<h2>Ex 5</h2>
Simply multiply the equation by 6, in order to get rid of the denominators:
<h2>Ex 6</h2>
We have to rearrange the equation in the form
So that 
is the radius, and 
will be the diamater. We have to complete the squares: rewrite the equation as
You can see that 
is the beginning of 
Similarly, 
is the beginning of 
So, if we add 16 and 9 to both sides, the equation becomes
So, now we know that 
, and thus the radius is 13, so the diameter is 26.
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
