Mathematics: Post your doubts here!

[whitecorp]

please could some one help me with this :?

make partial fractions

2(4x^2 + 1) /(2x+1)(2x-1)

ps the degree of power of numerator= denominator

You will have to do long division before using partial fractions to split things up. Hope this helps. Peace.

 

[whitecorp]

Pls guys help me, how to get c (y-intercepy}............ thanx....!!!
ans-:lgc=-o.6
c=0.251

I am not sure what you mean by y-intercept here,
but for lgc =-0.6, assuming that you are solving for c, then c=e^-0.6 =0.549, which is definitely not 0.251.

Hope this helps. Peace.

 

[whitecorp]

Hey thank you sooo much !
Yeah, my ans is : 2x + 3y - 6z = 8

Btw, do you do S1 and M1 as well ?

No problem. Here are the workings for 11(ii):



Hope it helps. Peace.

(I am not exactly sure what is M1 and S1, but to me its just mathematics, so I guess it shouldn't be a problem at all).

 

[hussamh10]

here is the answer to the question, i hope you understood it!

Thanks i got it

 

[bamteck]

whitecorp thank you man !

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w05_qp_4.pdf

Can you help me with M1, for no. 5 please

 

[SalmanPakRocks]

Find the binomial expansion of (3x-2)^4
I'm totally new to this so please write all the steps.

 

[Soldier313]

Find the binomial expansion of (3x-2)^4
I'm totally new to this so please write all the steps.

(3x - 2)^4

a. (-2 (-3x/2 + 1) )^4 ......i factorised it.....because as far as i know, the other number in the bracket needs to be +1 for the formula below to work

(-2)^4 (-1.5x + 1)^4

b. 16 (-1.5x + 1)^4

c. then from the data booklet : we substitute the values we have above into this equation

so we keep our 16 on the side for now and multiply it later.....

d. x = -1.5x in this case
n = 4
e.



go on in this way (just substitute) if they ask you to give the answer upto power 4, 5 etc


finally we use our 16 now

f. 16 (1 - 1.5x)^4 = 16 ( 1 - 6x + 13.5x^2 - 13.5x^3 .......... )

hope you understood.... inshaAllah......

 

[bamteck]

M1 Urgent Help Needed Please

7. A particle of mass m kg moves up a line of greatest slope of a rough plane inclined at 21◦ to the
horizontal. The frictional and normal components of the contact force on the particle have magnitudes
F N and R N respectively. The particle passes through the point P with speed 10 m s−1 , and 2 s later it
reaches its highest point on the plane.

(i) Show that R = 9.336m and F = 1.416m, each correct to 4 significant figures.

(ii) Find the coefficient of friction between the particle and the plane.

After the particle reaches its highest point it starts to move down the plane.

(iii) Find the speed with which the particle returns to P.

Please help me for the (iii) only !

[Quoted from Nov 2004 p4]

 

[iKhaled]

need help with question 9 a) may june 2011 paper 12

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_12.pdf

 

[iKhaled]

M1 Urgent Help Needed Please

7. A particle of mass m kg moves up a line of greatest slope of a rough plane inclined at 21◦ to the
horizontal. The frictional and normal components of the contact force on the particle have magnitudes
F N and R N respectively. The particle passes through the point P with speed 10 m s−1 , and 2 s later it
reaches its highest point on the plane.

(i) Show that R = 9.336m and F = 1.416m, each correct to 4 significant figures.

(ii) Find the coefficient of friction between the particle and the plane.

After the particle reaches its highest point it starts to move down the plane.

(iii) Find the speed with which the particle returns to P.

Please help me for the (iii) only !

[Quoted from Nov 2004 p4]

the distance moved by the particle is s= 1/2(10)(2) = 10m

while the particle is going down the friction force will act opposite to the motion therefore:

F = ma
mgsinθ - F = ma
10msin21.1 - 1.416m = ma
a = 10sin21.1 - 1.416
a= = 2.184 ms^-2

v^2 = u^2 + 2as
v^2 = 0 + 2(2.184)(10)
v = 6.61 ms^-1

i hope this makes sense to u!

 

[Shahed Shubeir]

can anyone explain me question 3 , may/june 2003 , paper 4 ( mechanics1)

 

[Taiyaba]

Please help me with this one

Find the sum of all odd numbers between 0 and 500 which are divisible by 7.

And this one

Show that the sum 1+3+5+(2n-1) is always a perfect square.

Thnx

 

[bamteck]

the distance moved by the particle is s= 1/2(10)(2) = 10m

while the particle is going down the friction force will act opposite to the motion therefore:

F = ma
mgsinθ - F = ma
10msin21.1 - 1.416m = ma
a = 10sin21.1 - 1.416
a= = 2.184 ms^-2

v^2 = u^2 + 2as
v^2 = 0 + 2(2.184)(10)
v = 6.61 ms^-1

i hope this makes sense to u!

Thank you very much mate ! Much appreciated !

 

[bamteck]

Please help me with this one

Find the sum of all odd numbers between 0 and 500 which are divisible by 7.

And this one

Show that the sum 1+3+5+(2n-1) is always a perfect square.

Thnx

the largest multiple of 7 less than 500 = 497 (7*71)
so the sum of all multiples of 7 less than 500 =
7 * sum(n=1...71)n = 7 (71)(72)/2 = 497*36 = 17892
the largest multiple of 14 less than 500 = 490 (14*35)
the sum of all multiples of 14 less than 500 =
14 * sum(n=1...35)n = 14(35)(36)/2 = 8820

so the sum of the odd multiples of 7 less than 500 is
17892 - 8820 = 9072

 

[bamteck]

need help with question 9 a) may june 2011 paper 12

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_12.pdf

Would love to help you but I'm bad at functions ! Arrggggggghhh !

 

[Taiyaba]

the largest multiple of 7 less than 500 = 497 (7*71)
so the sum of all multiples of 7 less than 500 =
7 * sum(n=1...71)n = 7 (71)(72)/2 = 497*36 = 17892
the largest multiple of 14 less than 500 = 490 (14*35)
the sum of all multiples of 14 less than 500 =
14 * sum(n=1...35)n = 14(35)(36)/2 = 8820
so the sum of the odd multiples of 7 less than 500 is
17892 - 8820 = 9072

Thnk u bt cn u plz xplain the second one also

 

[bamteck]

Thnk u bt cn u plz xplain the second one also

I don't know !

 

[Taiyaba]

I don't know !

It's ok

 

[NokiaN95638]

Help on P2 questions needed!!! (here @= theta)
Q. The Parametric equation of a curve are
X=2@+cos@ y= @+sin@

0≤@≤2π


1. Find dy/dx in terms of @. (no problem on this one)
2. Show that, at points on the curve where the gradient is 3/4 the parameter @ satisfies an equation of the form:
5sin(@+α)=2 , where the value of α is to be stated.
3. Solve the equation in part(ii) to find 2 possible values of @.

 

[Khunkar]

can anyone give your solution on:- 9709/O/N/11, question no. 2
thanks in advance.