Noo question djjddrf f f f f f
1 answer:
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Answer:
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
Last Question
Because this may not be the question you want, I'll omit the first step which should be done with latex. The key to all of these except C is to invert the second fraction (turn the second fraction upside down) and multiply.
A
=(6*-2) / (5*3) = - 12/15 = -4/5
B
= (7*5) / (4*3) = 35 / 12 = 2 11/12
C
Impossible to do 4/0 cannot be done.
D
= (-2*-4)/(3*5) =
8/15