...and I was wondering how long you all have been here? I will be out celebrating tomorrow. Any well-wishes?
atheist
2007-08-07 10:22:16 · 19 answers · asked by AuroraDawn 7 in Society & Culture ➔ Religion & Spirituality
Thanks so much everyone! I think that the only reason I have lasted this long is that I had to take a 6 month hiatus when my mom was sick with cancer. Clink and cheers to year number 2!
2007-08-07 10:49:46 · update #1
Aug. 9 2006. So close to my one year anniversary!
2007-08-07 10:32:08 · answer #1 · answered by Animal Girl 4 · 0⤊ 0⤋
Cool, and congrats. Ive been coming on here since September 2006. Got 2 accounts but both same name.
2007-08-07 17:31:54 · answer #2 · answered by Woody 3 · 0⤊ 0⤋
Congrats! I've only been on here since late March of this year. I'm trying not to make controversial comments because I want to make that anniversary, and second year...
2007-08-07 17:31:54 · answer #3 · answered by lil_lil85 3 · 0⤊ 0⤋
Hi, and congratulations, or should that be commiserations??
I opened my first account in July 06, but only used it for a couple of months. I came back on about 4 months ago and am on my 3rd account.
Have a great day out tomorrow!
An atheist
2007-08-07 17:28:07 · answer #4 · answered by Grotty Bodkin is not dead!!! 5 · 0⤊ 2⤋
Happy Birthday
We'll start a new calender the year can be 1 AD in your honour.
2007-08-07 17:34:20 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Since Feb. of this year. Sounds like a good enough reason to hoist a few suds!
2007-08-07 20:02:25 · answer #6 · answered by grumpy geezer 6 · 0⤊ 0⤋
I first came here just a few months after Y!A started. But it wasn't until recently that I joined R&S.
Happy anniversery!
2007-08-07 17:31:06 · answer #7 · answered by DEPRESSED™ 3 · 0⤊ 0⤋
Happy anniversary! I've been here since March, 2007.
2007-08-07 17:26:57 · answer #8 · answered by Daisy Indigo 6 · 0⤊ 1⤋
about 4 months
2007-08-07 17:25:42 · answer #9 · answered by . 6 · 0⤊ 1⤋
Congrats!!
I've been here since Apr 2006
2007-08-07 17:36:11 · answer #10 · answered by genaddt 7 · 0⤊ 0⤋