F=ar^t if the half-life is 96 years...
.5=r^96
ln(.5)=96lnr
ln(.5)/96=lnr
r=e^((ln.5)/96)
f=800e^(336(ln.5)/96)
f=70.71
So about 71mg will remain after 336 years.
Multiply the amount in the account by the interest rate:
2500 x 0.034 = 85.00
Answer: $85.00
D) Subtraction because in order to find the distance in between the marker you will need to subtract 14 from 29. So the operation will you need to find the answer is subtraction.
1. -3x -20
2. -6x-16
3. -14-x
