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Mathematics 7 Online
OpenStudy (anonymous):

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OpenStudy (ddcamp):

L×W = 48 L = 3W This implies that 3W×W = 48 3W²=48 W²=16 W=4 Plugging that into our first equation: L×4 = 48 L = 12 So now we have the outer perimeter of our field: |dw:1383365015722:dw|

OpenStudy (anonymous):

where is the 12 coming from, 4x3? the questin never gives a width measurement which is where im confused

OpenStudy (ddcamp):

A rectangle has an area of 48m^2. The length of the lot is three times the width. If the length is L and the width is W, we have two equations we can work with. Area = 48m² = length times width = L×W The length is three times the width → L = 3W Substituting 3W for L in the first equation, we find: 48m² = (3W)×W 48m² = 3W² 16m² = W² 4m = W We can then use that value for W in one of the original equations: L = 3W → L = 3(4) → L = 12

OpenStudy (anonymous):

ok ty

OpenStudy (anonymous):

so i understand that portion, how do i form the equation?

OpenStudy (ddcamp):

|dw:1383365605373:dw|

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