Umm, it is just finding points on a line, and I go for two points always, when x is 0 and when y is 0.lawl then how did u get the answer
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Umm, it is just finding points on a line, and I go for two points always, when x is 0 and when y is 0.lawl then how did u get the answer
dude.. and guys at random... post krne se pehle dekkh liya karo ke kisi aur ne swal ka jawab tou nhi dde dya.. besides i was asking fr sumone else... thnkoo anyways buddy
u need not go bak three pages... just the top of this page actuallydidn't feel like reading 3 pages
<--- is too lazy
le mai kiya hare kerri rocket science ay! xDUmm, it is just finding points on a line, and I go for two points always, when x is 0 and when y is 0.
Can you give the very first step?i always have trouble in trigonometry ... i reached into 3sinxcosx=1 .. i checked the mark scheme x=0.365 radians .. when verifying it turns to be correct but however i don't know how I am supposed to reach this value when sin and cos are multiplied .. PLZ HELP URGENTLY thanks further
3tanx.(cos^2)x=1Can you give the very first step?
Well, see, i got x=0.365 as one of the answers, but I am not sure my way is correct, very complicated as well as, not the only answer! Was there a range?3tanx.(cos^2)x=1
1.206 .. anyways nvm abt it im sure there is another solution for this .. thank youWell, see, i got x=0.365 as one of the answers, but I am not sure my way is correct, very complicated as well as, not the only answer! Was there a range?
Well i got this value too!1.206 .. anyways nvm abt it im sure there is another solution for this .. thank you
well i found it 3sinxcosx=1 .. u gotta pretend that sinxcosx = half sin 2 theta .. since since sin 2 theta = 2sincosx .. so we actually get 1.5 sin 2x = 1 .. sin 2x = 2/3 .. then x= 20.90 degrees and it turns to be correct that if u took math a2 u will know this formulaWell, see, i got x=0.365 as one of the answers, but I am not sure my way is correct, very complicated as well as, not the only answer! Was there a range?
Well i got this value too!
Okay here is the way.
P.S, i dint use more than Pure 1 knowledge.!
3tanx(cos^2)= 1
3 sinxcosx = 1
since , sinx = √1-cos^2 [ from sin^2 + cos^2 = 1]
3 √1-cos^2 cosx = 1
√1-cos^2 = 1/3cosx
1 - cos^2 = (1/3cosx)^2
1 - cos^2 = 1/9cosx^2
(9cosx^2)1 - cos^2 = 1
9cos^2 - 9cos^4 = 1
9cos^4 - 9cos^2 + 1 = 0
then using quaradictic formula u get.
cos^2(x) = 0.87267 or cos^2 = 0.12732
cos x = √0.87267 or cos x = √0.12732
x = cos^-1 (√0.87267) or x = cos^-1 (√0.12732)
do uk the common perpendicular method to find the normal of a plane from 2 other normals which is perpendicular to it ?Aoa wr wb
Can someone please help me with 7 ii (using the scalar product method) and 7 iii of this paper.....?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_31.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_ms_31.pdf
(PS: Please do provide a detailed explanation, vectors isn't exactly my cup of tea )
JazakAllah khair and thanx a ton!
Thanx a lot, yes i did get it.....ok for question 7(ii)
we got the equation of AB r = i +2j + 2k + t( 2i + 2j -2k) so any point lies on this line will have x= 1 +2t y = 2 + 2t z= 2-2t we can get the position vector of p from here. it says that OP (position vector) is perpendicular to the line AB which means that..
[(1+2t)i + (2+2t)j + (2-2t)k]X [2i + 2j -2k] = O from here u can find the value of t which will be t= -1/6 then substitute it to find x y and z and this will be the position vector of P
did u get it ?
sorry, er what do you mean?do uk the common perpendicular method to find the normal of a plane from 2 other normals which is perpendicular to it ?
which math book do u use?Thanx a lot, yes i did get it.....
sorry, er what do you mean?
Pure Mathematics 2 and 3 by Hugh Neil and Doughlas Quadling.which math book do u use?
owww......yes i do
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