Since there are no figures given, I will give an example.
You are given a silo that is shaped as
a closed cylinder with a conical end. The diameter of the silo is 4 ft, the
length of the cylindrical part is 6 ft, and the entire length of the silo is
10.5 ft. Suppose that you are asked to find the total volume of
grains that can be stored in the silo.
Given:
Cylinder part
D = 4 ft
H = 6 ft
Cone part
H = 10.5 – 6 = 4.5ft
D = 4ft
Required:
Volume of silo
Solution:
V of cylinder = πr²H
V of cylinder = π(4/2)²(6)
V of cylinder = 75.4 ft³
V of cone = πr²H/3
V of cone = π(4/2)²(4.5/3)
V of cone = 18.85 ft³
Total volume = 94.25 ft³
Answer:
<u>77.4 degrees and 12.6 degrees</u>
Step-by-step explanation:
Let the angles be x and x - 64.8.
Therefore, on the basis of knowledge that complementary angles add up to 90 degrees :
x + x - 64.8 = 90
2x = 154.8
x = <u>77.4 degrees</u>
x - 64.8 = 77.4 - 64.8 = <u>12.6 degrees</u>
9514 1404 393
Answer:
h(-5) = 129
Step-by-step explanation:
Put -5 where x is and do the arithmetic.
h(-5) = 4(-5)² -5(-5) +4
h(-5) = 4(25) +25 +4 = 129
Answer:
B. (-9/2, -11/2)
Step-by-step explanation:
-12+3 / 2 = -9/2
-3+-8 / 2 = -11/2
Answer:
Area of the sector = 57.26295cm²
Step-by-step explanation:
Radius of the circle=9cm
π= 3.142
Angle at B= 81° ( opposite angle of a quadrilateral)
Area of the sector = πr² * 81/360
Area of the sector = 3.142 * 9*9 * 0.225
Area of the sector= 3.142*81*0.225
Area of the sector = 57.26295cm²
