Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 10 - Section 10.2 - Integrated Review - Operations on Polynomials: 25

Answer

$y^{33}z^{39}$

Work Step by Step

$(y^{11}z^{13})^{3}$ $=(y^{11})^{3}(z^{13})^{3}$ $=(y^{11\times3})(z^{13\times3})$ $=(y^{33})(z^{39})$ $=y^{33}z^{39}$
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