1. **Identify irrational numbers from the list:** - A rational number can be expressed as a fraction of two integers. - An irrational number cannot be expressed as a simple fraction and has a non-repeating, non-terminating decimal expansion. A. $\frac{5}{11}$ is a fraction of integers, so it is rational. B. $\sqrt{\frac{5}{11}}$ is the square root of a non-perfect fraction, so it is irrational. C. $\sqrt{\frac{25}{11}} = \frac{\sqrt{25}}{\sqrt{11}} = \frac{5}{\sqrt{11}}$; since $\sqrt{11}$ is irrational, this is irrational. D. $\frac{5\pi}{11}$ involves $\pi$, which is irrational, so this is irrational. E. $\sqrt{\frac{11}{5}}$ is the square root of a non-perfect fraction, so it is irrational. F. $5.15$ is a terminating decimal, so it is rational. **Answer for 42:** B, C, D, E are irrational. 2. **Which number is irrational?** A) $9.6$ is a terminating decimal, so rational. B) $\frac{77}{9}$ is a fraction of integers, so rational. C) $\sqrt{25} = 5$ is rational. D) $\sqrt{13}$ is the square root of a non-perfect square, so irrational. **Answer for 43:** D 3. **Simplify $2^{-3}$ and find equivalent expressions:** Recall: $a^{-n} = \frac{1}{a^n}$ - A) $\frac{1}{2^{-3} \times 2^2} = \frac{1}{\frac{1}{2^3} \times 2^2} = \frac{1}{\frac{2^2}{2^3}} = \frac{1}{2^{-1}} = 2^1 = 2$ which is not equal to $2^{-3}$. - B) $\frac{2^0}{2^2} = \frac{1}{4} = 2^{-2}$ not equal to $2^{-3}$. - C) $\frac{2^6}{2^2 \times 2} = \frac{2^6}{2^3} = 2^{6-3} = 2^3$ not equal to $2^{-3}$. - D) $(2^3)^{-1} = 2^{-3}$ which is exactly the expression. **Answer for 44:** D