Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.3 Center of Mass - Exercises - Page 483: 2

$(1, \frac{7}{12})$

$M_{x}$ = $4(2)+2(2)+5(-1)+1(0)$ = $7$ $M_{y}$ = $4(1)+2(-3)+5(2)+1(4)$ = $12$ the total mass of the system is $12$ so the coordinates of the center of mass are $(\frac{M_{y}}{M},\frac{M_{x}}{M})$ = $(1, \frac{7}{12})$