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moongazer Group Title

question below :)

  • 2 years ago
  • 2 years ago

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  1. moongazer Group Title
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    |dw:1341855893671:dw|

    • 2 years ago
  2. moongazer Group Title
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    @myininaya could you help me ?

    • 2 years ago
  3. moongazer Group Title
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    |dw:1341856207758:dw|

    • 2 years ago
  4. Wired Group Title
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    \[\LARGE{\frac{\sqrt{(95t)^{3}}}{\sqrt[3]{(27s^{3}t^{-4}}){^2}}*(\frac{3s^{-2}}{4\sqrt[3]{t}})^{-1}}\] Hard to read, but is that the equation?

    • 2 years ago
  5. moongazer Group Title
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    No, The 5 there should be s

    • 2 years ago
  6. moongazer Group Title
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    did you get it?

    • 2 years ago
  7. Wired Group Title
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    Both s's?

    • 2 years ago
  8. moongazer Group Title
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    \[\sqrt{(9st)^3}\]

    • 2 years ago
  9. Wired Group Title
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    Is this correct? \[\LARGE{\frac{\sqrt{(9st)^{3}}}{\sqrt[3]{(27s^{3}t^{-4}}){^2}}*(\frac{3s^{-2}}{4\sqrt[3]{t}})^{-1}} \]

    • 2 years ago
  10. moongazer Group Title
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    yup :)

    • 2 years ago
  11. moongazer Group Title
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    could you help me? because I think I am getting the wrong answer

    • 2 years ago
  12. Wired Group Title
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    Blah, gotta use pen and paper for this one lol, getting some crazy answer.

    • 2 years ago
  13. moongazer Group Title
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    My teacher always gives us crazy equations :)

    • 2 years ago
  14. Wired Group Title
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    \[\LARGE{(\frac{3s^{-2}}{4\sqrt[3]{t}})^{-1} = \frac{(3s^{-2})^{-1}}{(4\sqrt[3]{t})^{-1}}} = \frac{4\sqrt[3]{t}}{3s^{-2}} = \frac{4s^{2}\sqrt[3]{t}}{3} \]

    • 2 years ago
  15. Wired Group Title
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    \[\Large{\sqrt{(9st)^{3}} = \sqrt{(9st)^{2}*(9st)} = 9st*\sqrt{9st}}\]

    • 2 years ago
  16. moongazer Group Title
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    I understand it, please continue :)

    • 2 years ago
  17. Wired Group Title
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    \[\Large{\sqrt[3]{(27s^{3}t^{-4})^{-2}} = \sqrt[3]{\frac{1}{(27s^{3}t^{-4})^{2}}}}\] Ack, gotta go!

    • 2 years ago
  18. moongazer Group Title
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    just continue it tomorrow if you want to :) what answer did you get ?

    • 2 years ago
  19. Wired Group Title
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    Didn't, that's all I did so far. Does it match up with what you have?

    • 2 years ago
  20. moongazer Group Title
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    I did another method tonight. But I already did that kind of method although but I didn't finish answering it. I'll just try answering this with that kind of way and the other way so I can check my answers :)

    • 2 years ago
  21. ganeshie8 Group Title
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    final answer i am getting as : \(4st^4\sqrt{st}\)

    • 2 years ago
  22. Wired Group Title
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    \[\Large{\sqrt[3]{\frac{1}{27^{2}(s^{3})^{2}(t^{-4})^{2}}} = \sqrt[3]{\frac{t^{8}}{9^{3}(s^{2})^{3}}} = \frac{t^{2}}{9s^{2}}\sqrt[3]{t^{2}}}\]

    • 2 years ago
  23. Wired Group Title
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    \[\LARGE{\frac{9st*\sqrt{9st}}{\frac{t^{2}}{9s^{2}}*\sqrt[3]{t^{2}}}*\frac{4s^{2}\sqrt[3]{t}}{3}} = \LARGE{\frac{108s^{5}\sqrt{9st}}{t\sqrt[3]{t}}}\] No idea if that's right or not. Didn't check every step.

    • 2 years ago
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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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