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Marina CMI [18]
2 years ago
6

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

Mathematics
1 answer:
SIZIF [17.4K]2 years ago
4 0

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}}

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