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...Coolnessjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj jjjjk klj kjfsdj jklfjskjg kfjsgkjf jk jdfskj kjksdfjgkl;fsdjkg jk jk jksjf k j j jk jkj sdj kjg kj k j kjfsdkj g; jkljgsdf jgkfd jgkjgk s;jiowrutiowu tui u ivcn,nb,.sn ,cvxn j kgjjn bn nxcm,n xm hs;w;q a a jg j kgsdfjkljg j j jkgjklg j k kj gjk j kg kl sfdkj kjsj ksj klgjs l; ksdljldfsj k ks jsdfjl; sjs s ;jls l jl l lsgjdjgs lksdfjg sldjgkl l;jgsdl ls;jgflkj js erchandise; Acclaim acquired the rights to Turok when it purchased Acclaim Comics (né Valiant) in 1994. Suffering from cash flow problems and falling sales, Turok became Acclaim's best hope for a financial turnaround. Iguana pushed the Nintendo 64's graphics capabilities to its limits, and were forced to compress or cut elements to fit the game on its 8-megabyte cartridge. Bugs delayed the game's release from holiday 1996 to 1997. Critical reception of Turok was highly positive. Becoming one of the most popular games for the console on release, Turok won praise for its graphics and evolution of the genre. Complaints centered on graphical slowdowns caused by multiple enemies appearing onscre layer controls a Native American warrior, Turok, who must stop the evil Campaigner from conquering the universe with an ancient and powerful weapon. As Acclaim's first exclusive title for the Nintendo 64, Turok was part of a strategy to develop outline that surveys key articles in Wikipedia, divided into 12 major branches...
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...(melting, vaporization & vapor pressure) 3 General Properties of Gases Properties of Gases • There is a lot of “free” space in gases • Gases can be expanded infinitely • Gases occupy containers uniformly & completely • Gases diffuse & mix rapidly Gas properties can be modeled using math which depends on: • V = volume of the gas (L or dm3) • T = temperature (K) • n = amount (mol) • P = pressure (atm) 5 FHSC1114 Physical Chemistry 4 6 1 Centre for Foundation Studies, UTAR Boyle’s Law Unit of Pressure & Volume • Unit of Pressure: The effect of pressure on gas volume 1 atm = 760 mm Hg = 101325 Pa (Nm-2) = 101.325 kPa = 1.013 bar = 760 torr P 1 V inversely proportional PV constant • Unit of Volume: PV1 P V2 1 2 1 dm3 = 1 L 1 m3 = 103 dm3 = 106 cm3 at constant n and T 7 Example 1: Boyle’s Law 8 Solution: A sample of CO2 has a pressure of 55 mm Hg in a volume of 125 mL. The sample is compressed to the new pressure 78 mm Hg. What is the new volume? Boyle’s Law: P1V1 P2V2 at cons tan t n and T P1 = 55 mm Hg , V1 = 125 mL P2 = 78 mm Hg , V2 = ? 9 10 Charles’s Law Example 2: Charles’s Law The effect of temperature on gas volume A sample of CO2 has volume of 25.0ml at room temperature. What is the final volume of gas if it is heated at 37 °C ? V T Directly proportional...
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...zecimilor; f-cifra sutimilor; g-cifra miimilor. 2. Fracţii -Fracţii zecimale finite: a, b = ab ; 10 a, bc = abc ; 100 -Fracţii zecimale periodice:- ab − a abc − a ; a, (bc) = ; 9 99 abc − ab abcd − ab mixte: a, b(c ) = ; a, b(cd ) = ; 90 990 simple: a, (b) = 3.. Rapoarte şi proporţii a a a⋅n se numeste raport ∀b ≠ 0; = = k, n ∈ Q* , b b b⋅ n k se numeşte coeficient de proporţionalitate ; Proprietatea fundamentală a proporţiilor: ac = ⇒a⋅ d = b⋅ c bd 4. Proporţii derivate: ⎧ ⎪ ⎪ a c ⎪ = ⇒⎨ b d ⎪ ⎪ ⎪ ⎩ b d sau = a c a = a±b c a a+c = b b+d d c a b sau = = b a c d c a±b c = sau ±d b a a−c = sau sau b b−d ±d d a2 c2 . = 2 b d2 2 5. Sir de rapoarte egale: a a + a 2 + a 3 + .... + a n ; a1 a = 2 = ......... = n = 1 b1 b2 bn b1 + b 2 + b 3 + ..... + b n (a 1 , a 2 , a 3 ,...... a n )şi (b 1 , b 2 , b 3 ,.... b n ) sunt direct a a1 a 2 = = .. = n = k . b1 b2 bn proporţionale ⇔ (a 1 , a 2 , a 3 ,...... a n )şi (b 1 , b 2 , b 3 ,.... b n ) sunt invers proporţionale ⇔ a1 ⋅ b1 = a 2 ⋅ b2 = .. = a n ⋅ bn 6. Modulul numerelor reale Proprietăţi: a ⎧ ⎪ a, ⎪ ⎨ 0, ⎪ ⎪− a, ⎩ def a〉0 a=0 a 〈0 1. a ≥ 0, ∀a ∈ R ; 2. a = 0, 3. a = −a, 4. a = b, 5. a ⋅b = a ⋅ b 6. a a = b...
