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
<em>Assuming [x] means the closest integer to x</em>
Answer:
Step-by-step explanation:
We can see that for any integer 
, there will always be a ![[x]](https://tex.z-dn.net/?f=%5Bx%5D)
, so there will always be a ![[x]+1](https://tex.z-dn.net/?f=%5Bx%5D%2B1)
. So, we don't need to worry about the domain impacting the range.
can be any integer from 
to 
, and as 
(at least in terms of a function's range and domain), the range of is equal to the range of , which is .
So, the answer is 
and we're done!
Answer:
Step-by-step explanation:
Worth of furnishings = $1580
Tax rate = 4.5%
Tax = 4.5% × $1580 = 0.045 × $1580 = $71.10
Amount paid = $1580 + $71.10 = $1651.10
Since the decorator's fee was 12% of the purchases. This will then be calculated as:
y = kx
where,
y = decorators fee
k = percentage of decorators fee on purchases
x = amount paid
y = 12% × $1651.10
= 0.12 × $1651.10
= $198.132
Step-by-step explanation:
2^-6 = 1/2^6 = 1/64
1/64 * 1/13 = 1/832
