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grigory [225]
3 years ago
10

What is the perimiter of ABC with vertices A (-2,9), B(7-3), and c(-2,-3) in the coordinate plane?

Mathematics
1 answer:
Serggg [28]3 years ago
6 0

Answer:

|AB| = √((7 - (-2))^2 + (-3 - 9)^2) = 15

|BC| = √((-2 - 7)^2 + (-3 - (-3))^2) = 9

|AC| = √((-2 - (-2))^2 + (-3 - 9)^2) = 12

Perimeter = 15 + 9 + 12 = 36

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Answer:

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Step-by-step explanation:

vertex form is

y = a(x - b)^2 + c     where a is a constant and (b,c) is the vertex.

The maximum is at (-2, 8)  because x 8 = height and x =-2 is equn. of symmetry

So here we have

y = a(x - (-2))^2 + 8

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Now at the point (-6, 2):

2 = a(-6+2)^2 + 8

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16a = -6

1 = -3/8.

So our equation is y = 3/8(x + 2)^2 + 8

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Step-by-step explanation:

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