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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    @jjamz87

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  2. anonymous
    • one year ago
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    ik its choice a and b but ineed the restriction?

  3. DecentNabeel
    • one year ago
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    \[\frac{x^2+4x-45}{x^2+10x+9}=\frac{x-5}{x+1}\]

  4. mathstudent55
    • one year ago
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    The first step is to factor the numerator and denominator.

  5. anonymous
    • one year ago
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    is the restriction -9

  6. DecentNabeel
    • one year ago
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    \[\mathrm{Factor}\:x^2+4x-45:\quad \left(x+9\right)\left(x-5\right)\] \[=\frac{\left(x+9\right)\left(x-5\right)}{x^2+10x+9}\] \[\mathrm{Factor}\:x^2+10x+9:\quad \left(x+9\right)\left(x+1\right)\] \[=\frac{\left(x+9\right)\left(x-5\right)}{\left(x+9\right)\left(x+1\right)}\] \[=\frac{\left(x+9\right)\left(x-5\right)}{\left(x+9\right)\left(x+1\right)}\] \[\mathrm{Cancel\:the\:common\:factors:}\:x+9\] \[=\frac{x-5}{x+1}\] that is the answer

  7. DecentNabeel
    • one year ago
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    @jammy987 are you understand

  8. triciaal
    • one year ago
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    |dw:1439493910772:dw|

  9. triciaal
    • one year ago
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    you do not want factor to be 0 so restrictions are the values that will make the factor = 0

  10. triciaal
    • one year ago
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    restrictions x not = -1, not = 5, not = -9

  11. triciaal
    • one year ago
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    so option A is correct do not divide by 0

  12. mathstudent55
    • one year ago
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    You don't want the denominator to be zero. The numerator can be zero. These two factors cannot be zero: \(x + 9 \ne 0\) \(x + 1 \ne 0\) x - 5 .can be zero because a zero in the numerator is permitted.

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