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发表于 17-10-2010 08:08 PM
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还有这个
P and Q are points on (x-2)^2+y^2=1 and (x+2)^2 + (y-1)^2 = 4 respectively.
Find the maximum and minimum values of PQ |
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发表于 17-10-2010 08:58 PM
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给你提示自己试试看
1)T11 = 110 (a+(n-1)d)
S20 = 290 [n/2 (2a+(n-1)d)]
sum of th ...
peaceboy 发表于 17-10-2010 06:22 PM 
第一题的我有用你给我的提示做过
110=a+10d
290=20/2[2a+19d] >>>2
a=110-10d >>>>1
1into2..
290 =20/2[2(110-10d)+19d)]
290 =10(220-d)
290 =2200-10d
290-2200=-10d
d=191 >>>可是答案错哦。。
第二题。。
Sn=3n+4n^2
a=S1 T2=S2-S1
=7 =3(2)+4(2)^2-7
=15
d=15-7=8
然后?
第三提。。。
a=6,d=12
n/2[2(6)+(n-1)12] > 500
n/2(12n) >500
6n^2 >500
.....?? |
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发表于 17-10-2010 09:07 PM
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第一题的我有用你给我的提示做过
110=a+10d
290=20/2[2a+19d] >>>2
a=110-10d >>>>1
1into2..
...
hongji 发表于 17-10-2010 08:58 PM 
抱歉,第一我看错题目
是S11 = 110才对 |
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发表于 17-10-2010 09:11 PM
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回复 2022# hongji
第二
Sn = 3n+4n^2 .
S(n-1) = 3(n-1) + 4(n-1)^2
=3n-3 + 4(n^2-2n+1)
=3n-3 + 4n^2-8n+4
= 4n^2 -5n +1
Tn = Sn - S(n-1)
=3n+4n^2 -(4n^2 -5n +1)
=8n-1
To show the sequence is an AP
Tn - T(n-1) = a , a is a constant
和上面一样自己做看
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发表于 17-10-2010 09:14 PM
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第一题的我有用你给我的提示做过
110=a+10d
290=20/2[2a+19d] >>>2
a=110-10d >>>>1
1into2..
...
hongji 发表于 17-10-2010 08:58 PM 
6n^2 >500
n^2 > 500/6
n > (500/6)^(1/2) |
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发表于 17-10-2010 09:15 PM
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抱歉,第一我看错题目
是S11 = 110才对
peaceboy 发表于 17-10-2010 09:07 PM 
啊啊啊。。我做到了第一题
那么第2,3? |
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发表于 17-10-2010 09:18 PM
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还有这个
P and Q are points on (x-2)^2+y^2=1 and (x+2)^2 + (y-1)^2 = 4 respectively.
Find the maxi ...
evildevil7n 发表于 17-10-2010 08:08 PM 
method 1 :
expand and find dy/dx = 0
method 2 :
find the center of those circle , then the x value + radius is the maximum pnt , x value - radius = minimum pnt |
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发表于 17-10-2010 09:22 PM
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6n^2 >500
n^2 > 500/6
n > (500/6)^(1/2)
peaceboy 发表于 17-10-2010 09:14 PM 
如果答案>>(500/6)^2
=n>9.8
=10
明天就要考数学了。。 |
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发表于 17-10-2010 09:24 PM
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如果答案>>(500/6)^2
=n>9.8
=10
明天就要考数学了。。
hongji 发表于 17-10-2010 09:22 PM 
对.
加油吧,有什么问题尽量问~! |
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发表于 17-10-2010 09:37 PM
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find the sum of all the integers between 1000 and 2000 which are not divisble
a=5.d=5
用S2000-S1000? |
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发表于 17-10-2010 09:43 PM
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if g(r)=1/r^2,simplify g(r)-g(r+1),Hence,find the sum of the first n terms of the series
3/1^2 . 2^2 +5/2^2 . 3^2 +7/3^2 . 4^2 +...
我最不会就是这些== |
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发表于 17-10-2010 09:57 PM
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find the sum of all the integers between 1000 and 2000 which are not divisble
a=5.d=5
用S2000- ...
hongji 发表于 17-10-2010 09:37 PM 
找sum from 1000 to 2000
a=1, d=1 , n=1001
然后找sum from 1000 to 2000 which is divisible by 5
1000 , 1005 , 1010 , 1015, ......,2000 <<< 201 term
a=1000 , d= 5 , n = 201
用sum from 1000 to 2000 剪掉sum from 1000 to 2000 which is divisible by 5就是答案 |
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发表于 17-10-2010 10:08 PM
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本帖最后由 peaceboy 于 17-10-2010 10:09 PM 编辑
回复 2031# hongji
if g(r)=1/r^2,simplify g(r)-g(r+1),Hence,find the sum of the first n terms of the series
3/1^2 . 2^2 +5/2^2 . 3^2 +7/3^2 . 4^2 +...
g(r) = 1/r^2
g(r+1) = 1/(r+1)^2
g(r) - g(r+1)= 1/r^2 - 1/(r+1)^2
= [(r+1)^2 - r^2]/ r^2 (r+1)^2
= [r^2 +2r +1 -r^2] / r^2 (r+1)^2
= [ 2r +1 ] / r^2 (r+1)^2
3/1^2 . 2^2 +5/2^2 . 3^2 +7/3^2 . 4^2 +... + [ 2r +1 ] / r^2 (r+1)^2
sum [ 2r +1 ] / r^2 (r+1)^2 to n term = sum g(r) - g(r+1) to n
= 1/1 - 1/2^2
+1/2^2 - 1/3^2
+1/3^2 - 1/4^2
+... - ....
+1/n^2 - 1/(n+1)^2
= 1- 1/(n+1)^2
= [(n+1)^2 -1]/(n+1)^2
= (n^2+2n) / (n+1)^2 |
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发表于 18-10-2010 07:17 PM
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麻烦大家帮我解答~希望大家可以帮我做出完整的做法~谢谢^^~
1)If (a√2+b√3)2=59-10√6,find the values of a and b.
2)If x^1/2 - 2x^ -1/2 =a, find in terms of a.
(i) x^1/2 + 2x^-1/2
3)Simplify lg₂8x2y - 1/2 lg₂y⁴/16 + 3 lg₂x3y2
4)Solve the simultaneous equation
x+2y=5
(iii) 2^(x+y) =32
5)The cost of a car is expected to increase by 12%every year. If a standard model now (2009)costs RM45000,when will it cost at least RM500000? |
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发表于 18-10-2010 08:03 PM
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麻烦大家帮我解答~希望大家可以帮我做出完整的做法~谢谢^^~
1)If (a√2+b√3)2=59-10√6,find the values ...
wuhu 发表于 18-10-2010 07:17 PM 
确定下全部题目有没有打错... |
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发表于 18-10-2010 08:22 PM
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回复 2033# peaceboy
今天的考试有一题很多分的
明明老师有给我们做
我竟然忘啦 |
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发表于 18-10-2010 08:24 PM
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回复 peaceboy
今天的考试有一题很多分的
明明老师有给我们做
我竟然忘啦
hongji 发表于 18-10-2010 08:22 PM 
再接再厉... |
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发表于 18-10-2010 08:28 PM
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回复 peaceboy
今天的考试有一题很多分的
明明老师有给我们做
我竟然忘啦
hongji 发表于 18-10-2010 08:22 PM 
学校的考试任你错没关系,STPM不要犯就行了 |
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发表于 18-10-2010 08:32 PM
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再接再厉...
Allmaths 发表于 18-10-2010 08:24 PM 
我要像你们一样哈哈 |
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发表于 18-10-2010 08:34 PM
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学校的考试任你错没关系,STPM不要犯就行了
peaceboy 发表于 18-10-2010 08:28 PM 
我要一直进步 |
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