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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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KE8717504
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|dw:1347684283899:dw|
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@jim_thompson5910
jim_thompson5910
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what do you have so far
KE8717504
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I'm just not sure how to work this.
KE8717504
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or how to go about starting it?
jim_thompson5910
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Use the idea that
Area = Length * Width
or
A = LW
Solve for W
W = A/L
jim_thompson5910
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Then make the proper substitutions. Then divide and simplify.
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@jim_thompson5910 this is what i get when i do that. 2ab^2+5a^2+21ab+8b
KE8717504
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That is diving the Area by the length
jim_thompson5910
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simplify
(5a^2+21ab+4b^2)/(2a+8b)
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5a+b/2 ?
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would that make my width?
jim_thompson5910
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Can you factor 5a^2+21ab+4b^2 at all?
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yes into (a+4b)(5a+1b) right?
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but u still get the same answer right 5a+b/2 right?
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@jim_thompson5910
Algebraic!
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It's right.
jim_thompson5910
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Did you mean \[\Large \frac{5a+b}{2}\] ?
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@jim_thompson5910 yea
jim_thompson5910
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then yes, you are correct
KE8717504
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I should get a learners metal on that one :0
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Thanks !!