中文歌詞! (6 Viewers)

[wrong_turn]

we ar e doing industrial chem as well. i hope it is easier than this shit we have been doing. i have a 30% titrations test when i get back to school.

that really sux! i shouldn't have chosen chemistry at all! why didn't i do economics instead?!

 

[Jago]

haha, eco isnt as easy as it sounds. Fun, useful but not easy.

 

[withoutaface]

Bio was an awesome subject, just paid attention in class did sfa else and managed 91

 

[(^o^)]

shut up bout bio already!
*cries*

 

[withoutaface]

shut up bout bio already!
*cries*

Would you prefer I talk about maths?

 

[Jago]

Please do.

/end nerdism

 

[paper cup]

we ar e doing industrial chem as well. i hope it is easier than this shit we have been doing. i have a 30% titrations test when i get back to school.

that really sux! i shouldn't have chosen chemistry at all! why didn't i do economics instead?!

yeh I wish I did eco as well.
alas

 

[withoutaface]

Suppose a is a root of P(x) (all coefficients real), and let |m represent the conjugate of m.

ax^n+bx^(n-1)+....=0
|(ax^n+bx^(n-1)+...)=|0
Using |(z+w)=|z+|w
|(ax^n)+|(bx^(n-1))+...=0
Using |(zw)=|z|w
(|a)(|x)^n+(|b)(|x)^(n-1)+...=0
Since when Im(m)=0, m=|m
a(|x)^n+b(|x)^(n-1)+...=0
:. if x is a root of P(x) then so is |x
If there's enough demand I'll do pythagoras.

 

[wrong_turn]

hmmm i don't want to do uni maths on that note, i don't want to do maths altogether!

i chose the wrong subjects...

 

[withoutaface]

Take this square inside a square. The congruence proof involves angles on a straight line etc but it's really not that interesting, so we'll just assume all the triangles have sides a, b, c.

Now first we find the area of the square one way, and that is
A=(a+b)^2=a^2+2ab+b^2

And now the other

A=c^2+4*(1/2 ab)=c^2+2ab

Now we equate them:

c^2+2ab=a^2+2ab+b^2
c^2=a^2+b^2
QED.

 

[Jago]

Stop it, you're making 4u maths people wet...

 

[withoutaface]

Stop it, you're making 4u maths people wet...

That's year 8 maths dude.

 

[wrong_turn]

That's year 8 maths dude.

his point exactly. people who indulge in the complex of things don't know the basis of things.

 

[paper cup]

Suppose a is a root of P(x) (all coefficients real), and let |m represent the conjugate of m.

ax^n+bx^(n-1)+....=0
|(ax^n+bx^(n-1)+...)=|0
Using |(z+w)=|z+|w
|(ax^n)+|(bx^(n-1))+...=0
Using |(zw)=|z|w
(|a)(|x)^n+(|b)(|x)^(n-1)+...=0
Since when Im(m)=0, m=|m
a(|x)^n+b(|x)^(n-1)+...=0
:. if x is a root of P(x) then so is |x
If there's enough demand I'll do pythagoras.

pythagoras wasn't even discovered by pythagoras.

 

[Jago]

That's year 8 maths dude.

I didn't actually read what you typed...i just assumed.

 

[LMF^^]

Readin mathz wit ^ and * and / and all that crap just turnz you away, even when I did 4U I never looked fo help on the net becauz of all thoze symbolz.

 

[xiao1985]

ahahah nerd justin... here's a q for ya:

prove sqrt 2 exist...

 

[jm1234567890]

ahahah nerd justin... here's a q for ya:

prove sqrt 2 exist...

that makes no sence, you mean prove root 2 is irrational?

 

[xiao1985]

no no... prove sqrt 2 exist...
as in... u know 1 exist, cuz u can count it...
u know 1/2 exist, cuz u can divide 1 into two , end up in 2 things which if u put together u get 1...

but how do u know sqrt2 exist?! =p

 

[jm1234567890]

no no... prove sqrt 2 exist...
as in... u know 1 exist, cuz u can count it...
u know 1/2 exist, cuz u can divide 1 into two , end up in 2 things which if u put together u get 1...

but how do u know sqrt2 exist?! =p

well.... root 2 can be expanded into a taylor series which will converge to some number.

if that's what you mean