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melomori [17]
1 year ago
6

Help me please translations hw

Mathematics
2 answers:
RUDIKE [14]1 year ago
6 0

The locations of the vertices are shown below:

  1. (- 5, 4), (- 4, 4), (- 5, 2)
  2. (- 4, 1), (- 1, 2), (0, - 1), (- 2, - 5)
  3. (2, 5), (3, 4), (1, 2)
  4. (3, - 5), (- 2, - 3), (0, - 1)
<h3>What are the coordinates of the images resulting from a translation?</h3>

Herein we find four representations of polygons on Cartesian plane and we must determine the coordinates of the images generated by translation, a kind of rigid transformations. Translation is described by the following operation:

P'(x, y) = P(x, y) + T(x, y)

Where:

  • P(x, y) - Original point
  • P'(x, y) - Resulting point
  • T(x, y) - Translation vector

Case 1 (K(x, y) = (3, - 3), X(x, y) = (4, - 3), R(x, y) = (3, - 5), T(x, y) = (- 8, 7))

K'(x, y) = (3, - 3) + (- 8, 7) = (- 5, 4)

X'(x, y) = (4, - 3) + (- 8, 7) = (- 4, 4)

R'(x, y) = (3, - 5) + (- 8, 7) = (- 5, 2)

Case 2 (E(x, y) = (0, 2), X(x, y) = (3, 3), V(x, y) = (4, 0), Z(x, y) = (2, - 4), T(x, y) = (- 4, - 1))

E'(x, y) = (0, 2) + (- 4, - 1) = (- 4, 1)

X'(x, y) = (3, 3) + (- 4, - 1) = (- 1, 2)

V'(x, y) = (4, 0) + (- 4, - 1) = (0, - 1)

Z'(x, y) = (2, - 4) + (- 4, - 1) = (- 2, - 5)

Case 3 (G(x, y) = (- 4, 5), J(x, y) = (- 3, 4), U(x, y) = (- 5, 2), T(x, y) = (6, 0))

G'(x, y) = (- 4, 5) + (6, 0) = (2, 5)

J'(x, y) = (- 3, 4) + (6, 0) = (3, 4)

U'(x, y) = (- 5, 2) + (6, 0) = (1, 2)

Case 4 (V(x, y) = (0, 0), Y(x, y) = (- 5, 2), F(x, y) = (- 3, 4), T(x, y) = (3, - 5))

V'(x, y) = (0, 0) + (3, - 5) = (3, - 5)

Y'(x, y) = (- 5, 2) + (3, - 5) = (- 2, - 3)

F'(x, y) = (- 3, 4) + (3, - 5) = (0, - 1)

To learn more on rigid transformations: brainly.com/question/28004150

#SPJ1

hjlf1 year ago
3 0

Answer:

See attachments.

Step-by-step explanation:

A translation is a type of <u>transformation</u> that moves a figure left, right, up or down.  Every point on the original figure is translated (moved) by the same distance in the same direction.

  • If a figure is to be moved left, <u>subtract</u> the number of units it is to be moved from the x-values.
  • If a figure is to be moved right, <u>add</u> the number of units it is to be moved to the x-values.
  • If a figure is to be moved up, <u>add</u> the number of units it is to be moved to the y-values.
  • If a figure is to be moved down, <u>subtract</u> the number of units it is to be moved from the y-values.

<h3><u>Question 1</u></h3>

Coordinates of the given <u>pre-image</u>:

  • K = (3, -3)
  • R = (3, -5)
  • X = (4, -3)

Given <u>mapping rule</u>:

(x, y) \rightarrow (x-8, y+7)

This tells us to subtract 8 from the x-values and add 7 to the y-values.

\sf \implies K(3,-3) \rightarrow K'(3-8, -3+7)=K'(-5,4)

\sf \implies R(3,-5) \rightarrow R'(3-8, -5+7)=R'(-5,2)

\sf \implies X(4,-3) \rightarrow X'(4-8, -3+7)=X'(-4,4)

<h3><u>Question 2</u></h3>

Coordinates of the given <u>pre-image</u>:

  • E = (0, 1)
  • X = (3, 3)
  • V = (4, 0)
  • Z = (2, -4)

Given <u>mapping rule</u>:

(x, y) \rightarrow (x-4, y-1)

This tells us to subtract 4 from the x-values and subtract 1 from the y-values.

\sf \implies E(0,1) \rightarrow E'(0-4, 1-1)=E'(-4,0)

\sf \implies X(3,3) \rightarrow X'(3-4, 3-1)=X'(-1,2)

\sf \implies V(4,0) \rightarrow V'(4-4, 0-1)=V'(0,-1)

\sf \implies Z(2,-4) \rightarrow Z'(2-4, -4-1)=Z'(-2,-5)

<h3><u>Question 3</u></h3>

Coordinates of the given <u>pre-image</u>:

  • G = (-4, 5)
  • J = (-3, 4)
  • U = (-5, 2)

Given <u>mapping rule</u>:

(x, y) \rightarrow (x+6, y)

This tells us to add 6 to the x-values and the y-values remain unchanged.

\sf \implies G(-4,5) \rightarrow G'(-4+6, 5)=G'(2,5)

\sf \implies J(-3,4) \rightarrow J'(-3+6, 4)=J'(3,4)

\sf \implies U(-5,2) \rightarrow U'(-5+6, 2)=U'(1,2)

<h3><u>Question 4</u></h3>

Coordinates of the given <u>pre-image</u>:

  • F = (-3, 4)
  • V = (0, 0)
  • Y = (-5, 2)

Given <u>mapping rule</u>:

(x, y) \rightarrow (x+3, y-5)

This tells us to add 3 to the x-values and subtract 5 from the y-values.

\sf \implies F(-3,4) \rightarrow F'( -3+3, 4-5)=F'(0,-1)

\sf \implies V(0,0) \rightarrow V'( 0+3, 0-5)=V'(3,-5)

\sf \implies Y(-5,2) \rightarrow Y'( -5+3, 2-5)=Y'(-2,-3)

Learn more about transformations here:

brainly.com/question/28354239

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Queremos maximizar el precio de tal forma que los ingresos no disminuyan.

Ese maximo precio es: $14,040.6

Sabemos que actualmente el precio es:

p = $6,000

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C = 120

Actualmente los ingresos son el producto de esos dos números, es decir:

ingresos = $6,000*120 = $720,000

Ahora sabemos que por cada incremento de $700 en el precio, el número de clientes decrece en 10.

Entonces podemos escribir el número de clientes como una ecuación lineal.

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Entonces tenemos:

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C(p) = (-10/$700)*p + 205.71

Los ingresos serán dados por:

ingresos = C(p)*p = (-10/$700)*p^2 + 205.71*p

Y queremos maximizar p de tal forma que esto sea igual a lo que obtuvimos antes:

(-10/$700)*p^2 + 205.71*p = $720,000

Entonces debemos resolver la ecuación cuadratica:

(-10/$700)*p^2 + 205.71*p - $720,000 = 0.

Las soluciones son dadas por la formula de Bhaskara.

p = \frac{-205.71 \pm \sqrt{(205.71)^2 - 4*(-10/\$ 700)*\$ 720,000} }{2*(-10/\$ 700)} \\\\p = \frac{-205.71 \pm 195.45}{(-20/\$ 700)}

La solución de maximo valor es:

p = (-205.71 - 195.45)/(-20/$700) = $14,040.6

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