Since center of dilation is the origin, this is easy. Just divide all of the x and y coordinate values by 3. Place the new point on the graph, and draw the triangle.
R' = R(3,6)/3 = (3/3,6/3)=(1,2). So R'(1,2)
S' = S(-3,6)/3 = (-3/3,6/3)=(-1,2). So S'(-1,2)
T' = T(-6,-6)/3 = (-6/3,-6/3)=(-2,-2). So T'(-2,-2)
So you now know the location of the 3 new points. R' at (1,2), S' at (-1,2) and T' at (-2,-2). Simply draw those 3 points on your graph and connect the lines to make a new triangle.
Given that a species of beetles grows 32% every year.
So growth rate is given by
r=32%= 0.32
Given that 100 beetles are released into a field.
So that means initial number of beetles P=100
Now we have to find about how many beetles will there be in 10 years.
To find that we need to setup growth formula which is given by
where A is number of beetles at any year n.
Plug the given values into above formula we get:


now plug n=10 years

Hence answer is approx 1606 beetles will be there after 20 years.
Now we have to find about how many beetles will there be in 20 years.
To find that we plug n=20 years

Hence answer is approx 25791 beetles will be there after 20 years.
Now we have to find time for 100000 beetles so plug A=100000





33.174666862=n
Hence answer is approx 33 years.