Engineering Mathematics 4 Episode 13 pps

Engineering Mathematics 4 Episode 13 pps

Engineering Mathematics 4 Episode 13 pps

... second moment of area about axis QQ, I QQ D 3 84 C 48 10 2 D 51 84 cm 4 Radius of gyration, k QQ D  I QQ area D  51 84 48 D 10 .4cm 49 8 ENGINEERING MATHEMATICS the techniques introduced in Sections ... matrices (a)  2 1  74  and  30 7 4  and (b)  31 4 43 1 14 3  and  275 21 0 63 4  . (a) Adding the corresponding elements gives:  2 1  74  C  30 7 4 ...

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Engineering Mathematics 4 Episode 1 ppsx

Engineering Mathematics 4 Episode 1 ppsx

... mid-ordinate rule 44 1 53 .4 Simpson’s rule 44 3 Assignment 14 447 54 Areas under and between curves 44 8 54. 1 Area under a curve 44 8 54. 2 Worked problems on the area under a curve 44 9 54. 3 Further worked ... ax n 40 7 47 .3 Standard integrals 40 8 47 .4 Definite integrals 41 1 48 Integration using algebraic substitutions 41 4 48 .1 Introduction 41 4 48 .2 Algebrai...

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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 1.6 942 i.e. tan x D 1.6 942 , and x D tan 1 1.6 942 D 59 .44 9 ° or 59 ° 27  [Check: LHS D 4sin59. 44 9 °  20 °  D 4sin39 .44 9 ° D 2. 542 RHS D 5co...

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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 ... 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 1012 3 2x 2 8202818 3x 6303 6 9 4 4 4 4 4 4 4 y 10 1 4...

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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 ... 20 144 11 25 121 9 30 81 7 35 49 5 40 25  X D 145  Y D 180  X 2 D 46 01 XY Y 2 275 25 300 100 240 225 240 40 0 275 625 270 900 245 1225 200 1600  X...

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Engineering Mathematics 4 Episode 11 pps

Engineering Mathematics 4 Episode 11 pps

... figures (a)  2 1 4 e 2x dx D  4 2 e 2x  2 1 D 2[ e 2x ] 2 1 D 2[ e 4  e 2 ] D 2[ 54. 5982 7.3891] D 94 .42 (b)  4 1 3 4u du D  3 4 ln u  4 1 D 3 4 [ln 4 ln 1] D 3 4 [1.3863 0] D 1 . 040 Now try ... 0] D 1 3  1 5 D 2 15 or 0 .133 3 Problem 7. Evaluate:   4 0 4cos 4  dÂ, correct to 4 significant figures   4 0 4cos 4  d D 4   4 0 cos 2 Â 2...

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Engineering Mathematics 4 Episode 2 docx

Engineering Mathematics 4 Episode 2 docx

... sides by 5x gives: 5x  3 x  D 5x  4 5  Cancelling gives: 15 D 4x1 15 4 D 4x 4 i.e. x D 15 4 or 3 3 4 Check: LHS D 3 3 3 4 D 3 15 4 D 3  4 15  D 12 15 D 4 5 D RHS (Note that when there is ... 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...

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Engineering Mathematics 4 Episode 3 pptx

Engineering Mathematics 4 Episode 3 pptx

... (a) 1 7 e 3 .46 29 (b) 8.52e 1.2651 (c) 5e 2.6921 3e 1.1171  (a) 4. 55 848 (b) 2 .40 444 (c) 8.051 24  6. (a) 5.6823 e 2.1 347 (b) e 2.1127  e 2.1127 2 (c) 4 e 1.7295  1 e 3.6817  (a) 48 . 041 06 (b) 4. 0 748 2 (c) ... D 2u C 2a1 Substituting s D 144 , t D 4intos D ut C 1 2 at 2 gives: 144 D 4u C 1 2 a 4 2 i.e. 144 D 4u C 8a2 Multiplying equation (1) by 2 gives: 84 D...

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Engineering Mathematics 4 Episode 4 potx

Engineering Mathematics 4 Episode 4 potx

... 1 or n P r = n! .n − r /! Thus, 4 P 2 D 4 3 D 12 or 4 P 2 D 4! 4 2! D 4! 2! D 4 ð3 ð2 2 D 12 Problem 19. Evaluate: (a) 7 C 4 (b) 10 C 6 (a) 7 C 4 D 7! 4! 7 4 ! D 7! 4! 3! D 7 ð6 ð5 4 ð3 ð2 4 ð3 ð23 ð2 D ... D 8 .4 360 125/180 D 34 .5cm.) THE BINOMIAL SERIES 117 (3.039) 4 may be written in the form 1 Cx n as: 3.039 4 D 3 C0.039 4 D  3  1 C 0.03...

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Engineering Mathematics 4 Episode 5 pptx

Engineering Mathematics 4 Episode 5 pptx

... 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 .316 D 0.7281x x D 41 .316 0.7281 D 56. 74 m From equation (2), height of building, h D 1.0724x D 1.07 24 56. 74 D 60 .85 ... 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 ,secX = 41 9 = 4 5 9...

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