Mathematics: Post your doubts here!

[minie23]

Rutzaba or PhyZac

Please help : http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w09_qp_31.pdf
no. 5(i)

 

[Esme]

further more

The two planes are x+2y-2z =2
And 2x-3y +6z =3
Lets say I like x and I want x to remain as it is and want y and z to converge to x.
Now look at the two equations closely
To eliminate y we can multiply the entire first equation by 3
And the second eq by2 thus the both terms in y wud be plus six and minus six and wud be eliminated…
The new eqs are 3x+ 6y- 6z=6
And second wud be 4x – 6y +12z =6
Y is eliminated….
7x +6z=12
7x= 12-6z
X =(12-6z)/7 …………… here is x in terms of z
Now to eliminate the term in z we will take first eq multiply the it by 3.
And take the sec eq as it i.
3x+ 6y -6z =6
2x-3y +6z =3
Z is eliminated we will get 5x+ 3y= 9
X= (9-3y) /5 this is x in terms of y

Now x=x in terms of y = x in terms of z
(X-0)/1 + (9-3y)/5 = (12-6z) /7

We re arrange… to make coefficient of y and z 1
(X-0)/1 + -3 (y-3) /5 = -6 (z-2) /7
(X-0)/1 +(y-3)/ (-5/3) = (z-2) /- 7/6
By comparing this eqwuth the general formula we know that a=0 b=3 c=2
And I=1 j= -5/3 and k=-7/6
To simplify these points or to remove them from fractions you can multiply them with ---6… remember abc CANNOT be multiplied or divided but only b1 can be ie I j and k can be.
We will get -6, 10 , 7
Thus we have the eq of line
(0, 3, 2) +t (-6, 10 , 7) J hope you get it

Esme

Yes I got it thank you.
One question though, in the end you multiplied b1 by -6. so you got it as (-6,10,7). What if I decide to multiply it by 6. The signs would then change (6,-10,-7). Is that ok ?
Oh and I tried it by another method and I got the answer with that too.

 

[Esme]

Rutzaba Can you help with this too ??

 

[abruzzi]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w05_qp_1.pdf
Could anyone please help me with qn no. 9(iii)?
Thanks

 

[Rutzaba]

Yes I got it thank you.
One question though, in the end you multiplied b1 by -6. so you got it as (-6,10,7). What if I decide to multiply it by 6. The signs would then change (6,-10,-7). Is that ok ?
Oh and I tried it by another method and I got the answer with that too.

the choice is urz love

Rutzaba Can you help with this too ??
View attachment 23962

wch prt

 

[Rutzaba]

may be tommz... and i havent seen part a this kinda question... y dun gimme the link n lemme research yea?

 

[iKhaled]

Rutzaba or PhyZac

Please help : http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_31.pdf

no. 5(i)

cos 4θ - 4cos 2θ + 3 = 8sin^4 θ

cos 4θ = 1-2sin^2 2θ
cos 2θ = 1-2sin^2 θ

1-2sin^2 2θ -4(1-2sin^2 θ) + 3
1-2(sin 2θ x sin 2θ) -4(1-2sin^2 θ) + 3

sin 2θ = 2sinθcosθ

1-2(2sinθcosθ x 2sinθcosθ) -4(1-2sin^2 θ) +3
1-8sin^2 θ cos^2 θ -4 +8sin^2 θ +3
-8sin^2 θ cos^2 θ + 8sin^2 θ
-8sin^2 θ(cos^2 θ -1 )
8sin^2 θ (1-cos^2 θ)

cos^2 θ + sin^2 θ = 1

8sin^2 θ x sin^2 θ
= 8sin^4 θ

i hope you got it!

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_1.pdf
Could anyone please help me with qn no. 9(iii)?
Thanks

you should know that tan θ = y/x and the gradient (m) is equal to y/x so u can say from here that tan θ = gradient. you need to find the angle between line l and the gradient of the curve at p so first of all we will find the gradient of the curve at point p

y = 12x^-1
dy/dx = -12x^-2
dy/dx = -3 ( i think you know that dy/dx means also the gradient)

now lets find the gradient of the line y = -2x + k so -2 is the gradient of the line now tan θ = gradient so..

tan θ = -3 and tan θ = -2
θ = -71.6° and θ = -63.4°

difference between them is 71.6 - 63.4 = 8.2° and that's our answer!!

i hope you got it!

 

[cool Asviva]

Ways to be divided
3 5 1 x 3!
3 3 3 x 3!/3!
1 7 1 x 3!/2!

9C3 * 6C5 * 1C1 * 3! = 3024
9C3 * 6C3 * 3C3 * 3!/3! = 1680
9C1 * 8C7 * 1C1 * 3!/2! = 216

Total No. Of Ways = 4920..

If the pies are identical then it would be a whole different story..

i didnt get why permutation is used here! dnt we assume that pies are identical? please explain.

 

[A star]

i didnt get why permutation is used here! dnt we assume that pies are identical? please explain.

no no its given in the question

 

[unseen95]

please help me with question number 5(ii), when i put the values 1 and 1.4 both give negative result. Link to the question paper is http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_31.pdf

 

[freezingfires]

Rutzaba or PhyZac
Can you please help me with Q10 part iii of May/June 2012 paper 32.It is a vector question Im unable to solve it .I'll highly appreciate if you give a step by step solution.Thanks
Here is the link:http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

 

[haha101]

can someone please explain the concept of rate of change

 

[externityxzx]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_3.pdf
please help me with Q6. (ii) & (iii) i don't understand the question
vectors-9. (i) (ii) and intergration-10. (ii) (iii)
Thanks!!!

 

[externityxzx]

please help me with question number 5(ii), when i put the values 1 and 1.4 both give negative result. Link to the question paper is http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_31.pdf

move the right part to the left side
secx-(3-1/2x^2)
x=1 => -0.649
x=1.4 => 3.863

 

[abruzzi]

you should know that tan θ = y/x and the gradient (m) is equal to y/x so u can say from here that tan θ = gradient. you need to find the angle between line l and the gradient of the curve at p so first of all we will find the gradient of the curve at point p

y = 12x^-1
dy/dx = -12x^-2
dy/dx = -3 ( i think you know that dy/dx means also the gradient)

now lets find the gradient of the line y = -2x + k so -2 is the gradient of the line now tan θ = gradient so..

tan θ = -3 and tan θ = -2
θ = -71.6° and θ = -63.4°

difference between them is 71.6 - 63.4 = 8.2° and that's our answer!!

i hope you got it!

Got it!
Thank you

 

[VelaneDeBeaute]

Alright, I know this question was posted before (If someone did it, or know when, they can give me the link to its solution -if it had been posted before)
Otherwise, I can't seem to be getting the hang of it. This is M/J/08, Paper 3, Q.8.

 

[A star]

this thread is full of p3 what about s1or p1 any one

 

[VelaneDeBeaute]

this thread is full of p3 what about s1or p1 any one

Even if it is, there's no way one can access the same post again. (Just as what happened to me right now. I looked through about 20 pages since 344, but was unable to find the same question)

 

[A star]

Even if it is, there's no way one can access the same post again. (Just as what happened to me right now. I looked through about 20 pages since 344, but was unable to find the same question)

ikr thats y we have a print screen option