Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter R - Section R.3 - Exponents, Roots, and Order of Operations - R.3 Exercises - Page 30: 69



Work Step by Step

Using order of operations and the concepts of the absolute value of a number, the given expression, $ 6+\dfrac{2}{3}(-9)-\dfrac{5}{8}\cdot16 ,$ simplifies to \begin{array}{l}\require{cancel} 6+\dfrac{2}{\cancel{3}}(-\cancel{9}^3)-\dfrac{5}{\cancel{8}}\cdot\cancel{16}^2 \\\\= 6+(-6)-10 \\\\= 6-6-10 \\\\= -10 .\end{array}
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