Maths, Addmaths and Statistics: Post your doubts here!

[AhsanAfzal]

lol i was not talking about sourav

 

[Talha Irfan]

the last part....can anyone explain it

Step 1 :- Integrate the given equation to find velocity equation
Step 2: - Find when the particle was instantaneously at rest by putting v=0
Step 3 : - Again Integrate The VELOCITY equation to find distance equation
Step 4: - put t=5 and t=4then subtract the s at t=4 from t=5 ans
Step 5: - IF distance at 5 is less than 4, then put the instantaneous time to find the turning point (max. distance) of particle then subtract d at 5 from the distance that come
Step 6: - Now Subtract t=4 From t=5 (from step 5) to get answer - I HOPE NO MISTAKES ARE MADE

 

[Talha Irfan]

HELP ME WITH THIS
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)

 

[Talha Irfan]

PERMUTATION AND COMBINATION !!!
IF out of N points M are collinear, then what is the number of straight lines which could be drawn ????
With the answer, please give a short explanation too!!

 

[Snowysangel]

PERMUTATION AND COMBINATION !!!
IF out of N points M are collinear, then what is the number of straight lines which could be drawn ????
With the answer, please give a short explanation too!!

N/NpM? when u get the answer tag me

 

[Salman Rahman]

HELP ME WITH THIS
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)

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?

 

[shahzadi afia]

HELP ME WITH THIS
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)

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

 

[Suchal Riaz]

Saad Mughal. Finally this thread is our and there will be no one ask, "is it maths question?"
I have solved it but it was a very time taking question. Tell me if there is a better way to solve it.

 

[Saad Mughal]

Saad Mughal. Finally this thread is our and there will be no one ask, "is it maths question?"
I have solved it but it was a very time taking question. Tell me if there is a better way to solve it.
View attachment 28125

How did you do this?
I did this by,
-(x-1)(x-2)(x-k) = 0
Simplifying,
-x^3 + (3+k)x^2 - (3k+2)x + 2k = 0,
Then just use the remainder.
I completed it in 5 minutes. These questions take these long always.

 

[Suchal Riaz]

How did you do this?
I did this by,
-(x-1)(x-2)(x-k) = 0
Simplifying,
-x^3 + (3+k)x^2 - (3k+2)x + 2k = 0,
Then just use the remainder.
I completed it in 5 minutes. These questions take these long always.

How u got here: -(x-1)(x-2)(x-k) = 0 these all are factors of a polynomial so does it mean that their product will be zero?

 

[Saad Mughal]

How u got here: -(x-1)(x-2)(x-k) = 0 these all are factors of a polynomial so does it mean that their product will be zero?

Yes it will.
When you factorize x^2-4x-12 = 0
You write in the form, (x-4)(x+2) = 0,
Similarly, you can do this here.

 

[Suchal Riaz]

Yes it will.
When you factorize x^2-4x-12 = 0
You write in the form, (x-4)(x+2) = 0,
Similarly, you can do this here.

I still don't get it why it will equal to zero. :'(
If u factorial any polynomial it doesn't mean it will equal to zero unless it is itself equal to zero. Or if u out such a value in it that makes it equal to zero.

 

[Saad Mughal]

I still don't get it why it will equal to zero. :'(
If u factorial any polynomial it doesn't mean it will equal to zero unless it is itself equal to zero. Or if u out such a value in it that makes it equal to zero.

Look, by (x-1)(x-2)(x-k) = 0,
It means that,
x=1
x=2
x=k
Respectively, if you get this, then, you can change it back.
x-1 = 0
x-2 = 0
x-k = 0
These three are zero so their multiples will also be 0.
So, (x-1)(x-2)(x-k) = 0
These are kinda the basics of algebraic factorization that we learned in maths.

 

[Suchal Riaz]

Look, by (x-1)(x-2)(x-k) = 0,
It means that,
x=1
x=2
x=k
Respectively, if you get this, then, you can change it back.
x-1 = 0
x-2 = 0
x-k = 0
These three are zero so their multiples will also be 0.
So, (x-1)(x-2)(x-k) = 0
These are kinda the basics of algebraic factorization that we learned in maths.

But who told you that (x-1) is equal to zero?

 

[Saad Mughal]

But who told you that (x-1) is equal to zero?

Obviously when x=1,
x-1 = 0
Just subtract one from both sides,
Since 1 is a factor of the polynomial, it means that x=1, further, x-1 = 0, x-2=0, x-k=0.
Since, y=f(x)=0, you can easily say (x-1)(x-2)(x-k)=0.

 

[KWIKIW]

Can someone share with me notes on Permutation and combinations and also explain it in detail if possible?

 

[Saad Mughal]

Can someone share with me notes on Permutation and combinations and also explain it in detail if possible?

https://www.khanacademy.org/math/pr.../permutations_and_combinations/v/permutations

 

[Suchal Riaz]

Obviously when x=1,
x-1 = 0
Just subtract one from both sides,
Since 1 is a factor of the polynomial, it means that x=1, further, x-1 = 0, x-2=0, x-k=0.
Since, y=f(x)=0, you can easily say (x-1)(x-2)(x-k)=0.

Got it, I did not see that f(x) = 0

 

[Suchal Riaz]

https://www.khanacademy.org/math/pr.../permutations_and_combinations/v/permutations

This is useless. It does not tell how to use them with restriction and the CIE always give question with restriction.

 

[Saad Mughal]

This is useless. It does not tell how to use them with restriction and the CIE always give question with restriction.

I don't know. This is the only non-youtube video I know.