O-Level Additional Math

[Belle2011]

Hello,

May I ask where can I get books/resources on math olympiad questions suitable for lower sec?
Thank-you.

cheers,
Belle.

[mum_sugoku]

JadeDry:

Hello,


Can you please help me with the following question?

Make \"t\" the subject of the formula:

T= 2(pi) * (the square root of(( (t^2) + (k^2)) / 2gt))

I have checked the answer which is:
t = (+ or -) (1/ (2*pi)) *(The square root of((2gt(T*2)) - ((4*(pi squared))(k squared))

I do not understand how they have used the lower case \"t\" in the formula.

Is it correct?

Thanks in advance.

Ya by right \"t\" shouldn't be in there if it is the subject.. So I too think that answer is incorrect.

Anyway this is a seemingly tedious sum (I mean the working). You'll need to apply the standard formula x=[-b +- sq rt (b^2 -4ac) ]/2a to solve it.. Here's what I've got after working it out, not sure if it's correct though:

t ={ (T^2)g +- {sq rt[(T^4)(g^2) - 16(pi^4)(k^2)]} } / (4pi^2)

[Xiao Hu]

Hi FrekiWang, other maths guru here,


Need help on this Q20, taken from Cedar Girl's 2011 Prelim2 papers. I just couldn't figure it out. Answer in the answer key is 90 degrees.

http://i40.tinypic.com/t9wj9c.jpg\">

Thanks in advance,
Xiao Hu

[njcstudent]

Hi, I am a sec 1 student in an IP school. Can anyone please recommend any good IP assessment books? Thanks in advance.

[mum_sugoku]

Xiao Hu:

Hi FrekiWang, other maths guru here,


Need help on this Q20, taken from Cedar Girl's 2011 Prelim2 papers. I just couldn't figure it out. Answer in the answer key is 90 degrees.

http://i40.tinypic.com/t9wj9c.jpg\">

Thanks in advance,
Xiao Hu

rhombus -> all 4 sides are equal, and AB//DC, AD//BC

a) DL is common to both triangles -->side
AD=CD -->side
tri ABD=tri CBD (SSS), therefore <ADL=<LDC -->angle

therefore tri ALD= tri CLD (SSA)

b) <ADL=<LBX (corr <), and
<DLX= <LBX+90 (ext < of tri)
but <DLX also = <DAL + <ADL (ext < of tri), and <ADL=<LBX
==> <DAL =90
==><LCD=90 (since tri ALD=tri CLD)

[Xiao Hu]

mum_sugoku:

Xiao Hu:

Hi FrekiWang, other maths guru here,


Need help on this Q20, taken from Cedar Girl's 2011 Prelim2 papers. I just couldn't figure it out. Answer in the answer key is 90 degrees.

http://i40.tinypic.com/t9wj9c.jpg\">

Thanks in advance,
Xiao Hu

rhombus -> all 4 sides are equal, and AB//DC, AD//BC

a) DL is common to both triangles -->side
AD=CD -->side
tri ABD=tri CBD (SSS), therefore <ADL=<LDC -->angle

therefore tri ALD= tri CLD (SSA)

b) <ADL=<LBX (corr <), and
<DLX= <LBX+90 (ext < of tri)
but <DLX also = <DAL + <ADL (ext < of tri), and <ADL=<LBX
==> <DAL =90
==><LCD=90 (since tri ALD=tri CLD)

Hi mum_sugoku,
Thanks so much! Yes, on hindsight, now I know I should have proceeded by using the congruent triangle results from part a.

I really appreciate your help,
Xiao Hu.

[mum_sugoku]

Xiao Hu:

Hi mum_sugoku,
Thanks so much! Yes, on hindsight, now I know I should have proceeded by using the congruent triangle results from part a.

I really appreciate your help,
Xiao Hu.

You are welcome . BTW if I'm not wrong, when question asks you to \"hence\" solve the subsequent problem, you are expected to make use of previous result to work out the solution, else you may be penalised for not following instruction!!!

[Herbie]

How to solve 499^2

[koguma]

Herbie:

How to solve 499^2

Assume 499^2 = 499 * 499,
Since 499 = (500-1)
so solve (500-1)^2 using (x-y)^2 formulae in algebra topic

[Herbie]

koguma:

Herbie:

How to solve 499^2

Assume 499^2 = 499 * 499,
Since 499 = (500-1)
so solve (500-1)^2 using (x-y)^2 formulae in algebra topic

Thanks! Thanks