1. **State the problem:** We need to find which expressions are equivalent to $28^{-3}$ from the given options. 2. **Recall the laws of exponents:** - $a^m \cdot a^n = a^{m+n}$ - $\frac{a^m}{a^n} = a^{m-n}$ - $(a^m)^n = a^{m \cdot n}$ - $a^{-m} = \frac{1}{a^m}$ 3. **Check each option:** **A:** $\frac{28^{-5}}{28^{-2}} = 28^{-5 - (-2)} = 28^{-5 + 2} = 28^{-3}$ ✔ **B:** $28^0 \cdot 28^3 = 28^{0+3} = 28^3 \neq 28^{-3}$ ✘ **C:** $\left(\frac{1}{28^{-3}}\right)^{-1} = (28^3)^{-1} = 28^{-3}$ ✔ **D:** $(7^{-3} \cdot 4^{-3})^{-1} = (7^{-3})^{-1} \cdot (4^{-3})^{-1} = 7^3 \cdot 4^3 = (7 \cdot 4)^3 = 28^3 \neq 28^{-3}$ ✘ **E:** $28^{-2} \cdot 28^5 = 28^{-2+5} = 28^3 \neq 28^{-3}$ ✘ **F:** $\frac{7^{-3}}{4^3} = 7^{-3} \cdot 4^{-3} = (7 \cdot 4)^{-3} = 28^{-3}$ ✔ 4. **Answer:** The expressions equivalent to $28^{-3}$ are A, C, and F. --- 5. **Angel's statement:** Raising a fraction to the power of $-1$ is the same as finding its reciprocal. **Explanation:** Yes, this is true because by definition, $a^{-1} = \frac{1}{a}$. So raising a fraction to the power $-1$ flips it, giving the reciprocal. --- 6. **Find $n$ such that:** $$\frac{42^n}{42} = \frac{\_}{42^3}$$ Rewrite left side: $$\frac{42^n}{42} = 42^{n-1}$$ Rewrite right side as: $$\frac{42^k}{42^3} = 42^{k-3}$$ Since the right side numerator is unknown, assume numerator is $42^m$ so that: $$\frac{42^m}{42^3} = 42^{m-3}$$ Set equal: $$42^{n-1} = 42^{m-3}$$ For equality, exponents must be equal: $$n - 1 = m - 3$$ If numerator on right is $42^1$ (just $42$), then $m=1$: $$n - 1 = 1 - 3$$ $$n - 1 = -2$$ $$n = -1$$ **Answer:** $n = -1$. --- **Summary:** - Equivalent expressions to $28^{-3}$: A, C, F - Angel's statement is correct - $n = -1$ satisfies the equation