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IRINA_888 [86]
2 years ago
13

quadrilateral ABCD is dilated by a scale factor of 2 centered around (2,2). Which statement is true about the dilation?

Mathematics
1 answer:
Setler [38]2 years ago
6 0

Answer:

The vertices of the image will have the coordinates

A(-5, 1) → A'(2x-2, 2y-2) → A'(-12, 0)

B(-4, 3) → B'(2x-2, 2y-2) → B'(-10, 4)

C(1, 2) → C'(2x-2, 2y-2) → C'(0, 2)

D(-3, 0) → D'(2x-2, 2y-2) → D'(-8, -2)

Step-by-step explanation:

<em>Note: You missed to mention the vertices of a quadrilateral ABCD. So, I am assuming the following vertices. It would anyways clear your concept.</em>

<em />

Let us suppose

  • A(-5, 1)
  • B(-4, 3)
  • C(1, 2)
  • D(-3, 0)

We know that the rule of the dilation with the center of dilation at (2,2) by a scale factor of 2 is:

  • (x, y) → (2x-2, 2y-2)

Therefore, the vertices of the image will have the coordinates

A(-5, 1) → A'(2x-2, 2y-2) → A'(-12, 0)

B(-4, 3) → B'(2x-2, 2y-2) → B'(-10, 4)

C(1, 2) → C'(2x-2, 2y-2) → C'(0, 2)

D(-3, 0) → D'(2x-2, 2y-2) → D'(-8, -2)

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Answer:

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Answer:

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V(y)  = Voy - g*t

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Si en esta ecuación hacemos V(y) = 0 estamos en el punto donde el componente en el eje y de la velocidad del proyectil es cero, ese punto es el punto medio del recorrido.

0 =  Vo*sen 60⁰     - g*t

g*t  =  Vo* √3/2

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t = 16*√3  / 9,8

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El tiempo total del primer recorrido es entonces por simetría

t₁ = 2 * 2,8278           t₁  = 5,6556 seg

La distancia del primer impacto al suelo es:

x = Vox * t₁                        ( Vox es constante   Vx = Vo*cos 60⁰ )

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Aplicando los mismos criterios ahora para el segundo bote

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V = 24 m/s

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t =  1,8759

2*t  = 2*1,8759

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x₂  =  24* 0,6428*3,7518

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Answer:

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Step-by-step explanation:

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