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anonymous

  • 5 years ago

And you WANT another problem?

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  1. anonymous
    • 5 years ago
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    Let me attach the one I am doing now that doesn't have an answer attached to it.

  2. amistre64
    • 5 years ago
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    now that I know what im looking at yes lol

  3. anonymous
    • 5 years ago
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  4. amistre64
    • 5 years ago
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    thats the exact same one; only with a sqrt7 lol

  5. anonymous
    • 5 years ago
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    ok, there you go. Let me try to work it too and see what I come up with and then you can let me know where I'VE SCREWED UP

  6. anonymous
    • 5 years ago
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    I know it's the same one basically, but it doesn't matter, cuz I keep getting them wrong consistently!

  7. anonymous
    • 5 years ago
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    cheat like crazy: use the computer.

  8. amistre64
    • 5 years ago
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    we know the integral goes to \[s+s^{-1/2}\] tho so thats a start

  9. anonymous
    • 5 years ago
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    43 to go!

  10. amistre64
    • 5 years ago
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    when we use the bottom limit we get: \[1-2 = -1 right?\]

  11. amistre64
    • 5 years ago
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    \[s -2s^{-1/2}\] i meant to type

  12. amistre64
    • 5 years ago
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    \[s - \frac{2}{\sqrt{s}}+1 \rightarrow s=\sqrt{7}\] is?

  13. amistre64
    • 5 years ago
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    \[\sqrt{7} -\frac{2}{\sqrt[4]{7}}+1\]

  14. amistre64
    • 5 years ago
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    essentally its just: \[F(\sqrt{7})-F(1)\] \[F(s) = s-{2\over\sqrt{s}}\]

  15. anonymous
    • 5 years ago
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    ok, you went WAYYYYYY too fast for me! I just got F(x)! Let me work this and see if I can get what you got, ok?

  16. amistre64
    • 5 years ago
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    ok.....

  17. anonymous
    • 5 years ago
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    i just successfully got sqrt7 - 2/ 4th root 7 for F(sqrt7). That's where I am in the figuring. Now for F(1)...

  18. amistre64
    • 5 years ago
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    yay!!

  19. amistre64
    • 5 years ago
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    and for shorthand: 4rt(7) is easier to read thru

  20. anonymous
    • 5 years ago
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    ok, then here's my answer: Let me get the equation thing and do it that way.

  21. anonymous
    • 5 years ago
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    \[\sqrt{7}-2/\sqrt[4]{7}+1\]

  22. amistre64
    • 5 years ago
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    thats it

  23. anonymous
    • 5 years ago
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    Is that right?

  24. anonymous
    • 5 years ago
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    YAY!!! Couldn't have done it without you! Did I tell you I have been working on this particular problem (#31) since noon today?

  25. amistre64
    • 5 years ago
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    yeah, :)

  26. amistre64
    • 5 years ago
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    lets try one that is not 'exactly' the same lol

  27. anonymous
    • 5 years ago
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    ok, so does this mean, now that you have taught me how to send you my problems, that you are willing to look at any of them I cannot do on my own (as long as you are online and available with the time?)?

  28. amistre64
    • 5 years ago
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    yeah, ow that i aint gotta read your mind it should be easier ;)

  29. anonymous
    • 5 years ago
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    That particular problem will have all the same types with the same integrand. That's the point, to let ua do the same ones over and over to get the precise idea of how top do them. I will see if the next problem is different and let you know, ok?

  30. amistre64
    • 5 years ago
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    interactmath.com is just like your math program except its free. you can practice on it all you like.

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is replying to Can someone tell me what button the professor is hitting...

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