I'm an IGCSE Maths Teacher - Post your questions here.

[notnek01]

On May 9th and May 14th I will be posting solutions in this thread for IGCSE Maths 0580/22 and 0580/42.

 

[mak kings]

really?? ..it means i had done it right...i gt the same crct answer in the 1st attempt !! huh !!

 

[notnek01]

really?? ..it means i had done it right...i gt the same crct answer in the 1st attempt !! huh !!

Yes the answers in the mark scheme are definitely wrong. You can check them yourself :

Try putting -sin(50) into you calculator and then sin(330). The values will be different. And the same is true with sin(210).

 

[igcsestudent201396]

Could you please explain shear and stretch?!

 

[Rockzstar]

Hey
CAn u please solve the question of paper 42 may june 2010
the whole question 10 (a ,b ,c ,d ) is a complete doubt ... unable to solve it

Please helP me 0ut !

 

[notnek01]

Hey
CAn u please solve the question of paper 42 may june 2010
the whole question 10 (a ,b ,c ,d ) is a complete doubt ... unable to solve it

Please helP me 0ut !

Start with part (a). Which specific part can't you do? Post all your working/ideas.

 

[notnek01]

Could you please explain shear and stretch?!

It's very hard to explain this in a forum. I uploaded some revision notes on the topic in this thread: https://www.xtremepapers.com/community/threads/maths-shear-and-stretch-explain.24950

 

[SohaibJawad]

Q9, Paper 22, May June 2010...please help me with this question

 

[notnek01]

Q9, Paper 22, May June 2010...please help me with this question

Each length of wood is 200cm rounded to the nearest cm. So for each length of wood, the upper bound is 200.5 and the lower bound is 199.5.

There are 32 lengths of wood so you need to multiply the lower bound by 32 to find the lower bound of the total (199.5 x 32).

Finally, change the answer into metres.

 

[SohaibJawad]

Each length of wood is 200cm rounded to the nearest cm. So for each length of wood, the upper bound is 200.5 and the lower bound is 199.5.

There are 32 lengths of wood so you need to multiply the lower bound by 32 to find the lower bound of the total (199.5 x 32).

Finally, change the answer into metres.

Thank You

 

[SohaibJawad]

Plz help me with this question also. Q12 Paper 22 M/J 2010, Part (a)...according to my calculations the answer should be 401.1 but marking scheme says its 440 :\

 

[notnek01]

Plz help me with this question also. Q12 Paper 22 M/J 2010, Part (a)...according to my calculations the answer should be 401.1 but marking scheme says its 440 :\

Can you post your working to show me how you got 401.1?

 

[Rockzstar]

Start with part (a). Which specific part can't you do? Post all your working/ideas.

i cant understand how to start itself .... please let me know the solution to this question

 

[Relon]

Could you plz tell me if this calculator is allowed in Igcse ?

Casio fx-991es plus

 

[notnek01]

Could you plz tell me if this calculator is allowed in Igcse ?

Casio fx-991es plus

Yes it is allowed. And I recommend it

 

[ZaqZainab]

You are doing a great work Mister! And i wold be really happy if you helped me with one Little Question I have well prepared for my Boards but this question Pope outta nowhere!!! http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s10_qp_22.pdf Question 8 and 10
My Working for Question 8 is: a)sinx-cosx=0
sinx=y and cosx=y
so y-y=o
b) y-y=5
do we have to find the point where this statement is true really don't understand!
Question 10: P=(x+3)/x
Now i haven't done anything over here because i have got to move the x there to start is it not
P*x=x+3
Now what
Please Help I Would really appreciate it

 

[ZaqZainab]

The first one is equivalent to sin(x)=cos(x) so this is where the graphs of sin(x) and cos(x) intersect.

The second one is a bit trickier : you need to find the x-value where the difference between the curves is 0.5. For example, at x=90, sin(x)=1 and cos(x)=0 --> the difference between the curves is 1. So a difference of 0.5 will occur somewhere between the intersection point (difference = 0) and x=90 (difference = 1). You can use the graphs to find this.

I can't indestand the second one Like how do we use the graph to find the difference do i have to go trough ALL the points on the line :/ Please Help

 

[notnek01]

I can't indestand the second one Like how do we use the graph to find the difference do i have to go trough ALL the points on the line :/ Please Help

You don't need to go through every point but you do need to use trial and error to get close to a difference of 0.5.

E.g. At x=60, the difference between the curves is about 0.87-0.5 = 0.37

At x=90, the difference is 1 - 0 = 1.

So the answer is somewhere between x=60 and x=90.

Does that help?

 

[mak kings]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s08_qp_2.pdf
Variant2- Q11
i can only do it till here ,,cud u plz detail me the steps aftr it .!!
26^2=10^2-2*3x*x*cos(120)

 

[Batrisyia]

First of all,the answer to this:[ AA^-1 & A^-1 A ] Which one is the correct way to use and will the answer BE DIFF?

Here. Find P if Q=[2 -1 (4&0 below) ] and PQ= [ 6 -3 (8 & -2 below) ]