Engineering Mathematics 4 Episode 2 docx

Engineering Mathematics 4 Episode 2 docx

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[aa  2b  a  b 2 ] [2 C5b 2 ] 9. 24 p  [2 35p q 2 p C2q C3q] [11q  2p] In ... following: 1. 2x ł4x C 6x  1 2 C 6x  2. 2x ł4x C6x  1 5  3. 3a 2a ð4a C a [4a1 2a] 4. 3a 2a4a C a [a3 10a] 5. 2y C 4 ł6y C 3 4 5y  2 3y  3y C 12 ...

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Engineering Mathematics 4 Episode 8 docx

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....

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Engineering Mathematics 4 Episode 1 ppsx

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...

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Engineering Mathematics 4 Episode 3 pptx

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...

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Engineering Mathematics 4 Episode 4 potx

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...

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Engineering Mathematics 4 Episode 5 pptx

Engineering Mathematics 4 Episode 5 pptx

... equations (1) and (2) gives: 0. 344 3x 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 3x 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...

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Engineering Mathematics 4 Episode 6 ppsx

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 4sin59. 44 9 °  20 °  D 4sin39 .44 9 ° D 2. 5...

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Engineering Mathematics 4 Episode 7 ppsx

Engineering Mathematics 4 Episode 7 ppsx

... 9 C7 .2 7 .2 7 .2 7 .2 7 .2 y D5x 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 D5x 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...

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Engineering Mathematics 4 Episode 9 pptx

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...

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Engineering Mathematics 4 Episode 10 ppsx

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...

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