Triangle ABC has coordinates of A(-4,-2), B(4,8), and C(-7,-2). Determine the perimeter of ABC to the nearest tenth.

The distance between points is given by:
 d = root ((x2-x1) ^ 2 + (y2-y1) ^ 2)
 We then look for the distance between the vertices:
 For AB:
 AB = root ((4 - (- 4)) ^ 2+ (8 - (- 2)) ^ 2)
 AB = 12.80624847
 For AC:
 AC = root ((- 7 - (- 4)) ^ 2 + (- 2 - (- 2)) ^ 2)
 AC = 3
 For BC:
 BC = root ((- 7-4) ^ 2 + (- 2-8) ^ 2)
 BC = 14.86606875
 The perimeter will be:
 P = AB + AC + BC
 Substituting values:
 P = 12.80624847 + 3 + 14.86606875
 P = 30.67231722
 Round to the nearest tenth:
 P = 30.7 units
 Answer:
 the perimeter of ABC is:
 P = 30.7 units