Mathematics Assignment

This Mathematics assignment covers topics such as laws of indices, exponential decay, compound interest, depreciation, and arithmetic sequences. It includes solved problems and calculations for each topic. The assignment is suitable for college or university students studying Mathematics.

Mathematics Assignment
Student Name:
Instructor Name:
Course Number:
12th May 2019

1a Using the laws of indices, for numbers having the same bases their powers
are added and subtracted if they are multiplied and divided respectively.
2 a2 b−4
6 c−4 × c 2
b2 a2
¿ a2 b− 4
3 a2 b2 × c 2
c−4
= b−6 c6
3
b) A number raised to a given power when again raised to another power, then
the final powers is the product of the initial powers.
¿ ) 4 ( a2 b2
b6 )
=2 a12 b−20×( a2 b2
b6 )
=2 a12 a2×( b2 b−20
b6 )
=2 a14 b−24
= 2 a14
b24
2a ¿ 9 ¿2( j+1 )=91
Equating the powers because the bases are equal we have.
2(j+1) =1
2j+2 =1
2j= -1
j= - 1
2

b) 53 ( j−3) ×54 j=52 (2 j+6)
Equating the powers we have
= 3(j-3) +4j=2(2j+6)
= 3j-9+4j=4j+12
=3j=21
j=7
3 a) ac=b
b) 70.463=x
2.462=x
c) 43 x−2=65
(3x-2) log 4=log 65
0.6021(3x-2) =1.81291
1.8063x-1.2042=1.8129
1.8063x=1.2042+1.8129
1.8063x=3.0171
x = 1.670
4 a) It is an exponential decay.
When x=0, y=100 and when x=3, y=6.4
The values of y decreases with increase in x values hence
exponential decay.

b) Initial value occurs when x=0 i.e. y=100( 2
5 ¿ ¿0 = 100
y=100
c) y =100( 2
5 ¿ ¿n
Table 1: Data
x -3 -2 -1 0 1 2 3 4 5 6
y=100(
2
5 ¿ ¿n
1562.5 625 250 100 40 16 6.4 2.56 1.024 0.4096
d) When x=0, y=100 and when x=1, y=40
Percentage change = ( 100−40
1−0 ) × 100= 60 %
Percentage change= 60 % decrease
-4 -2 0 2 4 6 8
0
200
400
600
800
1000
1200
1400
1600
1800
A GRAPH OF Y AGAINST X
X
Y

e) When x=9, y=100(0.4)9 =0.0262
=0.03 (2d.p)
5 a) 100%-27%=83%=0.83
B=220(0.83 ¿ ¿n
B=220( 0.83 ¿ ¿n milligrams
Where B=amount of ibuprofen remaining
B0=amount of original ibuprofen
n=number of hours that elapse
b) B=220(0.83)6
= 71.93
= 72 milligrams
c) 220 1hr 182.6 1hr 151.558 1hr 125.79 1hr 104.41
Half life occur when B=110 milligrams. The half life (110mg) lies between
3hrs and 4hrs.
125.79 1 hr 104.411
Fraction of time for 110 in the range of 125.79 and 110
125.79-104.41=21.38
125.79-110=15.79
15.79
21.38 × 60 minutes= 44.3 minutes
3 hours + 44.3 minutes = (3×60) +44.3= 224.3minutes
Half life=224 minutes (nearest minute)
6i) Simple interest

3.5
100 ×5 ×35000=6125
ii) Compound interest
A= P (1+ r
100 ¿ ¿n
Where A=Amount , P=principal , r=rate and
n= number of times of calculating compound interest
n=5÷ 1
12=60 , r= 3.27
12 =0.2725
A = (1+ 0.2725
100 ) 60
A= P (1+ r
100 ¿ ¿n
=35000(1.002725)60
A=41 208, interest =41208-35000= 6207
iii) A= P (1+ r
100 ¿ ¿n =35000(1+ 3.35
100 )5
A=41229
Interest =41229-35000=6229
Recommendation; Invest at 3.27% compounded
monthly it will give highest interest of $6229
7. Calculating depreciation
i) Straight line method
Total depreciation in 3 years =125000 × 18
100 × 3=67500
ii) Reducing balance method

Year 1 depreciation=125000 × 18
100 =22500
125000-22500=102500
Year 2 depreciation =102500 × 18
100 =18450
102500-18450=84050
Year 3 depreciation 84050 × 18
100 =15129
Total depreciation=22500+18450+15129=56079
iii) Sum of years digits method
Sum of digits=1+2+3=6
Year 1 depreciation=125000 × 3
6 =62500
125000-62500=62500
Year 2 depreciation =62500 × 2
6 =20833.3
62500-20833.3=41666.7
Year 3 depreciation =41666.7× 1
6 =6944.5
Total depreciation=20833.3+6944.5+62500=90277.8
90277.8-56079=34198.8
Sum of the year’s digits gives the greatest depreciation
by 34198.8.
=$ 34 199
8 a) a=10, d=12-10=2

nth term = a+(n-1)
=10+2(n-1)
=10+ 2n-2
nth term = 8+2n
80th term =8+2(80) =168
b) a =100 ,r=0.25
Sn= a(1−rn )
1−r
When n=15
S15= 100(1−0.2515)
1−0.25 = 100(0.9999)
0.75 =133.33
Sum of the first 15 terms =133.33