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thompsonjodi9638
03.05.2021 •
Mathematics
A rectangular garden has perimeter 66 ft and area 216 ft squared. Find the dimensions of the garden.
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Ответ:
The perimeter is then
2L + 2W = 66
The area is
L*W = 216
Solve the first equation for L
2L = 66 - 2W
L = 66/2 - 2W/2 = 33 - W
Put this into the second equation
(33 - W)*W = 216
33W - W^2 = 216
rearranging
W^2 - 33W + 216 = 0
Using the quadratic formula
[-(-33) plus or minus sqrt((-33)*(-33) - 4 * 1 * 216)]/2
[33 plus or minus sqrt(1089 - 864)]/2
[33 plus or minus sqrt(225)]/2
[33 plus or minus 15]/2
(33 + 15)/2 = 48/3 = 16 = W
(33 - 15)/2 = 18/2 = 9 = W
Checking back into the original equations
L = 24 W = 9
Step-by-step explanation:
Ответ:
Half the perimeter would be the sum of one length and one width.area = length x width.
X + y = 33 feet
Rewrite as x = 33-y
X x y = 216 feet^2
Replace x:
(33-y) x y = 216
Expand:
33y -y^2 = 216
Subtract 216 from both sides
33y -y^2 -216 = 0
Using quadratic formula solve for y
Y = 9 and 24
The dimensions would be 9 for by 24 ft
Ответ:
step-by-step explanation:
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