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...SOLUTIONS MANUAL CRYPTOGRAPHY AND NETWORK SECURITY PRINCIPLES AND PRACTICE FOURTH EDITION WILLIAM STALLINGS Copyright 2006: William Stallings -2- © 2006 by William Stallings All rights reserved. No part of this document may be reproduced, in any form or by any means, or posted on the Internet, without permission in writing from the author. -3- NOTICE This manual contains solutions to all of the review questions and homework problems in Cryptography and Network Security, Fourth Edition. If you spot an error in a solution or in the wording of a problem, I would greatly appreciate it if you would forward the information via email to ws@shore.net. An errata sheet for this manual, if needed, is available at ftp://shell.shore.net/members/w/s/ws/S. W.S. -4- TABLE OF CONTENTS Chapter 1: Chapter 2: Chapter 3: Chapter 4: Chapter 5: Chapter 6: Chapter 7: Chapter 8: Chapter 9: Chapter 10: Chapter 11: Chapter 12: Chapter 13: Chapter 14: Chapter 15: Chapter 16: Chapter 17: Chapter 18: Chapter 19: Chapter 20: Introduction ..................................................................................................5 Classical Encryption Techniques ...............................................................7 Block Ciphers and the Date Encryption Standard ................................13 Finite Fields .................................................................................................21 Advanced Encryption Standard ...................
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Words: 2727 - Pages: 11
...Nguy n T t Thu B t ð ng Th c B T ð NG TH C I. LÝ THUY T 1. ð nh nghĩa : Cho a ; b ∈R. M nh ñ “ a > b” ; “a ≥ b” ; “a < b” ; “ a ≤ b” g i là b t ñ ng th c 2.Tính ch t : * a > b và b > c ⇒ a > c * a >b⇔ a +c >b+c * a > b và c > d ⇒ a + c > b + d ac > bc khi c > 0 * a>b⇒ ac < bc khi c < 0 * a >b≥0⇒ a > b * a ≥ b ≥ 0 ⇔ a 2 ≥ b2 * a > b ≥ 0 ⇒ an > bn 3. B t ñ ng th c v giá tr tuy t ñ i * | x |< a ⇔ −a < x < a ( V i a > 0 ) x > a * | x |> a ⇔ ( V i a > 0) x < −a 4. B t ñ ng th c gi a trung bình c ng và trung bình nhân ( Bñt Cauchy) a+b a) Cho a, b ≥ 0 , ta có ≥ ab . D u ‘=’ x y ra khi và ch khi a = b 2 H qu :*. Hai s dương có t ng không ñ i thì tích l n nh t khi 2 s ñó b ng nhau *. Hai s dương có tích không ñ i thì t ng nh nh t khi 2 s ñó b ng nhau a+b+c 3 b) Cho a, b, c ≥ 0 , ta có ≥ abc . D u ‘=’ x y ra khi và ch khi a = b = c 3 5. Phương pháp ch ng minh b t ñ ng th c I. Phương pháp bi n ñ i tương ñương ð ch ng minh BðT d ng A ≥ B ta thư ng dùng các cách sau : Cách 1 : Ta ch ng minh A − B ≥ 0 . ð là ñi u này ta thư ng s d ng các h ng ñ ng th c ñ phân tích A − B thành t ng ho c tích c a nh ng bi u th c không âm. Chú ý : M t s k t qu ta thư ng hay s d ng * x 2 ≥ 0 ∀x và x 2 = 0 ⇔ x = 0 ; | x |≥ 0 ∀x và | x |= 0 ⇔ x = 0 * a 2 + b 2 + c 2 ≥ 0 . ð ng th c x y ra ⇔ a = b = c = 0 . Ví d 1 : Cho hai s th c a, b . Ch ng minh r ng : a 2 + b 2 ≥ 2ab . Gi i : Ta có a 2 + b 2 − 2ab = (a − b)2 ≥ 0 ⇒ a 2 + b 2 ≥ 2ab . ð ng th c có ⇔ a = b . Ví d 2 : Cho ba s th c a, b, c . Ch ng minh...
Words: 8140 - Pages: 33