The question is:
y = c e^kx
find c and k
at x = 0, y = 100
so
100 = c (1)
so c = 100
at x = 2, y = 4000
so
4000 = 100 e^2k
ln 40 = 2 k
3.69 = 2 k
so
k = 1.84
so
y = 100 e^(1.84 x)
PLEASE HELP wiTH ALL OF THEse. I DON'T KNOW WHERE TO START---THANKS!
1. At 2pm the number of bacteria in a colony was 100, by 4pm it was 4000. Write an exponential function to model the population y of bacteria x hours after 2pm.
2.Using the information in problem 1, how many bacteria were there at 7pm that day?
3.Radioactive iodine is used to determine the health of the thyroid gland. It decays according to the equation y = ae-0.0856t, where t is in days. Find the half-life of this substance.
4. An anthropologist finds there is so little remaining Carbon-14 in a prehistoric bone that instruments cannot measure it. This means that there is less than 0.5% of the amount of Carbon-14 the bones would have contained when the person was alive. How long ago did the person die?
5 answers
2) I think you can do that now
3)
At t = 0, y = a
what is t when y = a/2
a/2 = a e^-.0856 T
.5 = e^-.0856 T
ln .5 = -.0856 T
you take it from there
3)
At t = 0, y = a
what is t when y = a/2
a/2 = a e^-.0856 T
.5 = e^-.0856 T
ln .5 = -.0856 T
you take it from there
Look up the half life of C 14. You have it.
Then how many half lives is .5% = .005
(1/2)^n = .005
n ln .5 = ln .005
n * -.693 = -5.3
so we know that the thing is at least n years old
age >/= 7.65 half lives
Then how many half lives is .5% = .005
(1/2)^n = .005
n ln .5 = ln .005
n * -.693 = -5.3
so we know that the thing is at least n years old
age >/= 7.65 half lives
I love you Damon!!!
name the sets of numbers to which -7 belongs.