Answer:
b(b/a)^2
Step-by-step explanation:
Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then
b = a - (p% × a) = a(1-p%)
b/a = 1 - p%
p% = 1 - b/a = (a-b)/a
Let the worth of the car on December 31, 2012 be c
then
c = b - (b × p%) = b(1-p%)
Let the worth of the car on December 31, 2013 be d
then
d = c - (c × p%)
d = c(1-p%)
d = b(1-p%)(1-p%)
d = b(1-p%)^2
d = b(1- (a-b)/a)^2
d = b((a-a+b)/a)^2
d = b(b/a)^2 = b^3/a^2
The car's worth on December 31, 2013 = b(b/a)^2 = b^3/a^2
Answer:
Step-by-step explanation:
<u>Step 1: Combine</u>
Adding a negative number is same as subtracting a positive number.
Answer:
