College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Chapter P Test - Page 79: 7

Answer

$\textbf{(a)}\hspace{0.7cm} \dfrac{a^2}{b}$ $\textbf{(b)}\hspace{0.7cm}\dfrac{y^4}{4x^6}$ $\textbf{(c)}\hspace{0.7cm} 18x^{5/8}$ $\textbf{(d)}\hspace{0.7cm}-3\sqrt{5}$ $\textbf{(e)}\hspace{0.7cm}3\sqrt{2}x^{3/2}y^2$ $\textbf{(f)}\hspace{0.7cm} \dfrac{1}{4x^{10}y}$

Work Step by Step

$\textbf{(a)}\hspace{0.7cm} \dfrac{a^3b^2}{ab^3}=\dfrac{a^3}{a}.\dfrac{b^2}{b^3}=a^{(3-1)}b^{(2-3)}=a^2b^{-1}=\dfrac{a^2}{b}$ $\textbf{(b)}\hspace{0.7cm} (2x^3y^{-2})^{-2}=(2)^{-2}\times (x^3)^{-2}\times (y^{-2})^{-2}\\\hspace{3.5cm}=\dfrac{1}{2^2}\times x^{-6}\times y^4=\dfrac{y^4}{4x^6}$ $\textbf{(c)}\hspace{0.7cm} (2x^{1/2}y^2)(3x^{1/4}y^{-1})^{2}=(2x^{1/2}y^2)(3)^{2}(x^{1/4})^{2}(y^{-1})^{2}\\\hspace{5.2cm}=(2x^{1/2}y^2)9x^{1/8}y^{-2}\\\hspace{5.2cm}=(2\times 9)x^{(1/2+1/8)}y^{(2-2)}=18x^{5/8}$ $\textbf{(d)}\hspace{0.7cm} \sqrt{20}-\sqrt{125}=\sqrt{4\times5}-\sqrt{25\times5}=\sqrt{4}\times\sqrt{5}-\sqrt{25}\times\sqrt{5}\hspace{3.7cm} =2\sqrt{5}-5\sqrt{5}=-3\sqrt{5}$ $\textbf{(e)}\hspace{0.7cm} \sqrt{18x^3y^4}=\sqrt{18}\sqrt{x^3}\sqrt{y^4}=\sqrt{2\times9}x^{3/2}y^2=3\sqrt{2}x^{3/2}y^2$ $\textbf{(e)}\hspace{0.7cm} \bigg(\dfrac{2x^2y}{x^{-3}y^{1/2}}\bigg)^{-2}=\bigg(2x^{(2-(-3))}y^{(1-1/2)}\bigg)^{-2}=\bigg(2x^{5}y^{1/2}\bigg)^{-2}\hspace{3.7cm} =2^{-2}x^{-10}y^{-1}=\dfrac{1}{4x^{10}y}$
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