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

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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
User Henderso
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