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1 Mar 2013: 10 7 20 10 19.

29 Oct 2013: . 1 1 00 3 20 1 0.  ,. . ... 1 1 10 3 20 1 0.  ,. . 1 1 11 3 11 1 1.

18 Feb 2013: 20). Show. that this test is a size α uniformly most powerful test for testing H′0 : σ2 σ20 against. ... Improved No difference WorsePlacebo 18 17 15Half dose 20 10 20Full dose 25 13 12.

1 Nov 2013: . 1 1 00 3 20 1 0.  ,. . ... 1 1 10 3 20 1 0.  ,. . 1 1 11 3 11 1 1.

16 Oct 2013: 2. Between 10 and 20 there are 4 primes; does it ever happen again that there are 4 primesbetween two consecutive multiples of 10 (apart from between 0 and 10)?

26 Feb 2013: α (deg) 5 10 15 20 25 30 35 40 45sin 2α 0.1736 0.3420 0.5 0.6428 0.7660 0.8660 0.9397 0.9848 1range (m) 4860

21 Feb 2013: MATHEMATICAL TRIPOS PART II 2013CODING AND CRYPTOGRAPHY. EXAMPLE SHEET 4. The first 18 examples are minimal to cover the course; you are also encouraged to try questions19–20. ... 20 Consider the linear recurrence. F xn = a0xnd a1xnd1. ad1xn1.

4 Dec 2013: 2/3. 20. Show that |pG(1)| = (1)|G|pG(1) is the number of acyclic orientations of G.(An orientation of G is an assignment of a direction to each

26 Feb 2013: 20. Let K = Q(ζp2 ) where p is an odd prime number.

26 Oct 2013: What time was dawn? I I conceal a 10 kopeck or 20 kopeck coin and you guess itsvalue.

18 Mar 2013: So c = 0 and after further changeof coordinates, F = X0X2 X21 , and so V is the image of φ = (Y 20 : Y0Y1 : Y 12) : P1 P2. ... 20 2bU0U1 cU. 21 , then since the discrimnant of FP is 4(B.

11 Mar 2013: 12 TOM FISHER. degrees 10, 10, 20, 20, 30, 30. We use Theorem 4.4 to decide which of these arecovariants. ... We alreadynoted in the proof of Proposition 5.2 that the integer weight discrete covariantsform a free K[c4,c6]-module of rank 6 generated in

13 Aug 2013: 4 1. )(20). for p 3. Hence the topology of the orbital space is S3/Zp. ... Figure 2: The function F̃ (x). 20.

12 Sep 2013: c) This follows from the definitions — see [20, I.2.8(1), I.7.12]. ... Springer-Verlag, New York, 1975. GraduateTexts in Mathematics, No. 21. [20] Jens Carsten Jantzen.

19 Jun 2013: 20)v 7 1. 3(2ζ3 1)(u. 7 4u5 u3 3u),v 7 1. ... 3This depends on the relationship between the embeddings (20) and the Weil pairing.

21 Apr 2013: Let be the free subgroup of SL2(Z) generated by the matrices A := ( 1 20 1 ) andB := ( 1 02 1 ). One has that Γ(2) = {I}. Given a positive integer N, ... Thereis no basis of S3(Φ0(5)) satisfying Atkin–Swinnerton-Dyer congruence relations forp. 20

18 Mar 2013: This is an isomorphism, inverse (Y0 : Y1) 7 (Y 20 : Y0Y1 : Y 22 ).Let φ: V W be an morphism of (projective or affine) varieties.

15 Mar 2013: 2. 2. Kleisli Bicategories. The Kleisli formulation of a monad on a category C, given in [20], has notplayed a prominent role in mathematics. ... Some Reasons for Generalising Domain Theory.Mathematical Structures in Computer Science 20, 2010, 239-265.

3 Apr 2013: Corollary 2.20. Suppose that k is algebraically closed. Then g1 contains subregular nilpotent elements. ... 4r 8 2r 4r 4 4 2r 4E6 4 10 12 16 18 24 6 8E7 4 12 16 20 24 28 36 8 12E8 4 16 24 28 36 40

20 Feb 2013: 173.8 Transfer. 19. 4 Raising the level 20. 5 Construction of a special automorphic representation 26. ... 1] IndU2P20 | |. 1/2, P 20 being the Borel subgroup in U2.

18 Dec 2013: 20. Note that γ(s) R, in fact r(γ(s)) r(p). For C large enough, we havethat Ct r is timelike in the region r r(p), thus, with

25 Feb 2013: n 1)z τ/2 z) = e(τ/2 z)ϑ(z). (5.20). ... 20 alexei kovalev. Compact connected orientable surfaces are classified up to a homeomorphism bytheir Euler characteristic χ (or, equivalently, by the genus).

13 Jan 2013: λ1. 101.  λ2120.  λ3210.  =1 1 20 2 11 0 0. ... as φ is linear. Now apply φ1 to get (). Lemma 1.20.