2007-02-21 08:57:46 · 11 answers · asked by Anonymous in Entertainment & Music ➔ Celebrities
Would you even care if Britney is just an ordinary person? Issues over celebrities are too over rated. Somebody already made money and we ordinary people still the same. You cannot just judge people without really knowing people. Especially, expect all the issues about celebrities you may hear. THEY ARE CELEBRITIES.. But hey, everyone hates Britney and the funny thing is everytime I ask somebody about Britney Spear as an entertainer, people always deny that they liked Britney. I just wonder how she became a DIVA if NOBODY liked her? Celebrities may all take drugs but that's their life.. and that's their money!
2007-02-21 09:12:16 · answer #1 · answered by Jay Ares 1 · 0⤊ 0⤋
Federmine
2007-02-21 09:00:22 · answer #2 · answered by Anonymous · 0⤊ 0⤋
no person at here might understand for advantageous besides. and that i became examining in a mag, each and every time she has have been given drug examined, she handed it. So, no, i do no longer subject that she is a drug addict.
2016-11-24 22:33:57 · answer #3 · answered by pirozzi 4 · 0⤊ 0⤋
Publicity
Jx
2007-02-21 09:01:06 · answer #4 · answered by kirroyale3 3 · 1⤊ 1⤋
Everything under the sun.
2007-02-21 09:00:49 · answer #5 · answered by Anonymous · 0⤊ 0⤋
probaly everything. too bad she will get off easier than a "regular" person
2007-02-21 09:25:31 · answer #6 · answered by CATWOMAN 6 · 0⤊ 0⤋
prolly just weed, she doesn't seem like someone that goes hardcore
2007-02-21 09:05:04 · answer #7 · answered by David 3 · 0⤊ 0⤋
WHO CARES THAT LIL SLUT IS SO WIERD SHE SHAVED HER HEAD BALD AND SHE GOT TATTO'S ON HER HEAD AND CHECKED HERSELF INO REHAB......... SO SHE DOES ALMOST EVERY FREAKIN DRUG THAT RETARD
2007-02-21 09:10:55 · answer #8 · answered by Anonymous · 0⤊ 0⤋
probably everything u can think of. i can believe she hasn't already killed herself!
2007-02-21 09:05:42 · answer #9 · answered by Anonymous · 0⤊ 0⤋
all of them
2007-02-21 09:00:32 · answer #10 · answered by jester c 2 · 0⤊ 0⤋