Hi there!
5.
Use the property:
[tex]\frac{x^a}{x^b} = x^{a - b}[/tex]
Now, solve by variable.
[tex]\frac{x^2}{x} = x^{2 - 1} = x\\\\\frac{y^5}{y^4} = y^{5 - 4} = y[/tex]
Rewrite:
[tex]= x * y * z^3 = \boxed{\text{ D. } xyz^3}}[/tex]
6.
Solve the inside of the parenthesis, and each term separately.
Use the property:
[tex]x^0 = 1[/tex]
[tex](7 + 3^2)^0 = 1 \\\\(8^0 + 3)^2 = (1 + 3)^2 = 4^2 = 16\\\\1 + 16 = \boxed{ \text{ C. } 17}[/tex]
7.
Solve like above:
[tex](4 + 2^2)^0 = 1\\\\(7^0 + 4)^{-3} = (1 + 4)^{-3} = 5^{-3}[/tex]
The negative exponent indicates taking the reciprocal. Using the property:
[tex]x^{-a} = \frac{1}{x^a}[/tex]
Therefore:
[tex]5^{-3} = \frac{1}{5^3} = \boxed{ \text{ D. }\frac{1}{125}}[/tex]
