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

[notnek01]

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)

I think you mean 10x^2 and not 10^2 right?

cos(120) = -0.5 (using your calculator) so it becomes:

26^2 = 10x^2 - 6x^2 *(-0.5)

--> 26^2 = 10x^2 + 3x^2

--> 26^2 = 13x^2

Can you finish it from here?

 

[notnek01]

0580/21 MJ 2010 Q21 a) How do we know which matrices can be worked out? what is the criteria.. im having difficulties with matrices

You can only multiply two matrices A x B if the number of columns in A is equal to the number of rows in B.

A good method: Write out the orders next to each other and if the inside numbers are the same then the multiplication is possible.


E.g. A is a matrix of order 2 x 3 and B is a matrix of order 3 x 2.

__________________________________________________

Is AB possible?

Write out the orders next to each other: 2 x 3, 3 x 2

The inside numbers are the same (3) so this multiplication is possible.

___________________________________________________

Is BA possible?

Write out the orders next to each other: 3 x 2, 2 x 3

The inside numbers are the same (2) so this multiplication is possible.

___________________________________________________

Is A^2 possible?

Write out the orders next to each other: 2 x 3, 2 x 3

The inside numbers are different (3 and 2) so this multiplication is not possible.

____________________________________________________

Does this help?

 

[notnek01]

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) ]

AA^-1 & A^-1 A will both be equal to the identity matrix (I).

But for this question you need to be careful. If you multiply PQ on the left by Q^(-1) you get:

Q^(-1) PQ

But this doesn't simplify.


If you multiply PQ on the right by Q^(-1) you get:

P Q Q^(-1) = P I = P


Does this make sense? You wouldn't get a question like this in an IGCSE paper.

 

[notnek01]

teacher i am not able to learn that transformation and symmetry
can you give me some suggestions

Transformations and symmetry are two very big topics so you need to be a lot more specific.

I can help you with questions or if there is an area of these topics that you don't understand.

If you want my revision notes please email me at [email protected]

 

[notnek01]

Yes thanks a lot so in other words the no of columns in the left hand matrix must equal to the no of rows in right hand matrix
can you also help me with may June 2011 paper 22 question 19 b and c please

b)

hh(x) = h(h(x)) = h(2x-3) = 2(2x-3) - 3 = ...


c)

g(x+1) = 2^(x+1)

fg(x+1) = f(g(x+1)) = f(2^(x+1)) = (2^(x+1))^2 = ...

 

[Batrisyia]

AA^-1 & A^-1 A will both be equal to the identity matrix (I).

But for this question you need to be careful. If you multiply PQ on the left by Q^(-1) you get:

Q^(-1) PQ

But this doesn't simplify.


If you multiply PQ on the right by Q^(-1) you get:

P Q Q^(-1) = P I = P


Does this make sense? You wouldn't get a question like this in an IGCSE paper.

Nah, i dont get what ur doing at all. It is a QUESTION from pp fyi. Some old papers. But, for my first question, its NOT ALL-THE-TIME same. I mean the product would be different. I know its an identity matrix but it depends on the question.

 

[notnek01]

Nah, i dont get what ur doing at all. It is a QUESTION from pp fyi. Some old papers. But, for my first question, its NOT ALL-THE-TIME same. I mean the product would be different. I know its an identity matrix but it depends on the question.

Do you know which paper it's from? It could be part of an older syllabus where you were expected to do this kind of manipulation.

The reason that you don't understand my post is because you're not used to this kind of matrix manipulation, probably because it's not part of the current IGCSE syllabus.

If the inverse of A exists then AA^-1 & A^-1 A are ALL-THE-TIME same.

 

[mak kings]

http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_s12_qp_22.pdf ...17c .....in this y dont v use the perception v hav that wen lines r parallel so their gradients r also parallel..therefore cant v use the same gradient value from the first equation??????????

 

[mak kings]

http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_w12_qp_23.pdf ---Q17


HI SIR, I cnnt do thw working at all !!!! plz plz lay down fr me the detailed steps dn working ..plz ...id b so thankful o ur help!!!!!!!!!!!! and yeh I evn saw the mark scheme ....and they showd with omitting the angle values...bt iam using them ..so cud plz show the DETAILED STEPS with all the angle values......itd b veryhelpul of u!

 

[notnek01]

http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_s12_qp_22.pdf ...17c .....in this y dont v use the perception v hav that wen lines r parallel so their gradients r also parallel..therefore cant v use the same gradient value from the first equation??????????

Your method is correct - parallel lines have equal gradients.

But first you need to write both equations in the form y=mx+c:

5x = 4y + 10 --> 4y = 5x - 10 --> y = (5/4) x - 10/4

2y = kx - 4 --> y = (k/2) x - 2

So (5/4) = (k/2) ...

 

[notnek01]

http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_w12_qp_23.pdf ---Q17


HI SIR, I cnnt do thw working at all !!!! plz plz lay down fr me the detailed steps dn working ..plz ...id b so thankful o ur help!!!!!!!!!!!! and yeh I evn saw the mark scheme ....and they showd with omitting the angle values...bt iam using them ..so cud plz show the DETAILED STEPS with all the angle values......itd b veryhelpul of u!

There are a few ways that you can do this question. If you don't like the method below, please tell me and I'll give you another one.

