## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 2 - Equations, Inequalities, and Problem Solving - 2.2 Using the Principles Together - 2.2 Exercise Set - Page 95: 98

#### Answer

$1$

#### Work Step by Step

Cancelling common factors between the numerator and the denominator,, then the given expression, $\dfrac{2}{7}\cdot\dfrac{7}{2}$, simplifies to \begin{array}{l}\require{cancel} \dfrac{\cancel{2}}{\cancel{7}}\cdot\dfrac{\cancel{7}}{\cancel{2}} \\\\= 1.\end{array}

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