anonymous
  • anonymous
Freddie is at chess practice waiting on his opponent's next move. He notices that the 4-inch-long minute hand is rotating around the clock and marking off time like degrees on a unit circle. Part 1: How many radians does the minute hand move from 3:35 to 3:55? (Hint: Find the number of degrees per minute first.) Part 2: How far does the tip of the minute hand travel during that time? Part 3: How many radians on the unit circle would the minute hand travel from 0° if it were to move 3π inches? Part 4: What is the coordinate points associated with this radian measure?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
First step is to find how many degrees there are per minute. 60 minutes to 360 degrees. So how many degrees per minute?
anonymous
  • anonymous
6 degrees per minute
anonymous
  • anonymous
Right. And we're trying to figure out how many degrees there are for 20 minutes. \[\large \sf 20 \times 6 =120\] so we're solving for 120 degrees. Do you know what 120 degrees looks like in radian form?

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anonymous
  • anonymous
ummmm 2pi/3
anonymous
  • anonymous
Right. So part 1 done. Next it to find the distance the tip traveled. For this we'll use the formula \[\large \sf Arc~=~r \times \theta\]
anonymous
  • anonymous
lol i dont get his part sry
anonymous
  • anonymous
Well wanna use a different formula? We're looking for the distance the tip traveled. In other words, the arc it traveled around the circumference of the clock. |dw:1436398679715:dw|
anonymous
  • anonymous
yes ok
anonymous
  • anonymous
And we'll use the formula \[\large \sf Arc~=~r \times \theta\] to find the arc length.
anonymous
  • anonymous
This is the radian style formula by the way
anonymous
  • anonymous
wait whats the unknowns
anonymous
  • anonymous
So \(\large \sf \theta=\frac{2\pi}{3}\)
anonymous
  • anonymous
yes is r=4?
anonymous
  • anonymous
And r is stated in the top part of your question. Yeah, r is 4
anonymous
  • anonymous
ok so theta= 4pi?
anonymous
  • anonymous
\[\large \sf \theta=\frac{2\pi}{3}\]
anonymous
  • anonymous
i mean arc sorry
anonymous
  • anonymous
\[\large \sf Arc~=~4 \times \frac{2\pi}{3}\]\[\large \sf Arc~=~\frac{8\pi}{3}\]
anonymous
  • anonymous
oh yea sry i multiplied wrong
anonymous
  • anonymous
Anyways, Part 3 uses the same formula. The only difference is what we're solving for. \[\large \sf 3\pi~=~4 \times \theta\]

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