Given depreciation=
- 20% a year for the first five years
- 8% a year for the following ten years
Expressed in a piecewise continuous function for a machine that costs P on initial purchase
for 0<=t<=5
V(t)=P(1-0.2)^t
and for 5<t<=15
V(t)=P(1-0.2)^5*(1-0.08)^(t-5) because rate is 0.08 from year 6 and up.
=V(5)(1-0.08)^(t-5)
(36)
Comparing with the given equations, and putting P=10,000, we get
r1=(1-0.2)=0.8
r2=(1-0.08)=0.92
=> r1+r2=0.8+0.92=1.72
[there is no physical meaning to this quantity, probably just to make correction easier]
(37) need number of years, t, to have the value of machine equal to or less than $2000.
After the fifth year, value of machine
V(5)=10000(0.8)^5=3276.8
To arrive at V(t)=2000, we form the equation
V(t)=2000, or
3276.8(0.92)^(t-5)=2000, and solve for t
0.92^(t-5)=2000/3276.8=1/1.6384
take log on both sides
(t-5)=log(1/1638.4)/log(0.92)=5.92
=>
t=5.92+5=10.921
Therefore after 10.921 years, the value of the machine will be at $2000 or below.
(38)
For this problem, we divide
2x^4-3x^3+4x^2-5x+6 by (x-3) to get
a quotient of 2x^3+3x^2+13x+34 with a remainder of 108.
However, using the factor theorem, we only have to substitute (x-3)=0, or x=3 into the expression to find the remainder, namely
R=2x^4-3x^3+4x^2-5x+6
=2(3)^4-3(3)^3+4(3)^2-5(3)+6
=162-81+36-15+6
=108 as before.
