2006-08-06 17:46:07 · 19 answers · asked by shaolinmantis77 4 in Travel ➔ United States ➔ Other - United States
i bet Philippines!!!! boracay's beach is the best. Lots of romantic places in the Philippines...i guarantee you..
2006-08-06 17:52:30 · answer #1 · answered by karisma 2 · 0⤊ 0⤋
Depends on what kind of vacation you're looking for. Relaxing? Exciting? Warm/hot weather? Temperate weather? And how much can you afford and how far are you willing to travel to?
In the meantime: www.frommers.com
2006-08-07 00:50:23 · answer #2 · answered by Tygirljojo 4 · 0⤊ 0⤋
The panhandle of Florida, between Pensacola and Panama City Beach.
2006-08-07 00:50:17 · answer #3 · answered by Chris 3 · 0⤊ 0⤋
It depends on what you are looking for. Everyone is different. It also depends on your budget. You need to decide how much money you want to spend and then decide what you want to do. Then go from there.
My favorite cities in the world are Singapore and San Diego, California, USA.
For scenery, San Diego is also good. Hawaii is nice.
For culture, Japan is wonderful.
2006-08-07 03:46:38 · answer #4 · answered by Adam 7 · 0⤊ 0⤋
Depens on what kind of vacation...romantic or party? I always wanted to goto Europe, or Africa, right near the beeeeccchhh
2006-08-07 00:49:39 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Australia, if u want a real vacation
2006-08-07 00:51:05 · answer #6 · answered by Anonymous · 0⤊ 0⤋
The ocean. Hawaii, Hilton Head, Malibu, wherever! something about the sound and smell of the ocean is deeply relaxing and just the thot of it makes me............xxxzzz...sorry.....went to sleep there for a sec.
2006-08-07 00:53:59 · answer #7 · answered by bookluffer 3 · 0⤊ 0⤋
I wouldn't mind Australia myself. Thats a long way from Colorado.
2006-08-07 00:50:26 · answer #8 · answered by Stratobratster 6 · 0⤊ 0⤋
That question is very general.
Some people like New York, Hawaii, London, Rome, etc.
2006-08-07 00:49:33 · answer #9 · answered by NA A 5 · 0⤊ 0⤋
Mac.chu picchu in Peru.
2006-08-07 00:53:52 · answer #10 · answered by Dinker 2 · 0⤊ 0⤋