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lukranit [14]
2 years ago
11

Find the perimeter of △ABC with vertices A(−5, 3), B(4, −5), and C(−5, −5). Round your answer to the nearest hundredth.

Mathematics
1 answer:
sertanlavr [38]2 years ago
8 0

The  perimeter of △ABC with vertices A(−5, 3), B(4, −5), and C(−5, −5) is 29.04 units

<h3>Perimeter of a triangle</h3>

The perimeter of a triangle is the sum of the sides of the triangle.

Given the coordinate points A(−5, 3), B(4, −5), and C(−5, −5)

Determine the distance AB, AC and BC using the distance formula

AB = √(-5-3)²+(4+5)²

AB = √64+81
AB = √145

AB = 12.04 units

AC = √(-5-3)²+(-5+5)²

AC = √64

AC = 8 units

BC = √(-5-4)²+(-5+5)²

BC = √81
BC = 9 units

Take the sum

Perimeter of triangle ABC = 9 + 8 + 12.04

Perimeter of triangle ABC = 29.04 units

Hence the  perimeter of △ABC with vertices A(−5, 3), B(4, −5), and C(−5, −5) is 29.04 units

Learn more on perimeter of triangle here: brainly.com/question/24382052

#SPJ1

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3 years ago
Lizzie rolls two dice. What is the probability that the sum of the dice is:
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A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\

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Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

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Probability of an event E can be formulated as:

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P(A)  = \dfrac{12}{36} = \dfrac{1}{3}

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)}  Count 1

Possible cases for sum 3: {(1,2), (2,1)}  Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)}  Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)}  Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

P(B)  = \dfrac{15}{36} = \dfrac{5}{12}

C. Divisible by 3 and less than 7:

P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

P(A \cap B) = \dfrac{7}{\text{36}}

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