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Dodo1

  • 3 years ago

curve problems?

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  1. Dodo1
    • 3 years ago
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  2. Luigi0210
    • 3 years ago
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    Do you know what the graph looks like?

  3. Dodo1
    • 3 years ago
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    No, I have no idea!

  4. Luigi0210
    • 3 years ago
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    Well it might be a good idea to sketch a graph

  5. Dodo1
    • 3 years ago
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    would it possible to solve the problem without a graph?

  6. Dodo1
    • 3 years ago
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    would it be

  7. Luigi0210
    • 3 years ago
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    Well yes but it be a bit harder since we don't know which graph is higher

  8. Dodo1
    • 3 years ago
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    oh i see... how do you draw a graph

  9. Luigi0210
    • 3 years ago
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    wait, are you finding total area or net area?

  10. Dodo1
    • 3 years ago
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    it says find the area between the curves?

  11. Luigi0210
    • 3 years ago
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    If I'm not mistaken then the graphs should look like this when graphed |dw:1366855841011:dw|

  12. Dodo1
    • 3 years ago
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    sin and cos graph?

  13. Luigi0210
    • 3 years ago
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    Uhm no, the ends of the graph go off into infinity after they have intersected

  14. Dodo1
    • 3 years ago
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    Oh.

  15. Luigi0210
    • 3 years ago
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    Now, do you notice anything about the two graphs?

  16. Dodo1
    • 3 years ago
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    -they get smaller and smaller? -they intersects

  17. Luigi0210
    • 3 years ago
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    Well yes but also they are reflections of one another

  18. Dodo1
    • 3 years ago
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    Yes,

  19. Luigi0210
    • 3 years ago
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    So that means that whatever area on graph has the other has the opposite area. Such as -x and x.

  20. Dodo1
    • 3 years ago
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    Yes,

  21. Luigi0210
    • 3 years ago
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    So that means the area would be zero

  22. Dodo1
    • 3 years ago
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    the answer is zero?

  23. Dodo1
    • 3 years ago
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    i thought i need more calculations

  24. Luigi0210
    • 3 years ago
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    Well you could if you want but the answer would still be zero. These two graphs are going to cancel each others area out

  25. Dodo1
    • 3 years ago
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    i entered zero into the answer but it tells me that it was wrong

  26. Luigi0210
    • 3 years ago
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    Well let's solve it right now, do you know how to set it up as an integral?

  27. Dodo1
    • 3 years ago
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    No, I have no idea!!

  28. Dodo1
    • 3 years ago
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    @Luigi0210

  29. Luigi0210
    • 3 years ago
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    Sorry I was trying to evaluate the problem

  30. Luigi0210
    • 3 years ago
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    but anyways we would set up the problem like this: \[\int\limits_{0}^{9} (x^3-14x^2+45x)-(-x^3+14x^2-45x) dx\]

  31. Zarkon
    • 3 years ago
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    I'm betting that they want you to set up two different integrals...based on the two enclosed regions.

  32. Dodo1
    • 3 years ago
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    so between...= minus. i got it

  33. Luigi0210
    • 3 years ago
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    Setting up two equations would take a bit longer

  34. Zarkon
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=area+between+x^3-14x^2%2B45x+and+-x^3%2B14x^2-45x

  35. Dodo1
    • 3 years ago
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    whats next step?

  36. Luigi0210
    • 3 years ago
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    The answer I got was 253/2..

  37. Luigi0210
    • 3 years ago
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    *243/2

  38. Dodo1
    • 3 years ago
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    the answer was wrong but thanks for ur help!

  39. Luigi0210
    • 3 years ago
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    I'm sorry I couldn't help you solve it..

  40. Dodo1
    • 3 years ago
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    No thanks for your time!

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