一些關於數學的問題(一定要有步驟)
回答 (5)
2006-11-26 3:24 am
✔ 最佳答案
5[4y-3(y+1)]=y over 25[4y-3(y+1)]=y/2
5(4y-3y-3)=y/2
5*2*(y-3)=y
10(y-3)=y
10y-30=y
10y=y+30
10y-y=30
9y=30
y=30/9
y=10/3
2006-11-26 5:56 am
5 (4y - 3(y+1)) = y/2
5 (4y - 3y -3) = y/2
5(y - 3) = y/2
10y - 30 = y
9y = 30
y = 10/3
5 (4y - 3y -3) = y/2
5(y - 3) = y/2
10y - 30 = y
9y = 30
y = 10/3
2006-11-26 2:35 am
５(4y-3y-3)=y over2
5(y-3)=y over 2
5y-15=y over 2
5y=30y over2
5y=15
y=3
2006-11-25 18:37:16 補充：
我似係
5(y-3)=y over 2
5y-15=y over 2
5y=30y over2
5y=15
y=3
2006-11-25 18:37:16 補充：
我似係
2006-11-26 2:29 am
20y-15y-15=y/2
10y-30=y
y=30/9
10y-30=y
y=30/9
收錄日期: 2021-04-23 16:13:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061125000051KK03946