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Mathematics 16 Online
OpenStudy (anonymous):

MEDALS AND FAN How do i slove this;

OpenStudy (anonymous):

\[2\pi =2pi r \left(\begin{matrix}30 \\360\end{matrix}\right) \]

OpenStudy (usukidoll):

could you provide a screenshot of this? this just looks weird. O_O

OpenStudy (usukidoll):

and what are we solving for? for r? for ???

OpenStudy (aric200):

It looks fine to me

OpenStudy (anonymous):

What is the radius of a circle in which a 30° arc is 2pi inches long? 24 inches 12 inches 2√3 inches

OpenStudy (usukidoll):

that's a bit better. looking for the radius of the circle..

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

a 30 deg arc is 1/12 of a circle (since there are 360 deg in a circle) thus the circumference of this circle is 12x2 pi = 24 pi the circumference of a circle = 2 pi R, so we know that 2 pi R = 24 pi or R=12

OpenStudy (anonymous):

i hope i helped :)

OpenStudy (usukidoll):

oh yeah the circumference formula \[C = 2 \pi r\] \[\frac{C}{2\pi} = r\] a full circle is 360 degrees and we are given 30 degrees as the arc so \[\frac{360}{30} = 12\] so 12 must be r \[C = 2 \pi (12) =24 \pi\] so to check that \[\frac{24 \pi}{2\pi} = r\] \[12=r \] (my explanation is a bit messed up... I haven't done this in years )

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