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: 70

Answer

$23$

Work Step by Step

Using order of operations, the given expression, $ 7-\dfrac{3}{4}(-8)+12\cdot\dfrac{5}{6} ,$ simplifies to \begin{array}{l}\require{cancel} 7-\dfrac{3}{\cancel{4}}(-\cancel{8}^2)+\cancel{12}^2\cdot\dfrac{5}{\cancel{6}} \\\\= 7-(-6)+10 \\\\= 7+6+10 \\\\= 13+10 \\\\= 23 .\end{array}
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