Mathematics Assignment: Expression Reduction, Vectors & Carcass Weight

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Added on  2023/06/07

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Homework Assignment
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Solution
Q1)
(a -2b)2 + 2(a + b)2
[a(a-2b) -2b(a-2b)] + [2(a(a + b) + b(a + b))]
[a2 -2ab -2ab + 4b2] + [2(a2 + ab + ab + b2)]
a2 -4ab + 4b2 + 2a2 + 4ab + 2b2
3a2 + 6 b2
3(a2 + 2 b2)
Q2) 2* a+b
2*(¿2
1 ) ¿ + (¿3
2 ) ¿ = (¿1
4 )¿
Q3)
4Ph
4 =4m4
h = 16 m
P
Q4)
A(1, 6) and B(3, 24)
F(x) = y = b*ax
Y2/y1 = ax2 –x1
24/6 = a3-1
4 = a2
a = 2
and
b = y1/ax1 = 6/21
= 3
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F(x) = 3*2x
Q5)
x -2 -1 0 1
f(x) 4 2.5 1 -0.5
g(x) -6 -3 0 3
h(x) 0 2 2 3
Therefore,
C = g(x)
B = h(x)
A = f(x)
Q6)
a = e
ln (0.5)
t = e
ln (0.5)
4 = 0.8409
8 = b*0.84093
b = 13.454
f(x) = 13.454 *0.8409x
using the equation above
f(x) = 13.454 *0.8409x
when x = -1
f(x) = 13.454 *0.8409-1
= 8
when x = 11
f(x) = 13.454 *0.840911
= 2
when y = 0.5
0.5 = 13.454 *0.8409x
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= 0.5/13.454 = 0.8409x
0.037163668 = 0.8409x
Apply logarithm to base 10
Log(0.037163668) = xLog(0.8409)
X = Log(0.037163668)/ Log(0.8409)
X = 19
Filling the table below with the values calculated respectively
x -1 3 11 19
y 8 8 2 0.5
Tasks, on PC (Preferably Maple)
Task 1
Qa)
year
(x)
Average
carcass
weight in
kilograms
.
(y)
x2 xy
0 41 0 0
1 41.4 1 41.4
2 42.2 4 84.4
3 42.7 9 128.1
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4 42.4 16 169.6
5 44.2 25 221
6 44.4 36 266.4
7 45.7 49 319.9
8 46 64 368
Total Total Total Total
36 390 204 1598.8
y=a x +nb
x y=a x2 +b x
390 = 36a + 9b
1598.8 = 204a + 36b
Using elimination method
b = 40.75
a = 0.65
Qb) a = 0.65 shows Average carcass weight in kilograms and year are directly correlated
Task 2
Qa)
Y = 22.41*x0.663
Y = S.A
X = Volume
When,
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X = 0.2 dm3
Y = 22.41*0.20.663
Y = 7.7095 dm2
Qb) when volume is increased by 10%
110%
= 110/100 * 0.2 = 0.22 dm3
X = 0.22 dm3
Y = 22.41*0.220.663
Y = 8.2124 dm2
Percentage change of the surface area
¿ 8.21247.7093
7.7095 100
= 6.52%
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