Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.6 Exponents - R.6 Exercises - Page R-25: 44

Answer

$ \displaystyle \frac{12^{3}}{y^{8}}$

Work Step by Step

$\displaystyle \frac{12^{3/4}\cdot 12^{5/4}\cdot y^{-2}}{12^{-1}\cdot(y^{-3})^{-2}}$=$\qquad$ .... group like terms =$\displaystyle \frac{12^{3/4}\cdot 12^{5/4}}{12^{-1}}\cdot\frac{y^{-2}}{(y^{-3})^{-2}} \qquad$ ....... use $a^{m}\cdot a^{n}=a^{m+n}$ and $(a^{m})^{n}=a^{mn}$ =$\displaystyle \frac{12^{3/4+5/4}}{12^{-1}}\cdot\frac{y^{-2}}{y^{-3(-2)}}$ =$\displaystyle \frac{12^{2}}{12^{-1}}\cdot\frac{y^{-2}}{y^{6}}\qquad$ ....... use $\displaystyle \frac{a^{m}}{a^{n}}=a^{m-n}$ $=12^{2-(-1)}\cdot y^{-2-6}$ $=12^{3}y^{-8}$ $= \displaystyle \frac{12^{3}}{y^{8}}$
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