Ind the center of mass of the given system of point masses lying on the \( x \)-axis. \[ \begin{array}{r}m_{1}=6, m_{2}=3, m_{3}=4 \\ x_{1}=-5, x_{2}=0, x_{3}=4\end{array} \] \( x=\square \)

The center of mass \( x \) for a system of point masses is calculated using the formula: \[ x = \frac{\sum m_i x_i}{\sum m_i} \] For the given masses, we can plug in the values: 1. Calculate the total mass: \[ \sum m_i = m_1 + m_2 + m_3 = 6 + 3 + 4 = 13 \] 2. Calculate the weighted sum of positions: \[ \sum m_i x_i = m_1 x_1 + m_2 x_2 + m_3 x_3 = (6)(-5) + (3)(0) + (4)(4) = -30 + 0 + 16 = -14 \] Now, substituting these sums into the center of mass formula: \[ x = \frac{-14}{13} \] Thus, the center of mass of the system is: \[ x \approx -1.08 \]