I need to write an inequality for this problem : The width of a rectangular field is 20ft less than the length. If the perimeter is to exceed 180ft, how wide and long should the field be? Any help would be great!
2007-01-16 08:03:02 · 5 answers · asked by kelseakaboom 2 in Science & Mathematics ➔ Mathematics
If length=L, then width=L-20
Perimeter=2*length + 2 * width
Perimeter= 2L+2*(L-20) = 4L-40
Perimeter > 180
So, 4L-40 > 180
L > 55.
2007-01-16 08:10:14 · answer #1 · answered by wiganpieeater 1 · 1⤊ 0⤋
Let the length of the field be l ft,
then the width w should be w=(l-20).
Then the perimeter 2l +2w =2l+2(l-20)=4l-40.
It is given that, 4l-40>180.
This leads to 4l >220 or l > 55ft
Then w > 55-20 or w >35 ft.
Answer: length > 55 and width > 35.
The answer is easily verified to satisfy the conditions of the problem.
2007-01-16 08:19:40 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Write the length to be w + 20 for starters. So, now you have a rectangle with sides equal to w and w + 20.
You know that the perimeter is equal to double the length plus double the width. So, you can write:
p = 2*w + 2(w + 20) = 4w + 40
You know that p > 180. So, you can write it as:
4w + 40 > 180
Solve for w to get w > 35.
So, the length must be at least 55 while the width must be at least 35.
2007-01-16 08:10:43 · answer #3 · answered by Rev Kev 5 · 0⤊ 0⤋
Let
L = length
L - 20 = Width
Perimeter Formula
2L+ 2W = P
2L + 2(L - 20) > 180
2L + 2L - 40 > 180
4L - 40 > 180
4L - 40 + 40 > 180 + 40
4L > 220
4L / 4 > 220 / 4
L > 55
- - - - - - - s-
2007-01-16 08:25:32 · answer #4 · answered by SAMUEL D 7 · 0⤊ 0⤋
x=length
x-20=width
2x+2(x-20)>180
2x+2x-40>180
4x>220
x>55
length greater than 55, width greater than 35
2007-01-16 08:10:44 · answer #5 · answered by kelsey 7 · 1⤊ 0⤋