MedVision ad

Semester 1 Chatter Thread (2009) (1 Viewer)

Status
Not open for further replies.

Ben Netanyahu

Banned
Joined
Nov 15, 2008
Messages
1,758
Location
Tel Aviv, Israel
Gender
Male
HSC
N/A
yeah but see, the thing is

a) i dont know you
b) foreverpink is a tooliehead
c) bkm and i shall forever be mortal enemies. i am the organic chemist and she is not

sincerely
jim c ronshaw
 

Ben Netanyahu

Banned
Joined
Nov 15, 2008
Messages
1,758
Location
Tel Aviv, Israel
Gender
Male
HSC
N/A
i gave it to you about 20mins ago FP, you imbecile. the clock is messed up on bos.

furthermore, what statistical analysis would one use on excel to prove a correlation? anova?
 

ledzeppelin

Member
Joined
Aug 10, 2004
Messages
877
Location
Mosman
Gender
Male
HSC
2006
i gave it to you about 20mins ago FP, you imbecile. the clock is messed up on bos.

furthermore, what statistical analysis would one use on excel to prove a correlation? anova?
=correl gives you the correlation coefficient which indicates the strength and direction of any correlation
 

jb_nc

Google "9-11" and "truth"
Joined
Dec 20, 2004
Messages
5,391
Gender
Male
HSC
N/A
seriuzly halp me pleaSe :(
Possible derivation:
d/dx((x^2 y^2 (x^2-y^2))/(x^2+y^2))
| Factor out constants:
= | y^2 (d/dx((x^2 (x^2-y^2))/(x^2+y^2)))
| Use the quotient rule, d/dx(u/v) = (( du)/( dx))/v-(u ( dv)/( dx))/v^2, where u = x^2 (x^2-y^2) and v = x^2+y^2:
= | y^2 ((d/dx(x^2 (x^2-y^2)))/(x^2+y^2)-(x^2 (x^2-y^2) (d/dx(x^2+y^2)))/(x^2+y^2)^2)
| Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x^2 and v = x^2-y^2:
= | (y^2 (x^2 (d/dx(x^2-y^2))+(x^2-y^2) (d/dx(x^2))))/(x^2+y^2)-(x^2 y^2 (x^2-y^2) (d/dx(x^2+y^2)))/(x^2+y^2)^2
| Differentiate the sum term by term:
= | (y^2 (x^2-y^2) (d/dx(x^2)))/(x^2+y^2)-(x^2 y^2 (x^2-y^2) (d/dx(x^2)+d/dx(y^2)))/(x^2+y^2)^2+(x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)
| The derivative of x^2 is 2 x:
= | -(x^2 y^2 (x^2-y^2) (d/dx(x^2)))/(x^2+y^2)^2-(x^2 y^2 (x^2-y^2) (d/dx(y^2)))/(x^2+y^2)^2+(x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)
| The derivative of x^2 is 2 x:
= | -(x^2 y^2 (x^2-y^2) (d/dx(y^2)))/(x^2+y^2)^2+(x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| The derivative of y^2 is zero:
= | (x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| Differentiate the sum term by term and factor out constants:
= | (x^2 y^2 (d/dx(x^2)-d/dx(y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| The derivative of x^2 is 2 x:
= | -(x^2 y^2 (d/dx(y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)+(2 x^3 y^2)/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| The derivative of y^2 is zero:
= | (2 x y^2 (x^2-y^2))/(x^2+y^2)+(2 x^3 y^2)/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
 

