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natali 33 [55]
3 years ago
10

Describe a sequence of transformations that maps ABC to A'B'C'.

Mathematics
1 answer:
Sphinxa [80]3 years ago
7 0

The correct answer is " (1) Reflect ABC across the x-axis and call this new triangle A'B'C'. (2) Translate A'B'C' 2 units right and 6 units up so that its image is A''B''C''. "


It is assumed that the points are

A in ABC is (1,9), B in ABC is (3,12), and C in ABC is (4,4). 


A'' in A''B''C'' is (3,-3), B'' in A''B''C'' is (5,-6), and C'' in A''B''C'' is (6,2).


Both triangles are congruent. Since they are congruent, there are no contractions nor dilations occured. After rotating clockwise, we get A"C"B". So we need to reflect it to get A"B"C"


hope this helped :D

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Find the perimeter of quadrilateral PQRS with the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3).
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Answer:

P=16.53\ units

Step-by-step explanation:

we know that

The perimeter of quadrilateral PQRS is equal to the sum of its four length sides

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

we have

the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3)

step 1

Find the distance PQ

P(2,4), Q(2,3)

substitute in the formula

d=\sqrt{(3-4)^{2}+(2-2)^{2}}

d=\sqrt{(-1)^{2}+(0)^{2}}

d=\sqrt{1}

dPQ=1\ unit

step 2

Find the distance QR

Q(2,3), R(-2,-2)

substitute in the formula

d=\sqrt{(-2-3)^{2}+(-2-2)^{2}}

d=\sqrt{(-5)^{2}+(-4)^{2}}

dQR=\sqrt{41}\ units

step 3

Find the distance RS

R(-2,-2), and S(-2,3)

substitute in the formula

d=\sqrt{(3+2)^{2}+(-2+2)^{2}}

d=\sqrt{(5)^{2}+(0)^{2}}

dRS=5\ units

step 4

Find the distance PS

P(2,4), S(-2,3)

substitute in the formula

d=\sqrt{(3-4)^{2}+(-2-2)^{2}}

d=\sqrt{(-1)^{2}+(-4)^{2}}

dPS=\sqrt{17}\ units

step 5

Find the perimeter

P=PQ+QR+RS+PS

substitute the values

P=1+\sqrt{41}+5+\sqrt{17}

P=6+\sqrt{41}+\sqrt{17}

P=16.53\ units

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3 years ago
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