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 - 1.7 Multiplication and Division of Real Numbers - 1.7 Exercise Set: 120

Answer

$-\dfrac{3}{2}$

Work Step by Step

The given expression, $ \left( \dfrac{-3}{5} \right)\div\dfrac{6}{15} $, simplifies to \begin{array}{l}\require{cancel} \left( \dfrac{-3}{5} \right)\cdot\dfrac{15}{6} \\\\= \dfrac{-3(15)}{5(6)} \\\\= \dfrac{-\cancel{3}(\cancel{5}\cdot3)}{\cancel{5}(\cancel{3}\cdot2)} \\\\= \dfrac{-3}{2} \\\\= -\dfrac{3}{2} .\end{array}
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