A culture contains 2000 bacteria initially and doubles every 60 minutes. Assuming that the rate of growth is proportional to the number of bacteria, find a function that models the number P(t) of bacteria after t minutes.

The function

P(t) = 2000e^0.0011552453t or 2000e^(In2/60)t

Step-by-step explanation:

A culture contains 2000 bacteria initially and doubles every 60 minutes. Assuming that the rate of growth is proportional to the number of bacteria, find a function that models the number P(t) of bacteria after t minutes.

The function is gotten as:

P(t) = Poe^kt

Where P(t) = Number of the bacteria after t minutes

P(o) = Initial number of the bacteria

e = Euler number

k = Growth rate

t = time

We have to find k

We are told that the culture contains 2000 bacteria initially and doubles every 60 minutes.

Hence,

n(60) = 2 × 2000

n(60) = 4000

This:

4000 = 2000 × e^k × 60

4000/2000 = e^60k

2 = e ^60k

In 2 = In e^60k

In 2 = 60k

k = In2/60

k = 0.011552453

Hence, the function P(t) = 2000e^0.0011552453t or 2000e^(In2/60)t

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq