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: 12



Work Step by Step

Substituting $p=-4$ and $r=16,$ the given expression, $ \dfrac{\sqrt{r}}{8p+2r} ,$ evaluates to \begin{array}{l}\require{cancel} \dfrac{\sqrt{16}}{8(-4)+2(16)} \\\\= \dfrac{4}{-32+32} \\\\= \dfrac{4}{0} \\\\= \text{undefined} .\end{array}
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