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SephI

  • one year ago

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.

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

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

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

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

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

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

  8. mayankdevnani
    • one year ago
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    correct

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

  10. mayankdevnani
    • one year ago
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    @SephI

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

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

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

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

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