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5x^2/(x^2)^(1/3)
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12 May 2017, 03:49
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62% (01:33) correct 38% (01:37) wrong based on 226 sessions
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5x^2/(x^2)^(1/3)
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12 May 2017, 03:59
Bunuel wrote: \(\frac{5x^2}{\sqrt[3]{x^2}\)
A. \(5\sqrt[3]{x}\)
B. \(5x\sqrt[3]{x}\)
C. \(5x\sqrt[3]{x^2}\)
D. \(5x^2\sqrt{x^2}\)
E. \(125x^4\) \(\frac{5x^2}{\sqrt[3]{x^2}\) = \(\frac{5x^2}{x^{2/3}}\) = \(5 x^{22/3}\) = \(5 x^{4/3}\) = \(5x^{1+1/3}\) = \(5x\sqrt[3]{x}\)
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Re: 5x^2/(x^2)^(1/3)
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13 May 2017, 02:10
shashankism wrote: Bunuel wrote: \(\frac{5x^2}{\sqrt[3]{x^2}\)
A. \(5\sqrt[3]{x}\)
B. \(5x\sqrt[3]{x}\)
C. \(5x\sqrt[3]{x^2}\)
D. \(5x^2\sqrt{x^2}\)
E. \(125x^4\) \(\frac{5x^2}{\sqrt[3]{x^2}\) = \(\frac{5x^2}{x^{2/3}}\) = \(5 x^{22/3}\) = \(5 x^{4/3}\) = \(5x^{1+1/3}\) = \(5x\sqrt[3]{x}\) What is wrong with cubing numerator and denominator? It will be 125 * (x)power 6 / (x)power 2 Ans will be 125 * (x)power 4



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Re: 5x^2/(x^2)^(1/3)
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13 May 2017, 02:28
rocko911 wrote: shashankism wrote: Bunuel wrote: \(\frac{5x^2}{\sqrt[3]{x^2}\)
A. \(5\sqrt[3]{x}\)
B. \(5x\sqrt[3]{x}\)
C. \(5x\sqrt[3]{x^2}\)
D. \(5x^2\sqrt{x^2}\)
E. \(125x^4\) \(\frac{5x^2}{\sqrt[3]{x^2}\) = \(\frac{5x^2}{x^{2/3}}\) = \(5 x^{22/3}\) = \(5 x^{4/3}\) = \(5x^{1+1/3}\) = \(5x\sqrt[3]{x}\) What is wrong with cubing numerator and denominator? It will be 125 * (x)power 6 / (x)power 2 Ans will be 125 * (x)power 4 Hey Rocko911, Why do you want to take cube of the given expression.. Its no where written to find cube.
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Re: 5x^2/(x^2)^(1/3)
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13 May 2017, 02:33
rocko911 wrote: shashankism wrote: Bunuel wrote: \(\frac{5x^2}{\sqrt[3]{x^2}\)
A. \(5\sqrt[3]{x}\)
B. \(5x\sqrt[3]{x}\)
C. \(5x\sqrt[3]{x^2}\)
D. \(5x^2\sqrt{x^2}\)
E. \(125x^4\) \(\frac{5x^2}{\sqrt[3]{x^2}\) = \(\frac{5x^2}{x^{2/3}}\) = \(5 x^{22/3}\) = \(5 x^{4/3}\) = \(5x^{1+1/3}\) = \(5x\sqrt[3]{x}\) What is wrong with cubing numerator and denominator? It will be 125 * (x)power 6 / (x)power 2 Ans will be 125 * (x)power 4 If you want to solve by cubing the expression then you will have to take cube root of the same.. \(\frac{5x^2}{\sqrt[3]{x^2}\) =\(\sqrt[3]{(\frac{5x^2}{\sqrt[3]{x^2})^3}\) =\(\sqrt[3]{(\frac{125x^6}{x^2})}\) =\(\sqrt[3]{125x^4}\) = \(5x\sqrt[3]{x}\)
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Re: 5x^2/(x^2)^(1/3)
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13 May 2017, 02:50
shashankism wrote: shashankism wrote: Bunuel wrote: \(\frac{5x^2}{\sqrt[3]{x^2}\)
A. \(5\sqrt[3]{x}\)
B. \(5x\sqrt[3]{x}\)
C. \(5x\sqrt[3]{x^2}\)
D. \(5x^2\sqrt{x^2}\)
E. \(125x^4\) \(\frac{5x^2}{\sqrt[3]{x^2}\) = \(\frac{5x^2}{x^{2/3}}\) = \(5 x^{22/3}\) = \(5 x^{4/3}\) = \(5x^{1+1/3}\) = \(5x\sqrt[3]{x}\) What is wrong with cubing numerator and denominator? It will be 125 * (x)power 6 / (x)power 2 Ans will be 125 * (x)power 4 If you want to solve by cubing the expression then you will have to take cube root of the same.. \(\frac{5x^2}{\sqrt[3]{x^2}\) =\(\sqrt[3]{(\frac{5x^2}{\sqrt[3]{x^2})^3}\) =\(\sqrt[3]{(\frac{125x^6}{x^2})}\) =\(\sqrt[3]{125x^4}\) = \(5x\sqrt[3]{x}\)[/quote] Why do we need to take cube root???? Please help



