½ ( pKa < sub > 1 </ sub > + pKa < sub > R </ sub >), where pKa < sub > R </ sub > is the side-chain pKa.

Consequently it allowed symbols like " ½ " for ". 5 " and the superscript < sup > 2 </ sup > for " to the power of 2 ".

Cadmium has 8 known meta states, with the most stable being < sup > 113m </ sup > Cd ( t < sub > ½ </ sub >

44. 6 days ), and < sup > 117m </ sup > Cd ( t < sub > ½ </ sub > = 3. 36 hours ).

This element also has 8 meta states, with the most stable being < sup > 150m </ sup > Eu ( T < sub > ½ </ sub >= 12. 8 hours ), < sup > 152m1 </ sup > Eu ( T < sub > ½ </ sub >= 9. 3116 hours ) and < sup > 152m2 </ sup > Eu ( T < sub > ½ </ sub >= 96 minutes ).

Aside from ever-popular R / C-controlled aircraft, several kitplane manufacturers offer ½, ⅔, and ¾-scale replicas capable of comfortably seating one ( or even two ) and offering high performance combined with more forgiving flight characteristics.

It produced two top-ten R & B hits, "( Baby Tell Me ) Can You Dance " and " No ½ Steppin.

* Jonathan Rowson vs Neil R McDonald, London Agency 1998, Slav Defense: Exchange Variation ( D13 ), ½ – ½

n + ½ ).

I. e., not only and must be substituted by α and β, respectively ( the first entity means " spin up ", the second one " spin down "), but also the sign + by the − sign, and finally r < sub > i </ sub > by the discrete values s < sub > i </ sub > (= ± ½ ); thereby we have and.

Over the real numbers this is equivalent to being able to define a symmetric scalar product, u. v = ½ ( uv + vu ) that can be used to orthogonalise the quadratic form, to give a set of bases

However we know that if an eclipse occurred at some moment, then there will occur an eclipse again S synodic months later, if that interval is also D draconic months, where D is an integer number ( return to same node ), or an integer number + ½ ( return to opposite node ).

For instance, in the above 2 × 2 matrix example, the coefficient − c < sub > 1 </ sub >= a + d of λ above is just the trace of A, trA, while the constant coefficient c < sub > 0 </ sub >= ad − bc can be written as ½ (( trA )< sup > 2 </ sup >− tr ( A < sup > 2 </ sup >)).

By summing 1 + ½ + ¼ +... ( a geometric series ) we see that the machine performs infinitely many steps in a total of 2 minutes.

: NO < sub > 2 </ sub >< sup >-</ sup > + ½ O < sub > 2 </ sub > → NO < sub > 3 </ sub >< sup >-</ sup >

The Jacobi polynomials with α = β are called the Gegenbauer polynomials ( with parameter γ = α + ½ ).

Qd8 + Kh7 ½ – ½

* E = ρ e + ½ ρ ( u < sup > 2 </ sup > + v < sup > 2 </ sup > + w < sup > 2 </ sup > ) is the total energy per unit volume, with e being the internal energy per unit mass for the fluid,

: p < sub > AO </ sub >= 2 < sup >- n </ sup >⋅(( 1 + f < sub > A </ sub >)/( 1 + f < sub > O </ sub >))< sup > ½ </ sup >

* Trq < sub > 1 </ sub >= ½ Trq < sub > in </ sub > + ½ Trq < sub > d </ sub > for the slower output

: 2 ( CaSO < sub > 4 </ sub >· ½ H < sub > 2 </ sub > O ) + 3 H < sub > 2 </ sub > O → 2 ( CaSO < sub > 4 </ sub >. 2H < sub > 2 </ sub > O ) + Heat