ledzeppelin

Member
Joined
Aug 10, 2004
Messages
877
Location
Mosman
Gender
Male
HSC
2006
Possible derivation:
d/dx((x^2 y^2 (x^2-y^2))/(x^2+y^2))
| Factor out constants:
= | y^2 (d/dx((x^2 (x^2-y^2))/(x^2+y^2)))
| Use the quotient rule, d/dx(u/v) = (( du)/( dx))/v-(u ( dv)/( dx))/v^2, where u = x^2 (x^2-y^2) and v = x^2+y^2:
= | y^2 ((d/dx(x^2 (x^2-y^2)))/(x^2+y^2)-(x^2 (x^2-y^2) (d/dx(x^2+y^2)))/(x^2+y^2)^2)
| Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x^2 and v = x^2-y^2:
= | (y^2 (x^2 (d/dx(x^2-y^2))+(x^2-y^2) (d/dx(x^2))))/(x^2+y^2)-(x^2 y^2 (x^2-y^2) (d/dx(x^2+y^2)))/(x^2+y^2)^2
| Differentiate the sum term by term:
= | (y^2 (x^2-y^2) (d/dx(x^2)))/(x^2+y^2)-(x^2 y^2 (x^2-y^2) (d/dx(x^2)+d/dx(y^2)))/(x^2+y^2)^2+(x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)
| The derivative of x^2 is 2 x:
= | -(x^2 y^2 (x^2-y^2) (d/dx(x^2)))/(x^2+y^2)^2-(x^2 y^2 (x^2-y^2) (d/dx(y^2)))/(x^2+y^2)^2+(x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)
| The derivative of x^2 is 2 x:
= | -(x^2 y^2 (x^2-y^2) (d/dx(y^2)))/(x^2+y^2)^2+(x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| The derivative of y^2 is zero:
= | (x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| Differentiate the sum term by term and factor out constants:
= | (x^2 y^2 (d/dx(x^2)-d/dx(y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| The derivative of x^2 is 2 x:
= | -(x^2 y^2 (d/dx(y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)+(2 x^3 y^2)/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| The derivative of y^2 is zero:
= | (2 x y^2 (x^2-y^2))/(x^2+y^2)+(2 x^3 y^2)/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
thanking you, but the first line is just xy(x^2-y^2), not (x^2 y^2 (x^2-y^2)) :(
 

