O-Level Additional Math

[nounou]

Hi, thank you very much for your help. I understand your workings.


Q: how to find 1+2+3+4 .... +2009+2010+2011+2012 ???? :?:

Please help. :thankyou: :lovesite:

koguma:

nounou:

Hi, need help on another Q. Kindly help. Thank you.

Evaluate (2012^2 - 2011^2) + (2010^2 - 2009^2) + ...
+ (4^2 - 3^2) + (2^2 - 1^2)

:? :?:

:thankyou:

Use this formulae : a^2 − b^2 = (a + b)(a − b)

2012^2 - 2011^2
= (2012+2011)(2012-2011)
= 2012+2011

2010^2 - 2009^2
= (2010+2009)(2010-2009)
= 2010+2009

4^2 - 3^2
= (4+3)(4-3)
= 4+3

2^2 - 1^2
= (2+1)(2-1)
= (2+1)

so you need to add 1+2+3+4 .... +2009+2010+2011+2012 to get the ans.

hopefully someone else can give a shorter working answer.

[Jtutor]

Hi nounou,


To solve for 1+2+3+…+2010+2011+2012, you can group them together as follow:
(1+2012)+(2+2011)+(3+2010)+…
There will be a total of 2012/2=1006 pairs.
Hence 1006 x 2013 = 2,025,078.

Hope it helps.

Cheers,
Jtutor

[nounou]

:thankyou: :udaman: :lovesite:

Jtutor:

Hi nounou,

To solve for 1+2+3+..............+2010+2011+2012, you can group them together as follow:
(1+2012)+(2+2011)+(3+2010)+...
There will be a total of 2012/2=1006 pairs.
Hence 1006 x 2013 = 2,025,078.

Hope it helps.

Cheers,
Jtutor

[Yamong]

Can anyone recommend physics tution arround bukit timah?

[htn]

Hi


Please help.

Secondary Two Maths.

Factorise

1)2r^2-5r-3

Expand and simplify the following expressions
1)(a+b)^2 - (a-b)^2

2)(x+2y)(x-6y)- (x-3y)(x-y)

TIA

[koguma]

htn:

Hi


Please help.

Secondary Two Maths.

Factorise

1)2r^2-5r-3

Expand and simplify the following expressions
1)(a+b)^2 - (a-b)^2

2)(x+2y)(x-6y)- (x-3y)(x-y)

TIA

This is basic algebra which is taught in Sec 1. Hope you understand the steps.

1)2r^2-5r-3
= (2r+1) (r-3)

Note : for the 2 qns below, since outside bracket the sign is \"-\", when you remove bracket you must change the sign of all terms inside the bracket

2)(a+b)^2 - (a-b)^2
Use (a + b) ^2 = a^2 + 2ab + b^2
Use (a − b) ^2 = a ^2 − 2ab + b ^2

(a+b)^2 - (a-b)^2
= a^2 + 2ab + b^2 – (a ^2 − 2ab + b ^2)
= a^2 + 2ab + b^2 – a ^2 + 2ab - b ^2
= 4ab

3)(x+2y)(x-6y)- (x-3y)(x-y)
= x^2 -6xy+2xy-12y^2 – [x^2 –xy-3xy+3y^2 ]
= x^2 -6xy+2xy-12y^2 – x^2 +xy + 3xy -3y^2
= -15y^2

[kuts]

Hi, Can anybody help solve this problem. Very urgent. Thanks in advance



Marks\t 0\t1\t2\t3\t4
No pupils\t2\t1\t3\t1\tP

Find the value of p if the median is 2.5 marks

[Skyed]

kuts:

Hi, Can anybody help solve this problem. Very urgent. Thanks in advance



Marks\t 0\t1\t2\t3\t4
No pupils\t2\t1\t3\t1\tP

Find the value of p if the median is 2.5 marks

P = 5

[kuts]

Skyed:

kuts:

Hi, Can anybody help solve this problem. Very urgent. Thanks in advance



Marks\t 0\t1\t2\t3\t4
No pupils\t2\t1\t3\t1\tP

Find the value of p if the median is 2.5 marks

P = 5

Thanks, btw, could u please explain me

[Skyed]

kuts:

Skyed:

[quote=\"kuts\"]Hi, Can anybody help solve this problem. Very urgent. Thanks in advance



Marks\t 0\t1\t2\t3\t4
No pupils\t2\t1\t3\t1\tP

Find the value of p if the median is 2.5 marks

P = 5

Thanks, btw, could u please explain me[/quote]Hi, median is the middle mark. Assuming you take 2.5 as a mark on it's own, the number of students who got 0, 1, 2, and 2.5 is a total of 7. Then because you need 2.5 to be the number in the middle, the number of people who got 3 marks, 4 marks and the 2.5 mark has also to be equal to 7. Therefore, 7 -1 -1 = 5.

To double check, you can add up all the values (2+1+3+1+5) = 12 and then +1 = 13 divided by 2 to find the median, which is 2.5.