we are given
[tex] A(t)=3900(\frac{1}{2})^{\frac{t}{14}} [/tex]
(a)
For finding initial amount , we can set t=0
and find A(t)
[tex] A(0)=3900(\frac{1}{2})^{\frac{0}{14}} [/tex]
[tex] A(0)=3900(\frac{1}{2})^{0} [/tex]
[tex] A(0)=3900 [/tex]
so, initial amount is 3900 grams..........Answer
(b)
We can plug t=40
and find A(t)
[tex] A(40)=3900(\frac{1}{2})^{\frac{40}{14}} [/tex]
[tex] A(40)=538 [/tex]
so, the amount remaining after 40 hours is 538 grams...........Answer
Answer:
To solve this problem, we just need to substitute the right value for the variable time.
The given expression is
[tex]A(t)=3900(\frac{1}{2} )^{\frac{t}{14} }[/tex]
Where the initial condition is determined when [tex]t=0[/tex].
[tex]A(t)=3900(\frac{1}{2} )^{\frac{0}{14} }=3900(1)=3900[/tex]
Therefore, the initial amount is 3900 grams.
Then, after 40 hours refers to [tex]t=40[/tex]
[tex]A(t)=3900(\frac{1}{2} )^{\frac{40}{14} } \approx 538.24[/tex]
Therefore, after 40 hours, there's 538.24 grams remaining, approximately.