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5x^2/(x^2)^(1/3)
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13 May 2017, 03:33
rocko911 wrote: shashankism wrote: rocko911 wrote: What is wrong with cubing numerator and denominator? It will be 125 * (x)power 6 / (x)power 2 Ans will be
125 * (x)power 4 If you want to solve by cubing the expression then you will have to take cube root of the same.. \(\frac{5x^2}{\sqrt[3]{x^2}}\) =\(\sqrt[3]{\Bigg (\frac{5x^2}{\sqrt[3]{x^2}}\Bigg )^3}\) =\(\sqrt[3]{(\frac{125x^6}{x^2})}\) =\(\sqrt[3]{125x^4}\) = \(5x\sqrt[3]{x}\) Why do we need to take cube root???? Please help In case that you want to get rid of cube root of \(\sqrt[3]{x}\)
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Re: 5x^2/(x^2)^(1/3)
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13 May 2017, 11:46
nguyendinhtuong wrote: rocko911 wrote: shashankism wrote: What is wrong with cubing numerator and denominator? It will be 125 * (x)power 6 / (x)power 2 Ans will be
125 * (x)power 4 If you want to solve by cubing the expression then you will have to take cube root of the same.. \(\frac{5x^2}{\sqrt[3]{x^2}}\) =\(\sqrt[3]{\Bigg (\frac{5x^2}{\sqrt[3]{x^2}}\Bigg )^3}\) =\(\sqrt[3]{(\frac{125x^6}{x^2})}\) =\(\sqrt[3]{125x^4}\) = \(5x\sqrt[3]{x}\) Why do we need to take cube root???? Please help In case that you want to get rid of cube root of \(\sqrt[3]{x}\)[/quote] yeah to get rid of \(\sqrt[3]{x}\) we are cubing the whole expression.... but as told by Shashankism, He cubed the whole problem AND took CUBE ROOT ALSO.......SO WHY BOTH? Why not only cubing the whole expression and get rid of \(\sqrt[3]{x}\) ??????????



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Re: 5x^2/(x^2)^(1/3)
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13 May 2017, 18:17
rocko911 wrote: yeah to get rid of \(\sqrt[3]{x}\) we are cubing the whole expression.... but as told by Shashankism, He cubed the whole problem AND took CUBE ROOT ALSO.......SO WHY BOTH? Why not only cubing the whole expression and get rid of \(\sqrt[3]{x}\) ?????????? He did so to make the expression more simple for calculate. You could make cube only of \(\sqrt[3]{x^2}\), however, you still have to go around to come to the answer.
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Re: 5x^2/(x^2)^(1/3)
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13 May 2017, 23:45
nguyendinhtuong wrote: rocko911 wrote: yeah to get rid of \(\sqrt[3]{x}\) we are cubing the whole expression.... but as told by Shashankism, He cubed the whole problem AND took CUBE ROOT ALSO.......SO WHY BOTH? Why not only cubing the whole expression and get rid of \(\sqrt[3]{x}\) ?????????? He did so to make the expression more simple for calculate. You could make cube only of \(\sqrt[3]{x^2}\), however, you still have to go around to come to the answer. No, i dont get it See how I did this.. 5x^2 / (x^2)^(1/3) 5x(x^2) / x^(2/3) CUBING NUMERATOR AND DENOMINATOR 125 x^(6) / x^(2) 125 x^4 what is wrong with this ?



