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 1 - Introduction to Algebraic Expressions - Review Exercises: Chapter 1: 30

Answer

$\dfrac{3}{16}$

Work Step by Step

Multiplying by the reciprocal of the divisor, the given expression, $ \dfrac{9}{16}\div3 $, simplifies to \begin{array}{l}\require{cancel} \dfrac{9}{16}\cdot\dfrac{1}{3} \\\\= \dfrac{9(1)}{16(3)} \\\\= \dfrac{\cancel{3}\cdot3}{16(\cancel{3})} \\\\= \dfrac{3}{16} .\end{array}
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