Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 1 - Review - Page 44: 71

Answer

$\dfrac{1}{5}$

Work Step by Step

Using order of operations, the expression $ \dfrac{\sqrt{25}}{4+3\cdot7} $ simplifies to \begin{array}{l} \dfrac{5}{4+21} \\\\= \dfrac{5}{25} \\\\= \dfrac{1}{5} .\end{array}
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