33.
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.
A) 10 in. B) 7 in. C) 8 in. D) 9 in.
2006-09-20 02:38:55 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
it is C) 8
64/2=32
32-8=24
24-8=16
2*8=16
makes the answer 8 with the length 24
2006-09-20 02:43:51 · answer #1 · answered by autbrat 2 · 0⤊ 0⤋
Let the width of the rectangle be x.
Then, according to the question,
length of the rectangle = 2x+8
Now, perimeter of a rectangle is given by 2(length + width)
that is, 2[(2x+8) + x]
which is given to be 64 in.
Thus, 2[(2x+8) + x] = 64
2(3x + 8) = 64
6x + 16 = 64
6x = 48
x = 8
Thus, width of the rectangle = 8 in.
You can also get the length as:
2x+8 = 2 * 8 + 8
=24 in.
2006-09-20 03:25:26 · answer #2 · answered by noesis 2 · 0⤊ 0⤋
C
2w + (2w+8)*2 = 64
8+8+16+8+16+8=64
2006-09-20 02:45:14 · answer #3 · answered by speed math 2 · 0⤊ 0⤋
The answer is C. 2y+2x=64, y=8+2x, so by substitution, x=8.
2006-09-20 02:47:29 · answer #4 · answered by jedd c 3 · 0⤊ 0⤋
the length is 2w+8
the width is w
so the perimeter equation would be
2(2w+8)+2w=64
4w+16+2w=64
6w+16=64
6w=48
w=8in
c
2006-09-20 02:45:18 · answer #5 · answered by Anonymous · 1⤊ 0⤋
L=2x+8 solve make a diagram
2006-09-20 02:41:32 · answer #6 · answered by Olivia 1 · 1⤊ 0⤋
6x + 16 = 64
6x = 48
x = 8
2006-09-20 02:58:43 · answer #7 · answered by Anonymous · 0⤊ 0⤋
l = 2w+8
l+l+w+w = 2l+2w = 64
l+w = 32
(2w+8)+w = 3w+8 = 32
3w = 24
w = 8
answer is (c)
2006-09-20 02:41:57 · answer #8 · answered by Anonymous · 1⤊ 0⤋