6067 bullying series part 2 bullying perpetrator
... discussion What’s your opinion about bullying? What makes a person tease and frighten another? What can we to protect ourselves from bullies? Have you ever been a victim of bullying? Which punishment should ... envied her -4 Ariella used to laugh at her bullying victim and to dissuade her classmates to seat next to her ... the text, is still in the Middle School -2 She was very swee...
Ngày tải lên: 27/08/2016, 17:14
Ngày tải lên: 12/02/2014, 15:20
... recycle-light-ends, good-use-of-raw-materials, vent/flare lights, lights-are-cheap-as-fuel, etc The use of design rationale information to support design can be used to improve the documentation ... on computers The remarkable benefits conferred by the computer have left us thirsting for more: more computers and more powerful systems Computer power is already astonishing State-of-the-ar...
Ngày tải lên: 11/08/2014, 09:21
6066 bullying series part 1 bullying victim
... they his -10 who 11 they 12 who B) Match the words on the left with their ... discussion What’s your opinion about bullying? What makes a person tease and frighten another? What can we to protect ourselves from bullies? Have you ever been a victim of bullying? ... 10 Throughout the text Kelsey makes a description of herself Write a full description of her
Ngày tải lên: 27/08/2016, 06:55
6068 bullying series part 3 bullying upstander
... -3 Nowadays it is accepted that teenagers defend others from bullying -4 Sidney has got some bullying stories to tell ... for discussion What you know about bullying? Why students bully other students? What can we to protect ourselves from bullies? Have you ever been a victim of bullying? Which punishment should ... -5 What did Sidney, the upstander, when she realized the situat...
Ngày tải lên: 28/08/2016, 07:03
Longman preparation series for the toeic answer keys part 2
Ngày tải lên: 28/10/2013, 22:15
Tài liệu Longman preparation series for the toeic test advanced part 2 ppt
Ngày tải lên: 24/12/2013, 11:17
Tài liệu Longman preparation series for the new toeic test part 2 ppt
... PRACTICE TEST ONE You will find the Answer Sheet for Practice Test One on page 25 5 Detach it from the book and use it to record your answers Play the audio for Practice Test One when you ... directions are given for each part You must mark your answers on the separate answer sheet Do not write your answers in the test book •t PART Directions: For each question...
Ngày tải lên: 26/01/2014, 16:20
preparation series for the new toeic test advanced course 4 Episode 2 Part 2 potx
Ngày tải lên: 21/07/2014, 21:21
preparation series for the new toeic test advanced course 4 Episode 1 Part 2 potx
Ngày tải lên: 21/07/2014, 21:21
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 2 ppsx
... x1 + x2 − x3 = 2x1 − 3x2 + x3 = −x1 + 2x2 − x3 = 1 To obtain an original basis, we form the augmented tableau e1 0 e2 e3 0 a1 a2 a3 b 1 1 −3 1 1 1 and replace e1 by a1 e2 by a2 , and e3 ... 2. 2 Examples of Linear Programming Problems 15 c x + c2 x2 + · · · + c n x n subject to the nutritional constraints a 11 x1 + a 12 x2 + · · · + a1n xn a 21 x1 + a 22 x2 + · · · + a2n...
Ngày tải lên: 06/08/2014, 15:20
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 1 pot
... no 10 11 12 13 14 15 16 17 18 19 20 Gauss-Southwell 00 −0 8 711 11 1 445584 2 087054 2 13 0796 2 16 3586 2 17 027 2 2 1 727 86 2 17 427 9 2 17 4583 2 17 4638 2 17 46 51 2 17 4655 2 17 4658 2 17 4659 ... 17 4659 2 17 4659 00 −0 37 025 6 −0 376 011 1 446460 2 0 529 49 2 06 023 4 2 06 023 7 2 16 56 41 2 16 5704 2 16 8440 2 17 39 81 2 17 40...
Ngày tải lên: 06/08/2014, 15:20
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 2 ppt
... Daniel [D1] and Faddeev and Faddeeva [F1] 9.5 The partial conjugate gradient method presented here is identical to the so-called s-step gradient method See Faddeev and Faddeeva [F1] and Forsythe ... definition of pk pT qk = k T k gk H k gk (21 ) and hence xT Hk+1 x = aT a bT b − aT b bT b 2 + x T pk T k gk Hk gk (22 ) Both terms on the right of (22 ) are nonnegative—the first by t...
Ngày tải lên: 06/08/2014, 15:20
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 3 pot
... 11 12 DFP DFP (with restart) Self-scaling 20 0 .33 3 20 0 .33 3 20 0 .33 3 20 0 .33 3 2. 7 32 7 89 93. 65457 93. 65457 2. 811061 836 899 × 10 2 56. 929 99 56. 929 99 5 627 69 × 10 2 37 6461 × 10−4 1. 620 688 1. 620 688 20 0600 ... 1.564971 939 804 × 10 2 810 1 23 × 10−4 16 920 5 × 10−5 3 7 23 85 × 10−7 96 .30 669 994 0 23 × 10−1 22 5501 × 10 2 30 1088 × 10 3 636 716 × 10 3...
Ngày tải lên: 06/08/2014, 15:20
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 4 pps
... minimize 2 2x1 + 2x1 x2 + x2 − 10x1 − 10x2 2 subject to x1 + x2 3x1 + x2 The first-order necessary conditions, in addition to the constraints, are 4x1 + 2x2 − 10 + 2x1 + 2x2 − 10 + 1 x1 + x2 + =0 2 ... extremize 2 x1 + x2 + x2 x3 + 2x3 subject to 2 x + x2 + x3 = The first-order necessary conditions are 1+ x1 = 2x2 + x3 + x2 = x2 + 4x3 + x3 = One solution to this set is easily seen t...
Ngày tải lên: 06/08/2014, 15:20