|By Longhorn (Longhorn) on Tuesday, October 07, 2003 - 02:46 pm: Edit|
ok, this is a QC
w.x.y.z is not equal to 0
which is greater?
w² + x² + y² + z²
(w + x + y + z)²
BTW.. ² means ^2, i.e. squared
ok... now I solve it like this..
the second bracket gives me all the w x y and z squared, plus 2(wx+wy+wz+xy+xz+yz).
by reduction, remove the w² + x² + y² + z² from both sides and so there's zero on one side and
2(wx+wy+wz+xy+xz+yz) on the other. Now how am I to know whether 2(wx+wy+wz+xy+xz+yz) >0 or <0? so I put in D. now D ain't the answer! A is the answer and that implies that 2(wx+wy+wz+xy+xz+yz) is negative, but how can I be sure of that?
|By 31337 (31337) on Tuesday, October 07, 2003 - 03:03 pm: Edit|
(w + x + y + z)² = (0)^2 since w + x + y + z is zero. w² + x² + y² + z² will always be positive because any thing squared is positive with the exception of 0 which is not the case here. So the answer is B.
|By Longhorn (Longhorn) on Tuesday, October 07, 2003 - 03:06 pm: Edit|
good thinking! thanks 31337
|By Captcalculus (Captcalculus) on Tuesday, October 07, 2003 - 03:21 pm: Edit|
B? do you mean A?
|By 31337 (31337) on Wednesday, October 08, 2003 - 02:32 pm: Edit|
thanks Longhorn . Oh sorry, the answer is A.
|By Longhorn (Longhorn) on Wednesday, October 08, 2003 - 02:37 pm: Edit|
btw, doesn't your username mean something like elite just like hackers and CS clans say it?
|By 31337 (31337) on Wednesday, October 08, 2003 - 03:01 pm: Edit|
hehe, yeah it does :P
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