Answer :
From the given figure we can see
The population in town A is increased by a constant rate because
[tex]\begin{gathered} 11000-10000=1000 \\ 12000-11000=1000 \\ 13000-12000=1000 \end{gathered}[/tex]Since the difference between every 2 consecutive terms is the same, then
The rate of increase of population is constant and = 1000 people per year
The form of the linear equation is
[tex]y=mx+b[/tex]m = the rate of change
b is the initial amount
Then from the information given in the table
m = 1000
b = 10,000
Then the equation of town A is
[tex]y=1000t+10000[/tex]Fro town B
[tex]\begin{gathered} R=\frac{10500}{10000}=1.05 \\ R=\frac{11025}{10500}=1.05 \\ R=\frac{11576}{11025}=1.05 \end{gathered}[/tex]Then the rate of increase of town by is exponentially
The form of the exponential equation is
[tex]y=a(R)^t[/tex]a is the initial amount
R is the factor of growth
t is the time
Since R = 1.05
Since a = 10000, then
The equation of the population of town B is
[tex]y=10000(1.05)^t[/tex]We need to find t which makes the population equal in A and B
Then we will equate the right sides of both equations
[tex]10000+1000t=10000(1.05)^t[/tex]Let us use t = 4, 5, 6, .... until the 2 sides become equal
[tex]\begin{gathered} 10000+1000(4)=14000 \\ 10000(1.05)^4=12155 \end{gathered}[/tex][tex]\begin{gathered} 10000+1000(5)=15000 \\ 10000(1.05)^5=12763 \end{gathered}[/tex][tex]\begin{gathered} 10000+1000(6)=16000 \\ 1000(1.05)^6=13400 \end{gathered}[/tex][tex]\begin{gathered} 10000+1000(30)=40000 \\ 10000(1.05)^{30}=43219 \end{gathered}[/tex]Since 43219 approximated to ten thousand will be 40000, then
A and B will have the same amount of population in the year 30
The answer is year 30