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
Sub in the values and solve for W
50=3W+14
50-14=3W
36=3W
12=W
The team won 12 games.
12+14=26 (games drawn and won)
35-26=9
Therefore they lost 9 games
Answer:
6x^2 y^3 is the answer
Step-by-step explanation:
Multiply everything given
<span><span>b=2g-2h-4
this is correct</span></span>
The speed from home base to airport was:

The speed from airport to home base was:

Average speed of plane was:

Average speed of wind was:

Answer: Average speed of plane was 160mph and average speed of wind was 10mph.
I hope I helped :D