Algebra 2 (1st Edition)

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

Chapter 12 Sequences and Series - Chapter Test - Page 843: 22

Answer

$\frac{46}{99}$

Work Step by Step

$0.4646...=46(0.01)+46(0.01)^2+...=\frac{a_1}{1-r}=\frac{46(0.01)}{1-0.01}=\frac{0.46}{0.99}=\frac{46}{99}$
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