Question

Can someone please help me with this again? 5-7

Thank you so much! ​

Answer 1

Hi there!

5.

Use the property:

`(x^a)/(x^b) = x^(a - b)`
Now, solve by variable.

`(x^2)/(x) = x^(2 - 1) = x\\\\(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) = (1)/(x^a)`

Therefore:

`5^(-3) = (1)/(5^3) = \boxed{ \text{ D. }(1)/(125)}`