Mathematics: Post your doubts here!

[syed1995]

just supporting you as your comb seems normal with wacky combos i have seen

Oh I never that that the combination was crazy.. It is quite normal.. Pure sciences + Maths. I just said it was difficult.

 

[A star]

no i wasnt talking bout her combo just saying i have seen other combos which dont make sense like physics eco and acounts i mean whats the use of physics

 

[Rutzaba]

Oh I never that that the combination was crazy.. It is quite normal.. Pure sciences + Maths. I just said it was difficult.

no i wasnt talking bout her combo just saying i have seen other combos which dont make sense like physics eco and acounts i mean whats the use of physics

Yup, qualified for the aforementioned group.

sweet hearts ur convo is intercepting with the questions and there answers ... pleesh continue on ur wallls thankyee ^_^

 

[syed1995]

no i wasnt talking bout her combo just saying i have seen other combos which dont make sense like physics eco and acounts i mean whats the use of physics

Don't underestimate physics!

 

[syed1995]

sweet hearts ur convo is intercepting with the questions and there answers ... pleesh continue on ur wallls thankyee ^_^

Sowwie!

 

[Rutzaba]

Esme imma solve u the whole question but first i must tell you wat i am doing...
lets take a set of three variables that are a point on the line of which we have to find the equation...
a, b, c
then u know wat b1 is? its is substraction of one point from another on the line... suppose it is of coordinates i, j, k
then the equation of line would be sumthing like (a,b,c) +t (i, j, k) get this part...?
wat we have to do in this equation is to eliminate one variable either x or y or z and make one of them the subject of formula in terms of each variable... such as if we are making x as the subject of formula then... x in terms of x, in terms of y and in terms of z.
the general equation from wch we derive the equation of line is
(x-a)/i +(y-b)/j + (z-c)/k

 

[SalmanPakRocks]

No off-topics, please.

 

[Rutzaba]

No off-topics, please.

you just acted like punjab police! xD coming after the enemies have surrendered.... by the way chek inbox i solved ur q...

 

[Rutzaba]

Esme imma solve u the whole question but first i must tell you wat i am doing...
lets take a set of three variables that are a point on the line of which we have to find the equation...
a, b, c
then u know wat b1 is? its is substraction of one point from another on the line... suppose it is of coordinates i, j, k
then the equation of line would be sumthing like (a,b,c) +t (i, j, k) get this part...?
wat we have to do in this equation is to eliminate one variable either x or y or z and make one of them the subject of formula in terms of each variable... such as if we are making x as the subject of formula then... x in terms of x, in terms of y and in terms of z.
the general equation from wch we derive the equation of line is
(x-a)/i +(y-b)/j + (z-c)/k

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​

 

[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