Answer:
6
....................................
I've attached a plot of one such cross-section (orange) over the region in the x-y plane (blue), including the bounding curves (red). (I've set

for this example.)
The length of each cross section (the side lying in the base) has length determined by the horizontal distance

between the y-axis

and the curve

. In terms of

, this distance is

. The height of each cross section is twice the value of

, so the area of each rectangular cross section should be

.
This means the volume would be given by the integral
Answer:

Step-by-step explanation:
Step 1:-
using logarithmic formula 
so given 
now simplify
= 
<u>Answer:</u>-
![log(x^{3}y^{2})= [tex]3 log x+2 log y](https://tex.z-dn.net/?f=log%28x%5E%7B3%7Dy%5E%7B2%7D%29%3D%20%5Btex%5D3%20log%20x%2B2%20log%20y)
We are given these three people age below:
- Jim's age
- Carla's age
- Tomy's age
We define the age of Jim as any variable, because the problem doesn't give any specific age. I will define Jim's age as x.
Next, Carla is 5 years older than Jim. That means if Carla is 5 years older, her age would be x+5.
Then Tomy is 6 years older than Carla. That means the age would be 6+(x+5).
The sum of their three ages is 31 years old. That means we add all these ages and equal to 31.

Combine like terms and solve for x.

Then we substitute the value of x in ages to find these three people ages.
- Jim's age = x = 5
- Carla's age = x+5 = 5+5 = 10
- Tomy's age = 6+(x+5) = 6+(5+5) = 6+10 = 16.
Answer
- Jim's age = 5
- Carla's age = 10
- Tomy's age = 16
Answer;
The generator was not fair.
Explanation;
The number generator is not fair. This is because the correct percentage of black socks was not chosen at all.