I took the liberty of finding for the complete question.
And here I believe that the problem asks for the half life of Curium. Assuming
that the radioactive decay of Curium is of 1st order, therefore the
rate equation is in the form of:
A = Ao e^(-kt)
where,
A = amount after t years = 2755
Ao = initial amount = 3312
k = rate constant
t = number of years passed = 6
Therefore the rate constant is:
2755/3312 = e^(-6k)
-6k = ln (2755/3312)
k = 0.0307/yr
The half life, t’, can be calculated using the formula:
t’ = ln 2 / k
Substituting the value of k:
t’ = ln 2 / 0.0307
t’ = 22.586 years
or
t’ = 22.6 years
Answer:
Where is it I don't see it?
Step-by-step explanation:
You have to solve for 9x first
The first option
Reflect triangle UTS over the line y=2 and dilate it by a scale factor of 2 from point S
Answer:
aₙ = -2aₙ₊₁
Step-by-step explanation:
According to the sequence given 16, -8, 4, ...
a1 = 16
a2 = -8
a3 = 4
From the values, we can conclude thst;
a1 = -2(-8)
Since a2 = -8, then;
a1 = -2(a2)
Similarly
a2 = -2(4)
a2 = -2a3
The subsequent sequence are;
a3 = -2a4
The nth term will ne;
aₙ = -2aₙ₊₁
Hence the required recursive function is aₙ = -2aₙ₊₁
