Given a regular octagon, in how many ways can we color one diagonal red and another diagonal blue so that the two colored diagonals cross (in the interior)? Consider rotations and reflections distinct.

Rotation = 1.

Reflection = 5 or 4.

Step-by-step explanation:

A regular octagon is the octagon that has eight, 8 sides and all these eight sides are equal.

So, if we are going to consider rotation and reflection distinct in order to color one diagonal red and another diagonal blue so that the two colored diagonals cross (in the interior) we can do that by following the steps Below;

=> Make sure that the first diagonal with the red color is drawn first.

=> to another make fixing to another end.

=> Determine the possibilities.

The numbers of times can be gotten by using Burnside Lemma which gives Rotation = 1 And Reflection = 5 or 4.

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation:

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it