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anonymous
 3 years ago
I have done this problem many times and it is still wrong. I know where it's wrong, at least, but do not know the answer. I look here for some different guidance from what I have been getting, hoping it will help.
( x^(2/3) / 4y(2) ) ^(1/2)
x^(2/3)(1/2) = x^(2/6) = x^(1/3)
4^(1/2) = 2
And this is allegedly where I am making a mistake:
y^2 * 1/2 = y^(2/2) = y^1 = 1/y
So for my final answer I have
x^(1/3)

2/y
But this is wrong, I guess. Any help is appreciated!
anonymous
 3 years ago
I have done this problem many times and it is still wrong. I know where it's wrong, at least, but do not know the answer. I look here for some different guidance from what I have been getting, hoping it will help. ( x^(2/3) / 4y(2) ) ^(1/2) x^(2/3)(1/2) = x^(2/6) = x^(1/3) 4^(1/2) = 2 And this is allegedly where I am making a mistake: y^2 * 1/2 = y^(2/2) = y^1 = 1/y So for my final answer I have x^(1/3)  2/y But this is wrong, I guess. Any help is appreciated!

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Henryblah
 3 years ago
Best ResponseYou've already chosen the best response.0Is it 4(y)^2 or (4y)^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Neither. There were no brackets or grouping symbols in the denominator. It was (is) literally: (4y^2)^1/2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Should I ask this question in a different section? It is prealgebra as I understand it. Should it be in an even lower section?

phi
 3 years ago
Best ResponseYou've already chosen the best response.0You are not really wrong, but you have not simplified your result. you got \[ \frac{x^\frac{1}{3}}{\frac{2}{y}}\] fractions with fractions as their top or bottom are ugly. if you multiply top and bottom by y/2 you get \[ \frac{x^\frac{1}{3} \cdot \frac{y}{2}}{\frac{2}{y}\cdot \frac{y}{2}} = x^\frac{1}{3} \cdot \frac{y}{2} =\frac{x^\frac{1}{3}y}{2} \]
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