the length of a picture frame is 3 in. greater than the width. The perimeter is less than 52 in. Describe the dimensions of the frame.
Solve the problem by writing an inequality.
2007-08-25 10:29:07 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
l=3+w
2l+2w<52, 2(3+w)+2w<52, 6+2w+2w<52
6+2w+2w<52
4w<46
w<23/2=11.5
then l <14.5
2007-08-25 10:35:54 · answer #1 · answered by Kenneth H 3 · 0⤊ 1⤋
Let W = width in inches. Then length is (W+3).
The parameter (assuming rectangular) is 2W + 2(W+3).
2W + 2(W+3) < 52
2W + 2W + 6 < 52
2W + 2W < 46
4W < 46
W < 23/2
Now for length:
(W+3) < 23/2+3 = 29/2
So you can say the width is less than 23/2 and the length is less than 29/2.
2007-08-25 10:39:05 · answer #2 · answered by Dave P 2 · 1⤊ 1⤋
p<52-3>w
2007-08-25 10:33:16 · answer #3 · answered by Anonymous 2 · 0⤊ 1⤋
L+3=w
P<52
2L+2W=P
2L+2(L+3)<52
4L+6<52
4L<46
L<11.5
W<14.5
2007-08-25 10:44:43 · answer #4 · answered by James 3 · 0⤊ 1⤋
2(x+3) + 2x < 52
2x + 6 + 2x < 52
4x <52 - 6
4x < 46
x < 11.5 (The short side)
x + 3 < 14.5 (The Long side)
2007-08-25 10:38:39 · answer #5 · answered by lenpol7 7 · 1⤊ 1⤋
x=width
x+3 = length
2x+2(x+3)<52
4x+6<52
4x<46
x<11.5
length <14.5
2007-08-25 10:37:16 · answer #6 · answered by chasrmck 6 · 1⤊ 0⤋
2(w+3) + 2w<52
2w+6+2w<52
4w+6<52
4w<46
w<11.5
2007-08-25 10:35:56 · answer #7 · answered by DanYell 3 · 1⤊ 1⤋
14.5x11.5x14.5x11.5
x=width
x+3 = length
2x+2(x+3)=52
4x+6=52
4x=46
x=11.5
length =14.5
2007-08-25 10:33:02 · answer #8 · answered by Anonymous · 0⤊ 1⤋
let length be l and w be the width.
l-w>3
2l+2w<52
2007-08-25 10:34:14 · answer #9 · answered by aviral17 3 · 0⤊ 1⤋
2(x+3)+2x<52
maybe... I am mathematically challenged
2007-08-25 10:33:59 · answer #10 · answered by Count Chocula 5 · 0⤊ 1⤋