|By Lisasimpson (Lisasimpson) on Saturday, October 02, 2004 - 04:01 pm: Edit|
find a,b for the parabola y = ax^2 + bx + 3 if the vertex is (2,4)
i have 4 invites left to give, so the first person to solve it can have how many they want and then anybody who verifies the answer can have one too
|By Aspirer42 (Aspirer42) on Saturday, October 02, 2004 - 04:08 pm: Edit|
For another Gmail invite, prove that
(x^n) + (y^n) = (z^n) has no solutions for n > 2, when x, y, and z are non-zero integers.
Show all work.
|By Xasuke (Xasuke) on Saturday, October 02, 2004 - 05:40 pm: Edit|
I'll do the parabola one.. I don't want an invite.
y = ax^2 + bx + 3
plugging in the given vertex (2, 4):
4 = 4a + 2b +3
ok.. and using x=-b/2a for the x cord. of the vertex:
4=4a -8a +3
y = -.25x^2 + x + 3
Right now, I'm too lazy to think about the other problem.. Pretty much prove the Pythagorean theorem?
|By Lisasimpson (Lisasimpson) on Saturday, October 02, 2004 - 05:53 pm: Edit|
could that be a trick prob aspirer? cuz of x y and x are all 1 then the problem will work no matter what n is..unless i'm just reading it wrong or something.
|By Cherrybarry (Cherrybarry) on Saturday, October 02, 2004 - 07:26 pm: Edit|
where's andrew wiles when you need him?!!!
|By Thermodude (Thermodude) on Saturday, October 02, 2004 - 07:27 pm: Edit|
well...posting a 187 page proof on CC would take a while anyways....too bad Fermat just couldn't have written down his proof somewhere...
|By Ubercollegeman (Ubercollegeman) on Sunday, October 03, 2004 - 01:06 am: Edit|
I'd solve it, Aspirer, but the proof is too long for the contents of this message box to contain...
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