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Carmit has ugly hair and she looks like a man.
Ashley thinks that she's all that and everyone knows that I'M all that.
Jessica is just ugly.
Kimberly thinks she is all perfect and she thinks she stands out the most but everyone knows that it's all about ME!
Melody always tries to sing over top of me and she is annoying.
Asia is not talented at all.

I'm the best and they are messing up my reputation...should I just leave the Pussycat Dolls for good?

2007-07-31 08:58:26 · 9 answers · asked by Anonymous in Entertainment & Music Polls & Surveys

9 answers

why DONCHYA

2007-07-31 09:00:37 · answer #1 · answered by Brownie 5 · 0 0

Well, young lady, seems to me that since you are so perfect, beautifully, and talented, you can't/don't need to be part of a team.
It sounds that you would be doing both yourself and the other members a favor and go solo. It must be a great feeling being such a Deva.

2007-07-31 16:43:06 · answer #2 · answered by GERALD S. MCSEE 4 · 0 0

Yes! my pet,leave the pussycat dolls for good and sail away with me to a far off land in paradise,where we can live happily ever after.

2007-07-31 16:02:58 · answer #3 · answered by TEC 7 · 0 0

wait amy winehouse is in the pussycat dolls? never knew this. i like both the solo work and the pussycat dolls. altho the solo work is more unique.

2007-07-31 16:26:33 · answer #4 · answered by Anonymous · 0 0

Yes leave them and come with me. Us against the world

2007-07-31 19:32:28 · answer #5 · answered by Anonymous · 0 0

Whatever floats your boat.

2007-07-31 16:03:55 · answer #6 · answered by Anonymous · 0 0

that`s totally up to you....you seem to have formed a pretty good opinion of yourself !!!! you sure it`s all true,,or just ego ?? good luck..

2007-07-31 16:03:12 · answer #7 · answered by Anonymous · 0 0

Thanks for the two points. (Usually I don't do this, but it was such a stupid question, I had to)

2007-07-31 16:02:10 · answer #8 · answered by maT 4 · 0 0

that's up to you, but you are hot.

2007-07-31 16:00:45 · answer #9 · answered by Anonymous · 0 0

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