anonymous
  • anonymous
Rectangle A has an area of 4 - x2. Rectangle B has an area of x2 + 2x - 8. In simplest form, what is the ratio of the area of Rectangle A to the area of Rectangle B? Show your work.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
@Hero
mayankdevnani
  • mayankdevnani
\[\huge \frac{rect.A}{rect.B}=\frac{4-x^2}{x^2+2x-8}\]
mayankdevnani
  • mayankdevnani
|dw:1369724169976:dw|

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mayankdevnani
  • mayankdevnani
\[\huge \frac{4-x^2}{(x+4)(x-2)}\]
anonymous
  • anonymous
Thank you! :) I knew it was something like that
mayankdevnani
  • mayankdevnani
can you solve it??? @SephI
anonymous
  • anonymous
Yes, I would expand 4 - x^2 and cancel the x + 4
mayankdevnani
  • mayankdevnani
correct
mayankdevnani
  • mayankdevnani
plz show your work how would you expand \(\large 4-x^2\)
mayankdevnani
  • mayankdevnani
@SephI
anonymous
  • anonymous
Okay Okay 4 - x^2 -(x - 2)(x + 2) Then my negative xs cancel
anonymous
  • anonymous
Would this be correct? http://prntscr.com/171rge
mayankdevnani
  • mayankdevnani
@SephI \[\huge 4-x^2=2^2-x^2=a^2-b^2=(a+b)(a-b)\] \[\huge (2+x)(2-x)\]
mayankdevnani
  • mayankdevnani
\[\huge \frac{(2+x)(2-x)}{(x+4)(x-2)}\]

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