This is a typical radioactive decay problem which uses the general form:
A = A0e^(-kt)
So, in the given equation, A0 = 192 and k = 0.015. We are to find the amount of substance left after t = 55 years. That would be represented by A. The solution is as follows:
A = 192e^(-0.015*55)
<em>A = 84 mg</em>
No because 6 out of 6 is one hole in one 6 is only 1/6 of one hole
You can see that the function is constantly equal to 3 if x is less than -2 (included). So, we have
if [/tex] x\leq -2[/tex]
Similarly, the function is constantly 1 if x is less than -2 (excluded). So, we have
if [/tex] x> -2[/tex]
N being the number, n-4=-21. n=-17