1. **State the problem:** Find the area of a square with side length $\frac{2}{3}y$ cm. 2. **Formula for the area of a square:** $$\text{Area} = (\text{side length})^2$$ 3. **Apply the formula:** $$\text{Area} = \left(\frac{2}{3}y\right)^2$$ 4. **Square the fraction and the variable:** $$\left(\frac{2}{3}\right)^2 y^2 = \frac{2^2}{3^2} y^2 = \frac{4}{9} y^2$$ 5. **Final answer:** $$\boxed{\frac{4}{9} y^2 \text{ cm}^2}$$ --- 1. **State the problem:** Check if the product of $\left(-\frac{3}{5}x\right)$ and $\left(\frac{5}{3}x - 7\right)$ has one term after applying the distributive property. 2. **Apply the distributive property:** $$\left(-\frac{3}{5}x\right) \times \left(\frac{5}{3}x - 7\right) = \left(-\frac{3}{5}x\right) \times \frac{5}{3}x + \left(-\frac{3}{5}x\right) \times (-7)$$ 3. **Multiply the first terms:** $$-\frac{3}{5}x \times \frac{5}{3}x = -\frac{3}{5} \times \frac{5}{3} \times x \times x = -\cancel{\frac{3}{5}} \times \cancel{\frac{5}{3}} x^2 = -1 \times x^2 = -x^2$$ 4. **Multiply the second terms:** $$-\frac{3}{5}x \times (-7) = \frac{21}{5}x$$ 5. **Combine the terms:** $$-x^2 + \frac{21}{5}x$$ 6. **Conclusion:** The product has two terms, not one. So, Latoya is incorrect. **Summary:** The product expands to $-x^2 + \frac{21}{5}x$, which has two terms, not one.