Mathematics: Post your doubts here!

[iKhaled]

ohkay makes my work easy wait ..

thanks man!!

 

[biba]

Hi all, I would please like to know the solution of question 4 in this paper. I just reached that point where I should leave the question or I will cause myself problem. Thanks in advance

cosec2@ = sec@ + cot @
so,
1/sin2@ =(1/cos@) + (cos@/sin@)
1/sin2@ = (sin@ + cos^2@)/(cos@sin@)
1/sin2@ = [sin@ +(1- sin^2@)]/ (cos@sin@)
1/sin2@ = (sin@+1- sin^2@)/(cos@sin@)
i am multiplying right hand side with (2/2) so,

1/sin2@ = 2(sin@+1-sin^2@)/2sin@cos@
1/sin2@ = 2(sin@+1-sin^2@) / sin2@
sin2@ gets cancelled out therefore we are left with:
1 = 2(sin@+1-sin^2@)
1 = 2sin@ +2 - 2sin^2@
2sin^2@ - 2sin@ -1 = 0
now u can find the angle

 

[biba]

I KNOW RITE !!!
i am sleep deprived
you know CIE ! always in the torturing mood xD
its like they get peace when they see us in pain xD

true that

 

[MustafaMotani]

yes i got all of the question but i don't know how to deal with the 6 roots thingy i cant get this question

sorry the pic is little blur but still readable..
I assume yu have basic idea of functions..

 

[daviruss]

hey plzz solve this plzz( W=-1+i , find W^3

 

[MustafaMotani]

hey plzz solve this plzz( W=-1+i , find W^3

w^2 = 1 + 2i -1 = 2i
w^3 = w^2 x w
2i x (1 +i)
2i - 2

 

[daviruss]

w^2 = 1 + 2i -1 = 2i
w^3 = w^2 x w
2i x (1 +i)
2i - 2

thx BRO owe u one the the real number was -1

 

[MustafaMotani]

thx BRO owe u one the the real number was -1

oh... but you got it!! ryt.. ?

 

[daviruss]

oh... but you got it!! ryt.. ?

yup i did thx you man

 

[daviruss]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w11_qp_33.pdf


number 10 plzz

 

[biba]

can anyone help me with q's 4 ii of M/J 2010 P31

sin3xsinx = 1/2(cos2x)- 1/2(cos4x)
so u integrate 1/2(cos2x)- 1/2(cos4x) ,u wil get :
[1/4(sin2x) - 1/8sin4x]
now put in the limits for x.. u wil get ur answer

 

[LindaKim123]

MustafaMotani

can u please solve http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_31.pdf
question 7 i) ?

 

[fahim bin faruk]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

can someone help me out with no.9 (ii)???

 

[Sanis]

w= -1 + i (on diagram (-1,1))
w^2 = -2i (on diagram (0,-2))

midpoint = ((-1+0)/2, (1-2)/2)
(-1/2 , -1/2) = -0.5 -0.5i

cool now i got this part .. how do I calculate the radius?

 

[gary221]

ppl, a lil help pls...
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_31.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_ms_31.pdf

ques 10.
thnx!

 

[19islandprincess96]

MustafaMotani

can u please solve http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_31.pdf
question 7 i) ?

1/x + 1/y* dy/dx = derivative of ln (xy)

-3y^2 * dy/dx = derivative of y^3

so

1/x + 1/y* dy/dx - 3y^2 * dy/dx = 0

1/x + dy/dx (1/y - 3y^2) = 0

dy/dx (1/y - 3y^2) = -1/x

dy/dx [(1-3y^3)/y] = -1/x

dy/dx = y/-x(1-3y^3)

dy/dx = y/x(3y^3 - 1)

Hope this is clear!

 

[Esme]

Best of luck for 2mrw 2 awl of u !!! THNX 4 helping out at this eleventh hour!!!

So yes i m signing out 4 2day!!!
Pray 4 me!

Especially regards to
Esme
iKhaled
PhyZac
knowitall10
Rutzaba
LMGD33
Alice123
SararaIH
VampBhums
..... n all others! (Excuse me if i missed any1s name here!!!)

Thank you so much for remembering me among all these amazing people and tagging me. Feels good
Insha'Allah the paper will be good and easy for everyone.
Pray for all of us sis

 

[gary221]

ppl, a lil help pls...
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_ms_31.pdf

ques 10.
thnx!

@ anyone?

 

[Esme]

@ anyone?

Q10i
da/dt = KV
da/dt = k(4/3πr^3)
da/dt = 4/3πkr^3

A=4πr^2 V = 4/3πr^3

da/dr = 8πr dv/dr = 4πr^3

dr/dt = dr/da x da/dt
dr/dt = (1/8πr) x 4/3πkr^3 substitute r and dr/dt to find K then u will end up with the given equation

Posted by iKhaled previously

 

[Esme]

ppl, a lil help pls...
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_ms_31.pdf

ques 10.
thnx!

Posted by PhyZac previously:

Q110ii)
okay first, move value to make two integration
(let ! be the integration sign)
dr/dt = 0.08r^2
1/r^2 dr = 0.08 dt
! 1/r^2 dr = !0.08 dt
! r^-2 dr = 0.08t +c
-1/r = 0.08t + c
sub values given in question
-1/5 = 0.08(0) + c
c = -1/5

-1/r = 0.08t - 1/5
r = -1 / (0.08 - 1/5)
r = -5 /0.4t - 1


Now for part iii
i am not sure
but what i think is the answer must be positive.
to do this the value shudnt be smaller than 0, (try calculating)
and even not more than 2.5
i got 2.5 by
0.4t-1 =0
0.4t =1
t= 2.5