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SephI Group Title

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.

  • one year ago
  • one year ago

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  1. SephI Group Title
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    @Hero

    • one year ago
  2. mayankdevnani Group Title
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    \[\huge \frac{rect.A}{rect.B}=\frac{4-x^2}{x^2+2x-8}\]

    • one year ago
  3. mayankdevnani Group Title
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    |dw:1369724169976:dw|

    • one year ago
  4. mayankdevnani Group Title
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    \[\huge \frac{4-x^2}{(x+4)(x-2)}\]

    • one year ago
  5. SephI Group Title
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    Thank you! :) I knew it was something like that

    • one year ago
  6. mayankdevnani Group Title
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    can you solve it??? @SephI

    • one year ago
  7. SephI Group Title
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    Yes, I would expand 4 - x^2 and cancel the x + 4

    • one year ago
  8. mayankdevnani Group Title
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    correct

    • one year ago
  9. mayankdevnani Group Title
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    plz show your work how would you expand \(\large 4-x^2\)

    • one year ago
  10. mayankdevnani Group Title
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    @SephI

    • one year ago
  11. SephI Group Title
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    Okay Okay 4 - x^2 -(x - 2)(x + 2) Then my negative xs cancel

    • one year ago
  12. SephI Group Title
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    Would this be correct? http://prntscr.com/171rge

    • one year ago
  13. mayankdevnani Group Title
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    @SephI \[\huge 4-x^2=2^2-x^2=a^2-b^2=(a+b)(a-b)\] \[\huge (2+x)(2-x)\]

    • one year ago
  14. mayankdevnani Group Title
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    \[\huge \frac{(2+x)(2-x)}{(x+4)(x-2)}\]

    • one year ago
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