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

[AlphaWolf]

How do u draw a curve in a graph without messing it up, can u do it setions??? And in a cumulative frequency do can you draw them as straight lines.

 

[notnek01]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s12_qp_42.pdf
Question 9 tomorrow our exam so can u help plz

It's hard to help you with a drawing question on a forum.

Which specific part can't you do?

 

[notnek01]

How do u draw a curve in a graph without messing it up, can u do it setions??? And in a cumulative frequency do can you draw them as straight lines.

I like to draw a curve like an artist will do a sketch with a pencil. Go slowly between points and sketch the curve with a sharp pencil. Erase anything that you're not happy with.

A cumulative frequency graph should be a curve. Don't draw straight lines.

 

[inquisitiveness]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w11_qp_43.pdf q.12 b iii

 

[inquisitiveness]

and q.11 part (iv) and (v) for summer 2011 paper 43..

 

[notnek01]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w11_qp_43.pdf q.12 b iii

Any term in this sequence is equal to the term before it plus two times the term two before it.

So the term after 3413, 6827 is

6827 + 2*3413 = 13653.


Now call x the term before 3413, 6827.

Then 6827 = 3413 + 2x

--> x = (6827-3413)/2 = 1707

 

[notnek01]

and q.11 part (iv) and (v) for summer 2011 paper 43..

iv)

121 + 122 + .... + 200 = (1 + 2 + ... + 200) - (1 + 2 + ... + 120) = 200(201)/2 - 120(121)/2 = 12840

v)

2 + 4 + 6 + ... + 800 = 2(1 + 2 + 3 + ... + 400) = 2[(400)(401)/2] = 160400

 

[AlphaWolf]

oh ok i will try. However that graph was from my textbook.

 

[inquisitiveness]

w

iv)

121 + 122 + .... + 200 = (1 + 2 + ... + 200) - (1 + 2 + ... + 120) = 200(201)/2 - 120(121)/2 = 12840

v)

2 + 4 + 6 + ... + 800 = 2(1 + 2 + 3 + ... + 400) = 2[(400)(401)/2] = 160400

why are you putting 2 infront of 400 in the calculation part..i understand that 2(1+2+3+....400)but wouldnt even numbers mean that there are only 400 digits and not 800?so why not 400(400+1)/2

 

[inquisitiveness]

Any term in this sequence is equal to the term before it plus two times the term two before it.

So the term after 3413, 6827 is

6827 + 2*3413 = 13653.


Now call x the term before 3413, 6827.

Then 6827 = 3413 + 2x

--> x = (6827-3413)/2 = 1707

hm so if any term in this sequence is equal to the term before it plus two times the term two before it.....so for first three terms its 2, 3, 7,.......if we want to find 3rd term it would be 3+(2*3) which is not equal to 7..?how?

 

[$$AK$$]

How to do sequence questions in general?
Is there any formulas to learn?

 

[Mohammed salik]

m/j 2011 v42 Q9 C. thanx ALOT

 

[notnek01]

hm so if any term in this sequence is equal to the term before it plus two times the term two before it.....so for first three terms its 2, 3, 7,.......if we want to find 3rd term it would be 3+(2*3) which is not equal to 7..?how?

The term two before 7 is 2 not 3.

3 + (2*2) = 7.

 

[notnek01]

w
why are you putting 2 infront of 400 in the calculation part..i understand that 2(1+2+3+....400)but wouldnt even numbers mean that there are only 400 digits and not 800?so why not 400(400+1)/2

2(1+2+3+...400) is a sum of 400 numbers.

e.g.

2(1 + 2) = 2 + 4
2(1 + 2 + 3)= 2 + 4 + 6
...
2(1 + 2 + 3 + ... + 400) = 2 + 4 + 6 + ... + 800


400(400+1)/2 would give you 1 + 2 + 3 + ... 400

This is not the sum of the even numbers between 1 and 800.

 

[notnek01]

How to do sequence questions in general?
Is there any formulas to learn?

Useful formulas:

Arithmetic sequences with a constant first (common) difference: nth term = a + (n-1)d

where d is the common difference and a is the first term.

e.g. 3, 7, 11, 15, 19, ...

The common difference is 4 so the nth term is 3 + (n-1)*4 = 4n - 1


Quadratic sequences with a constant second difference: nth term = a + (n-1)d + 0.5(n-1)(n-2)c

where a is the first term, d is the difference between the first two terms and c is the constant second difference.

But my preferred method...

A quadratic sequence has nth term of the form an^2+bn+c. a is the second difference divided by 2 and you can find b and c by substituting values and solving the equations.

E.g. 4, 6, 10, 16, 24, ...

First differences are 2, 4, 6, 8, ... so second difference is constant (which means it's a quadratic sequences) and is equal to 2.

So a=2/2 = 1

Then the nth term must be n^2+bn+c. Substitute n=1 and n=2:

n=1 : 1 + b + c = 4
n=2 : 4 + 2b + c = 6

Solve these equations to find b and c.

 

[IGCESs monster]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w11_qp_42.pdf ##########

Can you draw or tell me how to do it

http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_w11_qp_42.pdf

 

[notnek01]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w11_qp_42.pdf ##########

Can you draw or tell me how to do it

http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_w11_qp_42.pdf

Which question?

 

[Mohammed salik]

m/j 2011 v42 Q9 C. thanx ALOT Plz

 

[notnek01]

m/j 2011 v42 Q9 C. thanx ALOT

Substitute different values of n:

n=1: The number of lines in the first diagram is 3 so

a(1^3) + b(1^2) + 1 = 3 --> a + b + 1 = 3 --> a + b = 2

n=2: The number of lines in the first two diagrams is 3 + 9 = 12 so

a(2^3) + b(2^2) + 2 = 12 --> 8a + 4b + 2 = 12 --> 8a + 4b = 10 --> 4a + 2b = 5

Can you solve the two equations to find a and b?

 

[Mohammed salik]

Substitute different values of n:

n=1: The number of lines in the first diagram is 3 so

a(1^3) + b(1^2) + 1 = 3 --> a + b + 1 = 3 --> a + b = 2

n=2: The number of lines in the first two diagrams is 3 + 9 = 12 so

a(2^3) + b(2^2) + 2 = 12 --> 8a + 4b + 2 = 12 --> 8a + 4b = 10 --> 4a + 2b = 5

Can you solve the two equations to find a and b?

oHHHHHHHHH Greattt! Thnx Alot!