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5 Dec 2006: All 25 signatories are members of the University of Cambridge Programme for Industry (CPI). By using a combination of academics from within the University and

27 Nov 2006: according to whether the chain features: (udsc) a light squark (b) a sbottom (t) a stop because there (one hopes) might be some ability to distinguish these chains (some of the ... What do you think would be a good thing to start doing now that I have

10 Oct 2006: n}.How large can we make a family A of subsets of [n] subject to thecondition that for all A, A′ A, there is some B B with B A A′? ... n}.How large can we make a family A of subsets of [n] subject to thecondition that for all A, A′ A, there is some

28 Nov 2006: TNF-treatment induces a rapid activation of IKK which in turnphosphorylates IB leading to the ubiquitination and protea-some-mediated degradation of IB. ... In the light of our results,we speculate that some of these effects might be due to thetransfer

3 Feb 2006: f(z) =a(z z0) bc(z z0) d. ,. for some a,b,c,d,z0 C. ... 1 < |z| < 2}, {0 < |z| < 1}, {0 < |z| < }. 8. (i) Let R and S be some Riemann surfaces, f : R S a continuous map, and p a pointin R.

9 Mar 2006: Then γ(b) = γ(a), so exp h(b) = 1and we must have h(b) = 2N πi for some integer N. ... f (w) =. n=0. an(w z)n. that converges to f (w) on some disc B(z, R).

1 Jun 2006: that P1BT = cP2 for some c k. Then fixing A2 and varying B we. ... Then Φ1 = cBT Φ2B for. some B GLm(k) and c k.

16 May 2006: A V (G), |A| = m = |b(A)| f (m). for some function f. ... But |B| > |X(<k)|, so B contains some k-set, whencetr B k.

21 Oct 2006: f(z) =a(z z0) bc(z z0) d. ,. for some a,b,c,d,z0 C. ... 1 < |z| < 2}, {0 < |z| < 1}, {0 < |z| < }. 8. (i) Let R and S be some Riemann surfaces, f : R S a continuous map, and p a pointin R.

31 Aug 2006: Dab2/Disabled2 (a cargo adaptor for proteins like megalin)binds to both a and b appendages, and it is likely that some ofthe other accessory proteins that bind to both a ... Nevertheless some acces-sory adaptors like b-arrestin (for internalisation of

19 Apr 2006: 2). Occasional mistakes in the primers do occurand in some rare cases strand displacement activity of PfuTurbok has introduced unwantedinsertions (data not shown). ... Over-production of proteins in Escherichia coli: mutant hosts that allow synthesis of

20 Feb 2006: Then γ(b) = γ(a), so exp h(b) = 1and we must have h(b) = 2N πi for some integer N. ... f (w) =. n=0. an(w z)n. that converges to f (w) on some disc B(z, R).

13 May 2006: for some a b. If a > b then the first binary quartic has singularreduction above (x : z) = (0 : 1). ... ]for some a b c. If b > c then the first ternary cubic has singularreduction at (x : y : z) = (0 : 0 : 1).

16 Jun 2006: b a t i o n for 4 - 5 w k at 17C. ... Some lines of the surface lattice and the choice of unit cell used are indicated in B.

9 Oct 2006: F [A, B] = (det A)p(det B)qF. for some integers p, q. ... We must show that φ is properly equivalent to a Weierstrassmodel πn(0, 0, 0, A, B) for some unique A, B K.

13 Jan 2006: in determining ρ: we could find dl by some othermeans, resample the patch in B-scan B at this unique shift, then correlate once to find ρ. ... f (dl) = Ae(dlc). 2. 2σ2 B. Since this does not decay to zero, it captures, to some extent, the shape of the

10 May 2006: Earlierproofs of Theorem B for some geodesic and magnetic flows using Fourier analysis canbe found in [12, 18]. ... Systems. 4 (1984) 67–80.[12] V. Guillemin, D. Kazhdan, Some inverse spectral results for negatively curved 2-manifolds,.

8 Oct 2006: GEOMETRY AND GROUPS. These notes are to remind you of the results from earlier courses that we will need at some pointin this course. ... d(xn, ) 0 as n. A subsequence of the sequence (xn) is given by taking only some of the terms in the sequence.

21 Mar 2006: neous cubic polynomial in the generalized Weierstrass normal form XZ24Y 3AX2Y BX3,for some complex constants A, B, A3 27B2 6= 0. ... You will need to recall appropriate results from several topics of the course (and some previousexample(s)).

22 Nov 2006: Questions 7–10 are more challenging than others and some parts certainly go beyond limitsof the examination. ... neous cubic polynomial in the generalized Weierstrass normal form XZ24Y 3AX2Y BX3,for some complex constants A, B, A3 27B2 6= 0.