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 answer is B
Step-by-step explanation:
0.77777777777777 and goes on forever since anything/9 is 0.(whatever that number is repeated)
'is' in mathematics means equal
'of' in mathematics means multiplication
percent means over 100
84 = 74% * x
84 = 74/100 * x
84 = 37/50 * x
Multiply 50/37 on both sides, so we can isolate 'x'
85 * 50/37 = x
x = 4250/37
x ≈ 114.864865
84 is 74 percent of <span>114.86</span>.
Answer:
21 cm
Step-by-step explanation:
Call the triangle ABC, with the right angle at B, the hypotenuse AC=25, and the given leg AB=10. The altitude to the hypotenuse can be BD. Since the "other leg" is BC, we believe the question is asking for the length of DC.
The right triangles formed by the altitude are all similar to the original. That means ...
AD/AB = AB/AC . . . . . . ratio of short side to hypotenuse is a constant
Multiplying by AB and substituting the given numbers, we get ...
AD = AB²/AC = 10²/25
AD = 4
Then the segment DC is ...
DC = AC -AD = 25 -4
DC = 21 . . . . . centimeters
