Question

I need help with this math question all parts please

Answer 1

we have the function

`P(t)=(60(1+0.4t))/(0.01t+3)`

Part A

For t=0

substitute

`\begin{gathered} P(t)=(60(1+0.4(0)))/(0.01(0)+3) \\ \\ P(t)=(60(1))/(3) \\ \\ P(t)=20 \end{gathered}`

The initial population was 20 insects

Part B

For t=5 years -------> Convert to months

t=60 months

substitute

`\begin{gathered} P(t)=(60(1+0.4(60)))/(0.01(60)+3) \\ \\ P(t)=(60(1+24))/(0.6+3) \\ \\ P(t)=(1500)/(3.6) \\ \\ P(t)=416.67 \end{gathered}`

The answer is 417 insects

Part C

Determine horizontal asymptote

`\begin{gathered} P(t)=(60\left(1+0.4t\right))/(0.01t+3) \\ \\ rewrite \\ P(t)=(60+24t)/(0.01t+3) \end{gathered}`

The horizontal asymptote is given by the ratio of

`(24)/(0.01)=2,400`

P(t)=2,400 ---------> horizontal asymptote

that means

as the value of time t increases ------> the value of the population tends to 2,400 insects

the population cannot be greater than 2400 insects

The range for the function P(t) is the interval [20, 2400)

All real numbers greater than or equal to 20 insects and less than 2400 insects

Part D

Using a graphing tool