This document provides study material and solved assignments for Maths. It covers topics like law of sine, amplitude and period of waveform, determinant of matrix, constructing histograms, and more.
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TASK 1 1.1 In ∆ABC, A = 530, B = 610and length of a = 12.6cm So, b and c can be calculated by following way - using law of sine - a=b=c Sin ASin BSin C from, given ∆ - A + B + C = 180 530+ 610+ C = 180 C = 660 therefore, 12.6=b=c Sin530Sin610Sin 660 12.6=b=c 0.790.870.91 so, b =13.87 cm while, c = 14.51 cm B Here, perpendicular = 15 unit hypotenuse = 17 unit then, base can be determined by -
b2= h2 –p2 = 172– 152 = 289 – 225 = 64 or,b =√64 = 8 unit Sin Ø= perpendicular / hypotenuse Sin Ø= 15/17 unit TASK 2 a) Amplitude and period of waveform y = 4 cos(2θ + 450) Amplitude of given equation is 4 and, period of above equation = 2 π / 2θ = π / θ a) Amplitude and period of waveform y = 6 sin (t - 300) Amplitude of given equation is 6 and, period of above equation = 2 π / t c) To Prove - sin2x (sec x + cosec x)= 1 + tan x cosx tanx taking LHS side of given equation- sin2x (sec x + cosec x) cosx tanx »sin2x (1/cosx + 1/sinx ) cox . sinx/cosx »sin2x (sin x + cos x)/ sinx cosx cox . sinx/cosx »sinx + cos x cox » tanx + 1 = RHS Hence Proved 2
When base of equation is same, then power is also equal - therefore, 2x2+ 2x – 3= 3 or,2x2+ 2x – 6 = 0 or,x2+ x – 3 = 0 Solving this quadratic equation - x =-b ± √ (b2– 4ac) 2a x =-1 ± √ (12– 4 x 1 x -3) 2 x 1 =-1 ±√13 3 therefore, x = (-1 + √13)/ 2 or, (-1 - √13)/ 2 4
Task 3 22x-1 A =∫∫dx. dy 1-x+2 22x-1 A =∫ [y]dx. 1-x+2 2 A =∫[2x – 1 + x – 2]dx 1 2 A =∫[3x – 3]dx 1 2 A = 3∫[x – 1]dx 1 2 A = 3 [ x2/2- x ]1 A = 3/2 5
a21a22a23 a31a31a33 then, determinant of any 3x3 matrix can be calculated by using the given formula A=a11( a22x a33– a23x a31) – a12( a21x a33– a23x a31) +a13(a21x a31- a22x a31) Using this formula, to find determinant of given matrix is given by can be calculated as 123 A =0-41 03-1 then,determinant of A = [ 1 ( -4 x -1 – 1 x3)- 2 ( 0 x -1 – 1 x 0 ) + 3 (0 x 3 – 1 x 0) ] = [ 1 (4 – 3) – 2 ( 0 – 0) + 3 ( 0 – 0 )] = 1 10
Lo3 Task 1 Task 2 Scenario 1 RevenueNumber of customers (£1000) JanuaryJuly Less than 52722 5 – 103839 10 – 154069 15 – 202241 20 – 301320 30 – 4045 Firstly, organised the above table into grouped frequency in following manner - RevenueNumber of customers (£1000) JanuaryJuly 0 to 52722 5 to 103839 10 to 154069 15 to 202241 20 to 301320 11
30 to 4045 As in the above mentioned table, group interval is not regular interval therefore, for further calculation, it is required to transform the table into equal grouped frequency in following way - Equal class interval - RevenueNumber of customers JanuaryJuly 0 to 106561 10 to 2062110 20 to 301320 30 to 4045 Now, for constructing the histogram of each monthly data, separate the table Monthly revenue of January - RevenueNumber of customers January 0 to 1065 10 to 2062 20 to 3013 30 to 404 Monthly revenue of July - RevenueNumber of customers July 12
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0 to 1061 10 to 20110 20 to 3020 30 to 405 a) Histogram of January month - Now, to calculate mode of January month data, use the below formula - c Mode = l +f1– f0xh 2 f1– f0– f2 here,f1represents the highest frequency data = 65 13 0 to 1010 to 2020 to 3030 to 40 0 10 20 30 40 50 60 706562 13 4 Hostogram Number of customers January Revenue Number of customers
f0defines the previous frequency fromf1= 0 f2refers to next frequency = 62 h represents the class difference = 10 and,l defines as lowest bound interval of modal class = 0, Put these values in above mentioned formulae to find mode in following way – Mode (z) = 0 +65 – 0x 10 2 x 65 – 0 – 62 = 0 +65 x 10 130 – 62 = 9.55 a) Histogram of July month - It has been depicted from this histogram that group 10 to 20 consists the maximum frequency, therefore, considering this group as modal class, mode can be calculated as - Mode = l +f1–f0xh 14 0 to 1010 to 2020 to 3030 to 40 0 20 40 60 80 100 120 61 110 20 5 Histogram Number of customers July Revenue Number of customers
2 f1– f0- f2 here,f1= 110,f0= 61,f2= 20, h = 10 and l = 10, then, Mode (z) = 10 +110 – 61x 10 2x110 – 61 – 20 = 10 +49 x 10 220 – 81 = 13.5 15
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Task 3 Given - Force (N) x Time (s) yx2y2xy 11.40.56129.960.31366.384 18.70.35349.690.12256.545 11.70.55136.890.30256.435 12.30.52151.290.27046.396 14.70.43216.090.18496.321 18.80.34353.440.11566.392 19.60.31384.160.09616.076 107.23.061721.521.405644.549 Then, using formula - ∑Y = a0N + a1∑X and, ∑XY = a0∑X+ a1∑X2 From, above table -∑X = 107. 2,∑XY = 44.549, N = 7 and∑X2= 1721.52,∑Y = 3.06 put these values in given, 3.06 = 7 a0+ 107.2 a1…Eq(i) 44.549 = 107.2 a0+ 1721.52 a1… Eq(ii) On solving both equation, 16
a0= 0.865and a1= - 0.028 then, equation of regression line of time on force – Y = a0+ a1X so,Y = 0.865 – 0.028 X Lo4 Task 1 Determine surface area of a line 350mm rotating around x axis at a distance of 440 mm using pappus theorem Task 2 M (x) = x3- 3x2- 4 Value of bending equation m, with the interval 0.5 from the range 1 to 4 can be determined by Variable xBending moment M 0.5-4.63 1-6 1.5−2.85 2-8 3.526.63 412 17
From this graph, it has estimated that bending moment of zero will be 2.4 near about. 18 00.511.522.533.544.5 -10 -5 0 5 10 15 20 25 30 Bending moment M