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anonymous
 one year ago
@Miracrown Tyler has a rectangular garden that measures 10 m wide by 13 m long. He wants to increase the area to 208 m2 by increasing the width and length by the same amount. What will be the length (longer dimension) of the new garden?
A.
14 m wide
B.
15 m wide
C.
16 m wide
D.
17 m wide
anonymous
 one year ago
@Miracrown Tyler has a rectangular garden that measures 10 m wide by 13 m long. He wants to increase the area to 208 m2 by increasing the width and length by the same amount. What will be the length (longer dimension) of the new garden? A. 14 m wide B. 15 m wide C. 16 m wide D. 17 m wide

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Miracrown
 one year ago
Best ResponseYou've already chosen the best response.1What do you think for this one?

hwyl
 one year ago
Best ResponseYou've already chosen the best response.0start by learning to write down the formula for area of the shape you are being asked to solve.

Miracrown
 one year ago
Best ResponseYou've already chosen the best response.1(10+x)(13+x)= 208 is the equation for area that is quadratic in x solve for x

hwyl
 one year ago
Best ResponseYou've already chosen the best response.0we have \(area = length \times width \rightarrow A = L \times W \) tentatively, the dimensions are L = 13 m and W = 10 m, so currently you have \( A = 13m \times 10m = 130~ m^2\) then you are told that that its new product (area) is increased to \(208~m^2 \) the question is, how much dimension should be added to both the length and width so that the new area reflects it considering that this dimension is the same for the length and the width.

hwyl
 one year ago
Best ResponseYou've already chosen the best response.0understanding the problem is the key, not just throwing formulaes like a cookie crumb.

hwyl
 one year ago
Best ResponseYou've already chosen the best response.0the word is ADDED is a mathematical procedure for addition \(A = (L+d) \times (W + d) \rightarrow (13~m+d~m) \times (10~m + d~m) = 208~m^2 \)

hwyl
 one year ago
Best ResponseYou've already chosen the best response.0the good thing about physical dimensions is that you always consider their units as they are a measurement, so if you try to solve and leave out the measurements (when you have many and different measurements given) then you run into trouble of not having the correct description.
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