jb_nc

Google "9-11" and "truth"
Joined
Dec 20, 2004
Messages
5,391
Gender
Male
HSC
N/A
thanking you, but the first line is just xy(x^2-y^2), not (x^2 y^2 (x^2-y^2)) :(
d/dx((x y(x) (x^2-y(x)^2))/(x^2+y(x)^2))
| Use the quotient rule, d/dx(u/v) = (( du)/( dx))/v-(u ( dv)/( dx))/v^2, where u = x y(x) (x^2-y(x)^2) and v = x^2+y(x)^2:
= | (d/dx(x y(x) (x^2-y(x)^2)))/(x^2+y(x)^2)-(x y(x) (x^2-y(x)^2) (d/dx(x^2+y(x)^2)))/(x^2+y(x)^2)^2
| Differentiate the sum term by term:
= | (d/dx(x y(x) (x^2-y(x)^2)))/(x^2+y(x)^2)-(x y(x) (x^2-y(x)^2) (d/dx(x^2)+d/dx(y(x)^2)))/(x^2+y(x)^2)^2
| The derivative of x^2 is 2 x:
= | -(x y(x) (x^2-y(x)^2) (d/dx(y(x)^2)))/(x^2+y(x)^2)^2+(d/dx(x y(x) (x^2-y(x)^2)))/(x^2+y(x)^2)-(2 x^2 y(x) (x^2-y(x)^2))/(x^2+y(x)^2)^2
| Use the chain rule, d/dx(y(x)^2) = ( du^2)/( du) ( du)/( dx), where u = y(x) and ( du^2)/( du) = 2 u:
= | -(2 x y(x)^2 (x^2-y(x)^2) (d/dx(y(x))))/(x^2+y(x)^2)^2+(d/dx(x y(x) (x^2-y(x)^2)))/(x^2+y(x)^2)-(2 x^2 y(x) (x^2-y(x)^2))/(x^2+y(x)^2)^2
| The derivative of y(x) is y'(x):
= | (d/dx(x y(x) (x^2-y(x)^2)))/(x^2+y(x)^2)-(2 x y(x)^2 (x^2-y(x)^2) y'(x))/(x^2+y(x)^2)^2-(2 x^2 y(x) (x^2-y(x)^2))/(x^2+y(x)^2)^2
| Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = y(x) (x^2-y(x)^2):
= | (x (d/dx(y(x) (x^2-y(x)^2)))+y(x) (x^2-y(x)^2) (d/dx(x)))/(x^2+y(x)^2)-(2 x y(x)^2 (x^2-y(x)^2) y'(x))/(x^2+y(x)^2)^2-(2 x^2 y(x) (x^2-y(x)^2))/(x^2+y(x)^2)^2
| Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = y(x) and v = x^2-y(x)^2:
= | (x (y(x) (d/dx(x^2-y(x)^2))+(x^2-y(x)^2) (d/dx(y(x)))))/(x^2+y(x)^2)+(y(x) (x^2-y(x)^2) (d/dx(x)))/(x^2+y(x)^2)-(2 x y(x)^2 (x^2-y(x)^2) y'(x))/(x^2+y(x)^2)^2-(2 x^2 y(x) (x^2-y(x)^2))/(x^2+y(x)^2)^2
| Differentiate the sum term by term and factor out constants:
= | (x y(x) (d/dx(x^2)-d/dx(y(x)^2)))/(x^2+y(x)^2)+(x (x^2-y(x)^2) (d/dx(y(x))))/(x^2+y(x)^2)+(y(x) (x^2-y(x)^2) (d/dx(x)))/(x^2+y(x)^2)-(2 x y(x)^2 (x^2-y(x)^2) y'(x))/(x^2+y(x)^2)^2-(2 x^2 y(x) (x^2-y(x)^2))/(x^2+y(x)^2)^2
| The derivative of x^2 is 2 x:
= | -(x y(x) (d/dx(y(x)^2)))/(x^2+y(x)^2)+(x (x^2-y(x)^2) (d/dx(y(x))))/(x^2+y(x)^2)+(y(x) (x^2-y(x)^2) (d/dx(x)))/(x^2+y(x)^2)-(2 x y(x)^2 (x^2-y(x)^2) y'(x))/(x^2+y(x)^2)^2+(2 x^2 y(x))/(x^2+y(x)^2)-(2 x^2 y(x) (x^2-y(x)^2))/(x^2+y(x)^2)^2
| Use the chain rule, d/dx(y(x)^2) = ( du^2)/( du) ( du)/( dx), where u = y(x) and ( du^2)/( du) = 2 u:
= | -(2 x y(x)^2 (d/dx(y(x))))/(x^2+y(x)^2)+(x (x^2-y(x)^2) (d/dx(y(x))))/(x^2+y(x)^2)+(y(x) (x^2-y(x)^2) (d/dx(x)))/(x^2+y(x)^2)-(2 x y(x)^2 (x^2-y(x)^2) y'(x))/(x^2+y(x)^2)^2+(2 x^2 y(x))/(x^2+y(x)^2)-(2 x^2 y(x) (x^2-y(x)^2))/(x^2+y(x)^2)^2
| The derivative of y(x) is y'(x):
= | (x (x^2-y(x)^2) (d/dx(y(x))))/(x^2+y(x)^2)+(y(x) (x^2-y(x)^2) (d/dx(x)))/(x^2+y(x)^2)-(2 x y(x)^2 y'(x))/(x^2+y(x)^2)-(2 x y(x)^2 (x^2-y(x)^2) y'(x))/(x^2+y(x)^2)^2+(2 x^2 y(x))/(x^2+y(x)^2)-(2 x^2 y(x) (x^2-y(x)^2))/(x^2+y(x)^2)^2
| The derivative of y(x) is y'(x):
= | (y(x) (x^2-y(x)^2) (d/dx(x)))/(x^2+y(x)^2)-(2 x y(x)^2 y'(x))/(x^2+y(x)^2)+(x (x^2-y(x)^2) y'(x))/(x^2+y(x)^2)-(2 x y(x)^2 (x^2-y(x)^2) y'(x))/(x^2+y(x)^2)^2+(2 x^2 y(x))/(x^2+y(x)^2)-(2 x^2 y(x) (x^2-y(x)^2))/(x^2+y(x)^2)^2
| The derivative of x is 1:
= | -(2 x y(x)^2 y'(x))/(x^2+y(x)^2)+(x (x^2-y(x)^2) y'(x))/(x^2+y(x)^2)-(2 x y(x)^2 (x^2-y(x)^2) y'(x))/(x^2+y(x)^2)^2+(2 x^2 y(x))/(x^2+y(x)^2)-(2 x^2 y(x) (x^2-y(x)^2))/(x^2+y(x)^2)^2+(y(x) (x^2-y(x)^2))/(x^2+y(x)^2)
 

