A mountain bike has a diameter of 26 inches. To the nearest foot how far does the tire travel when it makes 32 revolutions?

Respuesta :

Hello!

You have to find the circumference

You use the equation

[tex]2 \pi r[/tex]

r is radius which is half the diameter

This gives us 81.68

Multiply this by 32

81.68 * 32 = 2613.76

This is 217.8133333 feet

The answer is 217.8133 feet

Hope this helps!

The distance it covers in 32 revolution is 217.9 ft.

What is the circumference of the circle?

The Circumference (or) perimeter of circle = 2πR. where, R is the radius of the circle.

The diameter of the tire of the bike = 26 in

The radius (r) = 13 in

To find the distance it travels in 32 revolutions.

To find the revolution we need to find the circumference.

The circumference of a circle = 2πr

The circumference of a circle = 2× 3.14 ×13 in

The circumference of a circle = 81.71 in (approx)

The distance it covers in 32 revolution is = 81.71 ×32 in

The distance it covers in 32 revolution is = 2614.85 in

We know that 1 ft = 12 inch

2614.85 in = 2614.85 ÷ 12 ft = 217.9 ft (approx)

Hence, the distance it covers in 32 revolution is 217.9 ft.

Learn more about circumference here;

https://brainly.com/question/15524201

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