数学Paper 1讨论专区

笨蛋一个

不好意思，写错，U=Universal

cgcy

原帖由 jien87 于 27-6-2009 04:40 PM 发表
prove
2)(AuC)n(AuB")=A
4)(A-B)u(B-A)=(A-B)-(AnB)

2)(AuC)n(AuB")=A
    I think you had copied the wrong ques. The supposed ques is (AuB)n(AuB")=A
            (AuB)n(AuB')= Au (BnB')
                                   = Au 0 ,   0=empty set
                                    = A
4) (A-B)u(B-A)=(A-B)-(AnB)
    This do the same with ques 4. The supposed ques is (A-B)u(B-A)=(AuB)-(AnB).
    But i am still thinking of the solution. If anyone know the solution, you may post it up to share

~HeBe~_@

我也觉得是
(A-B)u(B-A)=(AuB)-(AnB)

Please refer to http://cforum2.cari.com.my/redir ... o=lastpost#lastpost

白羊座aries

If a function f is defined by
f(x)=x^2 for 0<x<=1
f(x)=2-x for 1<x<=3
and f(x+3) = f(x) for all values of x ,
sketch the graph of f for -4<=x<=6

evaluate f(13) , f(20) ,intergrate 0,6 f(x) dx

我完全不明白periodic functions

你说对了。报歉，写错问题。谢咯

(A-B)u(B-A)=(AUB)-(AnB)
(A-B)U(B-A)=
=(AnB"U(BnA"
=[(AnB"UB]n[(AnB"UA"]
=[(AUB)n(B"UB)]n[(AUA"n(B"UA"]
=[(AUB)nE]n[En(B"UA"]
=(AUB)n(BnA)"
=(AUB)-(AnB)

E=universe set

(A-B)u(B-A)=(AUB)-(AnB)
(A-B)U(B-A)=
=(AnB"U(BnA"
=[(AnB"UB]n[(AnB"UA"]
=[(AUB)n(B"UB)]n[(AUA"n(B"UA"]
=[(AUB)nE]n[En(B"UA"]
=(AUB)n(BnA)"
=(AUB)-(AnB)

E=universe set

Elsiewang

Determine the value of k such that √3+ki ∕1+√3i is a real number, and find this real number.
它是不是要化成 ax^2 + bx + c?
请各位帮忙！谢谢！

[ 本帖最后由 Elsiewang 于 3-7-2009 10:38 PM 编辑 ]

harry_lim

不是。。。找complex number的
a + bi的

meiling

原帖由 Elsiewang 于 3-7-2009 10:34 PM 发表
Determine the value of k such that √3+ki ∕1+√3i is a real number, and find this real number.
它是不是要化成 ax^2 + bx + c?
请各位帮忙！谢谢！

你要let imagery number be 0
也就是
√3+ki ∕1+√3i=x+yi
where y=0

[ 本帖最后由 meiling 于 4-7-2009 10:53 PM 编辑 ]

~HeBe~_@

楼上 已经 给提示了。
应该不成问题。
若还是做不到，再请教。
加油~

bell_25

这里有个问题想要问。。
Given f(x) = (1 / x-4) - (4 / x-1)
find the point of inflexion by giving your answer to one decimal place.

我找到
d^2/dx^2[f(x)] = [ 2 / (x-4)^3 ] - [ 8 / (x-1)^3 ] = 0
-6x^3 + 90x^2 - 378x + 510 = 0
然后我就不知道要怎样找value of x to get the point of inflexion 了。。
请指点。。谢谢。。

bell_25

这里有个问题想要问。。
Given f(x) = (1 / x-4) - (4 / x-1)
find the point of inflexion by giving your answer to one decimal place.

我找到
d^2/dx^2[f(x)] = [ 2 / (x-4)^3 ] - [ 8 / (x-1)^3 ] = 0
-6x^3 + 90x^2 - 378x + 510 = 0
然后我就不知道要怎样找value of x to get the point of inflexion 了。。
请指点。。谢谢。。

Elsiewang

从√(8-6i)怎样做到(3-i)？

Elsiewang

还是在卡。。。。。

Elsiewang

不要卡了。。。。。

Elsiewang

不要再卡了。。。

笨蛋一个

原帖由 bell_25 于 9-7-2009 01:40 AM 发表
这里有个问题想要问。。
Given f(x) = (1 / x-4) - (4 / x-1)
find the point of inflexion by giving your answer to one decimal place.

我找到
d^2/dx^2[f(x)] = [ 2 / (x-4)^3 ] - [ 8 / (x-1)^3 ] = 0
- ...

按calculator...

原帖由 Elsiewang 于 14-7-2009 12:48 PM 发表
从√(8-6i)怎样做到(3-i)？

let √(8-6i)=x-yi，square两边，然后找x和y...

芭樂

有什么好介绍的Math练习？？
要Chaoter By Chapter 的
能的话不要Q&A的

末日审判

帮帮我下，
1.The sum of the first four terms of a G.P is five times the sum of the sum of the first two terms.Find the value of the common ratio,r,given that r>1.

2.The 1st term of a G.P. exceeds the second term by 1 and the sum of the first three terms is 13/5.If there are positive as well as negative terms in the G.P.,find the sum of the first five terms.

[ 本帖最后由 末日审判 于 24-7-2009 10:21 PM 编辑 ]