Answer:
a)

b) The projected population after 6 years is of 103 stray cats.
c) The number of years required for the stray-cat population to reach 700 is 16.64.
Explanation:
The population N(t) after t years, following an exponential growth moel, is given by:

In which N(0) is the initial population and r is the growth rate.
In 1999 the town had 35 stray cats, and the relative growth rate was 18% per year.
This means that

(a) Find the function that models the stray-cat population n(t) after t years.


(b) Find the projected population after 6 years.
This is N(6).



The projected population after 6 years is of 103 stray cats.
(c) Find the number of years required for the stray-cat population to reach 700.
This is t for which N(t) = 700. So








The number of years required for the stray-cat population to reach 700 is 16.64.