A regular pentagon is cut out of a flat piece of cardboard. The pentagon is then placed back into the pentagonal hole. How many ways can the pentagon be placed back into the hole, if the pentagon can be rotated and/or reflected?

5

Step-by-step explanation:

If you place the hole in the cardboard so there's a vertex at the top, and always read the letters clockwise then the first 5 answers are

ABCDE

BCDEA

CDEAB

DEABC

EABCD

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0