The population of rabbits on an island are growing exponentially. Initially, there were 850 rabbits. Four months later, there are 1000 rabbits on the island. (a) Use the exponential growth equation, Q(t) = Qoekt to find the growth constand k. Let t be time in months. Round k to 4 decimal digits. 350 850 ,000 O0D rncreau (b) How fast is the population of rabbitsgrowing at t = 12? Write your answer is a full sentence in context of the problem. 850040b 12 27. :1,383.5 a ato o6 1,384 animals at 12 m (c) How fast is the growth rate changing at 12 months. Your answer must have proper units.

a) k = 0.0414

b) the population of rabbits after 12months is approximately 1148rabbits. It can be deduced that the number of rabbits keep growing at a slow pace

c) There is about one third increase in the initial growth of the rabbits after 12months

Step-by-step explanation:

Given the exponential growth equation Q(t) = Qoe^kt where;

Q(t) is the population of the rabbits at time t

Qo is the initial population of the rabbit

k is the growth constant

t is the time

If there are 850rabbits initially, this means @ t = 0, Qo = 850

Therefore, Q(t) = 850e^k(0)

Q(0)= 850

If there are 1000 rabbits 4 months later, 1000 = Qoe^4k

1000 = 850e^4k

e^4k = 1000/850

e^4k = 1.18

Applying ln to both sides, we have

lne^4k = ln1.18

4k = ln 1.18

k = ln 1.18/4

k = 0.0414 (to 4dp)

b) To know how far the rabbit is growing after 12months i.e @ t = 12

Using our exponential growth equation Q(t) = Qoe^kt

Given Qo = 850, k = 0.0414 t = 12

Substituting this values in the equation we have;

Q(12) = 850e^0.0414(12)

Q(12) = 850×1.35

Q(12) = 1147.8

This means that the population of rabbits after 12months is approximately 1148rabbits. It can be deduced that the number of rabbits keep growing at a slow pace

c) There is about one fifth of the initial population increase in growth rate every 4 months which is about one third increase in the initial growth of the rabbits after 12months

the image of (6, 9) under a dilation is (4, 6). the scale factor is

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3