Question

Can someone please help me with this again? 5-7Thank you so much! ​

Answer 1

Hi there!

5.

Use the property:

$\frac{x^a}{x^b} = x^{a - b}$
Now, solve by variable.

$\frac{x^2}{x} = x^{2 - 1} = x\\\\\frac{y^5}{y^4} = y^{5 - 4} = y$

Rewrite:
$= x * y * z^3 = \boxed{\text{ D. } xyz^3}}$

6.

Solve the inside of the parenthesis, and each term separately.

Use the property:
$x^0 = 1$

$(7 + 3^2)^0 = 1 \\\\(8^0 + 3)^2 = (1 + 3)^2 = 4^2 = 16\\\\1 + 16 = \boxed{ \text{ C. } 17}$

7.

Solve like above:
$(4 + 2^2)^0 = 1\\\\(7^0 + 4)^{-3} = (1 + 4)^{-3} = 5^{-3}$

The negative exponent indicates taking the reciprocal. Using the property:
$x^{-a} = \frac{1}{x^a}$

Therefore:
$5^{-3} = \frac{1}{5^3} = \boxed{ \text{ D. }\frac{1}{125}}$