Mathematics Assignment: Expression Reduction, Vectors & Carcass Weight

This document provides solutions to a mathematics assignment consisting of multiple questions. The first question involves simplifying an algebraic expression. The following questions cover vector coordinate determination, and solving for variables. The assignment further includes tasks related to mathematical modeling, specifically analyzing the relationship between average carcass weight and year, and calculating surface area based on volume. The solution provides step-by-step calculations and explanations for each problem, utilizing mathematical formulas and techniques. Desklib offers a variety of solved assignments and past papers for students seeking assistance with their studies.

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)
4∗P∗h
4 =4∗m∗4
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

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

= 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

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,

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
¿ 8.2124−7.7093
7.7095 ∗100
= 6.52%