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:
I need the values of either X or Y to solve this. I can solve for what X is though.
Step-by-step explanation:
A: X = -12
B: X =78
C: X = 12.13
I hope this helps you, but since both X and Y are unknown variables, you can't solve it, only simplify (which it already is.)
399,713 rounded to the nearest thousand would be <span>400,000
Hope I Helped!!! :-)
</span>
Answer:
10:30 a.m
Step-by-step explanation:
just got it right on edeg
Answer:
$1171.05
Step-by-step explanation:
1500 × (1-(6 / 100))⁴
