1. **State the problem:** We have three masses: $m_1$, $m_2$, and $m_3$. Given $m_1 = 2.7 m_2 = 2.7 m_3$, with $m_1$ at the origin $(0,0)$, $m_2$ at $(2.7, 9)$, and $m_3$ at $(2.7, -9)$. We need to find the $x$-coordinate of the center of mass. 2. **Formula for center of mass:** The $x$-coordinate of the center of mass for multiple masses is given by: $$x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$$ 3. **Assign values:** Let $m_2 = m$, then $m_1 = 2.7 m$ and $m_3 = m$. 4. **Calculate numerator:** $$\sum m_i x_i = m_1 x_1 + m_2 x_2 + m_3 x_3 = 2.7 m \times 0 + m \times 2.7 + m \times 2.7 = 0 + 2.7 m + 2.7 m = 5.4 m$$ 5. **Calculate denominator:** $$\sum m_i = 2.7 m + m + m = 4.7 m$$ 6. **Calculate $x_{cm}$:** $$x_{cm} = \frac{5.4 m}{4.7 m} = \frac{\cancel{m} 5.4}{\cancel{m} 4.7} = \frac{5.4}{4.7} \approx 1.149$$ 7. **Final answer:** The $x$-coordinate of the center of mass is approximately $1.149$ (rounded to three decimal places).