Engineering Mathematics 4 Episode 2 docx
... x 2y 2 (ii) 3a b 2 (i) x 2 4xy C4y 2 (ii) 9a 2 6ab C b 2 7. 3a C2[a 3a 2 ] [4 a] 8. 2 5[aa 2b a b 2 ] [2 C5b 2 ] 9. 24 p [2 35p q 2 p C2q C3q] [11q 2p] In ... following: 1. 2x ł4x C 6x 1 2 C 6x 2. 2x ł4x C6x 1 5 3. 3a 2a ð4a C a [4a1 2a] 4. 3a 2a4a C a [a3 10a] 5. 2y C 4 ł6y C 3 4 5y 2 3y 3y C 12 ...
Ngày tải lên: 13/08/2014, 08:21
Engineering Mathematics 4 Episode 8 docx
... ð 123 .69 ° , by De Moivre’s theorem D 21 97 6 7 42 . 14 ° D 21 97 6 3 82. 14 ° since 7 42 . 14 Á 7 42 . 14 ° 360 ° D 3 82. 14 ° D 21 97 66 22 . 14 ° since 3 82. 14 ° Á 3 82. 14 ° 360 ° D 22 . 14 ° Problem 2. ... Cj5 4 D [ p 74 6 144 .46 ° ] 4 D p 74 4 6 4 ð 144 .46 ° D 547 6 6 577. 84 ° D 547 6 66 21 7. 84 ° or 547 6 66 21 7 ° 15 in polar form....
Ngày tải lên: 13/08/2014, 08:21
Engineering Mathematics 4 Episode 1 ppsx
... of (a) 2 3 ð 2 4 2 7 ð 2 5 and (b) 3 2 3 3 ð 3 9 From the laws of indices: (a) 2 3 ð 2 4 2 7 ð 2 5 D 2 3C4 2 7C5 D 2 7 2 12 D 2 7 12 D 2 5 D 1 2 5 D 1 32 (b) 3 2 3 3 ð 3 9 D 3 2 3 3 1C9 D 3 6 3 10 D ... š3(c)š 1 2 (d) š 2 3 In Problems 4 to 8, evaluate the expressions given. 4. 9 2 ð 7 4 3 4 ð 7 4 C 3 3 ð 7 2 147...
Ngày tải lên: 13/08/2014, 08:21
Engineering Mathematics 4 Episode 3 pptx
... MATHEMATICS (b) 5 e 0 .25 e 0 .25 e 0 .25 C e 0 .25 D 5 1 .2 840 2 541 0.77880078 1 .2 840 2 541 C 0.77880078 D 5 0.50 52 24 6 2. 0 628 261 D 1 .2 24 6 , correct to 4 decimal places Problem 4. The instantaneous ... x 2 C 4x 32 D 0 [4, 8] 2. x 2 16 D 0 [4, 4] 3. x C 2 2 D 16 [2, 6] 4. 2x 2 x 3 D 0 1, 1 1 2 5. 6x 2 5x C 1...
Ngày tải lên: 13/08/2014, 08:21
Engineering Mathematics 4 Episode 4 potx
... Cx n , 2 1 C x 4 1 2 D 2 1 C 1 2 x 4 C 1 /2 1 /2 2! x 4 2 C 1 /2 1 /2 3 /2 3! x 4 3 CÐÐÐ D 2 1 C x 8 x 2 128 C x 3 10 24 ÐÐÐ = 2 Y x 4 − x 2 64 Y x 3 5 12 −··· This ... frequency, f 1 D 1 2 k 1 I 1 D 1 2 k 1 2 1 I 1 2 1 D 1 2 [1 C0. 04 k] 1 2 [1 0. 02 I] 1 2 D 1 2 1 C0. 04 1...
Ngày tải lên: 13/08/2014, 08:21
Engineering Mathematics 4 Episode 5 pptx
... equations (1) and (2) gives: 0. 344 3x C 120 D 1.0 724 x 0. 344 3x C 0. 344 3 120 D 1.0 724 x 0. 344 3 120 D 1.07 24 0. 344 3x 41 .316 D 0. 728 1x x D 41 .316 0. 728 1 D 56. 74 m From equation (2) , height of ... theorem: 41 2 D 9 2 CYZ 2 from which YZ D p 41 2 9 2 D 40 units. Thus, sin X = 40 41 ,tanX = 40 9 = 4 4 9 , cosec X = 41 40 = 1 1 40 ,s...
Ngày tải lên: 13/08/2014, 08:21
Engineering Mathematics 4 Episode 6 ppsx
... is: (a) 2 74. 7 m (b) 36 .4 m (c) 34. 3 m (d) 94. 0 m 32. (7, 141 ° ) in Cartesian co-ordinates is: (a) (5 .44 , 4. 41) (b) (5 .44 , 4. 41) (c) (5 .44 , 4. 41) (d) (5 .44 , 4. 41) 33. If tanA D 1 . 42 76, sec ... x D 6.3680 3.7588 D 1.6 9 42 i.e. tan x D 1.6 9 42 , and x D tan 1 1.6 9 42 D 59 .44 9 ° or 59 ° 27 [Check: LHS D 4sin59. 44 9 ° 20 ° D 4sin39 .44 9 ° D 2. 5...
Ngày tải lên: 13/08/2014, 08:21
Engineering Mathematics 4 Episode 7 ppsx
... 9 C7 .2 7 .2 7 .2 7 .2 7 .2 y D5x 2 C 9x C 7 .2 6.8 1 .45 7 .2 11 .2 x 22 .53 5x 2 20 31 .25 45 C9x 18 22 .5 27 C7 .2 7 .2 7 .2 7 .2 y D5x 2 C 9x C 7 .2 5 .2 1.55 10.8 Problem 5. Plot a graph of: y D 2x 2 and hence ... y D 2x 2 3x 4and y D 2 4x y D 2x 2 3x 4 is a parabola and a table of values is drawn up as shown below: x 2 10 12 3 2x 2 820...
Ngày tải lên: 13/08/2014, 08:21
Engineering Mathematics 4 Episode 9 pptx
... 22 .4 22 .8 21 .5 22 .6 21 .1 21 .6 22 .3 22 .9 20 .5 21 .8 22 .2 21.0 21 .7 22 .5 20 .7 23 .2 22. 9 21 .7 21 .4 22 .1 22 .2 22. 3 21 .3 22 .1 21 .8 22 .0 22 .7 21 .7 21 .9 21 .1 22 .6 21 .4 22 .4 22 .3 20 .9 22 .8 21 .2 22. 7 21 .6 22 .2 ... 48 places and the results are as shown. 2. 10 2. 29 2. 32 2 .21 2. 14 2. 22 2 .28 2. 18 2. 17 2...
Ngày tải lên: 13/08/2014, 08:21
Engineering Mathematics 4 Episode 10 ppsx
... Y Y 2 12 7 .25 6 .25 5 9 4. 25 9 .25 3 13 6 .25 5 .25 12 21 2. 75 2. 75 14 17 4. 75 1 .25 7 22 2. 25 3.75 3 31 6 .25 12. 75 28 47 18.75 28 .75 14 17 4. 75 1 .25 7 10 2. 25 8 .25 3 9 6 .25 9 .25 13 ... 35 49 5 40 25 X D 145 Y D 180 X 2 D 46 01 XY Y 2 275 25 300 100 24 0 22 5 24 0 40 0 27 5 625 27 0 900 24 5 122 5 20 0 1600 XY...
Ngày tải lên: 13/08/2014, 09:20