The 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.

a(t)=136(1.123)^4t

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

a.

60%

b.

59%

c.

41%

d.

40%

Explanation:

The function that represents the number of E.coli bacteria cells per 100 mL of water as the time t years elapses is:

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.

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

its c or d   take a lucky guess my dude