Answer:
1
Step-by-step explanation:
8/2(4)
8/8
1
Where the pictures of the graph
Answer:

Step-by-step explanation:






Step-by-step explanation:
I am not fully sure what your teacher is aiming for. it friends very much on what you were just discussing in class (which I don't know).
but the first thing coming to mind is a minus sign ("-"). squaring a negative number removed the minus and makes the result equal to squaring the same positive number.
just for the undoing the 1/2 :
that is, because a fraction as exponent specifies in its denominator the root to be calculated for the basic value or expression.
so, 1/2 means square root. and yes, square is the inverse function of a square root, and it "undoes" the square root.
in exponent calculation it just means that for exponent 1 to the power of exponent 2 we simply multiply both exponents. and so, 1/2 × 2 = 1
FYI - the numerator still represents an original "to the power of" operation.
so, e.g. 3/2 would mean put the basis to the power of 3 and then do the square root of that result. or the other way around. these operations are commutative (the sequence does not matter).