Mathematics: Post your doubts here!

[iKhaled]

can someone explain why x< 5/3 was not introduced in this question and the answer is only x< 1 ?

question 3 may/june 2003

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s03_qp_3.pdf

 

[Muhammad TAimoor]

i have no idea whether it wud or not Muhammad TAimoor

Me too!

 

[Rutzaba]

Pls help....
8 women and 5 men are standing in a line.
In how many arrangements will no two men standing next to each other?
thanks in advance

nhi yar. i was wrong it wud be done like this
5! multiplied by 8p6

 

[Rutzaba]

How to solve ths qustin...sinA-cosA=1/8. Find 1/secAcosecA .tel me frm whr ths questn is

huh? state the Q clearly

 

[Silent Hunter]

any mechanics revision notes? anybody ? thank you

 

[Jspake]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
No. 6 (ii)

Can anyone please explain how to get the coordinates of D by solving simultaneously?
Thanks

 

[Rutzaba]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
No. 6 (ii)

Can anyone please explain how to get the coordinates of D by solving simultaneously?
Thanks

Thats easy. All u need to have to make an eq of line is gradient and a set of coordinates

Now two things before i tell u the solution
1- parallel lines have same gradients
2- perpendicular lines use this formula m1 X m2 = -1
where m1is rhe gradient of straight line and m2 of the perp.
If u have one of these gradients u can find the other wth this frmula

 

[Rutzaba]

Now... using the gradient formula we found out that the gradient of Ab is 2/3
Lets take this as m1
so putting in the frmula 2/3 multiplied by m2 ( the gradient of line bc) = -1
So the gradient of line bc is -3/2 and u have coordinates of b put in it and make eq of bc put y=o to get coords of C
then u have 1 point C on the line Cd and you kno ab is parallel to cd. The gradient of cd wud thus be 2/3 wth hlp of this and coords of C make eq of cd. Then make eq of ad since its perp to ab its gradient wud be -3/2 you have coords of A. Now u cn make eq od ad.

Finally solve ad an cd to get d.

Sure there were ways less lengthier than this... but u specified that u needed simultaneously.

 

[Just visiting]

Can anyone please help me . I am studying for maths A2 . Should i study pure 2 before studying pure 3 or not.
An if possible plz tell me a good study source. I finished as but till now i find pure 3 horrible

 

[Jspake]

Now... using the gradient formula we found out that the gradient of Ab is 2/3
Lets take this as m1
so putting in the frmula 2/3 multiplied by m2 ( the gradient of line bc) = -1
So the gradient of line bc is -3/2 and u have coordinates of b put in it and make eq of bc put y=o to get coords of C
then u have 1 point C on the line Cd and you kno ab is parallel to cd. The gradient of cd wud thus be 2/3 wth hlp of this and coords of C make eq of cd. Then make eq of ad since its perp to ab its gradient wud be -3/2 you have coords of A. Now u cn make eq od ad.

Finally solve ad an cd to get d.

Sure there were ways less lengthier than this... but u specified that u needed simultaneously.

Thank you for your reply.
Just a slight error there.. the gradient of AB is 3/2.. so the rest is supposed to be changed i.e gradient of BC is -2/3 etc.

If you don't mind, could you please show me the other methods? This way is longer than I thought!
Thanks

 

[Rutzaba]

Thank you for yo r reply.
Just a slight error there.. the gradient of AB is 3/2.. so the rest is supposed to be changed i.e gradient of BC is -2/3 etc.

If you don't mind, could you please show me the other methods? This way is longer than I thought!
Thanks

No u are not ryt.... its gonna be positive 2/3 cux 2-(-2) wud becum 4 and 14 -8 is 6

 

[Rutzaba]

OOne of them is by property of rectangles if u kno 3 coordinates you can find 4rth.
One more is the midpoint of ac wud also be the midpoint of b d.

 

[Beaconite007]

Plz can some one explain me the conccept of how to do questions of integration using subsitution . THNKS

 

[salvatore]

Plz can some one explain me the conccept of how to do questions of integration using subsitution . THNKS

I assume you know how to find the integrals of the type (u)^n.

Now, to find the integrals of the more complicated forms i.e (ax+b)^n.. you have to use the substitution method in order to convert it into the form (u)^n, which will make it easier to find the integral.
I'll explain this in form of an example.

⌠ (x + 4)^5 dx
⌡

Let u = x+4
Hence, it can be written in the form u^5.

Remember: du = ( du/dx) dx
(du = Differential constant)
Therefore, (du/dx) = 1
du = 1 dx
Since du = 1 dx, we can say that du = dx

So the expression becomes ⌠ (u)^5 dx = ⌠(u)^5 du
.............................................⌡................... ⌡
(Dont mind the dots.. these are just to keep the integral sign in place)

The integral of (u)^5 = (u^6)/6 + C right?

Since u = x+4, the expression becomes :

((x+4)^6)/6 + C... This is the answer!

 

[freakybandi]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
can anyone please explain qs3(ii), 9 and 10 to me? please need it ASAP!

 

[SilverCrest]

can someone help me with thiS
Q Find the value of the constant a for which the coefficient of x is 64 in the expansion (1 +ax)(2-x)^6
the answer is 4 but how?

 

[Rutzaba]

can someone help me with thiS
Q Find the value of the constant a for which the coefficient of x is 64 in the expansion (1 +ax)(2-x)^6
the answer is 4 but how?

simple... ((2^6)) + 6c1 (2^4) (-x^1) bus thats as much as we are going to need
That wud give u (64-192 x) ( 1+ax)
Then 64 -192x +64ax -192ax^2
we only need x 's coefficient ... that wud be 64a-192 wen this is supposed to b equals to 64
64a -192 = 64

Solve this ... the answer is 4

 

[Tabi Sheikh]

Any body have notes about trignometry?

 

[pakiboy]

is application of differential equations included in our syllabus?

 

[iKhaled]

need help in question 8 may june 2007, the second part please

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_3.pdf