Answer
$\frac{161}{495}$
Work Step by Step
$0.32525...=0.3+2.5(0.01)+2.5(0.01)^2+...=0.3+\frac{a_1}{1-r}=0.3+\frac{2.5(0.01)}{1-0.01}=\frac{0.322}{0.999}=\frac{161}{495}$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 12 Sequences and Series - Chapter Test - Page 843: 24
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
