Triangle abc has vertices at (-4,0) (-1,6) and (3,-1) what is the perimeter of triangle abc rounded to the nearest tenth.
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Here, Vertices: A(-4, 0) B(-1, 6) C(3, -1)

Length of AB = √(6 - 0)² + (-1 + 4)²
= √6² + 3²
= √45

Length of BC = √(-1-6)² + (3+1)²
= √-7² + 4²
= √65

Length of AC = √(0+ 1)² + (-4 - 3)²
= √1² + 7²
= √50

Perimeter = sum of all sides:
So, P = √45 + √65 + √50 = √160 = 4√10

In short, Your Answer would be 4√10 units

Hope this helps!