Four golf balls are packaged in a cylindrical container as shown below. The balls just touch the top, bottom, and sides of the c
ylinder. The diameter of each ball is 4cm.
What is the volume of the cylinder rounded to the nearest cubic centimeter?
What is the total volume of the four balls rounded to the nearest cubic centimeter?
What percent of the volume of the container is occupied by the four balls?
What is the volume of the cylinder rounded to the nearest cubic centimeter?
What is the total volume of the four balls rounded to the nearest cubic centimeter?
What percent of the volume of the container is occupied by the four balls?
1 answer:
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Answer:
12). LM = 37.1 units
13). c = 4.6 mi
Step-by-step explanation:
12). LM² = 23² + 20² - 2(23)(20)cos(119)°
LM² = 529 + 400 - 920cos(119)°
LM² = 929 - 920cos(119)°
LM =
=
= 37.08
≈ 37.1 units
13). c² = 5.4² + 3.6² - 2(5.4)(3.6)cos(58)°
c² = 29.16 + 12.96 - 38.88cos(58)°
c² = 42.12 - 38.88cos(58)°
c =
c =
c = 4.6386
c ≈ 4.6 mi
Answer:
432pi
Step-by-step explanation:
