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anonymous

  • 5 years ago

who can help me with calculus??!!

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  1. anonymous
    • 5 years ago
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    I can

  2. anonymous
    • 5 years ago
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    ok it has tpo do with integrals

  3. anonymous
    • 5 years ago
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    ok

  4. anonymous
    • 5 years ago
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    \[\int\limits\limits_{1}^{7} 1/\sqrt[3]{t} dt\]

  5. anonymous
    • 5 years ago
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    if \[f(x)=1/\sqrt[3]{t}\] \[F(x)=(3/2)\sqrt[3]{t}^2\] The result is F(7) - F(1) Calculate that for yourself

  6. anonymous
    • 5 years ago
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    ok thank you!

  7. anonymous
    • 5 years ago
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    hey you still there?

  8. Jit4won
    • 5 years ago
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    I will help you

  9. anonymous
    • 5 years ago
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    ok

  10. anonymous
    • 5 years ago
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    \[\int\limits_{3}^{b} 3x ^{2} dx = 37 \]

  11. anonymous
    • 5 years ago
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    find the value of b?

  12. Jit4won
    • 5 years ago
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    \[\int\limits_{3}^{b}\]\[\int\limits_{}^{}3x ^{2}dx = 3x^{3}/3 = x^3\]

  13. Jit4won
    • 5 years ago
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    substituting the limits we get \[b^3 - 3 ^3 \]

  14. Jit4won
    • 5 years ago
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    now \[b^3 - 3^3 = 37 \]

  15. Jit4won
    • 5 years ago
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    \[b^3 =27+37\]

  16. Jit4won
    • 5 years ago
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    \[b^3=64 \]

  17. Jit4won
    • 5 years ago
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    \[ b^3 = 4^3 \]

  18. anonymous
    • 5 years ago
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    so what happen to the 37?

  19. Jit4won
    • 5 years ago
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    b=4 is your answer

  20. Jit4won
    • 5 years ago
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    37 got added with 27

  21. Jit4won
    • 5 years ago
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    37+ 27 = 54

  22. Jit4won
    • 5 years ago
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    oh* 37+ 27= 64

  23. anonymous
    • 5 years ago
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    ahh i get it now thnaks!

  24. Jit4won
    • 5 years ago
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    You may email me or skype

  25. anonymous
    • 5 years ago
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    can i ask you another question? i have a final tomorrow lol

  26. Jit4won
    • 5 years ago
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    yeah

  27. Jit4won
    • 5 years ago
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    Do u use skype? so that i could instruct you

  28. anonymous
    • 5 years ago
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    yes i have skype

  29. Jit4won
    • 5 years ago
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    Jit4won

  30. Jit4won
    • 5 years ago
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    call me

  31. anonymous
    • 5 years ago
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    ok give me a sec

  32. anonymous
    • 5 years ago
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    just added you

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spraguer (Moderator)
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