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KE8717504

  • 3 years ago

consider the rectangle below. It's area is 5a^2+21ab+4b^2 and it's length is 2a+8b. find it's width w, in terms of a and b, by dividing it's area by it's length.

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  1. KE8717504
    • 3 years ago
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    |dw:1347684283899:dw|

  2. KE8717504
    • 3 years ago
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    @jim_thompson5910

  3. jim_thompson5910
    • 3 years ago
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    what do you have so far

  4. KE8717504
    • 3 years ago
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    I'm just not sure how to work this.

  5. KE8717504
    • 3 years ago
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    or how to go about starting it?

  6. jim_thompson5910
    • 3 years ago
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    Use the idea that Area = Length * Width or A = LW Solve for W W = A/L

  7. jim_thompson5910
    • 3 years ago
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    Then make the proper substitutions. Then divide and simplify.

  8. KE8717504
    • 3 years ago
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    @jim_thompson5910 this is what i get when i do that. 2ab^2+5a^2+21ab+8b

  9. KE8717504
    • 3 years ago
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    That is diving the Area by the length

  10. jim_thompson5910
    • 3 years ago
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    simplify (5a^2+21ab+4b^2)/(2a+8b)

  11. KE8717504
    • 3 years ago
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    5a+b/2 ?

  12. KE8717504
    • 3 years ago
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    would that make my width?

  13. jim_thompson5910
    • 3 years ago
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    Can you factor 5a^2+21ab+4b^2 at all?

  14. KE8717504
    • 3 years ago
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    yes into (a+4b)(5a+1b) right?

  15. KE8717504
    • 3 years ago
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    but u still get the same answer right 5a+b/2 right?

  16. KE8717504
    • 3 years ago
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    @jim_thompson5910

  17. Algebraic!
    • 3 years ago
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    It's right.

  18. jim_thompson5910
    • 3 years ago
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    Did you mean \[\Large \frac{5a+b}{2}\] ?

  19. KE8717504
    • 3 years ago
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    @jim_thompson5910 yea

  20. jim_thompson5910
    • 3 years ago
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    then yes, you are correct

  21. KE8717504
    • 3 years ago
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    I should get a learners metal on that one :0

  22. KE8717504
    • 3 years ago
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    Thanks !!

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