Basic College Mathematics (9th Edition)

Published by Pearson
ISBN 10: 0321825535
ISBN 13: 978-0-32182-553-7

Chapter 3 - Adding and Subtracting Fractions - Review Exercises - Page 257: 79

Answer

$\frac{1}{250}$

Work Step by Step

$(\frac{1}{4})^{2}.(\frac{2}{5})^{3}$ = $\frac{1}{4}\times\frac{1}{4} \times \frac{2}{5}\times\frac{2}{5}\times\frac{2}{5}$ = $\frac{1\times1\times2\times2\times2}{4\times4\times5\times5\times5}$ (multiply nominators and denominators) = $\frac{1\times1\times1\times1\times1}{1\times2\times5\times5\times5}$ = $\frac{1}{250}$
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