anonymous
  • anonymous
helpppppp
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@jjamz87
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anonymous
  • anonymous
ik its choice a and b but ineed the restriction?
DecentNabeel
  • DecentNabeel
\[\frac{x^2+4x-45}{x^2+10x+9}=\frac{x-5}{x+1}\]

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mathstudent55
  • mathstudent55
The first step is to factor the numerator and denominator.
anonymous
  • anonymous
is the restriction -9
DecentNabeel
  • DecentNabeel
\[\mathrm{Factor}\:x^2+4x-45:\quad \left(x+9\right)\left(x-5\right)\] \[=\frac{\left(x+9\right)\left(x-5\right)}{x^2+10x+9}\] \[\mathrm{Factor}\:x^2+10x+9:\quad \left(x+9\right)\left(x+1\right)\] \[=\frac{\left(x+9\right)\left(x-5\right)}{\left(x+9\right)\left(x+1\right)}\] \[=\frac{\left(x+9\right)\left(x-5\right)}{\left(x+9\right)\left(x+1\right)}\] \[\mathrm{Cancel\:the\:common\:factors:}\:x+9\] \[=\frac{x-5}{x+1}\] that is the answer
DecentNabeel
  • DecentNabeel
@jammy987 are you understand
triciaal
  • triciaal
|dw:1439493910772:dw|
triciaal
  • triciaal
you do not want factor to be 0 so restrictions are the values that will make the factor = 0
triciaal
  • triciaal
restrictions x not = -1, not = 5, not = -9
triciaal
  • triciaal
so option A is correct do not divide by 0
mathstudent55
  • mathstudent55
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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