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: 32

Answer

$\dfrac{27}{25}$

Work Step by Step

Multiplying the numerators and the denominators, the given expression, $ \dfrac{9}{10}\cdot\dfrac{6}{5} $, simplifies to \begin{array}{l}\require{cancel} \dfrac{9(6)}{10(5)} \\\\= \dfrac{9(\cancel{2}\cdot3)}{\cancel{2}\cdot5(5)} \\\\= \dfrac{27}{25} .\end{array}
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