withoutaface

Premium Member
Joined
Jul 14, 2004
Messages
15,098
Gender
Male
HSC
2004
Possible derivation:
d/dx((x^2 y^2 (x^2-y^2))/(x^2+y^2))
| Factor out constants:
= | y^2 (d/dx((x^2 (x^2-y^2))/(x^2+y^2)))
| Use the quotient rule, d/dx(u/v) = (( du)/( dx))/v-(u ( dv)/( dx))/v^2, where u = x^2 (x^2-y^2) and v = x^2+y^2:
= | y^2 ((d/dx(x^2 (x^2-y^2)))/(x^2+y^2)-(x^2 (x^2-y^2) (d/dx(x^2+y^2)))/(x^2+y^2)^2)
| Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x^2 and v = x^2-y^2:
= | (y^2 (x^2 (d/dx(x^2-y^2))+(x^2-y^2) (d/dx(x^2))))/(x^2+y^2)-(x^2 y^2 (x^2-y^2) (d/dx(x^2+y^2)))/(x^2+y^2)^2
| Differentiate the sum term by term:
= | (y^2 (x^2-y^2) (d/dx(x^2)))/(x^2+y^2)-(x^2 y^2 (x^2-y^2) (d/dx(x^2)+d/dx(y^2)))/(x^2+y^2)^2+(x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)
| The derivative of x^2 is 2 x:
= | -(x^2 y^2 (x^2-y^2) (d/dx(x^2)))/(x^2+y^2)^2-(x^2 y^2 (x^2-y^2) (d/dx(y^2)))/(x^2+y^2)^2+(x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)
| The derivative of x^2 is 2 x:
= | -(x^2 y^2 (x^2-y^2) (d/dx(y^2)))/(x^2+y^2)^2+(x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| The derivative of y^2 is zero:
= | (x^2 y^2 (d/dx(x^2-y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| Differentiate the sum term by term and factor out constants:
= | (x^2 y^2 (d/dx(x^2)-d/dx(y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| The derivative of x^2 is 2 x:
= | -(x^2 y^2 (d/dx(y^2)))/(x^2+y^2)+(2 x y^2 (x^2-y^2))/(x^2+y^2)+(2 x^3 y^2)/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
| The derivative of y^2 is zero:
= | (2 x y^2 (x^2-y^2))/(x^2+y^2)+(2 x^3 y^2)/(x^2+y^2)-(2 x^3 y^2 (x^2-y^2))/(x^2+y^2)^2
My eyes burn :(
 

jb_nc

Google "9-11" and "truth"
Joined
Dec 20, 2004
Messages
5,391
Gender
Male
HSC
N/A
d/dy((x y (x^2-y^2))/(x^2+y^2))
| Factor out constants:
= | x (d/dy((y (x^2-y^2))/(x^2+y^2)))
| Use the quotient rule, d/dy(u/v) = (( du)/( dy))/v-(u ( dv)/( dy))/v^2, where u = y (x^2-y^2) and v = x^2+y^2:
= | x ((d/dy(y (x^2-y^2)))/(x^2+y^2)-(y (x^2-y^2) (d/dy(x^2+y^2)))/(x^2+y^2)^2)
| Use the product rule, d/dy(u v) = v ( du)/( dy)+u ( dv)/( dy), where u = y and v = x^2-y^2:
= | (x ((x^2-y^2) (d/dy(y))+y (d/dy(x^2-y^2))))/(x^2+y^2)-(x y (x^2-y^2) (d/dy(x^2+y^2)))/(x^2+y^2)^2
| Differentiate the sum term by term:
= | (x (x^2-y^2) (d/dy(y)))/(x^2+y^2)-(x y (x^2-y^2) (d/dy(x^2)+d/dy(y^2)))/(x^2+y^2)^2+(x y (d/dy(x^2-y^2)))/(x^2+y^2)
| The derivative of y is 1:
= | -(x y (x^2-y^2) (d/dy(x^2)))/(x^2+y^2)^2-(x y (x^2-y^2) (d/dy(y^2)))/(x^2+y^2)^2+(x y (d/dy(x^2-y^2)))/(x^2+y^2)+(x (x^2-y^2))/(x^2+y^2)
| The derivative of x^2 is zero:
= | -(x y (x^2-y^2) (d/dy(y^2)))/(x^2+y^2)^2+(x y (d/dy(x^2-y^2)))/(x^2+y^2)+(x (x^2-y^2))/(x^2+y^2)
| The derivative of y^2 is 2 y:
= | (x y (d/dy(x^2-y^2)))/(x^2+y^2)-(2 x y^2 (x^2-y^2))/(x^2+y^2)^2+(x (x^2-y^2))/(x^2+y^2)
| Differentiate the sum term by term and factor out constants:
= | (x y (d/dy(x^2)-d/dy(y^2)))/(x^2+y^2)-(2 x y^2 (x^2-y^2))/(x^2+y^2)^2+(x (x^2-y^2))/(x^2+y^2)
| The derivative of x^2 is zero:
= | -(x y (d/dy(y^2)))/(x^2+y^2)-(2 x y^2 (x^2-y^2))/(x^2+y^2)^2+(x (x^2-y^2))/(x^2+y^2)
| The derivative of y^2 is 2 y:
= | -(2 x y^2)/(x^2+y^2)-(2 x y^2 (x^2-y^2))/(x^2+y^2)^2+(x (x^2-y^2))/(x^2+y^2)
 

