Mathematics: Post your doubts here!

[soccerprinc3]

who did the stats ppr today?

 

[cross]

who did the stats ppr today?

I did. How did it go for you?

 

[soccerprinc3]

I did. How did it go for you?

it was great! how'd you find it?

 

[Sasuke Uchiha]

Anybody has got the mechanics 1 (M1) paper 42 for May/June 2013
I needed the answer for number 6(ii)

 

[cross]

it was great! how'd you find it?

Kinda easy. Which is not good - high GT.

 

[abbey789]

did u ppl get the answers to maths p3 2013?
i have em... contact me...

Do you have the statistics paper 72 too?

 

[Rutzaba]

not yet

 

[Mustafa Malik]

Hello everyone
I am new to this part of Xtremepapers cuz i hv just finished my O levels n in these vacations i wanna study for A levels .Although i created a thread but there was no reply. Actually i need the name of the best books for Math A levels that most renown skools follow.

 

[Fatima18]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_12.pdf
Hi guys....For number 4 ii) I em getting 8 for K..However the marking says its 8.5...how to solve?thanx

 

[Rutzaba]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
Hi guys....For number 4 ii) I em getting 8 for K..However the marking says its 8.5...how to solve?thanx

remember that wen evr it says that u need to calculate the tangent the discriminant wud be equals to 0
b^2 -4ac =0
now we make y the subject of formula and substitute it into eq of curve
2y+ x =k
2y= k-x
y= (k-x)/2 substitute the value into curve
y ^2+2x =13
((k-x)/2 )^ 2 +2x= 13
((k^2 -2kx +x^2)/4)) +2x =13
(k^2 -2kx +x^2) +8x =52
you get
x^2 + (8-2k) x + (k^2-52)
a=1 b= 8-2k c= k^2 -52
b^2 -4ac
(8-2k)^2 - 4(1)(k^2 -52) =0
64- 32k +4k^2 -4k^2 +208 =0
-32k + 272=0
32k= 272
k= 8.5

 

[Fatima18]

remember that wen evr it says that u need to calculate the tangent the discriminant wud be equals to 0
b^2 -4ac =0
now we make y the subject of formula and substitute it into eq of curve
2y+ x =k
2y= k-x
y= (k-x)/2 substitute the value into curve
y ^2+2x =13
((k-x)/2 )^ 2 +2x= 13
((k^2 -2kx +x^2)/4)) +2x =13
(k^2 -2kx +x^2) +8x =52
you get
x^2 + (8-2k) x + (k^2-52)
a=1 b= 8-2k c= k^2 -52
b^2 -4ac
(8-2k)^2 - 4(1)(k^2 -52) =0
64- 32k +4k^2 -4k^2 +208 =0
-32k + 272=0
32k= 272
k= 8.5

Thank you so much !

 

[Rutzaba]

Thank you so much !

remember me in ur prayers ^_^

 

[Fatima18]

remember me in ur prayers ^_^

sure u too

 

[Hemdon]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_12.pdf
Question No 9.1st part.....i cant solve it.....and the mark scheme says : one limit is -1 and other is 3...
Can some one explain please...

 

[waleedsmz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_12.pdf

Question No 9.1st part.....i cant solve it.....and the mark scheme says : one limit is -1 and other is 3...
Can some one explain please...

The cosine function can take up any value from -1 to 1. But since it is squared ( k = 2 ) , then it can take any value from 0 to 1. Therefore, to find the first limit, you substitute with 0 , it gives you 3. The other limit, you substitute with 1, it gives you 3-4 =-1.

 

[amna saleem]

can sme1 hlp me wid tax slab plz!my xa =m on sunday math!

 

[amna saleem]

sm1 hlp me wid tax slab?!

 

[amna saleem]

i need fr tax slab!!

 

[Vaibhav Mor]

Hi I'm having trouble in trigonometric identities, can someone please help me with this question:

 

[Mess]

Hello,
could you help me please? I'm having a hard time with number 5 (iii). I have got 2sec^2X - 1 - 2 secXtanX. There is a sign problem. :$
Please help me.