Petrosian vs. Spassky, World Championship 1966 ( game 12 ) 1. Nf3 g6 2. c4 Bg7 3. d4 d6 4. Nc3 Nd7 5. e4 e6 6. Be2 b6 7. O-O Bb7 8. Be3 Ne7 9. Qc2 h6 10. Rad1 O-O 11. d5 e5 12. Qc1 Kh7 13. g3 f5 14. exf5 Nxf5 15. Bd3 Bc8 16. Kg2 Nf6 17. Ne4 Nh5 18. Bd2 Bd7 19. Kh1 Ne7 20. Nh4 Bh3 21. Rg1 Bd7 22. Be3 Qe8 23. Rde1 Qf7 24. Qc2 Kh8 25. Nd2 Nf5 26. Nxf5 gxf5 27. g4 e4 28. gxh5 f4 29. Rxg7 Qxg7 30. Rg1 Qe5 31. Nf3 exd3 32. Nxe5 dxc2 33. Bd4 dxe5 34. Bxe5 + Kh7 35. Rg7 + Kh8 36. Rg6 + Kh7 37. Rg7 + Kh8 38. Rg6 + Kh7 39. Rg7 + ½ – ½

Coins were issued in ½, 1 and 2 Quart denominations.

A lower bound for the required energy is the kinetic energy K = ½ mv < sup > 2 </ sup > where m is the final mass.

Judit played board 2 and finished the tournament with the highest score of 12½ – ½ to win the individual gold medal.

In all, 31 radioisotopes of neodymium have been detected, with the most stable radioisotopes being the naturally occurring ones: < sup > 144 </ sup > Nd ( alpha decay with a half-life ( T < sub > ½ </ sub >) of 2. 29 × 10 < sup > 15 </ sup > years ) and < sup > 150 </ sup > Nd ( double beta decay, T < sub > ½ </ sub > = 7 × 10 < sup > 18 </ sup > years, approximately ).

Thus the solution is 2, not ½ as would be the result of an incorrect analysis.

Despite the name, reverse Polish notation is not exactly the reverse of Polish notation, for the operands of non-commutative operations are still written in the conventional order ( e. g. "/ 6 3 " in Polish notation and " 6 3 /" in reverse Polish both evaluating to 2, whereas " 3 6 /" in reverse Polish notation would evaluate to ½ ).

Gold 1 yen were introduced in 1871, followed by copper 1 rin, ½, 1 and 2 sen in 1873.

Thrust and momentum from exhaust, per unit mass expelled, scales up linearly with its velocity ( momentum = mv ), yet kinetic energy and energy input requirements scale up faster with velocity squared ( kinetic energy = ½ mv < sup > 2 </ sup >).

In the late 18th century, coins were issued in denominations of ⅓, ½, 1, 3, 6, 7½, 10 and 15 groszy, 1, 2, 4, 6 and 8 złotych.

The ⅓ and ½ grosz were denominated as the solidus and polgrosz, whilst the 7½ and 15 groszy ( copper ) were denominated as 1 and 2 silver groschen.

In this knock-out format tournament he defeated Gata Kamsky 1½ – ½, Karpov 2 – 0, Viswanathan Anand 1½ – ½, and finally the World Champion Garry Kasparov 1½ – ½ to win the first prize of approximately 75, 000 USD.

( 7 ½ lbs ), and is typically in the range of 2. 7 – 4. 6 kg ( 5. 5 – 10 pounds ).

It has historically been read as stating that the angle subtended by the Sun's diameter is 2 degrees, but Archimedes states in The Sand Reckoner that Aristarchus had a value of ½ degree, which is much closer to the actual average value of 32 ' or 0. 53 degrees.

Graphs of y = b < sup > x </ sup > for various bases b: # Powers of ten | base 10 (< span style =" color: green "> green </ span >), # The exponential function | base e (< span style =" color: red "> red </ span >), # Powers of two | base 2 (< span style =" color: blue "> blue </ span >), and base ½ (< span style =" color: cyan "> cyan </ span >).

The first mine was located about ½ mile from Dogwood Flat and the No. 2 mine was about 1-1 / 4 mile up the run.

The town measures about 2 & ½ miles east to west and about one mile north to south at its widest point.

The coins occur in the denominations,, ⅛, ¼, ½, 1, 2, 5, and 10 marks, with a few 100-mark coins for very large transactions.

The ½ and 2 Mark coins are the only ones that could be potentially confused.

Bekenstein suggested ( ½ ln 2 )/ 4π as the constant of proportionality, asserting that if the constant was not exactly this, it must be very close to it.