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Re: 5x^2/(x^2)^(1/3)
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14 May 2017, 00:49
rocko911 wrote: nguyendinhtuong wrote: rocko911 wrote: yeah to get rid of \(\sqrt[3]{x}\) we are cubing the whole expression.... but as told by Shashankism, He cubed the whole problem AND took CUBE ROOT ALSO.......SO WHY BOTH? Why not only cubing the whole expression and get rid of \(\sqrt[3]{x}\) ?????????? He did so to make the expression more simple for calculate. You could make cube only of \(\sqrt[3]{x^2}\), however, you still have to go around to come to the answer. No, i dont get it See how I did this.. 5x^2 / (x^2)^(1/3) \(\quad =\frac{5x^2}{(x^2)^\frac{1}{3}}\) OK5 x(x^2) / x^(2/3) \(\quad =\frac{5 \times x \times x^2}{x^\frac{2}{3}}\) Wrong. You mean \(=\frac{5 \times x^2}{x^\frac{2}{3}}\) ?CUBING NUMERATOR AND DENOMINATOR 125 x^(6) / x^(2) .....Hey, this step is wrong. You can't left the cube root sign like this. You have to keep it like this: \(=\sqrt[3]{\Bigg ( \frac{5 \times x^2}{x^\frac{2}{3}} \Bigg ) ^3}=\sqrt[3]{\frac{125 \times x^6}{x^2}}\)125 x^4 .....As this step led to incorrect value. This step must be \(=\sqrt[3]{125x^4}\)what is wrong with this ? Your way is wrong as I explained above
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Re: 5x^2/(x^2)^(1/3)
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14 May 2017, 01:10
[quote="nguyendinhtuong"][quote="rocko911"][quote="nguyendinhtuong"][quote="rocko911"]
5×x2x23=5×x2x23 ?
CUBING NUMERATOR AND DENOMINATOR
125 x^(6) / x^(2) .....Hey, this step is wrong. You can't left the cube root sign like this. You have to keep it like this: =(5×x2x23)3−−−−−−−−⎷3=125×x6x2−−−−−−√3=(5×x2x23)33=125×x6x23
125 x^4 .....As this step led to incorrect value. This step must be =125x4−−−−−√3
cube root was in denominator and when i already did x^(2/3) from where did the cube and a cube root came from?????????? i dont get it,, can u explain by basic and step by step approach?



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Re: 5x^2/(x^2)^(1/3)
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14 May 2017, 01:20
rocko911 wrote: cube root was in denominator and when i already did x^(2/3) from where did the cube and a cube root came from?????????? i dont get it,, can u explain by basic and step by step approach?
I don't get your question. Let make another example. For example, simplify this \(\sqrt[3]{8x^3}\) The correct way is \(\sqrt[3]{8x^3}=\sqrt[3]{(2x)^3}=2x\) Your way is like this \(\sqrt[3]{8x^3}=8x^3\). You left out cube root and get wrong value. However, you could left out cube when solving equation. \(\sqrt[3]{8x^3}=1 \implies 8x^3=1\) Hope this clears your problem
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Re: 5x^2/(x^2)^(1/3)
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25 Apr 2018, 10:43
@Bunel can someone explain why the answer is not C? I dont fully understand how the math was done to get B.



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Re: 5x^2/(x^2)^(1/3)
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25 Apr 2018, 12:40
rocko911 wrote: shashankism wrote: Bunuel wrote: \(\frac{5x^2}{\sqrt[3]{x^2}\)
A. \(5\sqrt[3]{x}\)
B. \(5x\sqrt[3]{x}\)
C. \(5x\sqrt[3]{x^2}\)
D. \(5x^2\sqrt{x^2}\)
E. \(125x^4\) \(\frac{5x^2}{\sqrt[3]{x^2}\) = \(\frac{5x^2}{x^{2/3}}\) = \(5 x^{22/3}\) = \(5 x^{4/3}\) = \(5x^{1+1/3}\) = \(5x\sqrt[3]{x}\) What is wrong with cubing numerator and denominator? It will be 125 * (x)power 6 / (x)power 2 Ans will be 125 * (x)power 4 In this case you are not necessarily multiplying the expression by 1, so you have created a different term by cubing both the numerator and denominator and then simplifying. So taking the cube root of your solution 125x^4 should get you back to to the answer.




Re: 5x^2/(x^2)^(1/3) &nbs
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