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Page "An die Jugend" ¶ 52
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::, 2 Bulb
:: 1 GeV / c < sup > 2 </ sup > = 1. 783 kg
:: 1 amu = 931. 46 MeV / c < sup > 2 </ sup > = 0. 93146 GeV / c < sup > 2 </ sup >
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:: Fe ( s ) → Fe < sup > 2 +</ sup >( aq ) + 2 e < sup >–</ sup >
:: O < sub > 2 </ sub >( g ) + 4 H < sup >+</ sup >( aq ) + 4 e < sup >–</ sup > → 2 H < sub > 2 </ sub > O ( l )
:: 2 Fe ( s ) + O < sub > 2 </ sub >( g ) + 4 H < sup >+</ sup >( aq )2 Fe < sup > 2 +</ sup >( aq ) + 2 H < sub > 2 </ sub > O ( l )

:: and &
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:: and n
:: A ( m, n )
:: A ( m, n )
:: n is an integer
define method factorial ( n :: < integer >)
:: ( C < sub > 6 </ sub > H < sub > 4 </ sub >)( CO < sub > 2 </ sub > CH < sub > 3 </ sub >)< sub > 2 </ sub > + 2 C < sub > 2 </ sub > H < sub > 4 </ sub >( OH )< sub > 2 </ sub > → 1 / n
:: to show / solve < tt > G </ tt >, show / solve < tt > G < sub > 1 </ sub ></ tt > and … and < tt > G < sub > n </ sub ></ tt >
:: Let n = 0
:: Example: See also Martin v. Wilks, 490 U. S. 755, 784 n. 21, 104 L. Ed.
In modern notation this says that given quantities p, q, r and s, then p: q :: r: s if for any positive integers m and n, np < mq, np = mq, np > mq according as nr < ms, nr = ms, nr > ms respectively.
m n :: nat
:: n < sub > x </ sub >, n < sub > y </ sub >, n < sub > z </ sub > are positive integers.
:: is a residue modulo p < sup > n </ sup > if k ≥ n
:: is a nonresidue modulo p < sup > n </ sup > if k < n is odd
:: is a residue modulo p < sup > n </ sup > if k < n is even and A is a residue
:: is a nonresidue modulo p < sup > n </ sup > if k < n is even and A is a nonresidue.

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