Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 4 - Polynomials - Mid-Chapter Review - Mixed Review: 3

Answer

$x^{16}y^{40}$

Work Step by Step

Using the laws of exponents, the given expression, $ (x^2y^5)^8 $, is equivalent to \begin{array}{l} x^{2(8)}y^{5(8)} \\\\= x^{16}y^{40} .\end{array}
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