The answer is 5.23 ft/second
The way I got this answer was first finding the circumference of the merry-go-round using the equation

, where r represents the radius, which is the distance from the center of a circle.
So your equation should look like this

. I used the rounded term of 3.14 for

Once you multiply those terms, you should get the answer, 31.4. Now this is the total distance around the merry-go-round in feet. So, if it takes 6 seconds to make a full rotation and we are trying to find the amount offeet traveled in one second, we should divide the circumference (31.4) by 6 to give us our answer.
So, 31.4÷6=5.2333...ft/second
And I rounded the answer to the nearest hundredth to get 5.23 feet/second
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By definition, the arc length is given by:
arc = R * theta * ((2 * pi) / 360)
Where,
theta: angle in degrees
R: radio
We have then:
(Arc) QPT if <QZT = 120:
theta = 360-120 = 240 degrees
R = 13.5 units
Substituting values we have:
(Arc) QPT = R * theta * ((2 * pi) / 360)
(Arc) QPT = (13.5) * (240) * ((2 * pi) / 360)
(Arc) QPT = 56.55 units
Answer:
(Arc) QPT = 56.55 units
Answer:
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Step-by-step explanation: