Answer:
M = 5742π
Step-by-step explanation:
Given:-
- Find the mass of a solid with the density ( ρ ):
ρ ( r, θ , z ) = 1 + z / 81
- The solid is bounded by the planes:
0 ≤ z ≤ 81 - r^2
0 ≤ r ≤ 9
Find:-
Find the mass of the solid paraboloid
Solution:-
- The mass (M) of any solid body is given by the following triple integral formulation:
- We can write the above expression in cylindrical coordinates:
![M = \int\limits\int\limits_r\int\limits_z {r*p(r,theta,z)} \, dz.dr.dtheta \\\\M = \int\limits\int\limits_r\int\limits_z {r*[ 1 + \frac{z}{81}] } \, dz.dr.dtheta\\\\](https://tex.z-dn.net/?f=M%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%5Cint%5Climits_z%20%7Br%2Ap%28r%2Ctheta%2Cz%29%7D%20%5C%2C%20dz.dr.dtheta%20%5C%5C%5C%5CM%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%5Cint%5Climits_z%20%7Br%2A%5B%201%20%2B%20%5Cfrac%7Bz%7D%7B81%7D%5D%20%7D%20%5C%2C%20dz.dr.dtheta%5C%5C%5C%5C)
- Perform integration:
![M = \int\limits\int\limits_r{r*[ z + \frac{z^2}{162}] } \,|_0^8^1^-^r^2 dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{(81-r^2)^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{6561 -162r + r^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + 40.5 -r +\frac{r^2}{162} ] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{[ 121.5r-r^2 -\frac{161r^3}{162} ] } \, dr.dtheta\\\\](https://tex.z-dn.net/?f=M%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%7Br%2A%5B%20z%20%2B%20%5Cfrac%7Bz%5E2%7D%7B162%7D%5D%20%7D%20%5C%2C%7C_0%5E8%5E1%5E-%5Er%5E2%20dr.dtheta%5C%5C%5C%5CM%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%7Br%2A%5B%2081-r%5E2%20%2B%20%5Cfrac%7B%2881-r%5E2%29%5E2%7D%7B162%7D%5D%20%7D%20%5C%2C%20dr.dtheta%5C%5C%5C%5CM%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%7Br%2A%5B%2081-r%5E2%20%2B%20%5Cfrac%7B6561%20-162r%20%2B%20r%5E2%7D%7B162%7D%5D%20%7D%20%5C%2C%20dr.dtheta%5C%5C%5C%5CM%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%7Br%2A%5B%2081-r%5E2%20%2B%2040.5%20-r%20%2B%5Cfrac%7Br%5E2%7D%7B162%7D%20%5D%20%7D%20%5C%2C%20dr.dtheta%5C%5C%5C%5CM%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%7B%5B%20121.5r-r%5E2%20-%5Cfrac%7B161r%5E3%7D%7B162%7D%20%5D%20%7D%20%5C%2C%20dr.dtheta%5C%5C%5C%5C)
![M = 2*\int\limits_0^\pi {[ 121.5r^2-r^3 -\frac{161r^4}{162} ] } |_0^6 \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 121.5(6)^2-(6)^3 -\frac{161(6)^4}{162} ] } \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 4375-216 -1288] } \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 2871] } \, dtheta\\\\M = 5742\pi kg](https://tex.z-dn.net/?f=M%20%3D%202%2A%5Cint%5Climits_0%5E%5Cpi%20%7B%5B%20121.5r%5E2-r%5E3%20-%5Cfrac%7B161r%5E4%7D%7B162%7D%20%5D%20%7D%20%7C_0%5E6%20%5C%2C%20dtheta%5C%5C%5C%5CM%20%3D%202%2A%5Cint%5Climits_0%5E%5Cpi%20%7B%5B%20121.5%286%29%5E2-%286%29%5E3%20-%5Cfrac%7B161%286%29%5E4%7D%7B162%7D%20%5D%20%7D%20%20%5C%2C%20dtheta%5C%5C%5C%5CM%20%3D%202%2A%5Cint%5Climits_0%5E%5Cpi%20%7B%5B%204375-216%20-1288%5D%20%7D%20%20%5C%2C%20dtheta%5C%5C%5C%5CM%20%3D%202%2A%5Cint%5Climits_0%5E%5Cpi%20%7B%5B%202871%5D%20%7D%20%20%5C%2C%20dtheta%5C%5C%5C%5CM%20%3D%205742%5Cpi%20%20kg)
- The mass evaluated is M = 5742π
Answer:
The length of AC is;
C. 50
Step-by-step explanation:
By the midpoint of a triangle theorem, we have that a segment that spans across and intersects with the midpoints of two sides of a triangle is equal to half the length of the third side and parallel to the length of the third side
The given parameters are;
The midpoints of ΔACE are B, D, and F
The length of EC = 44
The length of DF = 25
Therefore, we have;
Given that DF is a midsegment of triangle ΔACE, then DF ║ AC and
the length of DF = (1/2) × AC the length of AC
∴ The length of AC = 2 × The length of DF
The length of DF = 25
∴ The length of AC = 2 × 25 = 50
The length of AC = 50
It is c because 8x2=16 and 8x4=32 and so on
I can barely see the majority of the questions
i cant do all but to multiply fractions, you just multiply the numerator and the denominator together
so 3 2/3 x 6
change 3 2/3 to an improper fraction
9/3 x 6/1 = 27/3 which is 9/1 reduced
10 x 1 2/3 is 10/1 x 5/3 = 50/3
-4/3
Explanation: All parallel lines have the same slope
