A DIRECTOR OF A UNIVERSITY LffiRARY has many problems. Besides financial problems, he must decide how to build collections, what resources in the collections can best serve faculty and students, what kind of bibliographical control must be provided, how to cope with the growth of collections and building problems that ensue, how to make more effective public services, how to provide instruction...

Moving disabled people ‘off benefits and into work’ has been an explicit aim of work-first welfare reform since 2008, increasingly punitively 2010. The this article is to demonstrate, for the first time, how Universal Credit (UC) fits with intensifies that strategy. Empirical data from 28 in-depth interviews 19 claimants (nine were interviewed twice) three focus groups 23 Jobcentre staff show U...

Introduction: Experiences of trance in Sintren dancer with a history psychosocial problems, it is possible that comorbidity mental disorders will appear. One the Primary Health Care programs dealing disorders. By providing counseling training to Staffs, hoped there be guidance and assistance from for community.Methods: Screening carried out recruit synthetic dancers who have psychological probl...

Article history: Received 15 April 2014 Received in revised form 29 September 2014 Accepted 28 October 2014 Available online 4 November 2014

Let F denote the class of all functions univalent in the unit disk and convex in the direction of the real axis. In the paper we discuss the functions of the class F which are n-fold symmetric, where n is positive even integer. For the class of such functions we find the Koebe set as well as the covering set, i.e. T f2F f ðDÞ and S f2F f ðDÞ. Moreover, the Koebe constant and the covering consta...

For every fixed graph H, we determine the H-covering number of Kn, for all n > n0(H). We prove that if h is the number of edges of H, and gcd(H) = d is the greatest common divisor of the degrees of H, then there exists n0 = n0(H), such that for all n > n0, C(H,Kn) = d dn 2h dn− 1 d ee, unless d is even, n = 1 mod d and n(n− 1)/d+ 1 = 0 mod (2h/d), in which case C(H,Kn) = d ( n 2 ) h e+ 1. Our m...