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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Ibolit asked Jun 9, 2023

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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

Mathematics
high-school

Ibolit asked Jun 9, 2023

by Ibolit

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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

Raghavendra Bankapur answered Jun 13, 2023

by Raghavendra Bankapur

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