Answer
$9v^{\frac{5}{2}}$
Work Step by Step
$=(-3)(-3)\cdot(v^{2}\cdot v^{\frac{1}{2}})$
$=+9\cdot v^{2+\frac{1}{2}},\qquad$ ...because $a^{m}\cdot a^{n}=a^{m+n}$
$=9v^{\frac{4+1}{2}}$
$=9v^{\frac{5}{2}}$
Algebra 1: Common Core (15th Edition)
by Charles, Randall I.
Published by
Prentice Hall
ISBN 10:
0133281140
ISBN 13:
978-0-13328-114-9
Chapter 8 - Polynomials and Factoring - Get Ready! - Page 483: 16
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