Answer
$$\left( {\overline x ,\overline y } \right) = \left( {\frac{5}{8},\frac{{13}}{{16}}} \right)$$
Work Step by Step
\[\begin{gathered}
{\text{}} \hfill \\
\boxed{\begin{array}{*{20}{c}}
{{m_i}}&3&4&2&1&6 \\
{\left( {{x_i},{y_i}} \right)}&{\left( { - 2, - 3} \right)}&{\left( {5,5} \right)}&{\left( {7,1} \right)}&{\left( {0,0} \right)}&{\left( { - 3,0} \right)}
\end{array}} \hfill \\
\end{gathered} \]
$$\eqalign{
& m{\text{ is the total mass of the system}} \cr
& m = {m_1} + {m_2} + {m_3} + {m_4} + {m_5} \cr
& m = 3 + 4 + 2 + 1 + 6 \cr
& m = 16 \cr
& \cr
& {\text{*The moment about the }}y{\text{ - axis is }} \cr
& {M_y} = {m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3} + {m_4}{x_4} + {m_5}{x_5} \cr
& {M_y} = \left( 3 \right)\left( { - 2} \right) + \left( 4 \right)\left( 5 \right) + \left( 2 \right)\left( 7 \right) + \left( 1 \right)\left( 0 \right) + \left( 6 \right)\left( { - 3} \right) \cr
& {M_y} = 10 \cr
& \cr
& {\text{*The moment about the }}x{\text{ - axis is }} \cr
& {M_x} = {m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3} \cr
& {M_x} = \left( 3 \right)\left( { - 3} \right) + \left( 4 \right)\left( 5 \right) + \left( 2 \right)\left( 1 \right) + \left( 1 \right)\left( 0 \right) + \left( 6 \right)\left( 0 \right) \cr
& {M_x} = 13 \cr
& \cr
& {\text{The center of mass }}\left( {\overline x ,\overline y } \right)({\text{or center of gravity}}){\text{ is}} \cr
& \overline x = \frac{{{M_y}}}{m} = \frac{{10}}{{16}} = \frac{5}{8} \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{{13}}{{16}} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{5}{8},\frac{{13}}{{16}}} \right) \cr} $$
