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(the answer has to be in positive exponents)

 Jul 21, 2017

Best Answer 

 #2
avatar+20848 
+2

simplify this expression?

\(\dfrac{2^{-1}a^2b^{-3}}{4a^3b^{-5}}\)

 

\(\begin{array}{|rcll|} \hline && \dfrac{2^{-1}a^2b^{-3}}{4a^3b^{-5}} \\\\ &=& \dfrac{2^{-1}a^2b^{-3}}{2^2a^3b^{-5}} \\\\ &=& \dfrac{2^{-1}}{2^2}\cdot \dfrac{a^2}{a^3} \cdot \dfrac{b^{-3}}{b^{-5}} \\\\ &=& \dfrac{1}{2^22^{1}}\cdot \dfrac{1}{a^3a^{-2}} \cdot \dfrac{b^{-3}b^{5}}{1} \\\\ &=& \dfrac{1}{2^{2+1}}\cdot \dfrac{1}{a^{3-2}} \cdot \dfrac{b^{-3+5}}{1} \\\\ &=& \dfrac{1}{2^{3}}\cdot \dfrac{1}{a^{1}} \cdot \dfrac{b^{2}}{1} \\\\ &=& \dfrac{1}{2^{3}}\cdot \dfrac{1}{a} \cdot \dfrac{b^{2}}{1} \\\\ &=& \dfrac{b^{2}}{2^{3}a} \\\\ &=& \dfrac{b^{2}}{8a} \\ \hline \end{array}\)

 

laugh

 Jul 21, 2017
 #1
avatar
0

(b2) / (8a)

 Jul 21, 2017
 #2
avatar+20848 
+2
Best Answer

simplify this expression?

\(\dfrac{2^{-1}a^2b^{-3}}{4a^3b^{-5}}\)

 

\(\begin{array}{|rcll|} \hline && \dfrac{2^{-1}a^2b^{-3}}{4a^3b^{-5}} \\\\ &=& \dfrac{2^{-1}a^2b^{-3}}{2^2a^3b^{-5}} \\\\ &=& \dfrac{2^{-1}}{2^2}\cdot \dfrac{a^2}{a^3} \cdot \dfrac{b^{-3}}{b^{-5}} \\\\ &=& \dfrac{1}{2^22^{1}}\cdot \dfrac{1}{a^3a^{-2}} \cdot \dfrac{b^{-3}b^{5}}{1} \\\\ &=& \dfrac{1}{2^{2+1}}\cdot \dfrac{1}{a^{3-2}} \cdot \dfrac{b^{-3+5}}{1} \\\\ &=& \dfrac{1}{2^{3}}\cdot \dfrac{1}{a^{1}} \cdot \dfrac{b^{2}}{1} \\\\ &=& \dfrac{1}{2^{3}}\cdot \dfrac{1}{a} \cdot \dfrac{b^{2}}{1} \\\\ &=& \dfrac{b^{2}}{2^{3}a} \\\\ &=& \dfrac{b^{2}}{8a} \\ \hline \end{array}\)

 

laugh

heureka Jul 21, 2017

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