Answer: D
Step-by-step explanation:
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
Answer:
1
Step-by-step explanation:
4-3=1 denominator never changes may i be marked brainliest?
He spent $5.75 on the red sand and $25.30 on the brown sand he spent a total of $31.05 on sand to fill his sand box! Hope this helps you :)
Answer: Associative property
Step-by-step explanation: it is the same equation but they switched the numbers around on the other side of the equal sign so therefore its associative property
