Pretty much every real and imaginary number except for 6
2007-09-07 13:09:15
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answer #1
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answered by Anonymous
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2⤊
0⤋
7
2007-09-07 20:08:50
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answer #2
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answered by B.Hound 4
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0⤊
0⤋
7
2007-09-07 20:07:36
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answer #3
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answered by Me 6
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1⤊
0⤋
The answer is 7.
Proof:
Given 2 + 2 â 5, we have
2 + 2 + 2 â 5 + 2
2 + 2 + (1 + 1) â 7
Then by the associative and commutitive laws of addition, (2+1) + (2+1) â 7
And finally, 3 + 3 â 7
In reality though, it's hard to start off with an explicit inequality like this and conclude that it only leads one other inequality. But going by the initial premise this is what we arrive at.
2007-09-07 20:08:46
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answer #4
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answered by Anonymous
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7⤊
0⤋
I think 3+3 is not 33
2007-09-07 20:16:34
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answer #5
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answered by ***BlueSky*** 2
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0⤊
0⤋
Ever heard of L'Hospital's Rule???
(You should have learned about it in elementary calculus.)
Anyway, http://mathworld.wolfram.com/LHospitalsRule.html will remind you the basics.
The real world effect of L'Hospital's rule is that with a little creative function creation, you can do some pretty good fuzzy math. The easiest case to manipulate is where lim(f) â 0 and lim(g) â 0. Then "0/0" can be made to converge to any arbitrary number.
2007-09-07 20:12:51
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answer #6
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answered by Anonymous
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0⤊
1⤋
+2 to the nth power. Now go make a sandwich and watch some TV....your brains have fallen out!
2007-09-07 20:12:50
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answer #7
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answered by dreamdress2 6
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0⤊
0⤋
It wouldn't be 5 either - nor 7.
2007-09-07 20:08:57
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answer #8
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answered by Cheese Fairy - Mummified 7
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0⤊
0⤋
the imaginary number of "i" or maybe merged into a binary format made up of ones and zeroes until paper after paper is filled
2007-09-07 20:12:16
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answer #9
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answered by midnitepoets 6
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0⤊
0⤋
7, of course
2007-09-07 20:14:56
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answer #10
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answered by Anonymous
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0⤊
0⤋