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izhangy

  • one year ago

what is 4^(3/2)? How do you show work for it?

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  1. Luigi0210
    • one year ago
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    It's the same as: \[\sqrt[2]{4^3}\]

  2. Luigi0210
    • one year ago
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    What's 4^3?

  3. izhangy
    • one year ago
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    oh alright, 4^3=64...

  4. klimenkov
    • one year ago
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    What is \(\sqrt 4\) ?

  5. Luigi0210
    • one year ago
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    Yes, now what's \[\sqrt{64}\]

  6. Luigi0210
    • one year ago
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    @klimenkov 's method would of worked too

  7. izhangy
    • one year ago
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    8

  8. Luigi0210
    • one year ago
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    And that's your answer

  9. izhangy
    • one year ago
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    alrighty what is 8^(2/3)?

  10. Luigi0210
    • one year ago
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    what's 8^2?

  11. izhangy
    • one year ago
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    64

  12. Luigi0210
    • one year ago
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    and what's \[\sqrt[3]{64}\]

  13. izhangy
    • one year ago
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    ohhhh 4

  14. Luigi0210
    • one year ago
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    Yup!

  15. izhangy
    • one year ago
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    thanks dood

  16. klimenkov
    • one year ago
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    I think i would be easier to compute \(8^{\frac13}\) first.

  17. izhangy
    • one year ago
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    ohh dood 1 more please

  18. izhangy
    • one year ago
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    what is √(8)/2^(1/2)?

  19. Luigi0210
    • one year ago
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    Is that 1/2 with the whole equation or just 2?

  20. izhangy
    • one year ago
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    sq root 8 divided by 2 to the half power

  21. Luigi0210
    • one year ago
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    Well the 1/2 makes that 2 a rad 2: \[\frac{\sqrt{8}}{\sqrt{2}}\] Now what does \[\sqrt{8}\] simplify into?

  22. izhangy
    • one year ago
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    2√(2)

  23. Luigi0210
    • one year ago
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    Yup! \[\frac{2\sqrt{2}}{\sqrt{2}}\] Can you finish it?

  24. izhangy
    • one year ago
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    2, right?

  25. Luigi0210
    • one year ago
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    Mhm :)

  26. izhangy
    • one year ago
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    thanks bro

  27. izhangy
    • one year ago
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    do you know how to calculate average rate of change?

  28. Luigi0210
    • one year ago
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    Anytime. Average rate of change? As in calculus?

  29. izhangy
    • one year ago
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    yes :D

  30. Luigi0210
    • one year ago
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    Sorry, I haven't had calculus for a while :/ just open a new question and call Nincompoop, he's great at calculus

  31. izhangy
    • one year ago
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    ah alrighty thanks dood.

  32. Luigi0210
    • one year ago
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    Yup, and if he's not on call dan815

  33. izhangy
    • one year ago
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    kk

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