I'm going to find a relationship between the area of a sector and the arc length:

Area = a/360 x pi x radius^2

Arc length = a/360 x 2 x pi x radius

You can see that that for area compared to arc length, there is an extra radius and a missing 2.

So radius x Arc length / 2 = Area


For this question, Arc length = 4r so

Area = (5r) x 4r / 2 = 20r^2 / 2 = 10r^2


Does that make sense?

 

[StudentXYZ]

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

I am also having a problem with this question. I solved part (a) and (b) and c(ii), (iii) using the formula:
number of diagonals=(number of sides/2)* (number of sides-3)
but I was not able to solve c(i) and d.

 

[notnek01]

I am also having a problem with this question. I solved part (a) and (b) and c(ii), (iii) using the formula:
number of diagonals=(number of sides/2)* (number of sides-3)
but I was not able to solve c(i) and d.

Where did you get that formula from? Did you work it out yourself?

It is easier to look at the sequence and see the pattern:

0, 2, 5, 9, 14, ...

The difference are 2, 3, 4, 5, ... so the sequence continues like this:

0, 2, 5, 9, 14, 20, 27, 35, 43, ...

And you can use this pattern to work answer a and b.


c)

You already know the formula: number of diagonals = n/2(n-3)

So comparing this with 1/p * n (n-q), you can see that p = 2 and q = 3.


If you didn't know the formula already, to find p and q, substitute some values of n into the formula e.g. for n=3 you know the number of diagonals is 0 so:

3/p (3-q) = 0

And you could do the same for e.g. n=4 and use your equations to find p and q.

 

[notnek01]

teacher can you provide me with some hand written answers of student so that i can hows the format and will be able to do it properly

I'm sorry, I won't be able to provide you with this.

 

[notnek01]

bu can you solve past paper and give me IF POSSIBLE

I'm sorry, I can't give you that either.

I do have revision notes which include many past paper questions with full solutions (not handwritten). If you want these, please email me at [email protected]

 

[zain1245780]

sir, can you plz explain me the answers of Q.6, Q.16b, Q17b, Q18, Q19, Q20a, Q20c. I have tried to do these but the answers in the mark scheme were different. so plz explain how to do these and for Q17b how will we come to know the type of transformation? thanks in advance.

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w12_qp_22.pdf

 

[StudentXYZ]

Where did you get that formula from? Did you work it out yourself?

It is easier to look at the sequence and see the pattern:

0, 2, 5, 9, 14, ...

The difference are 2, 3, 4, 5, ... so the sequence continues like this:

0, 2, 5, 9, 14, 20, 27, 35, 43, ...

And you can use this pattern to work answer a and b.


c)

You already know the formula: number of diagonals = n/2(n-3)

So comparing this with 1/p * n (n-q), you can see that p = 2 and q = 3.


If you didn't know the formula already, to find p and q, substitute some values of n into the formula e.g. for n=3 you know the number of diagonals is 0 so:

3/p (3-q) = 0



And you could do the same for e.g. n=4 and use your equations to find p and q.

Thank you very much. Actually I got the formula from the mark scheme. I thought we were suppose to memorize it. I didn't recognize this as a sequence question.
What about part d. I tried n+1=n/2(n-3)+30, but I can't get an answer

 

[zain1245780]

plz help me with these questions with explanation to their answers. Q5, Q13, Q15, Q20, Q18, Q19d. http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w12_qp_21.pdf

 

[notnek01]

Thank you very much. Actually I got the formula from the mark scheme. I thought we were suppose to memorize it. I didn't recognize this as a sequence question.
What about part d. I tried n+1=n/2(n-3)+30, but I can't get an answer

The only nth terms that you need to be able to find are for 1) arithmetic series and 2) simple geometric series.

1) E.g. 3, 7, 11, 15, 19, ... These have nth terms of the form an+b

2) E.g. 3, 9, 27, 81, 243, 729,... These have nth terms of the form a^n

The nth terms of all other types of sequences can usually be found either by using previous sequences or by substituting values like in this question. Some sequences e.g. 1, 4, 16, 25, 36, ... you are expected to recognise.


d)

You could use algebra here, but you will end up with a quadratic equation which seems a but unnecessarily just for one mark!

Have a look at the sequence again: 0, 2, 5, 9, 14, 20, 27, 35, 43, ...

The question wants you to find two consecutive terms in this sequence where the difference is 30.

E.g. n=5, n+1 = 6. A polygon with 5 sides has 5 diagonals and a polygon with 6 sides has 9 diagonals so the difference is 4.

You need to find where the difference is 30. Remember, the difference goes up by 1 each time.


For sequences questions like this, it's always best to look for patterns and use logic before using algebra.

 

[notnek01]

sir, can you plz explain me the answers of Q.6, Q.16b, Q17b, Q18, Q19, Q20a, Q20c. I have tried to do these but the answers in the mark scheme were different. so plz explain how to do these and for Q17b how will we come to know the type of transformation? thanks in advance.

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w12_qp_22.pdf

plz help me with these questions with explanation to their answers. Q5, Q13, Q15, Q20, Q18, Q19d. http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w12_qp_21.pdf

You've asked a huge amount of questions. Please choose a maximum of three that you want me to help you with. Once you can do them, you can ask some more.

If you have this many questions then it's probably best to sit down with your teacher and not get help here.