Seriously need help with this one guys calculus 3 class here

Seriously need help with this one guys, calculus 3 class here. Is anyone able to do this? There\'s 3 parts to this problem. I\'m totally lossed on it. If you\'re able to do it can you please also explain the process, please and thank you.

Solution

1)

no.of edges for n-cube is n* (2^n-1)

                                                                dimension of   cube

                                            1                 2                   3                   4                      5     ...............

no.of edges                         1               2*2                3*2²              4*2³                 5*2⁴     ..............

2)

for 4-dim cube we have 16 vertices

from one one vertices we can draw 3 diagonals

so 16*3=48

so diagonals can be drawn

from this face diaginals are perpendicular

those are drawn 2 from one vertices

16*2=32

thus 32 vertices are perpendicular to each other