2007-02-10 15:32:58 · 19 answers · asked by Anonymous in Entertainment & Music ➔ Polls & Surveys
YES, sinlge ppl feel guitly
not singles feel obligated to spend money.
its a hallmark holiday with no significane whatsoever
2007-02-10 15:35:58 · answer #1 · answered by Anonymous · 0⤊ 1⤋
Even though I'm looking forward to it. I do think it's a worthless holiday, because I show love for people everyday, not just one day a year. But, not to sound materialistic (even though it will) Valentine's Day is a great day to get lots of flowers and candy and stuff.... hehe
2007-02-10 23:43:04 · answer #2 · answered by Ashley 2 · 0⤊ 0⤋
Not for the florists and the jewelers. In Ohio they even invented another one. It's called Sweetest's Day, it's the last Saturday of October or close to that. Makes me want to gak up a big ole furball.
2007-02-11 08:26:34 · answer #3 · answered by Anonymous · 1⤊ 0⤋
All holidays are pointless
2007-02-10 23:37:49 · answer #4 · answered by Anonymous · 2⤊ 0⤋
Well, maybe not ever but it's just a way for business' to make money. Lovers love every day.
2007-02-10 23:38:28 · answer #5 · answered by crazywoman88 4 · 1⤊ 0⤋
Yea it is. What love is the love that is celebrated specifically on one day and not the others? i think it's a waste of a day =)
2007-02-10 23:38:22 · answer #6 · answered by ~Moi~ 2 · 0⤊ 0⤋
No, it is a holiday of love and cheer, scrooge. No holiday is pointless unless your a buhummbug.
2007-02-10 23:37:11 · answer #7 · answered by Anonymous · 1⤊ 2⤋
only if you dont have a partner/
when I was growing up, my parents gave us one flower and some candy and a card, cause we were their valentines, ive always loved valetines day
2007-02-10 23:36:34 · answer #8 · answered by BoTToms UP 5 · 2⤊ 0⤋
I believe Singles' Awareness Day is mainly geared toward perpetuating our fantabulous capitalist system. :)
2007-02-10 23:37:58 · answer #9 · answered by 1 2 · 1⤊ 0⤋
No, because it's not just about love, it's also a day to celebrate special friendships.
2007-02-10 23:36:37 · answer #10 · answered by Viviana 6 · 2⤊ 0⤋