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: 31

Answer

$\frac{388}{495}$

Work Step by Step

$0.78383...=0.7+8.3(0.01)+8.3(0.01)^2+...=0.7+\frac{a_1}{1-r}=0.7+\frac{8.3(0.01)}{1-0.01}=\frac{388}{495}$
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