2007-04-04 02:42:08 · 11 answers · asked by Bush Invented the Google 6 in Politics & Government ➔ Politics
To clarify: when did Bush announce that Pelosi is breaking the law and violating the Constitution by going to Syria? Takers??? Anyone? Bueller?
2007-04-04 02:58:50 · update #1
He has. It was on CNN this morning. He keeps insisting that his lack of diplomacy is working in the middle east. He also stated that Syria will get a mixed message from the Pelosi visit. However, he did not acknowledge a better plan for dealing with Syria.
2007-04-04 02:46:43 · answer #1 · answered by CHARITY G 7 · 0⤊ 0⤋
If she's going to show faith and support there's no problem with her visiting other nations.. it's only if she were trying to change foreign policy there that there is a problem.. being as that is the presidents right to regulate.
you are correct.. she broke no law.
2007-04-04 09:48:14 · answer #2 · answered by pip 7 · 0⤊ 0⤋
He has condemned her going to Syria, and, Republican spokespersons have also condemned the action by Pelosi.
2007-04-04 09:46:03 · answer #3 · answered by Anonymous · 0⤊ 1⤋
I was hoping that she was being fitted for a Burka...I wish she would just stay there...but they may not have her "fix" of Botox that she needs not to look like the Crypt Keeper...
2007-04-04 10:00:15 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Bush did speak out against it this AM. His usual saber rattling speech on axis of evil.
2007-04-04 09:51:16 · answer #5 · answered by Anonymous · 0⤊ 1⤋
He has. Come out from under your rock once and a while.
2007-04-04 09:52:39 · answer #6 · answered by Anonymous · 0⤊ 1⤋
Meh, I could really care less where she goes.
2007-04-04 09:44:43 · answer #7 · answered by Anonymous · 0⤊ 0⤋
He has. Hes said over and over again that it was a terrible idea.
2007-04-04 09:48:33 · answer #8 · answered by Anonymous · 1⤊ 1⤋
wouldn't it be nice if a suicide bomber struck her convoy
2007-04-04 09:47:22 · answer #9 · answered by phoenix 2 · 1⤊ 1⤋
He did.
2007-04-04 09:45:54 · answer #10 · answered by Anonymous · 1⤊ 1⤋