Question 1157061
A piece of charcoal is found to contain 39​% of the element B that it originally had.
Use the radioactive decay formula: A = Ao*2^(-t/h), where
A = resulting amt after t time
Ao = initial amt
t = time of decay
h = half life of substance
:
​(a) When did the tree from which the charcoal came​ die?
 Use 5800 years as the​ half-life of the element B.
Assuming initial amt is 1
2^(-t/5800) = .39
{{{-t/5800}}}ln(2) = ln(.39)
{{{-t/5800}}}= {{{ln(.39)/ln(2)}}}
{{{-t/5800}}} = -1.35845
t = -5800 * -1.3845
t = 7879 yrs
: 
​(b) Using a graphing​ utility, graph the relation between the percentage of the 
 element B remaining and time.
the equation for this: y = 100*2^(-x/5800)
{{{ graph( 300, 200, -2000, 15000, -20, 105, 100*2^(-x/5800), 50) }}}
:
​(c) Using​ INTERSECT, determine the time that elapses until half of the the 
 element B remains. Did that
:
​(d) Verify the answer found in part ​(a). confirmed, entered 7879 and got 39