Maths 32? Discussion 24 Hours

[aqibmansoor97]

anyone here giving Accounting P3??

me hopefully it's easier than today ru giving acct p3 we r the first batch to give pray cash flow doesn't come I think question would be from non profit partnership I am sure it comes every year 3 standard costing 4 investment appraisal 5 maybe ratios 6 it van be consignment accts

 

[farhan141]

So just out of curiosity, what was the answer to the part a of complex number question?

Something like √2 +- i

 

[Faiqa Ahmad]

IF P3 threshold drops in the lows 50's has happened before everyone is hopeful iA and s1 goes near perfect still a chance for A*

Really? Because my P3 was really bad... I had a little mishap in the middle (forgot to drink water the entire day so nearly fainted) and that's when my paper started to go downhill although I was good at P3. I even left some parts unanswered and couldn't even do the 'show' questions... Yep that's how bad my day was. Idk if there's still a chance for A* TBH.

 

[Abdullah Zahid]

Really? Because my P3 was really bad... I had a little mishap in the middle (forgot to drink water the entire day so nearly fainted) and that's when my paper started to go downhill although I was good at P3. I even left some parts unanswered and couldn't even do the 'show' questions... Yep that's how bad my day was. Idk if there's still a chance for A* TBH.

Then i guess aim for an A atleast sometimes aiming for A* makes your paper even worse A is not bad either

 

[asadalam]

Something like √2 +- i

Close,it was +- √2 + i

 

[asadalam]

Ill try to remember some of the answers if you want to know still

 

[Abdullah Zahid]

shortest distance was 2.5 confirm and +-√2+i was my complex ans

 

[Dark Destination]

How can +- come first when there is +- in the "middle" of the quadratic formula
Do you rember the question ? It was something like iZ^2+2Z+(something)=0

Cause the co-efficients of this quadratic equation are not real, and therefore the roots will be complex but won't be conjugate pairs of each other. They are only conjugate pairs if co-efficients are real.

 

[Dark Destination]

Please try to remember the question, things will be more clear I am sure about the iz^2+2z part it's some something like 2i at the end

I don't remember the question, but your answer is definitely wrong cause it shows +/- before imaginary part which means the complex numbers are conjugates of each other when they shouldn't be.
And 2 people have already mentioned above that their answer was different.

 

[asadalam]

You are absolutely right , the question was iz^2+2z-2i=0 I remember,
Shitt, I hope the thresholds are really low this time, did you solve the diff, equation by the way ?

In it least value of x was like 3(sqrt3)/2

 

[asadalam]

How the hell did you guys solve differential equation ?

Alright i'll try and recall,split them into 2 parts,on left was 1/x which would simple give lnx
Right was sin20/3+cos20 (0 is theeta )
Rearrange so that upper part was derivative of lower i.e -1/2 (-2sin20/3+cos20)
As upper is derivative it just becomes -1/2 ln(3+ cos20)
Then just simplify with e and all that.

 

[asadalam]

Holy hell, thats elegant !
Are you appearing for further maths too by any chance?

Haha no but my p3 went great alhamdulillah

 

[Albert Lianto]

I would just like to ask if it would be enough to prove that ABCD, already known as a parallelogram, could be proven a rhombus by proving that AC is perpendicular to BD, right? Because some of my friends proved it by proving AB=BC, any one of these two are okay, still, right?

I've got people saying proving that the diagonals are perpendicular may imply that ABCD is a kite, but if ABCD is already known to be a parallelogram, ABCD hence must be a rhombus, right? I wonder if proving it this way is sufficient..

 

[farhan141]

Close,it was +- √2 + i

Yep exactly that . Sorry my memory is weak

 

[farhan141]

To prove the parallelogram is a rhombus, you must show that all their sides are equal

That can be square as well. After proving sides equal u have to prove that the diagonals are not equal.

 

[farhan141]

Alright i'll try and recall,split them into 2 parts,on left was 1/x which would simple give lnx
Right was sin20/3+cos20 (0 is theeta )
Rearrange so that upper part was derivative of lower i.e -1/2 (-2sin20/3+cos20)
As upper is derivative it just becomes -1/2 ln(3+ cos20)
Then just simplify with e and all that.

Can u write the question on a paper or something? I dont remember it and now I'm freaking out

 

[Albert Lianto]

To prove the parallelogram is a rhombus, you must show that all their sides are equal

That can be square as well. After proving sides equal u have to prove that the diagonals are not equal.

I feel like if a parallelogram has its diagonals perpendicular, it is automatically a rhombus. That's why I didn't prove the sides are equal. I reckon using one of those two ways (either proving that the diagonals are perpendicular or proving that the length of all the sides are the same) are enough to prove ABCD is a rhombus. And technically, a square satisfies the conditions for a rhombus, so a square is essentially a rhombus, no? Square is like a subset in rhombus..

 

[farhan141]

Square has equal diagonals as well as sides while rombus has one longer diagonal and one shorter

 

[asadalam]

That can be square as well. After proving sides equal u have to prove that the diagonals are not equal.

Or that angles are not equal to 90,by finding dot product of vectors and showing its not zero,then it leaves only rhombus and square is eliminated as well

Can u write the question on a paper or something? I dont remember it and now I'm freaking out

Haha just tell me your ans of least value of x,that'll confirm if you did it right or not

 

[Master Minds]

Math P3 has been uploaded along with answers on https://justpastpapers.com/
So, let's share our expected score for this exam. I am expecting around 67