Two circles, one of radius 5 inches, the other of radius 2 inches, are tangent at point p. two bugs start crawling at the same time from point p, one crawling along the larger circle at $3\pi$ inches per minute, the other crawling along the smaller circle at $2.5\pi$ inches per minute. how many minutes is it before their next meeting at point p?

  40 minutes

Step-by-step explanation:

The circumference of the larger circle is ...

  C = 2πr = 2π(5 in) = 10π in

The bug navigates the circumference at 3π in/min, so will take

  time = distance/speed = (10π in)/(3π in/min) = 10/3 min

to travel once around.

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The circumference of the smaller circle is ...

  C = 2πr = 2π(2 in) = 4π in

The bug navigates this circumference at 2.5π in/min, so will take

  (4π in)/(2.5π in/min) = 8/5 min

to travel once around.

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The bugs will meet at a time that is the least common multiple of these times. Both can be expressed in 15ths of a minute as ...

  {50/15, 24/15}

Then the LCM of these will be ...

  (1/15)LCM(50, 24) = (1/15)(50×24)/GCD(50, 24) = 1200/30 = 40

It will be 40 minutes before the bugs next meet at point P.

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A graphing calculator can make use of the mod function to show when the bugs meet at point P (total is displacement of the two bugs from P is zero). It shows the meeting occurs after 40 minutes.

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer