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 5 - Polynomials and Factoring - 5.1 Introduction to Factoring - 5.1 Exercise Set: 65

Answer

$\dfrac{3}{25}$

Work Step by Step

The given expression, $ \dfrac{2}{5}\div\dfrac{10}{3} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{2}{5}\cdot\dfrac{3}{10} \\\\= \dfrac{2(3)}{5(10)} \\\\= \dfrac{\cancel{2}(3)}{5(\cancel{2}\cdot5)} \\\\= \dfrac{3}{25} .\end{array}
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