Log(2)/log(1.064) ≈ 11.17 . . . . hours
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The population can be given by
p(n) = p₀×1.064ⁿ . . . . where n is the number of hours
You want to find n whe p(n) = 2*p₀.
2p₀ = p₀×1.064ⁿ . . . . . . . . . . . . substitute the given information
2 = 1.064ⁿ . . . . . . . . . . . . . . . . . divide by p₀
log(2) = n×log(1.064) . . . . . . . . take logs to make it a linear equation
log(2)/log(1.064) = n . . . . . . . . divide by the coefficient of n
Answer:
3
Step-by-step explanation:
13-4=9
9/3=3
It represente the absolute value
Last angle is 122 if its a triangle
It is hard to comprehend your question. As far as I understand:
f(x,y) = e^(-x)
Find the volume over region R = {(x,y): 0<=x<=ln(6), -6<=y <= 6}.
That is all I understood. It would be easier to understand with a picture or some kind of visual aid.
Anyways, to find the volume between the surface and your rectangular region R, we must evaluate a double integral of f on the region R.
![\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy](https://tex.z-dn.net/?f=%5Ciint_%7BR%7D%5E%7B%20%7D%20e%5E%7B-x%7DdA%3D%5Cint_%7B-6%7D%5E%7B6%7D%20%5Cint_%7B0%7D%5E%7Bln%286%29%7D%20e%5E%7B-x%7Ddx%20dy)
Now evaluate,
![\int_{0}^{ln(6)}e^{-x}dx](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7Bln%286%29%7De%5E%7B-x%7Ddx)
which evaluates to, 5/6 if I did the math correct. Correct me if I am wrong.
Now integrate this w.r.t. y:
![\int_{-6}^{6}\frac{5}{6}dy = 10](https://tex.z-dn.net/?f=%5Cint_%7B-6%7D%5E%7B6%7D%5Cfrac%7B5%7D%7B6%7Ddy%20%3D%2010)
So,
![\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy = 10](https://tex.z-dn.net/?f=%5Ciint_%7BR%7D%5E%7B%20%7D%20e%5E%7B-x%7DdA%3D%5Cint_%7B-6%7D%5E%7B6%7D%20%5Cint_%7B0%7D%5E%7Bln%286%29%7D%20e%5E%7B-x%7Ddx%20dy%20%3D%2010)