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anonymous

  • 4 years ago

Can someone how to do 3^3/2 by hand?

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  1. anonymous
    • 4 years ago
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    explain**

  2. anonymous
    • 4 years ago
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    or any x^(y/z)

  3. UnkleRhaukus
    • 4 years ago
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    is that \[3^{3/2}\] or\[(3^3)/2\]

  4. anonymous
    • 4 years ago
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    the first one.. 3^(3/2)

  5. campbell_st
    • 4 years ago
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    so its really (3^3)^1/2 so do the inner part 1st (27)1/2 \[\sqrt{27}\] \[\sqrt{27}=\sqrt{9 \times3} =\sqrt{3^2 \times 3} = 3\sqrt{?}\]

  6. campbell_st
    • 4 years ago
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    oops ? = 3

  7. anonymous
    • 4 years ago
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    3sqrt3?

  8. sasogeek
    • 4 years ago
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    \[\ 3^\frac{3}{2}\sqrt{3^3}=\sqrt{9}=3 \]

  9. anonymous
    • 4 years ago
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    3^ 3/2 = surd 3^3 used, indices law.. surd 3^3 = surd 27 \[\sqrt{27}\] = \[\sqrt{(9)(3)}\] = \[3\sqrt{3}\]

  10. anonymous
    • 4 years ago
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    So what if it was 3^5/3?

  11. sasogeek
    • 4 years ago
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    \[\ \sqrt{27}=\sqrt{\text{9 x 3}} = \sqrt{9} \text{ x } \sqrt{3}= 9\sqrt{3} \]

  12. sasogeek
    • 4 years ago
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    I'm so slow at latex :(

  13. sasogeek
    • 4 years ago
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    \[\huge \sqrt[n]{A^b}=A^\frac{b}{n}\]

  14. anonymous
    • 4 years ago
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    for 3^(5/3) = \[\sqrt[3]{3^5}\] = \[\sqrt[3]{243}\] = \[\sqrt[3]{(27)(9)}\] = \[3 \sqrt[3]{9}\]

  15. sasogeek
    • 4 years ago
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    \[\huge note \ that \ \sqrt[2]{A}=\sqrt{A} \]

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