Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - Chapter Review - Page 842: 28

Answer

$\frac{8}{9}$

Work Step by Step

Following the example: $0.888...=8(0.1)+8(0.1)^2+...=\frac{a_1}{1-r}=\frac{8(0.1)}{1-0.1}=\frac{0.8}{0.9}=\frac{8}{9}$
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