Precalculus (6th Edition)

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

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises - Page 74: 30

Answer

$\color{blue}{-2^{1/2}my^{5/2}}$

Work Step by Step

RECALL: (1) $a^{m/n}=\sqrt[n]{a^m} = (\sqrt[n]{a})^m$ (2) $(ab)^m=a^mb^m$ (3) $(a^m)^n=a^{mn}$ Use rule (1) above to obtain: $-m\sqrt{2y^5}=-m(2y^5)^{1/2}$ Use rule (2) above to obtain: $=-m(2^{1/2})(y^5)^{1/2}$ Use rule (3) above to obtain: $= -m(2^{1/2})(y^{5(1/2)}) \\=-m(2^{1/2})y^{5/2} \\=\color{blue}{-2^{1/2}my^{5/2}}$
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