Regular hexagon fghijk shares a common center with square abcd on a coordinate plane. || . across which lines can the combined figure reflect onto itself? a. any of the perpendicular bisectors of the sides of the hexagon b. either diagonal of the square c. any of the perpendicular bisectors of the sides of the square d. there are no lines across which this figure can reflect onto itself

(C) Any of the perpendicular bisectors of the sides of the square

Step-by-step explanation:

In Regular Hexagon FGHIJK, we have 6 line of reflection across which the hexagon reflects onto itself. Those lines are:

3 perpendicular bisectors of sides i.e. perpendicular bisector of IJ , IH and GH

3 lines passing through vertices i.e. HK, IF and GJ.

While in Square, we have 4 line of reflection across which the square reflects onto itself. Those lines are:

2 perpendicular bisectors of sides AB and BC i.e. HK and perpendicular bisector of CD

2 digonals of square i.e. AC and BD

Also from figure we know that perpendicular bisector of CD and perpendicular bisector of IJ is the same line.

So, for combined figure we have to take common lines from both figures i.e. perpendicular of sides CD or IJ and line HK.

80π/3 or about 83.775

step-by-step explanation:

since a cone is 1/3 of a cylinder, we need to find the base's area. since area = πr^2, we get 16π. then, multiplying this by 5, we get 80π. dividing by 3, since the cone's volume is 1/3 of that of a cylinder's, we get 80π/3.

hope this !

12.5 feet

step-by-step explanation: