B. 120 in^3
since the formula is a= pi (6)^2 (10/3), 6 squared is 36 which when multiplied by (10/3) becomes 120. Hence pi is multiplied by 120 and we obtain 120pi
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:
<em>Answer: False</em>
Step-by-step explanation:
<em>Relations vs Functions</em>
For a given relation between two variables x and y to be a function, it must meet the following condition: every value of x must be related to one and only one value of y.
When the graph of a relation is given, we can easily tell the difference by using the vertical line test as follows:
Imagine a vertical line moving through the x-axis. If any line touches the graph of the relation at more than one point, then the relation is not a function.
If we place vertical lines through the x-axis, they would touch the function twice at some values of x.
Thus, the graph is not a function
Answer: False
-36s+54 i hope that was helpful
if it's a mix fraction, then you write 22 1/2. if improper fraction, then write 45/2
