Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.3 Center of Mass - Exercises - Page 483: 3

$(\frac{a+b}{3}, \frac{c}{3})$

$M_{x}$ = $0(m)+0(m)+c(m)$ = $cm$ $M_{y}$ = $a(m)+b(m)+0(m)$ = $(a+b)m$ the total mass of the system is $3m$ so the coordinates of the center of mass are $(\frac{M_{y}}{M},\frac{M_{x}}{M})$ = $(\frac{(a+b)m}{3m}, \frac{cm}{3m})$ = $(\frac{a+b}{3}, \frac{c}{3})$