Intermediate Algebra (12th Edition)

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

Chapter 2 - Chapters R-2 - Cumulative Review Exercises - Page 214: 11



Work Step by Step

Substituting $p=-4,$ and $q=\dfrac{1}{2},$ the given expression, $ -3(2q-3p) ,$ evaluates to \begin{array}{l}\require{cancel} -3\left(2\cdot\dfrac{1}{2}-3(-4)\right) \\\\= -3\left(1+12\right) \\\\= -3\left(13\right) \\\\= -39 .\end{array}
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