He number of e. coli bacteria cells in a pond of stagnant water can be represented by the function below, where a represents the number of e. coli bacteria cells per 100 ml of water and t represents the time, in years, that has elapsed.

based on the model, by approximately what percent does the number of e. coli bacteria cells increase each year?

a.
59%
b.
40%
c.
41%
d.
60%

The function that represents the number of E.coli bacteria cells per 100 mL of water as t years elapses, and is missing in the question, is:

   Option A. 59%

Explanation:

The base, 1.123, represents the multiplicative constant rate of change of the function, so you just must substitute 1 for t in the power part of the function::

Then, the multiplicative rate of change is 1.590, which means that every year the number of E.coli bacteria cells per 100 mL of water increases by a factor of 1.590, and that is 1.59 - 1 = 0.590 or 59% increase.

You can calculate it also using two consecutive values for t. For instance, use t =1 and t = 1

step-by-step explanation:

|ab| = 1.4 = 4

h = 1.4 = 4

a(abc) area = 4.4 = 16/2 = 8

area of abc = 8 cm^2

good luck ! ; )

greater than one

step-by-step explanation: