width ---- w
length ---- 3w - 2
2w + 2(3w-2) = 156
2w + 6w - 4 = 156
8w = 160
w = 20
park is 20 by 58
the perimeter of a dog park is 156 ft. If the length of the park is 2 ft. less than 3 times the width, what are the dimensions of the dog park?
6 answers
The dimensions of the dog part are 20ft by 56 ft.
Set up the problem
w + w + ( 3w - 2 ) + ( 3w - 2 ) = 156
Set up the problem
w + w + ( 3w - 2 ) + ( 3w - 2 ) = 156
P = 2L + 2W
156 = 2(3W - 2) + 2W
156 = 6W - 4 + 2W
160 = 8W
20 = W
~~~~~~~~~~~~~~~~`
L = 58
156 = 2(3W - 2) + 2W
156 = 6W - 4 + 2W
160 = 8W
20 = W
~~~~~~~~~~~~~~~~`
L = 58
sorry I post 56 instead of 58, the dimensions are 20ft by 58ft
How did you get the length to be 58. Where did that number come from?
ohhhhh nvrm i got
1 answer