A. I make trouble
B. Trouble makes me
C. I do not make trouble
D. None of the above
2007-06-12 03:42:58 · 7 answers · asked by it's me 5 in Arts & Humanities ➔ Other - Arts & Humanities
A
2007-06-12 03:46:18 · answer #1 · answered by annie78 3 · 0⤊ 0⤋
D
However, if you said that all troublemakers always make trouble, and you are a trouble maker, then it WOULD be logically correct to say that you are a troublemaker and you would make trouble.
As it is written, it could go either way, since troublemakers COULD make trouble or troublemakers could be behaving well and could NOT make trouble.
Logically, if A=B and B=C, then A MUST = C
whale
2007-06-12 10:50:32 · answer #2 · answered by WilliamH10 6 · 0⤊ 0⤋
D
there is no stipulation in the question of all or always. So some could make trouble but other could not make trouble. With you being only one it's like saying most people like liver and automatically assuming you like liver.
2007-06-12 10:55:08 · answer #3 · answered by rambo1214 3 · 1⤊ 0⤋
i ´m inclined to suggest a mixture of both A and B - insomuch as social identity is informed by, not only our actions, but also how those actions are perceived and interpreted by others - in this way society ´s interpretation of our actions, and its response to them can come to affect our understanding of self within that society. In a social situation therefore, a trouble maker is not only someone who actually makes trouble, but someone who is perceived (justly or unjustly) as being a trouble maker. In this sense trouble makes you just as much as you make trouble. fun - isn't it :-)
2007-06-12 11:09:47 · answer #4 · answered by paul c 1 · 0⤊ 1⤋
C
you'd like to think of yourself as trouble, but you have no idea what a trouble maker really is...don't put yourself in that catagory......
2007-06-12 10:52:48 · answer #5 · answered by doclakewrite 7 · 0⤊ 1⤋
D. None of the above ...
Because it is never safe, anyhow, anywhere, if you are one!
2007-06-12 10:49:42 · answer #6 · answered by arabianbard 4 · 0⤊ 1⤋
D
dear, oh dear, oh dear....
2007-06-12 11:09:59 · answer #7 · answered by Anonymous · 0⤊ 1⤋