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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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!

Pre-Algebra
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Is it 4(y)^-2 or (4y)^-2
Neither. There were no brackets or grouping symbols in the denominator. It was (is) literally: (4y^-2)^1/2
Should I ask this question in a different section? It is pre-algebra as I understand it. Should it be in an even lower section?

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  • phi
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 \[ \frac{x^\frac{1}{3} \cdot \frac{y}{2}}{\frac{2}{y}\cdot \frac{y}{2}} = x^\frac{1}{3} \cdot \frac{y}{2} =\frac{x^\frac{1}{3}y}{2} \]

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