+0

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

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+10

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

Pittassuaq  Dec 10, 2015

#2
+91469
+10

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

I did that the super long way ..

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

Melody  Dec 10, 2015
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#1
+91469
0

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

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

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

Melody  Dec 10, 2015
#2
+91469
+10

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

I did that the super long way ..

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

Melody  Dec 10, 2015

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