Intermediate Algebra (6th Edition)

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

Chapter 1 - Test: 11



Work Step by Step

$\frac{6(7-9)^{3}+(-2)}{(-2)(-5)(-5)}$ =$\frac{6(-2)^{3}+(-2)}{10(-5)}$ =$\frac{6(-8)+(-2)}{-50}$ =$\frac{-50}{-50}$ = 1
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