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:
Subtract from the dividend an easy multiple of the divisor
Record the partial quotient in a column to the right of the problem.
Repeat until the dividend has been reduced to zero or the remainder is less than the divisor.
Add the partial quotients to find the quotient.
Step-by-step explanation:
Hope this helps :)
The ratio of cm to mi is 1/30, and we are looking for x/110. Since 1/30 = x/110, 30x=110.
x = 110/30 = 11/3 = 3.7 cm
Answer:
a ( copper and nickel
b ( 1034.9
Step-by-step explanation:
Add all the values of melting points
2,651 + -37.89 = 2613.11
2613 + -458 = 2155.11
2155.11 + 1984.32 = 4139.43
Divide the number by the number of numbers: 4139.43/4 = 1034.8575
1034.8575 rounded to the nearest tenth is 1034.9, thats how you find the mean, or the average.
