Maths, Addmaths and Statistics: Post your doubts here!

[Talha Irfan]

u have a diagram?...and these are vectors right?...but need to show the figure, cause we cant know which is BC and AB and what are there directions, are they in opposite direction?

tbh; no diagram is given, yeah they are vectors like i said AB so from A to B and the figure is a regular octagon as i said

 

[Talha Irfan]

yeah! u have to show the diagram...or we might misplace the positions of the letters!..and get wrong vectors

tbh; no diagram is given, yeah they are vectors like i said AB so from A to B and the figure is a regular octagon as i said

 

[Suchal Riaz]

Do we need to know this determinant method?

 

[Talha Irfan]

View attachment 28134
Do we need to know this determinant method?

No, Not In Syllabus
However, we use one-way determinant method when solving the simultaneous equations through matix method

(a b c d)(x y) = (e f)

 

[Talha Irfan]

N/NpM? when u get the answer tag me

sure difficult haan

 

[nidz]

http://papers.xtremepapers.com/CIE/...ematics - Additional (4037)/4037_s06_qp_2.pdf

Q.7 (b) can't really understand how they used the method shown in the marking scheme to get the answer. :/

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/...ematics - Additional (4037)/4037_s06_qp_2.pdf

Q.7 (b) can't really understand how they used the method shown in the marking scheme to get the answer. :/

We dont know how much candies are there and are not sure that all candies are three less or more.
There are 6 types of candies. Order does not matter. If u have to choose out of six types of candies and they must not be all same, then there are only two ways u can do this. Firstly, choose three out of 6 flavours, one each. Three different flavours. This will be 6C3.
Secondly u can choose two of any same flavour and one other of any other flavour. In total u can choose out of 6 for 2 candies and as they are same u must take it as choosing one out of 6 as u are actually choosing just a single flavour for two candies. Amd for the other single candy the flavour must be only one of the five remaining. So it simplifies to 6C1 X 5C1. Add them together u get 50 answer.
If u have more ambiguity please ask.

 

[Hamza Hameed Khan]

Part 3 please anyone ?

 

[Saad Mughal]

Part 3 please anyone ?

15 tickets are there,
If all the boys get tickets, then only 3 of the 8 girls can get tickets.
Therefore,
Number of ways = 12C12 * 8C3 = 56.

 

[Talha Irfan]

15 tickets are there,
If all the boys get tickets, then only 3 of the 8 girls can get tickets.
Therefore,
Number of ways = 12C12 * 8C3 = 56.

Bro, Please Solve This If You Can : ABCDFEFGH is a regular octagon and AB = p and BC = q ; Express AH in terms of p and q and show that
AE + BH + CG + DF = 2(2+√2)(q - √2p)

 

[ijlalm]

relative velocity involving i and j component vectors..i need notes regarding it.

 

[AhsanAfzal]

relative velocity involving i and j component vectors..i need notes regarding it.

https://www.xtremepapers.com/commun...st-your-doubts-here.4773/page-272#post-507467

 

[KWIKIW]

How to find out which to apply, permutation or combination?

 

[Mohammad Farzanullah]

Bro, Please Solve This If You Can : ABCDFEFGH is a regular octagon and AB = p and BC = q ; Express AH in terms of p and q and show that
AE + BH + CG + DF = 2(2+√2)(q - √2p)

this is such a diificult question that my sir said to award Rs 100 to the one who solves it correctly!

 

[robinhoodmustafa]

Saad Mughal
Help. How to find the coefficient of x^3

 

[Saad Mughal]

Saad Mughal
Help. How to find the coefficient of x^3

Are you sure that the coefficient of x^3 is given? Isn't it x^2?

 

[robinhoodmustafa]

Are you sure that the coefficient of x^3 is given? Isn't it x^2?

Yeaaa.. It is x^2

 

[sj0007]

How to find out which to apply, permutation or combination?

if there is a selection to take place it is combination but if there is an arrangement or placing objects in an order it is permutation......i hope it wuz helpful....gud luck 4 ur exam!!

 

[Saad Mughal]

Yeaaa.. It is x^2

Then,
(1-3x)(1+6C1 ax + 6C2 a^2x^2+ ......)
Coefficient of x^2 = 1(6C2)(a^2) - 3(6C1)(a)
24 = 15a^2 - 18a
Simplifying,
5a^2 - 6a - 8 = 0
5a^2 - 10a + 4a - 8 = 0
(5a+4)(a-2) = 0
Since a is positive,
a -2 = 0
a =2

 

[Saad Mughal]

Bro, Please Solve This If You Can : ABCDFEFGH is a regular octagon and AB = p and BC = q ; Express AH in terms of p and q and show that
AE + BH + CG + DF = 2(2+√2)(q - √2p)

This is in our book as well. It is very brain draining since it involves using a number of right-angled triangles and assumptions to calculate the vectors. Questions like these don't come in CIE for obvious reasons.
http://mathforum.org/library/drmath/view/55029.html
I searched and found this solution. Hope it helps.