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4 Aug 2015: The problem is now completely specified bythree geometrical quantities, namely the non-dimensionalmean gap, H = kH, the corrugation amplitude, a = ka, andthe asymmetry parameter, g. ... K. Sia and G. M. Whitesides, Electrophoresis, 2003, 24,.

23 Apr 2015: Drosophila strains and microscopyThe following stocks were used: ubiECad-GFP –hereinafter. ECadGFP [24], sGMCA [12,25]; zipperCPTI002907 (available. ... MbsN300 (24 embryos/115 vertices) (B), ASGal4/UAS-ctMLCK (17 embryos/101 vertices) (C), and

28 Jan 2015: collaboration. 4. The laboratory is to be constructed on University land located within the Cambridge Biomedical Campus on Puddicombe Way between the Li Ka Shing Centre and the multi-storey car

15 Dec 2015: Oncology, Li Ka Shing. 2015. Richard James Gilbertson. Operations Research. 2007.

14 Feb 2015: Theessential techniques are described in Chapter 5, which introduces the concept of themultivector derivative [18, 24]. ... ArBs 〈AB〉|sr| (1.23)ArBs 〈AB〉sr, (1.24). which we call the interior and exterior products respectively.

4 Apr 2015: the discussion of Church's alternatives in Jech[7],chapter 11, section 4, problems 23 and 24. ... If for some v < d), ho(v,a) = w, rv{a) = 6 by 3-24; by 3-21 {pgp = gv} is fc-solid, and.

11 Apr 2015: a]k = {=(f) |f f ka & f is injective}. a (153). ... Write HF for the class of hereditarily finite sets, defined as the union ofall finite transitive sets.24 LEMMA Let k ω; then for any x, x Vk1.

18 Feb 2015: Anumber of other helpful introductory articles can be found, including those byHestenes [35, 36], Vold [24, 28], and Doran & Lasenby [37]. ... 3.24). Hence µRR̃ is equal to minus its reverse and so must be a bivector.

5 Feb 2015: CONTENTS. Editorial 2. The Master's Letter 4. The College Library: Past, Present and Future 7. Science in Britain, circa 1992 10. Some Historical Notes on the College Council 15. Book Reviews 19. Obituaries 27. College Sport 36. College Societies 49.

30 Nov 2015: φ|Rn{0} = 0 , tφ|Rn{0} = 0 and lim|x|. φ = 0 , (10.24). ... From (8.43) we have that. F1[sin ka] = i2(δ(x a) δ(x a)) (10.32).