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GaiginJen
Group Title
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!
 one year ago
 one year ago
GaiginJen Group Title
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!
 one year ago
 one year ago

This Question is Closed

Henryblah Group TitleBest ResponseYou've already chosen the best response.0
Is it 4(y)^2 or (4y)^2
 one year ago

GaiginJen Group TitleBest ResponseYou've already chosen the best response.1
Neither. There were no brackets or grouping symbols in the denominator. It was (is) literally: (4y^2)^1/2
 one year ago

GaiginJen Group TitleBest ResponseYou've already chosen the best response.1
Should I ask this question in a different section? It is prealgebra as I understand it. Should it be in an even lower section?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
You 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 \[ \frac{x^\frac{1}{3} \cdot \frac{y}{2}}{\frac{2}{y}\cdot \frac{y}{2}} = x^\frac{1}{3} \cdot \frac{y}{2} =\frac{x^\frac{1}{3}y}{2} \]
 one year ago
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