.
#(10)-a)# Justine is incorrect on the first claim because you can have aline like #IF# that is a line of symmetry and goes through two vertices. Also, the line that is perpendicular to #IF# is a line of symmetry but does not go through any vertices or the origin.
#(10)-b)# Justine is incorrect on the second claim. The angle of rotational symmetry for a regular hexagon is #60^@#. The central angle facing each side is #=360^@/6=60^@#. Every time you rotate the hexagon by #60^@# the hexagon is totally indistinguishable from the original one.
#(11)-PartA#
Point #P# travels a distance equal to the circumference of the circle in #30# seconds.
The circumference of the circle is:
#C=2pir# where #r# is the radius.
In this problem, #r=7# feet
#C=2pi(7)=14pi=44# feet
Point #P# travels at the rate of #44# feet each #30# seconds. To convert this to feet per minute, we divide #44# by #30# to get feet per second and multiply the result by #60# seconds to get feet per minute:
#speed=44/30*60=88# feet per minute
#(11) - Part B#
After #8.25# minutes, point #P# has traveled:
#88(8.25)=726# feet
Let's divide this distance by #C# to find out how many times point #P# has returned to its original position:
#726/44=16.5# times
After #8.25# minutes, #P# has gone around #16# times and passed its original position by #1/2# the circle, i.e. point #P# is now at #(-7,0)# assuming the merry-go-round is turning counterclockwise as shown below:
