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anonymous

  • one year ago

I'm having trouble with some partial fraction decomposition problems. I have to write out the form of the partial fraction of the function Q = limit(1 to 10) 1x/(x^2+6x+9) dx. Determine the numerival values of the coefficients, A and B ----A/denominator + B/denominator

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  1. triciaal
    • one year ago
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    start by finding the factors of the denominator

  2. triciaal
    • one year ago
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    |dw:1439752537963:dw|

  3. anonymous
    • one year ago
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    I did that and got (x+3)^2

  4. anonymous
    • one year ago
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    thats the answer

  5. anonymous
    • one year ago
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    Is it really?? I thought I had to reduce after that point

  6. anonymous
    • one year ago
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    And what about the limit? The homework program I'm using says that's wrong

  7. anonymous
    • one year ago
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    well it is wrong stop looking for handouts duhhh

  8. triciaal
    • one year ago
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    do you need to integrate the function and find value of 10 minus value of 1?

  9. anonymous
    • one year ago
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    I'm not, I just need help getting to the next point. I had factored it, then put it in the form \[\frac{ 1x }{ (x+3)(x+3) } = \frac{ A }{ (x+3) } + \frac{ B }{ (x+3) }\]

  10. anonymous
    • one year ago
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    Then I guess I'm supposed to find where x = 0 and solve for A then insert to find B, but I'm having a tough time just isolating just one, normally the factors wouldn't be the same thing

  11. anonymous
    • one year ago
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    I've just been struggling on this problem for a while now

  12. triciaal
    • one year ago
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    @zepdrix

  13. triciaal
    • one year ago
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    how do you know the denominator for A or B is not (x +3)^2

  14. triciaal
    • one year ago
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    |dw:1439753193971:dw|

  15. anonymous
    • one year ago
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    |dw:1439753340124:dw|

  16. anonymous
    • one year ago
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    Wow drawing is hard on this

  17. ganeshie8
    • one year ago
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    Heyy

  18. anonymous
    • one year ago
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    Hi

  19. ganeshie8
    • one year ago
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    you need to square the second one, try this \[\frac{ 1x }{ (x+3)(x+3) } = \frac{ A }{ (x+3) } + \frac{ B }{ (x+3)^{\color{red}{2}} }\]

  20. anonymous
    • one year ago
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    B = 1x - A?

  21. misty1212
    • one year ago
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    then \[A(x+3)^2+B(x+3)=x\] is the next step you need the square when you have repeated factors in the denominator

  22. misty1212
    • one year ago
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    oh wait i think i made a mistake there

  23. misty1212
    • one year ago
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    when you add you should get \[A(x+3)+B\] as your numerator, so \[A(x+3)+B=x\]

  24. misty1212
    • one year ago
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    multiply out, get \[Ax+3A+B=x\]telling you \(A=1\)

  25. ganeshie8
    • one year ago
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    and you can get the value of \(B\) easily, simply plugin \(x=-3\)

  26. misty1212
    • one year ago
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    and \[3A+B=0\]so \[3+B=0\\ B=-3\]

  27. misty1212
    • one year ago
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    or @ganeshie8 way works too

  28. anonymous
    • one year ago
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    Oh my goodness thank you guys so much!

  29. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

  30. anonymous
    • one year ago
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    How do I close this problem? First question I've asked on here...

  31. misty1212
    • one year ago
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    you should see an option in the bottom for "close question"

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