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Suppose a merry-go-round is spinning once every 3 seconds. If a point on the outside edge has a speed of 12.56 feet per second, what is the radius of the merry-go-round? (Use 3.14 for
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Suppose a merry-go-round is spinning once every 3 seconds. If a point on the outside edge has a speed of 12.56 feet per second, what is the radius of the merry-go-round? (Use 3.14 for

User Tawny
by
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1 Answer

5 votes

Answer:

r = 6 feet

Explanation:

Ler r represents the radius

Hence perimeter = edge speed times 3second

2 times 3.14 times r = 12.56 times 3

r = 6 feet

User Ritesh Chandora
by
4.7k points
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