x^3y^0x^-7=1x^-4?

Question

I don't understand this problem.

Answer 1

x^3 y^0 x^-7
=x^3 *x^-7 [as y^0=1]
=x^(3-7)
=x^ -4

Answer 2

I assume the first part is three terms multiplied together, this gives (x^3)(y^0)(x^-7)
But since anything to the zero power is 1, this means y^0 = 1
And so you get (x^3)(x^-7) and the exponents add:
(x^3)(x^-7) = x^(3-7) = x^(-4) as given.

Answer 3

x^3y^0x^-7=1x^-4
x^-4 = x^-4
x can be anything you like.

Remember y^0 =1. Anything ^0 = 1
x^3*x^-7 = x^(3-7) = x^-4

Answer 4

y^0 = 1 so you're left with (x^3)(x^-7). When multiplying powers, you add the exponents. (x^3)(x^-7) = x^-4.