Tài liệu Linux Device Drivers-Chapter 1 :An Introduction to

Tài liệu Linux Device Drivers-Chapter 1 :An Introduction to Device Drivers ppt

... Chapter 1 :An Introduction to Device Drivers As the popularity of the Linux system continues to grow, the interest in writing Linux device drivers steadily increases. Most of Linux is independent ... modularity makes Linux drivers easy to write, to the point that there are now hundreds of them available. There are a number of reasons to be interested i...

Ngày tải lên : 24/12/2013, 01:17

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Tài liệu Starting Your Own Business: An introduction to different aspects of Norwegian regulations pptx

... Authorisation 10 CHOICE OF BUSINESS ENTITY 11 Sole Proprietorship (Self-employed Business) 11 General Partnerships 12 Business with Limited Liability 12 Cooperatives (SA) 15 Norwegian branch ... Obligations and rights 18 More information 18 Special conditions 18 ACCOUNTING AND AUDITING 19 Legal Basis 19 Accounting 19 Required documentation 19 Accounting principles...

Ngày tải lên : 18/02/2014, 00:20

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Tài liệu BPT BẠC NHẤT 1 ẨN

... trªn trôc sè:à Bài gi i:ả 1: 3 01: 29 1: 2 81: 27 1: 26 1: 25 1: 24 1: 23 1: 2 21: 21 1:20 1: 19 1: 18 1: 17 1: 16 1: 15 1: 14 1: 13 1: 12 1: 11 1 :10 1: 09 1: 08 1: 07 1: 06 1: 05 1: 04 1: 03 1: 02 1: 01 1:00 0:590:58 0:570:560:55 0:54 0:53 0:52 0: 51 0:50 0:490:48 0:47 0:46 0:45 0:440:43 0:42 0: 41 0:400:390:380:370:360:35 0:340:330:32 0: 310 :300:29 0:28 0:270...

Ngày tải lên : 27/11/2013, 06:11

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Tài liệu An introduction to Neural Networks pptx

... feed-forward network was 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 Figure 4.3: Example of function approximation with a feedforward network. Top left: The original learning ... equal to one on one side of the dividing line which is dened by: w 1 x 1 + w 2 x 2 = ; (3 .18 ) and equal to zero on the other side of this l...

Ngày tải lên : 20/12/2013, 19:15

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Tài liệu Compilers and Compiler Generators an introduction with C++ pptx

... INC 10 1E 20 STA LETTERS ; END 12 19 21 LDA LETTERS + 1 ; INC(Total) 14 05 INC 15 1E 21 STA LETTERS + 1 17 35 00 BRN LOOP ; END 19 19 20 EXIT LDA LETTERS 1B 0F OTC ; Write(Letters) 1C 19 ... "z") 0B 38 12 BPZ * + 6 0D 19 00 LDA LETTERS ; THEN INC(Letters) 0F 05 INC 10 1E 00 STA LETTERS ; END 12 19 01 LDA LETTERS + 1 ; INC(Total) 14 05 INC 15 1E...

Ngày tải lên : 26/01/2014, 07:20

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Tài liệu Metals and Society: an Introduction to Economic Geology pptx

... Paleoproterozoic 0 .1 11 10 18 0 0 .10 205 43 McArthur River Australia Paleoproterozoic 0.2 9.2 4 .1 41 237 31 Mount Isa Australia Paleoproterozoic 6.8 5.9 14 8 12 4 15 Red Dog U.S. Mississippian 16 .6 4.6 83 16 5 ... Laterites 13 2 5.4 .1 Bauxite 13 2 5.4.2 Ni Laterites 13 6 5.5 Other Lateritic Deposits 13 8 5.6 Supergene Alteration 13 8 References . . 13 9 6 The Future of...

Ngày tải lên : 19/02/2014, 22:20

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Tài liệu Lab 4.2.1 Configuring ISDN BRI (U-Interface) ppt

... Ottawa(config)#interface bri 0 Ottawa(config-if)#isdn spid1 510 55 510 0000 01 55 510 00 Ottawa(config-if)#isdn spid2 510 55 510 010 0 01 55 510 01 Step 8 Review switch status a. Check the state of the ... (SPIDs) may have to be specified for the ISDN Switch to respond to the ISDN TE1 correctly. The SPIDs, supported by the Adtran simulator, are specified as isdn spid1 and is...

Ngày tải lên : 11/12/2013, 14:15

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Tài liệu ISDN Lab Scenario 1 -- OSPF Build to Specifications docx

... 0x7F3 10 .10 .12 .2 10 .10 .11 .1 1 81 0x800000 01 0xFBFD 10 .10 .13 .1 10 .10 .11 .1 232 0x800000 01 0x1F8 10 .10 .13 .2 10 .10 .11 .1 1 81 0x800000 01 0xF008 17 2 .19 .1. 4 10 .10 .11 .1 252 0x800000 01 0x96C4 19 2 .16 8 .1. 0 ... 10 .10 .11 .1 183 0x800000 01 0xFBFD 10 .10 .13 .2 10 .10 .11 .1 183 0x800000 01 0xF008 17 2 .19 .1. 4 10 .10 .11 .1 233 0x800000 01...

Ngày tải lên : 11/12/2013, 14:15

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Tài liệu Đề thi và đáp án môn Toán cao cấp ppt

... của to n không gian 3 ¡ . Câu II: a/ ( ) n 1 n 1 n 1 +∞ = + − − ∑ = n 1 2 n 1 n 1 +∞ = + + − ∑ là chuỗi số dương Nhận thấy rằng 2 1 2n n 1 n 1 n 2 n n 1 n 1 > ⇔ > + + − ∀ ≥ + + − ( *1) * ... 2 y x x 1= − − có 2 nghiệm: 1 2 1 5 1 5 x vµ x 2 2 − + = = nên 2 1 5 x x 1 0 x 2 + − − > ∀ > ⇒ 2 n n 1 0 n n 1 n 2− − > ⇔ > + ∀ ≥ (*2) *...

Ngày tải lên : 13/12/2013, 15:16

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