Mathematics: Post your doubts here!

[Aqsa Mohd]

how to get my self done wd p3 .. :S

 

[Snowberry]

Can someone help me?

Find the cartesian equations of the plane through (--1,1,2) and (2,0,-1) perpendicular to the plane x+y-2z=1.​

use the two points to obtain a direrction vector in the plane....and usee the given plane equation's normal as another vector in the required plane....cross these two vectors and there u go ....

Can someone please write it? I'm still stuck. What does 'cross these vectors' mean please?

 

[kinal]

than

Aoa wr wb!

yeah just as hassam said...like in the a part..u got

z = -2..

instead of z it's z^2 now
so z^2 = -2

similarly do for the other 2 roots...when u take the square root u obviously get..2 answer

thank you!

 

[Chooi1993]

How to get the initial value for q4 ?​

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_32.pdf​

 

[user]

how to get my self done wd p3 .. :S

aoa wr wb!
solve one paper ...try recent ones..check ur paper from the mark scheme and mark yourself....now make out which topics you werent really sure..revise them thoroughly...
go through your notes for all chapters...solve few more papers....mark urself..if you feel u r unsure about something revise that chapter again

 

[aaakhtar19]

How to get the initial value for q4 ?​

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_32.pdf​

I guess taking any initial value would solve the question I'll recommend to find the answer by calculator
and rounding it off to whole number and then using this as initial value

 

[user]

I guess taking any initial value would solve the question I'll recommend to find the answer by calculator
and rounding it off to whole number and then using this as initial value

smart

 

[Snowberry]

Can someone help me?

Find the cartesian equations of the plane through (--1,1,2) and (2,0,-1) perpendicular to the plane x+y-2z=1.​

use the two points to obtain a direrction vector in the plane....and usee the given plane equation's normal as another vector in the required plane....cross these two vectors and there u go ....

Can someone please write it? I'm still stuck. What does 'cross these vectors' mean please?

 

[user]

Can someone please write it? I'm still stuck. What does 'cross these vectors' mean please?

cross product/vector roduct/cross the vectors..al are the same thing...Check this: Vectors , to find out how to do that...

 

[abdullah12]

Can someone please explain j10 v31 Q9)i)

 

[aaakhtar19]

Can someone please explain j10 v31 Q9)i)

Page 100 same thread post 1990 already solved

 

[leadingguy]

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s06_qp_3.pdf

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s06_ms_3.pdf
question 4 part 2

i reached til the eq. 25 < 25cos(θ - 73.74) < 25

can any one help me out by just doing 2 to 3 more steps not the whole question plz

 

[kagome]

Help plz... O/N 2010 32 Q7ii

 

[smzimran]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_3.pdf

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_ms_3.pdf
question 4 part 2

i reached til the eq. 25 < 25cos(θ - 73.74) < 25

can any one help me out by just doing 2 to 3 more steps not the whole question plz

Why did you reach to this?
This is not an equation, it is an inequality!

25cos(θ - 73.74) = 15
Simply solve this but remember the range is negative also so also take the value of negative (360 - x) in the 4th quadrant

 

[iqraa nadeem]

Hi.. Can someone pls help with nov11/33.q10 =S
here's the link
https://docs.google.com/viewer?a=v&...h5Lf5D&sig=AHIEtbTfFEbI3bmoyd0Wnx3O3bppOVsZ_w

 

[leadingguy]

Hi.. Can someone pls help with nov11/33.q10 =S
here's the link
https://docs.google.com/viewer?a=v&q=cachel5TBi4TmzcJ:www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w11_qp_33.pdf november 2011 paper 33 9709&hl=en&gl=pk&pid=bl&srcid=ADGEESjynU7l4xl4mks2Y9oAnb1ghOQR75yUtAXuiAgJKP_3ljz5gDqaCPFYXCpxIS64HCC4oTVS4IwZIOBFrPlov_GWo38j47c-BwVoNuC1LpcY83kSTKful1Zt-ubLe1vbidh5Lf5D&sig=AHIEtbTfFEbI3bmoyd0Wnx3O3bppOVsZ_w

whats thiis lnk??

 

[Keitak]

Thanks for replying!
when you make X the subject does the answer you get have to be within the new range or the one specified in the question?

I get (2X-71.57)= 50.77 and 2x-71.57=(360-50.77)

2X=122.3 and 2X=380.8
X=61.2 X=190.4
in 2nd quadrant, only sin is +ve
then, do 360-190.4=169.6
then, 180-169.6=10.4

final answer should be in range specified in the question

why is it "360-50.7"
I dont undestand this part
"in 2nd quadrant, only sin is +ve
then, do 360-190.4=169.6
then, 180-169.6=10.4"

 

[iqraa nadeem]

MYy bad ..here's the link for nov11/33 i need help with q10
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf

 

[iqraa nadeem]

I also need help with nov10/33 q10 part b (ii)
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w10_qp_33.pdf

 

[Aahliya]

i had the same problem
what i did first was to make lny the subject ---> lny=(2x+1)/x
now differentiate each side ------> lny becomes 1/y (dy/dx)
differentiate (2x+1)/x using quotient rule ------- [2x-(2x+1)]/x^2 = -1/x^2
therefore------- 1/y (dy/dx)= -1/x^2,
take 1/y to the other side becomes ------− - y/x^2

Well !! That's the only way left to solve this question .. Nyz thanks