College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Summary, Review, and Test - Review Exercises - Page 789: 50



Work Step by Step

Let $0.\overline{6}=x$, then $6.\overline{6}=10x$. Thus $6.\overline{6}-0.\overline{6}=10x-x\\6=9x\\x=\frac{6}{9}=\frac{2}{3}$
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