Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.1 Introduction to Algebraic Expressions - R.1 Exercise Set - Page 938: 51

Answer

$10$

Work Step by Step

The given expression, $ \dfrac{7000+(-10)^3}{10^2\times(2+4)} ,$ evaluates to \begin{array}{l}\require{cancel} \dfrac{7000+(-1000)}{100\times(6)} \\\\= \dfrac{7000-1000}{600} \\\\= \dfrac{6000}{600} \\\\= 10 .\end{array}
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