數學方程x2,好急!!!!! 10點呀~

2008-06-13 8:49 pm
1.3
---(x分之3) - 2x=2(x+1)
x

2.A man runs from A to B.If he increases his speed by 10%,what is the percentage decrease of the time required?
2題都要列式,thx!!
更新1:

有小小補充: 1. ritaathena你第1條條式錯左- - 係x做分母,3做分子唔係掉轉-口- 2. Uncle Michael你第2條問題可唔可以冇x^2(x既2次方)- - 姐係第4個步驟同埋之後果d唔好咁計...因為我d程度未到冇深- -" thxx a lot~

回答 (5)

2008-06-13 10:12 pm
✔ 最佳答案
The answers are as follows:


圖片參考:http://ndc.hkcampus.net/~ndc-007/080613_1.jpg

=====

圖片參考:http://ndc.hkcampus.net/~ndc-007/080613_2.jpg
2008-06-14 1:00 am
第1.3條
3/x -2x=2(x+1)
3-2x^2=2x^2+2x
4x^2+2x-3=0
x=(-2+-開方(4+48))/8
x=(-2+-2*開方13)/8
x=-1+-開方13

2.公式s=vt--------1
當v增加10%,距離不變,則t=t1(速度快左,時間短左t>t1)
s=1.1*vt1 ------2
vt=1.1*vt1
t=1.1t1
t1=t/1.1
所以時間減少, (t-t1)/t*100%=0.1/1.1*100=9.09%

2008-06-13 17:01:07 補充:
第1.3條
3/x -2x=2(x+1)
3-2x^2=2x^2+2x
4x^2+2x-3=0
x=(-2+-開方(4+48))/8
x=(-2+-2*開方13)/8
x=(-1+-開方13)/4

打小左個/4
2008-06-14 12:15 am
第1題係一定要用x²咁計......
我都幫你唔到

2.
% decrease=1/1.1=10/11=(100/11)%
2008-06-13 9:53 pm
1.3

x/3-2x=2(x+1)
x-6x=6x+6
-11x=6
x= - 6/11

2. speed = distance / time
distance = speed x time
let k be the new time

distance = (1+10%)speed x k x time
so, k = 1/(1+10%) = 0.909
so, the time decrease 9.09%
參考: 自己
2008-06-13 9:04 pm
口口口口口口口口口 (x分之3) - 2x=2(x 1)x
口口口口口口口口口x分之3) - 2x/2=2(x 1)/2x
口口口口口口口口 (x分之3) - xx3 =x 1x3x
口口口口口口口口口口口口口 x-x=x 1x
口口口口口口口口口口口口 x-x x=2x 1x
口口口口口口口口口口口口口口x=2x 1x
口口口口口口口口口口口口口x 1=3x
口口口口口口口口口口口口x 1-x=3x-x
口口口口口口口口口口口口口1/2=2x/2
口口口口口口口口口口口口口0.5=x
口口口口口口口口口口口口口 x=0.5
口口口口口口口口口口口口口口 ==

2. 我唔識英文
參考: me


收錄日期: 2021-04-24 09:51:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080613000051KK00771

檢視 Wayback Machine 備份