Pls help find the amount at the end of 6 hours

A. 7,348

Step-by-step explanation:

P = le^kt

intitial population = 500

time = 4 hrs

end population = 3,000

So we have all these variables and we need to solve for what the end population will be if we change the time to 6 hours. First, we need to find the rate of the growth(k) so we can plug it back in. The given formula shows a exponencial growth formula. (A = Pe^rt) A is end amount, P is start amount, e is a constant that you can probably find on your graphing calculator, r is the rate, and t is time.

A = Pe^rt

3,000 = 500e^r4

now we can solve for r

divide both sides by 500

6 = e^r4

now because the variable is in the exponent, we have to use a log

ln(6) = 4r

we can plug the log into a calculator to get

1.79 = 4r

divide both sides by 4

r = .448

now lets plug it back in

A = 500e^(.448)(6 hrs)

A = 7351.12

This is closest to answer A. 7,348

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8