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 - Page 278: 125

Answer

$\dfrac{m^{8}}{n^{11}}$

Work Step by Step

Using the laws of exponents, the given expression, $ \dfrac{m^2n^{-8}}{m^{-6}n^3} ,$ is equivalent to \begin{array}{l}\require{cancel} m^{2-(-6)}n^{-8-3} \\\\= m^{2+6}n^{-8-3} \\\\= m^{8}n^{-11} \\\\= \dfrac{m^{8}}{n^{11}} .\end{array}
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