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
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
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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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More answers

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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