1. To solve this problem you must sum the volume of the cone and the volume of the hemisphere. This means that the volumen of the prop is:
Vt=Vc+Vh
Vt is the volumen of the prop.
Vc is the volumen of the cone.
Vh is the volume of the hemisphere.
2. The volume of the cone (Vc) is:
Vc=1/3(πr²h)
r=9 in
h=14 in
π=3.14
4. Then, you have:
Vc=(3.14)(9 in)²(14 in)/3
Vc=3560.76 in³/3
Vc=1186.92
5. The volume of the hemisphere (Vh) is:
Vh=2/3(πr³)
π=3.14
r=9 in
6. Then, you have:
Vh=(2)(3.14)(9 in)³/3
Vh=4578.12 in³/3
Vh=1526.04 in³
7. Finally, the volumen of the prop (Vt) is:
Vt=Vc+Vh
Vt=1186.92 in³+1526.04 in³
Vt=2713.0 in³
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What is the volume of the prop?
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The volume of the prop is 2713.0 in³
Since you haven't identified this figure, I'm going to assume that it's a rectangle.
The Perimeter of a rectangle of length L and width W is P = 2L + 2W.
Here you are given the Perimeter and the length, and are to find the width, W.
Solving the above equation for W, we get P - 2L = 2W.
Dividing by 2 (to isolate that W), we get
P
-- - L = W
2
Substitute P= 6 yds and L = 6 feet (or 2 yds), find W (in yards).
40n-15 can be broken down by
5(8n-3)
Answer:
B: ∠DKH and ∠ENH
Step-by-step explanation:
None
