Let’s analyze and solve the three questions shown in the image:

71. Evaluate: $-(10^0)^7$

• Step 1: Simplify $10^0$. Any number raised to the power of 0 is 1: $10^0 = 1$
• Step 2: Apply the exponentiation: $(10^0)^7 = 1^7 = 1$
• Step 3: Multiply by the negative sign: $-(10^0)^7 = -1$

Answer: B) -1

72. What is $\left(\frac{x}{y}\right)^3$ as a quotient of powers?

• Step 1: Use the property of exponents: $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$. $\left(\frac{x}{y}\right)^3 = \frac{x^3}{y^3}$

Answer: C) $\frac{x^3}{y^3}$

73. Simplify: $\frac{\left(-3x^2y\right)^4}{3(xy^2)^3}$

• Step 1: Expand the numerator: $\left(-3x^2y\right)^4 = (-3)^4 (x^2)^4 y^4 = 81x^8y^4$

• Step 2: Expand the denominator: $3(xy^2)^3 = 3 \cdot x^3 \cdot (y^2)^3 = 3x^3y^6$

• Step 3: Divide numerator by denominator: $\frac{81x^8y^4}{3x^3y^6} = \frac{81}{3} \cdot \frac{x^8}{x^3} \cdot \frac{y^4}{y^6}$ Simplify each term: $\frac{81}{3} = 27, \quad \frac{x^8}{x^3} = x^{8-3} = x^5, \quad \frac{y^4}{y^6} = y^{4-6} = y^{-2} = \frac{1}{y^2}$ Combine results: $\frac{81x^8y^4}{3x^3y^6} = \frac{27x^5}{y^2}$

Answer: C) $\frac{27x^5}{y^2}$

Would you like further details or clarification?

Related Questions:

1. How would the answers change if $10^0$ had an exponent other than 7 in question 71?
2. Can you rewrite $\left(\frac{x}{y}\right)^3$ in expanded form?
3. What happens if the denominator in question 73 has additional variables?
4. How do negative exponents simplify in similar problems?
5. Could you use a shortcut to solve question 73 without full expansion?

Tip: Always simplify exponents step by step to avoid mistakes, especially when handling multiple variables!