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anyanavicka [17]
3 years ago
13

Graph △XYZ with vertices X(2, 3), Y(−3, 2), and Z(−4,−3) and its image after the translation (x, y)→(x+3, y−1)

Mathematics
1 answer:
viktelen [127]3 years ago
5 0

Answer:

  • X'(5, 2)
  • Y'(0, 1)
  • Z'(-1, -4)

Step-by-step explanation:

The translation increases each x-coordinate by 3, moving the point 3 units to the right. It decreases each y-coordinate by 1, moving the point 1 unit down.

  (x, y) ⇒ (x+3, y-1)

  X(2, 3) ⇒ X'(5, 2)

  Y(-3, 2) ⇒ Y'(0, 1)

  Z(-4, -3) ⇒ Z'(-1, -4)

The red arrows show the translation of each point in the graph.

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The quadratic function h(t)=-16.1t^2 + 150 models a balls height, in feet, over time, in seconds, after it is dropped from a 15
katrin [286]
<h2>Hello!</h2>

The answer is: The first graphic representation.

<h2>Why?</h2>

We are given a quadratic equation, meaning that it could be two possible solutions for the exercise, however, we are talking about time, so we have to consider only the obtained positive values.

Let's make the equation equal to 0 in order to find the values of "t"

h(t)=-16.1t^2 + 150\\0=-16.1t^2 + 150\\16.1t^2=150\\t^2=\frac{150}{16.1}=9.32\\t=+-\sqrt{9.32}=+-3.05\\t1=3.05\\t2=-305

So, discarding the negative value, we can use the possitive value to find the correct graphic representation.

To find the correct graphic representation we must take into consideration the following:

- We must remember that the sign of the coefficient of the quadratic term (t^2) will define if the parabola opens downward or upward.

From the given quadratic (or parabola) equation we have:

a=-1\\b=0\\c=150

So, since the coefficient of the quadratic term is negative, the parabola opens downward.

- Since we are looking for a graphic that represents the change in height over time, we need to look for a graphic that shows only positive values for the x-axis (time)

- We are looking for a parabola which y-axis intercept is equal to 150.

Therefore, the graphic representation of the quadratic function that models a ball's height over time is the first graphic representation.

Have a nice day!

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Ainat [17]
For a smoothing constant of 0.2

Time period – 1 2 3 4 5 6 7 8 9 10

Actual value – 46 55 39 42 63 54 55 61 52

Forecast – 58 55.6 55.48 52.18 50.15 52.72 52.97 53.38 54.90

Forecast error - -12 -.6 -16.48 – 10.12 12.85 1.28 2.03 7.62 -2.9
The mean square error is 84.12

The mean forecast for period 11 is 54.38
For a smoothing constant of 0.8

Time period – 1 2 3 4 5 6 7 8 9 10

Actual value – 46 55 39 42 63 54 55 61 52

Forecast – 58 48.40 53.68 41.94 41.99 58.80 54.96 54.99 59.80

Forecast error - -12 6.60 -14.68 0.06 21.01 -4.80 0.04 6.01 -7.80The mean square error is 107.17

The mean forecast for period 11 is 53.56


Based on the MSE, smoothing constant of .2 offers a better model since the mean forecast is much better compared to the 53.56 of the smoothing constant of 0.8.
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