SELÇUKLULAR'A AİT NADİDE BİR TEKNİK MİNAİ: II. KILIÇ ARSLAN KÖŞKÜ ÖRNEKLERİ

ASFALT BETONUNDA RİJİTLİK MODÜLÜNÜN BELİRLENMESİ İÇİN BİR YAKLAŞIM AN APPROACH for DETERMINING the STIFFNESS MODULUS on ASPHALT CONCRETE

In this study, stiffness modulus parameters of asphalt concrete were determined experimentally for different temperature and exposure times. The stiffness modules were calculated according to Nijboer stiffness module. Basic physical properties and the quantity of bitumen of asphalt core samples were designated for determining the stiffness modules. The samples were exposed to 17 C (reference te...

Atypical Hemolytic Uremic Syndrome: Report of a Pediatric Case a Ti̇pi̇k Hemoli̇ti̇k Üremi̇k Sendrom: Bi̇r Olgu Sunumu

Atypical hemolytic uremic syndrome (HUS) is a rare disease in childhood. The morbidity and mortality of this disease is found high. In this report, a 13 years old boy who had five times of HUS attacks was presented. The family history revealed that his uncle was on hemodialysis for chronic renal failure. The patient had acute renal failure, hemolytic anemia, thrombocytopenia, and dark urine whe...

Mutation accumulation is still potentially problematic, despite declining paternal age: a comment on Arslan et al. (2017)

Final Report on Dual-tone Multiple Frequency Dtmf Detector Implementation Gg Uner Arslan for Ee382c Embedded Software Systems

Dual-tone Multi-frequency DTMF signals are used in touch-tone telephones as well as many other areas such a s i n teractive control, telephone banking, and pager system. As analog telephone lines are converted to digital, researchers became interested in digital DTMF detectors. There are many digital DTMF detecting algorithms, but most of them do not comply with the related International Teleco...

Electronic Companion — “ A Single - Product Inventory Model for Multiple Demand Classes ” by Hasan Arslan , Stephen C . Graves , and

Proposition 1. Consider two stocking policies s1 and s2 where for some j , s2 j = s1 j − 1, s2 j−1 = s1 j−1 + 1, and s2 i = s1 i , ∀ i = j j − 1. We have that: (i) ILi s2 i s 2 N ≥ ILi s1 i s1 N for i ∈ 1 j − 1 , and (ii) ∑j i=1 Bi i s 2 i s 2 N ≥ ∑j i=1 Bi i s 1 i s 1 N . Proof. From (6) and (7) we find that ILj s2 j s 2 N = ILj s1 j − 1 s1 j+1 s1 N = ILj s1 j s1 N − 1. Thus from (8) and (9), ...