Math Problem, Lounge
Name:
Anonymous
2006-11-04 18:52
If k > m and m is < (y x c)/6z, what number does z have to be if c = 15 and m = k/y?
Name:
Anonymous
2006-11-04 19:08
Nerd
Name:
Anonymous
2006-11-04 19:15
14
Name:
Anonymous
2006-11-04 19:40
m = GTFO
Name:
Anonymous
2006-11-04 19:57
we're not doing your homework
Name:
Anonymous
2006-11-04 20:10
>>1
Since you are multiplying three variables, this is a third grade problem with three solutions:
● 1/0
● Infinity
● 0.000...1
Name:
Anonymous
2006-11-04 20:24
and FATMOUSE
Name:
Anonymous
2006-11-04 23:39
Name:
Anonymous
2006-11-05 1:55
z < ((15(y^2)k)/6) + 0.000...1
Name:
pissed off jerk nuts
2006-11-05 2:46
go back to the fucking science and math board, you fucking waste of asshole tissue.
Jesus fuckin' christ! you people have some fucking nerve!
Name:
Anonymous
2006-11-05 4:50
0.000...1 = 0
Name:
Anonymous
2006-11-05 7:14