Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Prerequisites - Review Exercises - Page 62: 33


(a) $-8z^3$ (b) $\frac{1}{y^2}$

Work Step by Step

(a) $(-2z)^3=(-2)^3.z^3=-8z^3$ (b) $\frac{(8y)^0}{y^2}=\frac{8^0.y^0}{y^2}=\frac{(1)(1)}{y^2}=\frac{1}{y^2}$
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