plz help me to answer the question.if you do not know the answer,plz answer no~i need the step.Thanks for your help`~
1)Given A=P(1+rt),find the value of t when A=1240,P=1000 and r=0.06.
2)Given E=(n-2)x180,find the value of n when E=1080.
F.2 maths plz help me quick~
2007-09-25 2:59 am
回答 (3)
2007-09-25 3:11 am
✔ 最佳答案
1) A=P(1+rt)1240=1000(1+0.06t)
1240=1000+60t
240=60t
t=240/60
t=4
2) E=(n-2)x180
1080=(n-2)x180
1080=180n-360
1080+360=180n
n=1440/180
n=8
參考: me
2007-09-25 3:12 am
Given A=P(1+rt),find the value of t when A=1240,P=1000 and r=0.06
1240=1000(1+0.06t)
1240=1000+60t
1240-1000=60t
240=60t
t=4
Given E=(n-2)x180,find the value of n when E=1080
1080=(n-2)x180
6=n-2
n=8
1240=1000(1+0.06t)
1240=1000+60t
1240-1000=60t
240=60t
t=4
Given E=(n-2)x180,find the value of n when E=1080
1080=(n-2)x180
6=n-2
n=8
參考: 自己
2007-09-25 3:09 am
1)
A/P=1+rt
1.24=1+0.06t
t=0.24/0.06
t=4
2)
E/180=n-2
1080/180=n-2
n-2=6
n=8
A/P=1+rt
1.24=1+0.06t
t=0.24/0.06
t=4
2)
E/180=n-2
1080/180=n-2
n-2=6
n=8
參考: 我計架 希望幫到你
收錄日期: 2021-04-13 13:36:24
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070924000051KK02830