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Danithestudent
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#121
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#121
(Original post by _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
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_ELiTeCaRR_
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#122
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#122
(Original post by Danithestudent)
So let me guess, your first name is Jack and your surname is ****.

Thus being, you are NOTHING Weeeeee

Nothing is making any logical sense tonight.
lol let's not be too pedantic
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makesomenoise
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#123
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#123
(Original post by _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?
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_ELiTeCaRR_
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#124
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#124
(Original post by 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
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_ELiTeCaRR_
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#125
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#125
(Original post by 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?
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serious narb
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#126
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#126
(Original post by _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....
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Danithestudent
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#127
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#127
He's Xtina to my Britney Spears.
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makesomenoise
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#128
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#128
(Original post by sillynarb2)
nothing wrong with that dazya, come on....
As soon as you need another example, guess what
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_ELiTeCaRR_
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#129
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#129
(Original post by sillynarb2)
nothing wrong with that dazya, come on....
well i mean he has a girls name too " tina "
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makesomenoise
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#130
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#130
(Original post by Danithestudent)
He's Xtina to my Britney Spears.
And there's no Madonna!
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Danithestudent
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#131
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#131
(Original post by 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

Jk sorry couldn't help it.
I got + rep once because someone thought I was my avatar who just happened to be Britney
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serious narb
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#132
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#132
(Original post by Danithestudent)
Sillynarb2, do realise how much you resemble Kiera Knightley? you should go to one of those lookalike agencies

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
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username9816
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#133
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#133
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".
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_ELiTeCaRR_
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#134
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#134
(Original post by Danithestudent)
Sillynarb2, do realise how much you resemble Kiera Knightley? you should go to one of those lookalike agencies
omg you're so funny!
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Danithestudent
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#135
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#135
(Original post by sillynarb2)
i get told that a lot
I feel somewhat strangely attracted to you.......:cool:

*giggles*
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makesomenoise
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#136
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#136
(Original post by 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...
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_ELiTeCaRR_
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#137
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#137
(Original post by 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 P1 hehe integration
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Danithestudent
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#138
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#138
(Original post by 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
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serious narb
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#139
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#139
(Original post by Danithestudent)
I feel somewhat strangely attracted to you.......:cool:

*giggles*


always tempted tp get a bradd pitt avatar if people always think everyone looks like their avatars
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Danithestudent
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#140
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#140
(Original post by _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 bull****ting
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