Card Sorting F85144

1. **State the problem:** We need to sort the given expressions into three groups: those equivalent to $2\sqrt{5}$, those equivalent to $5$, and those that are neither. 2. **Recall important formulas:** - $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ - $(\sqrt{a})^{2} = a$ 3. **Evaluate each expression:** - $\sqrt{2} \cdot \sqrt{5} = \sqrt{2 \times 5} = \sqrt{10}$ (Neither) - $\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{5} = (\sqrt{2} \cdot \sqrt{2}) \cdot \sqrt{5} = \sqrt{4} \cdot \sqrt{5} = 2 \sqrt{5}$ (Equivalent to $2\sqrt{5}$) - $\sqrt{25} = 5$ (Equivalent to $5$) - $\sqrt{4} \cdot \sqrt{5} = \sqrt{4 \times 5} = \sqrt{20} = 2 \sqrt{5}$ (Equivalent to $2\sqrt{5}$) - $\sqrt{20} = \sqrt{4 \times 5} = 2 \sqrt{5}$ (Equivalent to $2\sqrt{5}$) - $(\sqrt{5})^{2} = 5$ (Equivalent to $5$) 4. **Sort the cards:** - Equivalent to $2\sqrt{5}$: $\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{5}$, $\sqrt{4} \cdot \sqrt{5}$, $\sqrt{20}$ - Equivalent to $5$: $\sqrt{25}$, $(\sqrt{5})^{2}$ - Neither: $\sqrt{2} \cdot \sqrt{5}$ **Final answer:** - $2\sqrt{5}$ group: $\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{5}$, $\sqrt{4} \cdot \sqrt{5}$, $\sqrt{20}$ - $5$ group: $\sqrt{25}$, $(\sqrt{5})^{2}$ - Neither group: $\sqrt{2} \cdot \sqrt{5}$