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y= (7x)-4/(sqrt3(-2) = 7*x-((4/sqrt3(-2)))

 Dec 10, 2015

Best Answer 

 #2
avatar+105509 
+10

y= (7x)-4/(sqrt3(-2) = 7*x-((4/sqrt3(-2)))

 

I did that the super long way ..

 

\(y= 7x-\frac{4}{\sqrt[3]{-2}}\\~\\ y= 7x+\frac{4}{\sqrt[3]{2}}\\~\\ y= 7x+\frac{4}{2^{1/3}}\times \frac{2^{2/3}}{2^{2/3}}\\~\\ y= 7x+ \frac{4\times 2^{2/3}}{2}\\~\\ y= 7x+ \frac{2\times 2^{2/3}}{1}\\~\\ y= 7x+ \sqrt[3]{32}\\~\\ \)

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 Dec 10, 2015
 #1
avatar+105509 
0

y= (7x)-4/(sqrt3(-2) = 7*x-((4/sqrt3(-2)))

 

I assume you want me to rationalize the denominator  ://

 

 

\(y= (7x)-\frac{4}{\sqrt[3]{-2}} \\~\\ y= 7x+\frac{4}{\sqrt[3]{2}} \\~\\ y= \frac{7x\sqrt[3]{2}+4}{\sqrt[3]{2}} \\~\\ y= \frac{7x*2^{1/3}+4}{2^{1/3}} \\~\\ y= \frac{2^{2/3}}{2^{2/3}}\times \frac{7x*2^{1/3}+4}{2^{1/3}} \\~\\ y= \frac{7x*2+4*2^{2/3}}{2} \\~\\ y= \frac{7x+2*2^{2/3}}{1} \\~\\ y= 7x+2^{5/3} \\~\\ y= 7x+\sqrt[3]{32} \\~\\ \)

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 Dec 10, 2015
 #2
avatar+105509 
+10
Best Answer

y= (7x)-4/(sqrt3(-2) = 7*x-((4/sqrt3(-2)))

 

I did that the super long way ..

 

\(y= 7x-\frac{4}{\sqrt[3]{-2}}\\~\\ y= 7x+\frac{4}{\sqrt[3]{2}}\\~\\ y= 7x+\frac{4}{2^{1/3}}\times \frac{2^{2/3}}{2^{2/3}}\\~\\ y= 7x+ \frac{4\times 2^{2/3}}{2}\\~\\ y= 7x+ \frac{2\times 2^{2/3}}{1}\\~\\ y= 7x+ \sqrt[3]{32}\\~\\ \)

Melody Dec 10, 2015

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