Alex_Mattucci
  • Alex_Mattucci
Please Help!!! Bernard is realizing that his dimensions are not fitting the way that he wants it to because he forgot about the deck. The deck is going to be a rectangular space that borders the pool. Also, for the pool to fit just right there must be a special product represented when either the area or volume expressions are determined. He wants you to design your own rectangular pool using polynomial expressions, including the deck to borders the pool. A. Include a diagram or sketch of your pool and deck. Your diagram must have all dimensions represented with polynomials and ensure that a special product will be used. B. Determine the polynomial expression to represent the area of land space that the pool will cover. C. Determine the area of the deck that borders the pool based on your dimensions.
Algebra
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SOLVED
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chestercat
  • chestercat
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Alex_Mattucci
  • Alex_Mattucci
@SolomonZelman Please help me with this question here!
anonymous
  • anonymous
Wow.. ok ill help you
Alex_Mattucci
  • Alex_Mattucci
OMG Thank You So much!

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anonymous
  • anonymous
yw
anonymous
  • anonymous
Width = W L= 3W + 3 D = 7 - 2L The dimensions are L X W or (3W + 3) (7-2L) B) 3W = 3 - L W = (3 - L) / 3 D = 7 - 2L D = 7 - 2(3W + 3) D = 7 - 6W - 6 D - 7 + 6 = -6W -(D -7+6) = 6W -(D - 1) = 6W (1 - D) / 6= W C) Area = L X W Volume = Area X D Substituting 3W+3 for "L" and 7 - 2L for "W" in the case of Area, you get the following: Volume = (3W + 3) (7 - 2L) D
anonymous
  • anonymous
or task 2, you have been given a lot of latitude. The three special products you might consider are: (a+b)(a+b) = a^2 + 2ab + b^2) (a-b)(a-b) = a^2 -2ab + b^2 (a-b)(a+b) = a^2 - b^2 Remember it is your design. It must have a deck surrounding it.
anonymous
  • anonymous
for*
anonymous
  • anonymous
A ) Width = W L= 3W + 3 D = 7 - 2L The dimensions are L X W or (3W + 3) (7-2L)
anonymous
  • anonymous
|dw:1449712947537:dw|
anonymous
  • anonymous
|dw:1449713257766:dw|
anonymous
  • anonymous
do you follow?
Alex_Mattucci
  • Alex_Mattucci
yes! Thank you so much for all your help!
anonymous
  • anonymous
Your welcome :)
anonymous
  • anonymous
i think thats everything is it?
anonymous
  • anonymous
:)
Alex_Mattucci
  • Alex_Mattucci
I am still not sure about B and C
anonymous
  • anonymous
ok why?
Alex_Mattucci
  • Alex_Mattucci
I looked at what you posted... Width = W L= 3W + 3 D = 7 - 2L The dimensions are L X W or (3W + 3) (7-2L) B) 3W = 3 - L W = (3 - L) / 3 D = 7 - 2L D = 7 - 2(3W + 3) D = 7 - 6W - 6 D - 7 + 6 = -6W -(D -7+6) = 6W -(D - 1) = 6W (1 - D) / 6= W C) Area = L X W Volume = Area X D Substituting 3W+3 for "L" and 7 - 2L for "W" in the case of Area, you get the following: Volume = (3W + 3) (7 - 2L) D ... I dont know what part in here is the answer to B and C
anonymous
  • anonymous
ohhh ok im sorry
Alex_Mattucci
  • Alex_Mattucci
no, it is not your fault. it is probably really obvious and i am missing it
Alex_Mattucci
  • Alex_Mattucci
oh wait, nvm. it was simply in front of me. Thank you so much for your help!
anonymous
  • anonymous
You are very welcome :D im glad i could help

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