Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.6 Rational Exponents - R.6 Exercises - Page 64: 81

Answer

$\color{blue}{-10y^2+y}$

Work Step by Step

RECALL: (1) $a(b+c) = ab + ac$ (2) $a^m \cdot a^n = a^{m+n}$ Use rule (1) above to obtain: $=y^{5/8} \cdot y^{3/8} -y^{5/8} \cdot 10y^{11/8}$ Use rule (2) above to obtain: $=y^{5/8+3/8} - 10y^{5/8+11/8} \\=y^{8/8} - 10y^{16/8} \\=y^1 -10y^2 \\=y-10y^2 \\=\color{blue}{-10y^2+y}$
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