SOLUTION: Hi I understand the formulas of circumference and diameter and whatnot, I have 2 homework questios and I don’t seem to fully understand it: 1. Anthony is biking around a circul

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Question 1114729: Hi I understand the formulas of circumference and diameter and whatnot, I have 2 homework questios and I don’t seem to fully understand it:
1. Anthony is biking around a circular track in jamming to some tunes. His bike tires are 2 feet in diameter, and the track is 1000 feet in diameter. How many times do Anthony’s tires go around one lap around the track?
2. In modern cataract surgery (eye surgery) the ophthalmologist( eye doctor) makes an incision that is 6% of the circumference of the cornea. If the diameter of the patients cornea is 8 mm what is the length of the incision?

I hope someone is willing to explain this to me, thanks.
Found 2 solutions by ikleyn, stanbon:
Answer by ikleyn(52797)   (Show Source): You can put this solution on YOUR website!
.
1.  Answer.   =  = 500 times.


2.  Answer.   =  = 1.51 millimeter (aproximately).




    To solve the problems, you need to know 


        a)  Circumference of the circle = , where  = 3.14 and D is the diameter.


        b)  1%  is  0.01  of the value;  6% = 0.06 of the value.

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Be aware:   The solutions by the other tutor for both  #1  and  #2  are   I N C O R R E C T !


Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
1. Anthony is biking around a circular track in jamming to some tunes. His
bike tires are 2 feet in diameter, and the track is 1000 feet in diameter. How many times do Anthony’s tires go around one lap around the track?
circumference of the tire:: pi*d = 2pi ft.
# of revolutions = 1000/(2pi) = 500/pi is approximately 159.15
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2. In modern cataract surgery (eye surgery) the ophthalmologist( eye doctor) makes an incision that is 6% of the circumference of the cornea. If the diameter of the patients cornea is 8 mm what is the length of the incision?
circumference = 8pi mm
6% of circumference = 1.508 mm
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Cheers,
Stan H.
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