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How many do u pull?

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_ELiTeCaRR_
go away you boring little person

Oh dear, talking to yourself! It's the first sign of madness you know!

The second is growing hair on the palms of your hands.

The third is looking for it
Danithestudent
So let me guess, your first name is Jack and your surname is Shit.

Thus being, you are NOTHING Weeeeee

Nothing is making any logical sense tonight.


lol let's not be too pedantic :smile:
_ELiTeCaRR_
go away you boring little person


Oh damn he called me boring, must cry...

How about no? How about you go away you little 13-year-old Maxim-reading despo?
XTinaA
Oh damn he called me boring, must cry...

How about no? How about you go away you little 13-year-old Maxim-reading despo?


no
XTinaA
Oh damn he called me boring, must cry...

How about no? How about you go away you little 13-year-old Maxim-reading despo?

i feel like having a banter - can't sleep

why do you have an avatar of a female when you're male?
what are you some kind of weirdo?
_ELiTeCaRR_
i feel like having a banter - can't sleep

why do you have an avatar of a female when you're male?
what are you some kind of weirdo?


nothing wrong with that dazya, come on....
He's Xtina to my Britney Spears.
sillynarb2
nothing wrong with that dazya, come on....


As soon as you need another example, guess what :biggrin:
sillynarb2
nothing wrong with that dazya, come on....


well i mean he has a girls name too " tina "
Danithestudent
He's Xtina to my Britney Spears.


And there's no Madonna! :wink:
sillynarb2
nothing wrong with that dazya, come on....
Sillynarb2, do realise how much you resemble Kiera Knightley? you should go to one of those lookalike agencies :smile:

Jk sorry couldn't help it.
I got + rep once because someone thought I was my avatar who just happened to be Britney
Danithestudent
Sillynarb2, do realise how much you resemble Kiera Knightley? you should go to one of those lookalike agencies :smile:

Jk sorry couldn't help it.
I got + rep once because someone thought I was my avatar who just happened to be Britney


i get told that a lot
Find the area of the finite region bounded by the x-axis and the curve with equation y = (x + 1)(x - 2).

Curve's equation: y = x^2 - x - 2

The curve crosses the x-axis at x = -1 and x = 2.

To calculate the area of the required finite region, find:

Definite integral (limits 2 to -1) x^2 - x - 2 dx.

= [(x^3)/3 - (x^2)/2 - 2x] with limits (2 to -1)

= (8/3 - 2 - 4) - (-1/3 - 1/2 + 2)

= -10/3 - 7/6
= -4.5

Therefore the area of the required finite region = 4.5

N:B The fact that the calculated area turned out to be -ve indicates that the finite region actually lies below the x-axis; hence explaining the -ve sign infront of the value "4.5".
Danithestudent
Sillynarb2, do realise how much you resemble Kiera Knightley? you should go to one of those lookalike agencies :smile:



omg you're so funny!
sillynarb2
i get told that a lot
I feel somewhat strangely attracted to you.......:cool:

*giggles*
bono
Find the area of the finite region bounded by the x-axis and the curve with equation y = (x + 1)(x - 2).

Curve's equation: y = x^2 - x - 2

The curve crosses the x-axis at x = -1 and x = 2.

To calculate the area of the required finite region, find:

Definite integral (limits 2 to -1) x^2 - x - 2 dx.

= [(x^3)/3 - (x^2)/2 - 2x] with limits (2 to -1)

= (8/3 - 2 - 4) - (-1/3 - 1/2 + 2)

= -10/3 - 7/6
= -4.5

Therefore the area of the required finite region = 4.5

N:B The fact that the calculated area turned out to be -ve indicates that the finite region actually lies below the x-axis; hence explaining the -ve sign infront of the value "4.5".


Umm, talk about off-topic...
bono
Find the area of the finite region bounded by the x-axis and the curve with equation y = (x + 1)(x - 2).

Curve's equation: y = x^2 - x - 2

The curve crosses the x-axis at x = -1 and x = 2.

To calculate the area of the required finite region, find:

Definite integral (limits 2 to -1) x^2 - x - 2 dx.

= [(x^3)/3 - (x^2)/2 - 2x] with limits (2 to -1)

= (8/3 - 2 - 4) - (-1/3 - 1/2 + 2)

= -10/3 - 7/6
= -4.5

Therefore the area of the required finite region = 4.5

N:B The fact that the calculated area turned out to be -ve indicates that the finite region actually lies below the x-axis; hence explaining the -ve sign infront of the value "4.5".


lol bono :smile::smile: P1 hehe integration
bono
Find the area of the finite region bounded by the x-axis and the curve with equation y = (x + 1)(x - 2).

Curve's equation: y = x^2 - x - 2

The curve crosses the x-axis at x = -1 and x = 2.

To calculate the area of the required finite region, find:

Definite integral (limits 2 to -1) x^2 - x - 2 dx.

= [(x^3)/3 - (x^2)/2 - 2x] with limits (2 to -1)

= (8/3 - 2 - 4) - (-1/3 - 1/2 + 2)

= -10/3 - 7/6
= -4.5

Therefore the area of the required finite region = 4.5

N:B The fact that the calculated area turned out to be -ve indicates that the finite region actually lies below the x-axis; hence explaining the -ve sign infront of the value "4.5".

bono's on drugsssss :biggrin:
Danithestudent
I feel somewhat strangely attracted to you.......:cool:

*giggles*


:biggrin:

always tempted tp get a bradd pitt avatar if people always think everyone looks like their avatars :biggrin:
_ELiTeCaRR_
omg you're so funny!

Really?! Do you think so? Well my friends and family always laughed at me, but I thought they were laughing AT me not with me. Do you think I could make a career outta it?! Hmm hmmm? We could do a double act the gorgeous sexy sadist and the funny looking masochist, but then I remembered you want to be a dentist and I want to be a journalist so it would have to be both sadists,


oh by the way I'm bullshitting :smile:

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