Chapter 8 - Further Applications of Integration - 8.3 Applications to Physics and Engineering - 8.3 Exercises - Page 607: 23

$M_y=14$ $M_x=10$ so the center of mass is $\left(1.4;1\right)$

$$M_y=m_1\cdot x_1+m_{2}\cdot x_2+m_{3}\cdot x_3=4\cdot 2+2\cdot (-3)+4\cdot 3=14$$ $$M_x=m_1\cdot y_1+m_{2}\cdot y_2+m_{3}\cdot y_3=4\cdot (-3)+2\cdot 1+4\cdot 5=10$$ so the center of mass is: $$\left(\frac{14}{4+2+4}=1.4;\frac{10}{4+2+4}=1\right)$$