anonymous
  • anonymous
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.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1347684283899:dw|
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
what do you have so far

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

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