In a coordinate plane, triangle ABC has vertices A ( - 4, 5), B ( - 4, - 4), and C (3,-4),. Find the perimeter of triangle ABC to the nearest tenth.

Answer: perimeter of triangle ABC to the nearest tenth = 27.4

Step-by-step explanation: Given that

vertices are

A ( - 4, 5), B ( - 4, - 4), and C (3,-4)

Length AB = root ([-4-5]^2 + [-4+4]^2)

Length AB = sqr (9^2) = 9

Length AC = root ([-4-5]^2 + [3+4]^2)

Length AC = sqr (9^2 + 7^2)

Length AC = sqr (130) = 11.4

Length BC = root ([-4+4]^2 + [3+4]^2)

Length BC = sqr (7^2) = 7

Perimeter = 9 + 11.4 + 7 = 27.4