G^−m ÷ g^n
1st g^−m=1/g^m, hence g^−m ÷ g^n = (1/g^m) /(g^n)==> 1/(g^m)(g^(n)
==> 1/(g^m+n) or g^(-m-n)
The angle between the hour and minute
Answer:
I believe if you have a letter with no value or the value is unknown you put it over 1
The x coordinates will stay the same. the y coordinates will be opposite.
