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)
Answer:
neither
Step-by-step explanation:
because the square root is not whole number and neither is the cube root of 165. so it is neither
Answer:
104
155
180
Step-by-step explanation:
an easy way is to solve what is in the parenthesis then multiply by the outside number
but if you need to use the distributive property:
1. 4 x (20 - -6)
(4 x 20) - (4 x -6) = 80 - -24 = 80 + 24 = 104
2. -5 x (24 - -7)
(-5 x 24) - (-5 x -7) = -120 - 35 = 155
3. 12 x (-30 + 45)
(12 x -30) + (12 x 45) = -360 + 540 = 180