Mathematics: Post your doubts here!

[The Sarcastic Retard]

Does anyone takes S2?

 

[The Sarcastic Retard]

Q9 (ii) check my mistake. Thanks.

 

[Copy Cat]

(iv) & (v) part

 

[The Sarcastic Retard]

View attachment 59541
Q9 (ii) check my mistake. Thanks.

View attachment 59540

DarkEclipse
qwertypoiu
Anum96
Anyone?????????????????

 

[Anum96]

View attachment 59541
Q9 (ii) check my mistake. Thanks.

View attachment 59540

DarkEclipse
qwertypoiu
Anum96
Anyone?????????????????

 

[Anum96]

View attachment 59537
View attachment 59538
Ok so i know how to solve it but the thing is here they have classed them as 0-9 then 10-19 then 20-34 and so on so the midpoints will be 4.5 and 14.5 and 27 etc...
Why can't you split the data like 0≤X<10 and 10≤X<20 and 20≤X<35 and so on??

The first data says <10 which means 10 is excluded therefore it should be read as 0-9 only.

 

[Rizwan Javed]

View attachment 59545
(iv) & (v) part

There're 6 colors. Select three colors from these 6. This can be done in 6C3 ways.

Now from those three colors that have been selected, there's one colour which is repeated. So select 1 colour from the 3 colous selected which is to be repeated. This can be done in 3C1 ways.

Now the selections have been made, so arrange them in the four places.

So the total arrangements possible will be:

6C3 * 3C1 * 4!/2! = 720 arrangements

 

[Rizwan Javed]

View attachment 59545
(iv) & (v) part

In the (v) part there are three different possibilities:
1. All the pegs are of different colors.
2. Three different colors.
3. Two different colors.

Deal these possibilities, one by one:

All different colours:

Select 4 colours from the 6 colours given. This will done in 6C4 ways. Then arrange these in the 4 places (4!).
So the total no. of ways will be : 6C4 * 4! = 360

Three different colours:
We've already done that in the previous part. There're 720 arrangements for this.

2 different colours:
Select two colours from the given 6. This will done in 6C2 ways. Now arrange the 4 pegs, keeping in mind that there are two pairs of pegs, each pair having a similar colour.
So the total possible arrangements will be: 6C2 * 4!/(2!*2!) = 90

Sum all of these cases, you'll get the required result:

360 + 720 + 90 = 1170 arrangements.

 

[Anum96]

Is there a mark for solving simultaneous equations? I mean, in the ms they give a method mark for "solving simultaneously". Does that mean if we skip the step and use the calc to do it directly, we will lose a mark?

Yes. working has a separate mark and answer has a separate one mark. Writing answer directly will give you just 1 mark.

 

[Anum96]

http://papers.gceguide.com/A Levels/Mathematics (9709)/9709_w15_qp_33.pdf
can anyone help me with 9b pls

|z| = 5
---> (0,0) = 5
Make a circle within center origin and radius 5

|z - 5| = |z|
----> (5,0) = 0
Mark the point and make the perpendicular..

Find the arguments and modulus and state the answer

 

[Farhan Ismail]

 

[Copy Cat]

There're 6 colors. Select three colors from these 6. This can be done in 6C3 ways.

Now from those three colors that have been selected, there's one colour which is repeated. So select 1 colour from the 3 colous selected which is to be repeated. This can be done in 3C1 ways.

Now the selections have been made, so arrange them in the four places.

So the total arrangements possible will be:

6C3 * 3C1 * 4!/2! = 720 arrangements

In the (v) part there are three different possibilities:
1. All the pegs are of different colors.
2. Three different colors.
3. Two different colors.

Deal these possibilities, one by one:

All different colours:

Select 4 colours from the 6 colours given. This will done in 6C4 ways. Then arrange these in the 4 places (4!).
So the total no. of ways will be : 6C4 * 4! = 360

Three different colours:
We've already done that in the previous part. There're 720 arrangements for this.

2 different colours:
Select two colours from the given 6. This will done in 6C2 ways. Now arrange the 4 pegs, keeping in mind that there are two pairs of pegs, each pair having a similar colour.
So the total possible arrangements will be: 6C2 * 4!/(2!*2!) = 90

Sum all of these cases, you'll get the required result:

360 + 720 + 90 = 1170 arrangements.

Thanks a ton!!!!

 

[Konstantino Nikolas]

Yes. working has a separate mark and answer has a separate one mark. Writing answer directly will give you just 1 mark.

Thank you.

 

[Konstantino Nikolas]

View attachment 59550

ln(2 + e^-x) = 2
2 + e^-x = e^2
e^-x = e^2 - 2
-xlne = e^2 - 2
-x = 5.39
x = -5.39

 

[The Sarcastic Retard]

ln(2 + e^-x) = 2
2 + e^-x = e^2
e^-x = e^2 - 2
-xlne = e^2 - 2
-x = 5.39
x = -5.39

No sir, u made a mistake.
-xlne = ln5.39
x = -1.68

 

[The Sarcastic Retard]

View attachment 59547View attachment 59548

Oh thanks! I spotted out my mistake.

 

[The Sarcastic Retard]

Are P3 vectors, differential equation and complex numbers tough????

 

[Konstantino Nikolas]

No sir, u made a mistake.
-xlne = ln5.39
x = -1.68

Oops! You're right.

Are P3 vectors, differential equation and complex numbers tough????

Differential equations are pretty easy. Vectors and complex numbers - the first sub-questions are generally easy but the further sub-questions require good thinking and practice.

 

[The Sarcastic Retard]

Oops! You're right.



Differential equations are pretty easy. Vectors and complex numbers - the first sub-questions are generally easy but the further sub-questions require good thinking and practice.

I dont find them that difficult, its just thinking a bit and that's all, but my marks goes only in this parts Hence I am poor at it.. Anyways...

 

[Konstantino Nikolas]

I dont find them that difficult, its just thinking a bit and that's all, but my marks goes only in this parts Hence I am poor at it.. Anyways...

Hahaha ... That's what I'm talking about. Practice more and you will hopefully get it. Complex numbers and vectors also require a bit of imagination, or rather, visualisation.