Maths question:
回答 (9)
2006-11-04 6:07 am
✔ 最佳答案
(2-t)/4-t=(2t-3)/5 (2-t)/4=(2t-3)/5+t
(2-t)/4=(2t-3)/5+5t/5
(2-t)/4=(2t-3+5t)/5
(2-t)/4=(7t-3)/5
4(7t-3)=5(2-t)
28t-12=10-5t
33t=22
t=22/33
t=2/3
2006-11-08 2:16 am
your answer is correct...
2006-11-04 5:58 pm
(2-t)/4-t=(2t-3)/5
(2-t)/4=(2t-3)/5+t
2-t=((2t-3)/5+t)(4)
2-t=(8t-12)/5+4t
2-t-4t=(8t-12)/5
2-5t=(8t-12)/5
10-25t=8t-12
22=33t
t=2/3
(2-t)/4=(2t-3)/5+t
2-t=((2t-3)/5+t)(4)
2-t=(8t-12)/5+4t
2-t-4t=(8t-12)/5
2-5t=(8t-12)/5
10-25t=8t-12
22=33t
t=2/3
2006-11-04 6:39 am
(2-t)/(4-t)=(2t-3)/5
(2-t)/(4-t) X5(4-t)=(2t-3)/5 X5(4-t)
5(2-t)=(2t-3)(4-t)
8t^2-6t-2=0
4t^2-3t-1=0
(4t+1)(t-1)=0
t=-1/4, -1
肯定！
2006-11-03 22:40:40 補充：
t=-1/4, 1不是-1，SORRY！
(2-t)/(4-t) X5(4-t)=(2t-3)/5 X5(4-t)
5(2-t)=(2t-3)(4-t)
8t^2-6t-2=0
4t^2-3t-1=0
(4t+1)(t-1)=0
t=-1/4, -1
肯定！
2006-11-03 22:40:40 補充：
t=-1/4, 1不是-1，SORRY！
2006-11-04 4:44 am
(2-t)/(4-t) = (2t-3)/5
10-5t = (2t-3)(4-t)
10-5t = 8t - 2t^2 -12 + 3t
2t^2 -16t + 22 = 0
t^2 - 8t + 11 = 0
where, a = 1, b = -8, c = 11 (use quadratic formula)
t = [8 +/- √64 - 4 (11)]/2
t = (8 +/- √20)/2
t = (8 +/- 2√5)/2
t = 4 +/- √5
10-5t = (2t-3)(4-t)
10-5t = 8t - 2t^2 -12 + 3t
2t^2 -16t + 22 = 0
t^2 - 8t + 11 = 0
where, a = 1, b = -8, c = 11 (use quadratic formula)
t = [8 +/- √64 - 4 (11)]/2
t = (8 +/- √20)/2
t = (8 +/- 2√5)/2
t = 4 +/- √5
2006-11-04 4:42 am
(2-t)/4-t=(2t-3)/5
5(2-t-4t)=4(2t-3)
10-5t-20t=8t-12
10-25t=8t-12
22=33t
t=3/2
2006-11-03 20:43:16 補充：
唔係呀,係2/3
5(2-t-4t)=4(2t-3)
10-5t-20t=8t-12
10-25t=8t-12
22=33t
t=3/2
2006-11-03 20:43:16 補充：
唔係呀,係2/3
2006-11-04 4:27 am
(2-t)/4-t=(2t-3)/5
(2t-3)(4-t)=5(2-t)
-2t²+11t-12=10-5t
-2t²+16t-22=0
t²-8t+11=0
t=(8±√20)/2
=4±√5
(2t-3)(4-t)=5(2-t)
-2t²+11t-12=10-5t
-2t²+16t-22=0
t²-8t+11=0
t=(8±√20)/2
=4±√5
2006-11-04 4:27 am
1.)(2-t)/4-t=(2t-3)/5
(2-t)/4=(2t-3)/5+t
(2-t)/4=(2t-3)/5+5t/5
(2-t)/4=(2t-3+5t)/5
by timing 20(which is the L.C.M. of both 5 and 2)on the whole equation
5(2-t)=(28t-12)
10-5t-28t=-12
10+12=28t+5t
22=33t
t=2/3
2006-11-03 20:29:44 補充：
should be'which is the L.C.M. of both 5 and 4'
(2-t)/4=(2t-3)/5+t
(2-t)/4=(2t-3)/5+5t/5
(2-t)/4=(2t-3+5t)/5
by timing 20(which is the L.C.M. of both 5 and 2)on the whole equation
5(2-t)=(28t-12)
10-5t-28t=-12
10+12=28t+5t
22=33t
t=2/3
2006-11-03 20:29:44 補充：
should be'which is the L.C.M. of both 5 and 4'
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https://hk.answers.yahoo.com/question/index?qid=20061103000051KK03510