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anonymous

  • one year ago

help me

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  1. anonymous
    • one year ago
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    @MadisonEmery

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  2. anonymous
    • one year ago
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    What do you think it is?

  3. anonymous
    • one year ago
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    c

  4. anonymous
    • one year ago
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    am i right

  5. anonymous
    • one year ago
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    i am pretty sure c is correct but let me check 1 last time

  6. anonymous
    • one year ago
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    what is the x at the bottom of the problem

  7. anonymous
    • one year ago
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    idk lol

  8. welshfella
    • one year ago
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    Note: as x approaches 1 the denominator approaches zero

  9. welshfella
    • one year ago
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    you can use l'hopitals theorem to do this find the derivative of the numerator and the denominator and plug in x = 1

  10. welshfella
    • one year ago
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    are you familiar with this?

  11. anonymous
    • one year ago
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    \[\frac{x^3+5 x^2+3 x-9}{x-1}=x^2+6 x+9 \]When x = 0, x^2+6 x+9=9

  12. anonymous
    • one year ago
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    omg to many people helping lol this is consufing

  13. welshfella
    • one year ago
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    c is not right

  14. anonymous
    • one year ago
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    ok then what is

  15. welshfella
    • one year ago
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    have you done l'hopitals rule in calculus?

  16. anonymous
    • one year ago
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    i have not

  17. welshfella
    • one year ago
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    oh well i can't think of any other way to do this. There probably is though...

  18. anonymous
    • one year ago
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    man i just need to answer this question is there aother way to help with this

  19. welshfella
    • one year ago
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    oh ok if we factor the numerators we get (x - 1)(x^2 + 6x + 9) ------------------ (x - 1) so the x-1 cancels out to give x^2 + 6x + 9

  20. welshfella
    • one year ago
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    now plug in x = 1 into x^2 + 6x + 9 and see what you get

  21. anonymous
    • one year ago
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    i got 16

  22. welshfella
    • one year ago
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    thats the answer

  23. welshfella
    • one year ago
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    ok?

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spraguer (Moderator)
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