ledzeppelin

Member
Joined
Aug 10, 2004
Messages
877
Location
Mosman
Gender
Male
HSC
2006
d/dy((x y (x^2-y^2))/(x^2+y^2))
| Factor out constants:
= | x (d/dy((y (x^2-y^2))/(x^2+y^2)))
| Use the quotient rule, d/dy(u/v) = (( du)/( dy))/v-(u ( dv)/( dy))/v^2, where u = y (x^2-y^2) and v = x^2+y^2:
= | x ((d/dy(y (x^2-y^2)))/(x^2+y^2)-(y (x^2-y^2) (d/dy(x^2+y^2)))/(x^2+y^2)^2)
| Use the product rule, d/dy(u v) = v ( du)/( dy)+u ( dv)/( dy), where u = y and v = x^2-y^2:
= | (x ((x^2-y^2) (d/dy(y))+y (d/dy(x^2-y^2))))/(x^2+y^2)-(x y (x^2-y^2) (d/dy(x^2+y^2)))/(x^2+y^2)^2
| Differentiate the sum term by term:
= | (x (x^2-y^2) (d/dy(y)))/(x^2+y^2)-(x y (x^2-y^2) (d/dy(x^2)+d/dy(y^2)))/(x^2+y^2)^2+(x y (d/dy(x^2-y^2)))/(x^2+y^2)
| The derivative of y is 1:
= | -(x y (x^2-y^2) (d/dy(x^2)))/(x^2+y^2)^2-(x y (x^2-y^2) (d/dy(y^2)))/(x^2+y^2)^2+(x y (d/dy(x^2-y^2)))/(x^2+y^2)+(x (x^2-y^2))/(x^2+y^2)
| The derivative of x^2 is zero:
= | -(x y (x^2-y^2) (d/dy(y^2)))/(x^2+y^2)^2+(x y (d/dy(x^2-y^2)))/(x^2+y^2)+(x (x^2-y^2))/(x^2+y^2)
| The derivative of y^2 is 2 y:
= | (x y (d/dy(x^2-y^2)))/(x^2+y^2)-(2 x y^2 (x^2-y^2))/(x^2+y^2)^2+(x (x^2-y^2))/(x^2+y^2)
| Differentiate the sum term by term and factor out constants:
= | (x y (d/dy(x^2)-d/dy(y^2)))/(x^2+y^2)-(2 x y^2 (x^2-y^2))/(x^2+y^2)^2+(x (x^2-y^2))/(x^2+y^2)
| The derivative of x^2 is zero:
= | -(x y (d/dy(y^2)))/(x^2+y^2)-(2 x y^2 (x^2-y^2))/(x^2+y^2)^2+(x (x^2-y^2))/(x^2+y^2)
| The derivative of y^2 is 2 y:
= | -(2 x y^2)/(x^2+y^2)-(2 x y^2 (x^2-y^2))/(x^2+y^2)^2+(x (x^2-y^2))/(x^2+y^2)
Where are you generating this from?
Thanks for the halp, but after figuring out what the final line of that actually says, I've actually got it in a simpler form than this, just not sure if it can go any further
I'd give some rep but must spread some around first apparently
 

jb_nc

Google "9-11" and "truth"
Joined
Dec 20, 2004
Messages
5,391
Gender
Male
HSC
N/A
Where are you generating this from?
Thanks for the halp, but after figuring out what the final line of that actually says, I've actually got it in a simpler form than this, just not sure if it can go any further
I'd give some rep but must spread some around first apparently
wolfram alpha search engine

it fuckin rules

just felt like posting walls of text though

if you post first line into the bar it will generate a latex form which is easy to read
 

withoutaface

Premium Member
Joined
Jul 14, 2004
Messages
15,098
Gender
Male
HSC
2004
matlab algebra toolbox needed imo.
Fuck matlab's a piece of shit. I hate it so hard.

Also I did the railcorp personality test. Do they actually cull people based on that shit, cause it's fucking hard to see how you could make yourself look like a balanced human being from the responses they asked you to choose.
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top