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GaiginJen

  • 2 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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  1. Henryblah
    • 2 years ago
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    Is it 4(y)^-2 or (4y)^-2

  2. GaiginJen
    • 2 years ago
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    Neither. There were no brackets or grouping symbols in the denominator. It was (is) literally: (4y^-2)^1/2

  3. GaiginJen
    • 2 years ago
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    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?

  4. phi
    • 2 years ago
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    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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