Answer:
![3.14r^2(h-\frac{1}{3}h_1)](https://tex.z-dn.net/?f=3.14r%5E2%28h-%5Cfrac%7B1%7D%7B3%7Dh_1%29)
Step-by-step explanation:
Let h be the cylinders height and r the radius.
-The volume of a cylinder is calculated as:
![V=\pi r^2h](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E2h)
-Since the cone is within the cylinder, it has the same radius as the cylinder.
-Let
be the height of the cone.
-The area of a cone is calculated as;
![V=\pi r^2 \frac{h}{3}\\\\=\frac{1}{3}\pi r^2h_1](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E2%20%5Cfrac%7Bh%7D%7B3%7D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2h_1)
The volume of the solid section of the cylinder is calculated by subtracting the cone's volume from the cylinders:
![V=V_{cy}-V_{co}\\\\=\pi r^2h-\frac{1}{3}\pi r^2 h_1, \pi=3.14\\\\=3.14r^2(h-\frac{1}{3}h_1)](https://tex.z-dn.net/?f=V%3DV_%7Bcy%7D-V_%7Bco%7D%5C%5C%5C%5C%3D%5Cpi%20r%5E2h-%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2%20h_1%2C%20%5Cpi%3D3.14%5C%5C%5C%5C%3D3.14r%5E2%28h-%5Cfrac%7B1%7D%7B3%7Dh_1%29)
Hence, the approximate area of the solid portion is ![3.14r^2(h-\frac{1}{3}h_1)](https://tex.z-dn.net/?f=3.14r%5E2%28h-%5Cfrac%7B1%7D%7B3%7Dh_1%29)
Answer:
50 pounds
Step-by-step explanation:
Dan and june mix two kind of feed for pedigreed dogs
Feed A worth is $0.26 per pound
Feed B worth is $0.40 per pound
Let x represent the cheaper amount of feed and y the costlier type of feed
x+y= 70..........equation 1
0.26x + 0.40y= 0.30×70
0.26x + 0.40y= 21.........equation 2
From equation 1
x + y= 70
x= 70-y
Substitutes 70-y for x in equation 2
0.26(70-y) + 0.40y= 21
18.2-0.26y+0.40y= 21
18.2+0.14y= 21
0.14y= 21-18.2
0.14y= 2.8
Divide both sides by the coefficient of y which is 0.14
0.14y/0.14= 2.8/0.14
y= 20
Substitute 20 for y in equation 1
x + y= 70
x + 20= 70
x= 70-20
x = 50
Hence Dan and june should use 50 pounds of the cheaper kind in the mix
Answer:
1680
Step-by-step explanation:
280 x 6 = 1680
The answer is 72 all u need to do plug in the numbers where the letters are
Answer:
See attached sheet
Step-by-step explanation: