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:
18
Step-by-step explanation:
2,700,000
count the zeros starting from the right
there's five so on the other number move five from the end then those two numbers use the bully and wimp method 5 and higher is included with everything else zeros
Answer:
75 degrees
Step-by-step explanation:
The shape is a trapezoid. When finding angle measurements in trapezoids:
Angle A + Angle C = 180 degrees
Angle B + Angle D = 180 degrees
Because you need to find the angle of B, we will use Angle B + Angle D = 180 degrees to solve this. Angle D is 105 degrees, so 105 + Angle B = 180. 180-105 equals 75, so angle B is 75 degrees.
