Engineering Mathematics 4 Episode 5 pptx

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 m Problem ... speed/time is shown in Fig. 20.3. 30 25 Graph of speed/time 20 15 Speed (m/s) 10 5 0 12 3 Time (seconds) 45 6 2 .5 4. 0 7.0 15. 0 5. 5 8. 75 10. 75...

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Engineering Mathematics 4 Episode 3 pptx

Engineering Mathematics 4 Episode 3 pptx

... 0.62e 4. 178 [(a) 5. 0988 (b) 0.0 640 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 .52 e 1.2 651 (c) 5e 2.6921 3e 1.1171  (a) 4. 55 848 (b) 2 .40 444 (c) 8. 051 24  6. ... C x 2 2! C x 3 3! C x 4 4! CÐÐÐ Hence e 0 .5 D 1 C0 .5 C 0 .5 2 21 C 0 .5 3 321 C 0 .5 4 4 321 C 0 .5 5 5 4 321...

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Engineering Mathematics 4 Episode 9 pptx

Engineering Mathematics 4 Episode 9 pptx

... in megapascals for 15 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 ... Table 40 .1. PRESENTATION OF STATISTICAL DATA 317 39 .5 40 .0 39.8 39 .5 39.9 40 .1 40 .0 39.7 40 .4 39.3 40 .7 39.9 40 .2 39.9 40 .0 40 .1 39.7 40 .5...

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Engineering Mathematics 4 Episode 12 pptx

Engineering Mathematics 4 Episode 12 pptx

... 150 kPa the constant is given by 3 ð 150  D 45 0 kPa m 3 or 45 0 kJ. Hence p v D 45 0, or p D 45 0 v Work done D  6 2 45 0 v dv D [ 45 0 ln v] 6 2 D 45 0[ln 6 ln 2] D 45 0 ln 6 2 D 45 0 ln 3 D 49 4 .4kJ Problem ... 1toC4 R.m.s. deviation D  1 4 1  4 1 y 2 dx D  1 5  4 1 2x 2  1 2 dx D  1 5  4 1 4x 4  4x 2 C 1dx D  1 5  4x 5 5  4x 3 3 C x...

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Engineering Mathematics 4 Episode 1 ppsx

Engineering Mathematics 4 Episode 1 ppsx

... of 4 5 8  3 1 4 C 1 2 5 4 5 8  3 1 4 C 1 2 5 D 4 3 C1 C  5 8  1 4 C 2 5  D 2 C 5 ð 5 10 ð 1 C8 ð 2 40 D 2 C 25  10 C16 40 D 2 C 31 40 D 2 31 40 Problem 4. Find the value of 3 7 ð 14 15 Dividing ... 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 problems on the area u...

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Engineering Mathematics 4 Episode 2 docx

Engineering Mathematics 4 Episode 2 docx

... 5 will divide into, is 5x. Multiplying both 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 ... x 2 D 25 is thus written as x = 5 Problem 15. Solve: 15 4t 2 D 2 3 ‘Cross-multiplying’ gives: 15 3 D 24t 2  i.e. 45 D 8t 2 45 8 D t 2 i...

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Engineering Mathematics 4 Episode 4 potx

Engineering Mathematics 4 Episode 4 potx

... /! 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 5 4 ð3 ð2 4 ð3 ð23 ð2 D 35 (b) 10 C 6 D 10! 6!10 ... annulus). d = 2. 25 cm d = 5 . 45 cm Figure 17. 15 Area of shaded part D area of large circle  area of small circle D D 2 4  d 2 4 D  4 D...

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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. 758 8 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. 54 2...

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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 ... equa- tion 4x 2 D4x C 15, i.e. 4x 2 C 4x  15 D 0. This is shown in Fig. 30.9, where the roots are x D2 .5 and x D 1 .5 as before. y 30 25 20 15 10 5 0 −3 −2 −112...

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Engineering Mathematics 4 Episode 8 docx

Engineering Mathematics 4 Episode 8 docx

... theorem: 7 Cj5 4 D [ p 74 6 144 .46 ° ] 4 D p 74 4 6 4 ð 144 .46 ° D 54 7 6 6 57 7. 84 ° D 54 7 6 66 217. 84 ° or 54 7 6 66 217 ° 15  in polar form. Since r 6  D r cos  C jr sin Â, 54 7 6 6 217. 84 ° D 54 7 6 cos ... Cj1.000 5 6 45 ° D 5 cos 45 °  Cj sin 45 °  D 5cos 45 °  Cj5sin 45 °  D 3 .53 6 j3 .53 6 4 6 120 ° D 4 cos 120 ° C j sin 120...

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