Answer:
f(10) = - 26
Step-by-step explanation:
To evaluate f(10), substitute x = 10 into f(x), that is
f(10) = 4 - 3(10) = 4 - 30 = - 26
1.7n +1
I think the is the answer
I hope this helps :)
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
Answer:
Step-by-step explanation:
This is a differential equation problem most easily solved with an exponential decay equation of the form
. We know that the initial amount of salt in the tank is 28 pounds, so
C = 28. Now we just need to find k.
The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is
. Thus, the change in the concentration of salt is found in
inflow of salt - outflow of salt
Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:

Therefore,
or just
and in terms of time,

Thus, our equation is
and filling in 16 for the number of minutes in t:
y = 24.834 pounds of salt