Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 7 - Exponents and Exponential Functions - Chapter Review - Page 476: 39



Work Step by Step

$(-2r^{-4})^2(-3r^2z^8)^{-1}$ $=(-2)^2r^{-8}(-3)^{-1}r^{-2}z^{-8}$ $=-\frac{4}{3}r^{-10}z^{-8}$ $=-\frac{4}{3r^{10}z^8}$
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