| 1. | uniquely American | ||
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Utterly fucked. Terribly sad. Unjust. Awful. Devastating.
Source: THE PRESIDENT: You work three jobs? MS. MORNIN: Three jobs, yes. THE PRESIDENT: Uniquely American, isn’t it? I mean, that is fantastic that you’re doing that. (Applause.) Get any sleep? (Laughter.) So guess what, that crappy food from China killed my cat!
Whoa, that's uniquely American, dude! Sorry as hell to hear that! |
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| 2. | unet | ||
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Your uniquely individualised network Unet is unique to your taste and considerations that are uniquely yours.
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| 3. | Marthie | ||
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Uniquely The Awesomeness! Marthie Is Uniquely The Awesomeness!
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| 4. | Fundamental Theorem of Arithmetic | ||
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The fundamental theorem of arithmetic states that {n: n is an element of N > 1} (the set of natural numbers, or positive integers, except the number 1) can be represented uniquely apart from rearrangement as the product of one or more prime numbers (a positive integer that's divisible only by 1 and itself). This theorem is also called the unique factorization theorem and is a corollary to Euclid's first theorem, or Euclid's principle, which states that if p is a prime number and p/ab is given (a does not equal 0; b does not equal 0), then p is divisible by a or p is divisible by b. Proof: First prove that every integer n > 1 can be written as a product of primes by using inductive reasoning. Let n = 2. Since 2 is prime, n is a product of primes. Suppose n > 2, and the above proposition is true for N < n. If n is prime, then n is a product of primes. If n is composite, then n = ab, where a < n and b < n. Therefore, a and b are products of primes. Hence, n = ab is also a product of primes. Since that has been established, we can now prove that such a product is unique (except for order). Suppose n = p sub1 * p sub2 * ... * p subk = q sub1 * q sub2 * ... * q subr, where the p's and q's are primes. If so, then p sub1 is divisible by (q sub1 * ... * q subr) by Euclid's first theorem. What is the relationship between p sub1 and one of the q's? If the r in q subr equals 1, then p sub1 = q sub1 since the only divisors of q are + or - 1 and + or - q and p > 1, making p = q. What about the other factors in the divisor? If p does not divide q, then the greatest common denominator of p and q is 1 since the only divisors of p are + or - 1 and + or - p. Thus there are integers m and n so that 1 = am + bn. Multiplying by q subr yieds q subr = amq subr + bnq subr. Since we are saying that p is divisible by q, let's say the q sub1 * q subr = cp. Then q subr = amq subr + bnq subr = amq subr + bcm = m(aq subr + bc). Therefore, p is divisible by q sub1 of q sub2 * ... * q subr. If p sub1 is divisible by q sub1, then p sub1 = q sub 1. If this does not work the first time... more...
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| 5. | C.O.L.O.U.R.S. | ||
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Cool Outrages Lovers Of Uniquely Raw Style. C.O.L.O.U.R.S what's that spell Cool Outrages Lovers Of Uniquely Raw Style.
Farnsworth Bentley C.o.l.o.u.r.s Man I got c.o.l.o.u.r.s. |
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| 6. | .you | ||
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Uniquely yours, attributable to you That is SO .you to try and justify a full stop in the middle of a sentence.
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| 7. | Freeking | ||
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similar to 'freaking', 'freakin' Adj. to describe really good, or very much, almost uniquely outstanding That's a freeking awesome coat
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