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标题: ppII第一套一道数学题求解 [打印本页]

作者: xiaoshang 时间: 2016-8-16 10:09:36 标题: ppII第一套一道数学题求解

The length of rectangle B is 10 percent less than the length of rectangle A,and the width of rectangle B is 10 percent greater than the width of rectangle A
Quantity A: the area of rectangle A
Quantity B: the area of rectangle B

感觉不难，但是我怎么算都是相等的啊，求解答

作者: hxb1014 时间: 2016-8-16 11:06:30

We denote the length of rectangle A by l, and the width by w. Then A = wl , B = w(1 + 10%) l(1 -10%) = 1.1 x 0.9 wl = 0.99A < A, hence we choose B

作者: hxb1014 时间: 2016-8-16 11:06:52

We denote the length of rectangle A by l, and the width by w. Then A = wl , B = w(1 + 10%) l(1 -10%) = 1.1 x 0.9 wl = 0.99A < A, hence we choose B

作者: xiaoshang 时间: 2016-8-16 15:34:37

hxb1014 发表于 2016-8-16 11:06
We denote the length of rectangle A by l, and the width by w. Then A = wl , B = w(1 + 10%) l(1 -10%) ...

谢谢谢谢，懂了，是我自己没转过弯来

作者: xiaoshang 时间: 2016-8-16 15:34:59

hxb1014 发表于 2016-8-16 11:06
We denote the length of rectangle A by l, and the width by w. Then A = wl , B = w(1 + 10%) l(1 -10%) ...

谢谢谢谢，懂了，是我自己没转过弯来

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