Mathematics: Post your doubts here!

[The Godfather]

In s1, why is variance = [(sum of x2 + sum of y2)/ (x+y)] - (mean)2

Sorry no S1 in my syllabus.

 

[The Godfather]

Hey ! Amm is math compolsury in cie ? Lol i know it is but still if i can do some other subjects instead of this math mental math will really make me mental please tell me what can i do

Why are you so afraid of mathematics?
It keeps our mind active and fresh.
Nothing is hard until you put your heart and all your grace in your work.
I would suggest you to take mathematics. No other suggestion from my side.

 

[loveyuan]

http://www.google.com.my/url?sa=t&r...8lfqolpMgSXZ9Ww4YfymVCg&bvm=bv.73231344,d.dGc

question number (3 a i )
anyone ? answer is 0.244

 

[Thought blocker]

http://www.google.com.my/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CBsQFjAA&url=http://www.utm.ac.mu/resource/documents/formatted%20probability%20and%20statistics%20-%20bis.pdf&ei=ru3tU92pCdfs8AXD1YGIDg&usg=AFQjCNG7F7V8lfqolpMgSXZ9Ww4YfymVCg&bvm=bv.73231344,d.dGc

question number (3 a i )
anyone ? answer is 0.244

DK Statistics.

 

[skittle_geek]

How do you solve Qn.7b..Please help

 

[Thought blocker]

How do you solve Qn.7b..Please help

Hope it helps.

 

[chishtyguy]

skittle_geek
My version of the solution

 

[skittle_geek]

Hope it helps.

Thanks so much ..it helped a lot

 

[Thought blocker]

Someone plz help me in this stupid confusing domain range chapter :/
If someone can help me understand ~Domain Range~ chapter and ~Vector Paper 3~ only, I would be really Thankful to him/her :/


Help needed!!!
I know that range of a function is the domain of the inverse function and vice versa but how to find domain in this question
part (iv) only
and PLZZ answer with deep knowledge ( I mean with more theory and working which is clearly understandable to solve rest of the questions):
Thanks and May Allah bless us all with knowledge and A/A*s in A Levels Ameen
View attachment 46644


kitkat <3 :P Angelina_25 shazmina fantastic girl Anyone :SSS
Plejj :'/

Here Rookaya did this. http://www.twiddla.com/1729925 Aly Emran

 

[Thought blocker]

Aly Emran
Part i) RANGE OF G INVERSE
Domain of g(x) = range of g inverse.
Domain of g(x) >= 4
Hence, Range of g inverse >=4
Part ii) DOMAIN OF G INVERSE
Range of g(x) = Domain of g inverse
First complete square :¬
- ( x - 4)^2 + 16 = g(x)
Now exchange g(x) with y.
y = - (x - 4)^2 +16
Now exchange x with y and vice-verse.
x = -(y - 4)^2 +16
Make y as subject.
y = 4 + root (16 - x)
Now exchange y with g inverse.
So g inverse is 4 + root (16 - x)
Now for g inverse, any value of x >= 16 will have undefined answer, yeah? so x <= 16
Another way is :¬
Due to the minus sign, it is the maximum point on the graph. So x <= 16 is range of g(x) and according to Range of g(x) = Domain of g inverse, domain of g inverse is x <= 16

Hope you got it.

 

[Miral Lovato]

Why are you so afraid of mathematics?
It keeps our mind active and fresh.
Nothing is hard until you put your heart and all your grace in your work.
I would suggest you to take mathematics. No other suggestion from my side.

Actually i would love to learn something im not good at but the problem is that i study by myself so maths cant be done without tutor thats why i asked .

 

[Thought blocker]

Actually i would love to learn something im not good at but the problem is that i study by myself so maths cant be done without tutor thats why i asked .

Look, I did Math on my own, yet got E in just 1 month preparation, you have 6+ months to prepare give your best shot sister.
Study from :- http://www.examsolutions.net/maths-revision/syllabuses/CIE/period-1/specification.php
All the best.

 

[Miral Lovato]

Look, I did Math on my own, yet got E in just 1 month preparation, you have 6+ months to prepare give your best shot sister.
Study from :- http://www.examsolutions.net/maths-revision/syllabuses/CIE/period-1/specification.php
All the best.

Okay thankoooo

 

[Thought blocker]

Oh Brillient
got it
but cant we find the mid point and see that is the graph a min or max and then decide greater then and less than?
Ps: Is it necessary to do inverse of g(x) in part 1?? because part 2 demands inverse so the examiner might not accept the inverse answer of part 1 i guess :/
plzz help me on this issue too
Thanks in advance

No its not like that, till I know.
Go to this site :- http://www.examsolutions.net/maths-revision/syllabuses/CIE/period-1/P1/module.php 2 chapter is function.
Everything in it.
All the best. ^_^

 

[Edison Weeho]

can anyone help ???

31/may/june 2010 Q7 iii)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf

 

[Thought blocker]

can anyone help ???

31/may/june 2010 Q7 iii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf

Not much with it! Saved answer here :¬

As the question states that the argument of z is least, z must represent the lowest possible point in the shaded region. Draw a tangent from the center to the lowest point on the circle (this doesn't have to be accurate) as shown by the blue line. This line represents |z|. The angle between the radius and the tangent is always 90 so a right-angled triangle is formed. The opposite side, with is equal to the radius is 1 unit. The magnitude of complex u is the hypotenuse and the adjacent represents |z|. you can easily find |u|, √(2^2+2^) = √8.
So, |z| =√(√8)^2 -1^2 = √8-1 =√7

 

[Lama AN]

Guys are we going to face such questions in Pure 1 ?
For functions ?

 

[Thought blocker]

Guys are we going to face such questions in Pure 1 ?
For functions ?

No. I have not faced a question like this. :/

 

[Edison Weeho]

Not much with it! Saved answer here :¬

As the question states that the argument of z is least, z must represent the lowest possible point in the shaded region. Draw a tangent from the center to the lowest point on the circle (this doesn't have to be accurate) as shown by the blue line. This line represents |z|. The angle between the radius and the tangent is always 90 so a right-angled triangle is formed. The opposite side, with is equal to the radius is 1 unit. The magnitude of complex u is the hypotenuse and the adjacent represents |z|. you can easily find |u|, √(2^2+2^) = √8.
So, |z| =√(√8)^2 -1^2 = √8-1 =√7

thank you very much for ur explanations ~

 

[MarcoReus]

If you haven't done Add Maths - how difficult is A Level Maths? Should I do Add.Maths before I start studying A Level Maths?