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 - Test: Chapter 1 - Page 78: 12

Answer

$\dfrac{3}{7}$

Work Step by Step

Cancelling the common factor between the numerator and the denominator, then the given expression, $ \dfrac{24}{56} $, simplifies to \begin{array}{l}\require{cancel} \dfrac{\cancel{8}\cdot3}{\cancel{8}\cdot7} \\\\= \dfrac{3}{7} .\end{array}
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