for this problem...The length of a rectangle is 8 in. more than twice its width. If the perimeter of the rectangle is 64 in., find the width of the rectangle. I am getting 8 inches. Does this sound correct to anybody?
2006-06-23 14:36:28 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
2[w+ {2w+8}]= 64
3w+8 = 32
w= 8
your are right
2006-06-23 14:40:12 · answer #1 · answered by Vivek 4 · 0⤊ 0⤋
Yes, it is correct. Let say that the width of the rectangule is denoted by x. Than the other side is 2x+8.
The perimeter is the addition of the length of all edges.
2*(x+2x+8) = 64 => 3x+8=32 => 3x=24 => x=8
The width of rectangle is 8.
2006-06-23 21:46:01 · answer #2 · answered by etropus 1 · 0⤊ 0⤋
8" is correct for the width, 24" for the length.
2006-06-23 22:00:14 · answer #3 · answered by davidosterberg1 6 · 0⤊ 0⤋
I think you should think a little more. Because it's very easy. See:
We will have this: x(width) so 2x+8 is length
So x+2x+8=64/2
So x will be 8. you are correct.
2006-06-23 21:43:11 · answer #4 · answered by phipha2006 4 · 0⤊ 0⤋
l = 2w + 8
p = 2(l + w)
64 = 2((2w + 8) + w)
64 = 2(2w + 8 + w)
64 = 2(3w + 8)
32 = 3w + 8
3w = 24
w = 8
l = 2w + 8
l = 2(8) + 8
l = 16 + 8
l = 24
w = 8
l = 24
The width is 8 inches so yes you are correct.
2006-06-23 22:27:46 · answer #5 · answered by Sherman81 6 · 0⤊ 0⤋
Let x = length, let y = width.
Perimeter = 2x + 2y
x = 2y + 8.
2x + 2y = 64
Subsituting x from the 1st equation into the second:
2(2y + 8) + 2y = 64
4y + 16 + 2y = 64
6y + 16 = 64
6y = 48
y = 8
You are correct.
2006-06-23 21:43:22 · answer #6 · answered by I Know Nuttin 5 · 0⤊ 0⤋
length =24 in
width = 8 in
so you are correct
2006-06-23 23:59:59 · answer #7 · answered by abhinav 2 · 0⤊ 0⤋
i got 12. 64=2w+2L; L=2w+8; therefore 64=2w+4w+16 Solve for w and get 8.
2006-06-23 21:43:41 · answer #8 · answered by bigdog2all2 1 · 0⤊ 0⤋
l-2b=8
2l+2b=64
l=32-b
32-b-2b=8
3b=24
b=8
2006-06-23 21:43:16 · answer #9 · answered by Vijay V 1 · 0⤊ 0⤋