Question

A population of rabbits is described by the function R(t) = 100(2t/5), where t is measured in months and R is measured in rabbits. Create a clear and properly labeled graph of R(t) on the domain 0 ≤ t ≤ 15 months.

Required:
a. Find ΔR on [1,2].
b. Find R(0).
c. Find R(10)
d. When will the population be 500 rabbits?

Answer 1

To find the average rate of change, initial value, and population at a specific time for the given population of rabbits, we can use the provided function. We can also determine when the population will reach a certain value.

Step-by-step explanation:

To create a graph of the function, we need to plot various points on the graph. We can choose different values of t within the given domain and calculate the corresponding values of R(t). For example, when t = 0, we have R(0) = 100(2(0)/5) = 0. When t = 1, we have R(1) = 100(2(1)/5) = 40. Using this method, we can find points to plot on the graph.

(a) To find ΔR on [1,2], we substitute the values of t into the function. R(1) = 40 and R(2) = 100(2(2)/5) = 80. ΔR is the difference between these values: ΔR = R(2) - R(1) = 80 - 40 = 40.

(b) To find R(0), we substitute t = 0 into the function: R(0) = 100(2(0)/5) = 0.

(c) To find R(10), we substitute t = 10 into the function: R(10) = 100(2(10)/5) = 400.

(d) To find when the population will be 500 rabbits, we set R(t) equal to 500 and solve for t. 500 = 100(2t/5). Dividing both sides by 100 gives us 5 = 2t/5. Multiply both sides by 5 gives us 25 = 2t. Dividing both sides by 2 gives us t = 12.5 months.

Answer 2

Answer and Step-by-step explanation: The graph is shown in the attachment.

a. ΔR on [1,2] is mathematically expressed as:

ΔR = R(2) - R(1)

which means difference of population of rabbits after 2 months and after 1 month.

`R(1) = 100((2)/(5).1 )`

`R(1) = 100((2)/(5) )`

R(2) =
`100((2)/(5).2 )`

`R(2) = 100((4)/(5) )`

`\Delta R = 100((4)/(5) )-100((2)/(5) )`

`\Delta R = 100[(4)/(5) - (2)/(5) ]`

`\Delta R=` 40

Difference of rabbits between first and second months is 40.

b. R(0) = 100(
`(2)/(5) .0`)

R(0) = 0

Initially, there no rabbits in the population.

c. R(10) =
`100((2)/(5).10 )`

R(10) = 400

In 10 months, there will be 400 rabbits.

d. R(t) = 500

`500=100((2)/(5).t )`

`(500)/(100)=(2)/(5).t`

`t = (500.5)/(100.2)`

t = 12.5

In 12 and half months, population of rabbits will be 500.