Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 1 - Review - Page 44: 67



Work Step by Step

$\frac{(8-10)^3-(-4)^2}{2+8(2)\div4}$ $\frac{(-2)^3-(-4)^2}{2+16\div4}$ $\frac{-8-16}{2+4}$ $\frac{-24}{6}$ $-4$
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