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14 Feb 2015: The most important quantities associated with a single 2-spinor κA are its flag-pole Ka = κAκA′, and the flagplane determined by the bivector Pab = κAκBA′B′ ABκA. ... P = 12ψσ1ψ̃ = κ(γ1(γ0 γ3))κ̃. (24). Since σ1 anticommutes with

12 Oct 2015: St Catharine’s. Fitzwilliam. 2022–23. Pembroke. Girton. 2006–07. Churchill. Pembroke. 2023–24. St John’s.

19 Feb 2015: Employing the one-particleidentity. ik1B1ik. 1B1, 5.24. we see that with no sum over a.

4 Nov 2015: Anatomy/Vet Anatomy CBC 17 52.94 11.76 8.24 16.47 0 4.71 0 5.88 0 0. ... Quayside Quayside 41 32.31 10.26 11.28 14.87 0 24.1 0 6.67 0.51 0.

16 Jun 2015: ApJ, 270, 342 (1983). 24. Mackay, C.D. and Astill, D.M., CCD Development at Cambridge, England'. ... 31. Mackay, C.D., CCDs in Astronomy'. Ann. Rev. Ast. And Astrophys., 24, 255, 1986.

4 Feb 2015: Electrophysiological recordings from immobilizedand dissected C. elegans were essentially performed as describedpreviously [24]. ... in muscle indicatesa latency of the order of 80 ms [24].

21 Oct 2015: 4. (a) Show that the symmetric group S4 has a subgroup of order d for each divisor d of 24,and find two non-isomorphic subgroups of order 4. ... 10. Let G be a group. If H is a normal subgroup of G and K is a normal subgroup of H, is Ka normal subgroup

4 Apr 2015: We sketch a proof of the following theorem of arithmetic. 24 METATHEOREM If Z is consistent, so is Z H. ... a (24). 25 In the rest of this section we study a construction that starting from any model of M0 yields one of M1 H.The latter model may be

5 Feb 2015: It is readily verified that Ka = K2a = a, or, as an operator equation,. ... Ka = Ka = (a) K. (4.52). It follows thatKej = ej, Kej = ej,.

22 Dec 2015: 855,515 95,186 – 95,186 (24,604) 70,582 – 190,043 2,116,140 2,099,231. ... 036 24,032 3,242 – 3,242 (4,480) (1,238) – 2,327 25,121 24,852.