Aculture contains 2000 bacteria initially and doubles every 40 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.

Hello there!

We know that the culture has 2000 bacteria initially (at t = 0 min), and it doubles when t = 40min, again at t = 80 min, and so on

And the rate of grouth is proportional to the population, then we can obtain an expression of the form:

P(t) = 2000*k(t) where k(t) is the factor that defines the growth of the population:

we know that:

p(0) = 2000*k(0) = 2000

p(40) = 2000*k(40) = 2000*2 = 4000

p(80) = 2000*k(80) = 2000*2*2 = 8000

etc

then we can see that k(t) = 2^(t/40)

where k(0) = 1, k(40) = 2, k(80) = 2*2, etc

then the function that models the population of the culture is:

p(t) = 2000*2^(t/40)

where t is in minutes.

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!