Asked by Anonymous
The length of a rectangle is 7 centimeters less than twice its width. Its area is 72 square meters. Find the dimensions of the rectangle.
Answers
Answered by
Reiny
width --- x
length --- 2x-7
x(2x-7) = 72
2x^2 - 7x - 72 = 0
(x-8)(2x+9) = 0
x = 8 or x = -9/2, rejecting the negative width
the rectangle is 8 by 9
length --- 2x-7
x(2x-7) = 72
2x^2 - 7x - 72 = 0
(x-8)(2x+9) = 0
x = 8 or x = -9/2, rejecting the negative width
the rectangle is 8 by 9
Answered by
Chol
Let l be 2x -7
w = x
A = lw
72 = (2x-7)x
72 = 2x^2 -7x
2x^2 -7x -72 =0
(2x+9)(x-8) = 0
x -8 = 0
x = 8
l = 2x-7
l = 2(8) -7
l = 16-7= 9
L = 9 m and w =8 m
Answered by
Ms. Sue
Aren't the dimensions?
width --- x
length --- 2x-0.07
width --- x
length --- 2x-0.07
Answered by
Reiny
good catch Ms Sue.
so
x(2x-.07) = 72
2x^2 - .07x - 72 = 0
x = (.07 ± √576.0049)/4
= 6.017 or a negative
width = 6.017
length = 11.965
check: area = 6.017(11.965) = 71.993 , not bad
so
x(2x-.07) = 72
2x^2 - .07x - 72 = 0
x = (.07 ± √576.0049)/4
= 6.017 or a negative
width = 6.017
length = 11.965
check: area = 6.017(11.965) = 71.993 , not bad
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