Answer:
1a ) N_6 = 50 mg
1b) N_3 = 70.7 mg
1c) N_1 = 89.1 mg
2) f(t) = 100e^(-0.1155t)
Step-by-step explanation:
Formula to get amount remaining is;
N_t = N_o × ½^(t/t_½)
We are given;
N_o = 100 mg
t_½ = 6 years
We have; N_t = 100(½^(t/6))
Thus;
1a) amount left after 6 years is;
N_6 = 100 × ½^(6/6)
N_6 = 50 mg
1b) amount left after 3 years is;
N_3 = 100 × ½^(3/6)
N_3 = 70.7 mg
1c) amount left after 1 year is;
N_1 = 100 × ½^(1/6)
N_1 = 89.1 mg
2) to find the function to model the amount left after t years, let's use the formula;
f(t) = f_o × e^(-rt)
Where r is decay constant
Thus;
f(t) = 100e^(-rt)
We recall that N_t = 100(½^(t/6))
Thus;
100(½^(t/6)) = 100e^(-rt)
Divide both sides by 100 to get;
(½^(t/6)) = e^(-rt)
In (½^(t/6)) = In e^(-rt)
Using power rule, we have;
(t/6) In ½ = -rt In e
-0.6931t/6 = - rt (because In e = 1)
t will cancel out to give;
r = 0.6931/6
r = 0.1155
Thus;
f(t) = 100e^(-0.1155t)