I know this is alot but I am giving all of the points I have on this questions I
ID: 2983128 • Letter: I
Question
I know this is alot but I am giving all of the points I have on this questions I am new and need help before 10pm. If someone already answer a question could you try another 1 will rate accordingly.
It is known that a certain kind of algae in the Herring Sea can double in population every 4 days. Suppose that the population of algae grows exponentially, beginning now with a population of 3,000,000.
Enter exact value answers below.
(a) How long it will take for the population to quadruple in size?
days
(b) What is the growth constant for this algae?
k =
(c) How long it will take for the population to triple in size?
days
The radioactive isotope plutonium-239 has a half-life of 24110 years.
The decay constant of plutonium-239 is
Enter the exact value expression as your answer.
Hint: Be very careful with positive and negative signs.
If you have 75 grams of plutonium-239 to begin with, the equation that describes the radioactive decay is
P(t) =
Match the right answer.
a. How many cm3 of the sulfate remain after 5 minutes?
b. When will half of the initial amount remain?
c. After how long will 5 cm3 remain?
Solve for t d. How much sulfate was injected?
e. State the differential equation satisfied by A(t).
Solve for t f. When is the sulfate being absorbed at a rate of 5 cm3 per minute? g. How fast is the sulfate being absorbed after 5 minutes?
Explanation / Answer
standard exponential growth formula: y=Ce^(kt)
C is initial population constant
e is 2.7182818
k is growth constant
t is the independant variable
you know that y(0) = 3,000,000
and that y(4) = 6,000,000; doubles every 2 days
3,000,000 = Ce^(k0)
3,000,000 = Ce^0
3,000,000 = C
now use 3,000,000 for C
6,000,000 = 3,000,000Ce^(k2)
2 = e^(k2)
ln(2) = 2k
ln(2)/2 = k, the growth constant, it equals about 0.3465
y=3,000,000e^(0.3465t)
now for time to triple
9,000,000 = 3,000,000e^(0.3465t)
3=e^(0.3465t)
ln(3)=0.3465t
t=ln(3)/0.3465 = 3.17 days