A water tank is contaminated with a colony of 20 000 e. Coli bacteria. After 3 days the bacteria has grown to 28 284 e. Coli. What is the doubling time of the bacteria using an equation

6

Step-by-step explanation:

The formula for doubling time is given as:

P(t) = Po(2)^t/k

Where:

P(t) = Population after time t = 28,284

Po = Initial Population = 20,000

t = time = 3 days

k = doubling time = ?

28,284 = 20,000 × (2)^3/k

Divide both sides by 20,000

28,284/20,000 = 20,000/20,000 × (2)^3/k

1.4142 = (2)^3/k

2^1/2 = (2)^3/k

Divide both sides by 2

1/2 = 3/k

Cross Multiply

k × 1 = 2 × 3

k = 6

Therefore, the doubling time(k) of the bacteria is 6

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!