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: 84


$\color{blue}{-15y^{19/10} - 10y^{13/10}}$

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: $=-5y \cdot 3y^{9/10} + (-5y) \cdot 4y^{3/10} \\=-5y \cdot 3y^{9/10} -5y \cdot 4y^{3/10} \\=-15y \cdot y^{9/10} - 20y \cdot y^{3/10}$ Use rule (2) above to obtain: $=-15y^{1+(9/10)} - 20y^{1+(3/10)} \\=-15y^{10/10+9/10} - 20y^{10/10+3/10} \\=\color{blue}{-15y^{19/10} - 10y^{13/10}}$
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