfrac{(-z)^{-6}}{5z^{3}} \\\\= \dfrac{1}{5z^{3}(-z)^{6}} \\\\= \dfrac{1}{5z^{3}z^{6}} \\\\= \dfrac{1}{5z^{3+6}} \\\\= \dfrac{1}{5z^{9}} .\end{array}
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