maths f.2 before 10:45!!

Let width be x cm,length be 5x m
5x+5x+x+x=48
12x=48
x=4
So its lenght is 5*4=20cm
width is 5cm

wait me.......
to be continuous

2007-12-17 22:52:04 補充：
Let 50-cent coins has x and 2 dollar coins has (22-x)0.5x+2(22-x)=260.5x+44-2x=261.5x=18x=12So 50-cent coins has 12 and 2 dollar coins has(22-12)=10

1. let x be its width, and 5x be its length.
2x+2(5x) = 48
12x =48
x =4
Therefore, the length and the width of the rectangle are 5(4) = 20cm and 4cm respectively.

2. let x be the number of 50-cent coins and (22 - x) be the number of 2 dollar coins.
0.5x + 2 (22 - x) = 26
0.5x + 44 - 2x =26
3 / 2 x = 18
x = 12

Therefore, he has 12 50-cent coins and 10 2 dollar coins.

2007-12-18 22:15:06 補充：
I hope that it may help you ^^, I have tried my best to help

1. let x be width
2(x+5x)=48 x=4cm(width) then length=4x5=20cm

2. let x be number of cent coins and y be number of dollar coins
0.5x+2y=26-----(1)
x+y=22-------(2)
from (2) y=22-x----(3)
sub(3)into (1) y= 10 x=12

1)Let x cm be its width.
Then its length is 5x cm.
2(5x+x)=48
6x=24
x=4
So its width is 4 cm. Its length=5*4 cm=20 cm

2)Let y be the number of 2 dollar coins he has.
Then he has (22-y) 50-cent coins.
2y+0.5(22-y)=26
2y+11-0.5y=26
1.5y+11=26
1.5y=15
y=10
So he has ten $2 dollar coins and (22-10)=12 50-cent coins.

1. length=20 width =4
2.10個 $2 coins 12個 50-cent coins

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