Acertain 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. if that car was worth a dollars on december 31, 2010 and was worth b dollars on december 31, 2011, what was the car worth on december 31, 2013 in terms of a and b ?

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

Let x = time, hours, from 1:00
Let y = water level, inches

Note that from 1:00 to 2:30, x = 2.5 - 1.0 = 1.5 hours.

The given data is
x:   0      1.5
y:  13   28.0

The change in x is 1.5 - 0 = 1.5 hours
he change in y is 28 - 13 = 15 inches

The rate of change is (15 inches)/(1.5 hours) = 10 inches/hour

10 inches per hour