Engineering Mathematics 4 Episode 6 ppsx

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 A ... 5) [(5.83, 59. 04 ° or (5.83, 1.03 rad)] 2. (6. 18, 2.35) [ (6. 61, 20.82 ° or (6. 61, 0. 36 rad)] 3. (2, 4) [ (4. 47, 1 16. 57 ° or (4. 47, 2.03 rad)] 4. (5 .4, 3.7)...

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Engineering Mathematics 4 Episode 1 ppsx

Engineering Mathematics 4 Episode 1 ppsx

... 40 20 1A4E 16 D 1 ð 16 3 C A ð 16 2 C 4 ð 16 1 C E ð 16 0 D 1 ð 16 3 C 10 ð 16 2 C 4 ð 16 1 C 14 ð 16 0 D 1 40 96 C 10 ð2 56 C 4 ð 16 C 14 ð1 D 40 96 C2 560 C 64 C 14 D 67 34 Thus, 1A4E 16 = 67 34 10 To convert ... to 6, determine the values of the expressions given: 1. 23 .6 C 14. 71 18.9  7 .42 1 [11.989] 2. 73. 84  113. 247 C8.21  0. 068 [31. 265 ]...

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Engineering Mathematics 4 Episode 7 ppsx

Engineering Mathematics 4 Episode 7 ppsx

... below. t 10 20 30 40 50 60 80 100 s 4. 9 7 .6 11.1 15 .4 20 .4 26 .4 40 .6 58.0 s 3 t 0.19 0.23 0.27 0.31 0.35 0.39 0 .47 0.55 240 ENGINEERING MATHEMATICS Stress s N/cm 2 8 . 46 8. 04 7.78 Temperature ... as shown below: I 2.2 3 .6 4. 1 5 .6 6.8 lg I 0. 342 0.5 56 0 .61 3 0. 748 0.833 P 1 16 311 40 3 753 1110 lg P 2.0 64 2 .49 3 2 .60 5 2.877 3. 045 A graph of lg P aga...

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Engineering Mathematics 4 Episode 10 ppsx

Engineering Mathematics 4 Episode 10 ppsx

... 171.1 xy x 2 y 2 45 9.2 43 .2 48 86  744 .8 25.7 21580 8.9 9 .4 8 4. 3 0 3733 46 .5 3.7 581  242 .1 24. 3 241 1 6 74. 8 62 .9 7 242  xy D2172  x 2 D 169 .2  y 2 D 40 441 Thus r D 2172 p 169 .2 40 441 D −0.830 This ... x 2 y 2 45 .3 52 .6 39.1 39.3 18.1 85 .6 32.8 39.1 27 .6 7 .6 7 .6 7 .6 5.9 22 .6 1 .6 8 .4 5.1 14. 1 79.7 39.1 162 .6 539.1 351 .6 8 26....

Ngày tải lên: 13/08/2014, 09:20

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Engineering Mathematics 4 Episode 2 docx

Engineering Mathematics 4 Episode 2 docx

... 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 C12  6. 2y C 4 ł6y C 3 4 5y  2 3y C ... ð10 3 (b) 0.375 m 2 (c) 240 ð10 3 Pa  60 ENGINEERING MATHEMATICS Check: LHS D 3 4  2 D 3 6 D 1 2 RHS D 4 3 4 C 4 D 4 12 C4 D 4 8 D 1 2 Hence the solution t D 4...

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Engineering Mathematics 4 Episode 3 pptx

Engineering Mathematics 4 Episode 3 pptx

... 0.0 64 0 37 (c) 40 .44 6] In Problems 5 and 6, evaluate correct to 5 decimal places: 5. (a) 1 7 e 3 . 46 29 (b) 8.52e 1. 265 1 (c) 5e 2 .69 21 3e 1.1171  (a) 4. 55 848 (b) 2 .40 444 (c) 8.051 24  6. (a) 5 .68 23 e 2.1 347 (b) e 2.1127  ... significant figures: (a) 1 4 ln 4. 7291 (b) ln 7. 869 3 7. 869 3 (c) 5.29 ln 24. 07 e 0.1 762 (a) 1 4 ln 4. 7291 D 1 4 1.5537 349 D...

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Engineering Mathematics 4 Episode 4 potx

Engineering Mathematics 4 Episode 4 potx

... 32ax 3 C 16x 4  2. Use the binomial theorem to expand 2  x 6  64 192x C 240 x 2  160 x 3 C 60 x 4  12x 5 C x 6  3. Expand 2x  3y 4  16x 4  96x 3 y C216x 2 y 2  216xy 3 C 81y 4  4. Determine ... 35 (b) 10 C 6 D 10! 6! 10 6 ! D 10! 6 !4! D 210 Problem 20. Evaluate: (a) 6 P 2 (b) 9 P 5 (a) 6 P 2 D 6! 6 2! D 6! 4! D 6 ð5 4 ð3 ð2 4 ð3 ð2 D...

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Engineering Mathematics 4 Episode 5 pptx

Engineering Mathematics 4 Episode 5 pptx

... tan 1 1. 762 9 D 60 .44 ° Measured from 0 ° , the two angles between 0 ° and 360 ° whose tangent is 1. 762 9 are 60 .44 ° and 180 ° C 60 .44 ° ,i.e. 240 .44 ° 1. 762 9 60 .44 240 .44 90° 270° 360 ° x 180° y ... 1.0724x 0. 344 3x C 0. 344 3120 D 1.0724x 0. 344 3120 D 1.07 24 0. 344 3x 41 .3 16 D 0.7281x x D 41 .3 16 0.7281 D 56. 74 m From equation (2), height of...

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Engineering Mathematics 4 Episode 8 docx

Engineering Mathematics 4 Episode 8 docx

... 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 66 217. 84 ° or 547 6 66 217 ° 15  in polar form. Since r 6  D r cos  C jr sin Â, 547 6 6 217. 84 ° D 547 6 cos 217. 84 ° C ... form: (a) 16 6 75 ° 2 6 15 ° (b) 10 6  4 ð 12 6  2 6 6   3 (a) 16 6 75 ° 2 6 15 ° D 16 2 6 75 °  15 °  D 8 66 60...

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Engineering Mathematics 4 Episode 9 pptx

Engineering Mathematics 4 Episode 9 pptx

... samples of tin were determined and found to be: 34. 61 , 34. 57, 34. 40, 34. 63 , 34. 63 , 34. 51, 34. 49, 34. 61 , 34. 52, 34. 55, 34. 58, 34. 53, 34. 44, 34. 48 and 34. 40 Calculate the mean and standard deviation from ... 40 .2 40 .6 40 .1 39.7 40 .2 40 .3         There is no unique solution, but one solution is: 39.3–39 .4 1; 39.5–39 .6 5; 39.7–39.8 9; 39.9 40 .0 17...

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