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ikadub [295]
3 years ago
5

Which statements are true about reflections? Check all that apply.

Mathematics
2 answers:
sukhopar [10]3 years ago
4 0

Answer:

<h3>a. An image created by a reflection will always be congruent to its pre-image </h3>

This statement is true because a reflection is a rigid transformation, that is, it doesn't change the shape and size of the original figure, it just moves it. That's why the reflectio is congruent to its pre-image.

<h3>b. An image and its pre-image are always the same distance from the line of reflection </h3>

This statement is also true, because the line of reflection is an axis that it's a reference for each figure. Also, when we reflect a figure, we are changing some coordinates to their opposite, for example, x=-2 changes to x'=2, which means the distance from the line of reflection is the same, two units.

<h3>c. If a point on the pre-image lies on the line of reflection, the image of that point is the same as the pre-image. </h3>

When we reflect a figure, and its reflected image has a points on the line of reflection, that means the pre-image also has its correspoinding point on the line of reflection, because actually that point won't change its coordinates is ON the line of reflection, that is, when the image reflects, the points remains fixed.

<h3>d. The line of reflection is perpendicular to the line segments connecting corresponding vertices. </h3><h3>f. The line segments connecting corresponding vertices are all parallel to each other</h3>

If we reflect a figure across a vertical axis, the coordinates that change are the horizontal ones.

If we reflect a figure across a horizontal axis, the coordinates that change are the vertical ones.

So, in the first case, if we unit the initial horizontal coordinates with the reflected one, they are gonna be united with horizontal lines, and all of them are gonna be parallels and perpendicular to the line of reflection, which in this case is vertical. That's why d and f are also true.

Gemiola [76]3 years ago
3 0

1,2,3,4,6                                                                                                                                                                   have a good day

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Norma-Jean [14]

Answer:

S(t) = a.sin (b.t) + d

a = -1.5, b = (2π/365), d = 3.47

S(t) = -1.5 sin (2πt/365) + 3.47

Step-by-step explanation:

Complete Question is presented in the attached image to this solution.

- Dingane has been observing a certain stock for the last few years and he sees that it can be modeled as a function S(t) of time t (in days) using a sinusoidal expression of the form

S(t) = a.sin(b.t) + d.

On day t = 0, the stock is at its average value of $3.47 per share, but 91.25 days later, its value is down to its minimum of $1.97.

Find S(t). t should be in radians.

S(t) =

Solution

S(t) = a.sin(b.t) + d.

At t = 0, S(t) = $3.47

S(0) = a.sin(b×0) + d = a.sin 0 + d = 3.47

Sin 0 = 0,

S(t=0) = d = 3.47.

At t = 91.25 days, S(t) = $1.97

But, it is given that T has to be in radians, for t to be in radians, the constant b has to convert t in days to radians.

Hence, b = (2π/365)

S(91.25) = 1.97 = a.sin(b×91.25) + d

d = 3.47 from the first expression

S(t = 91.25) = a.sin (91.25b) + 3.47 = 1.97

1.97 = a.sin (2π×91.25/365) + 3.47

1.97 = a sin (0.5π) + 3.47

Sin 0.5π = 1

1.97 = a + 3.47

a = -1.5

Hence,

S(t) = a.sin (b.t) + d

a = -1.5, b = (2π/365), d = 3.47

S(t) = -1.5 sin (2πt/365) + 3.47

Hope this Helps!!!

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Answer:

The point (-3, 2) lies outside of the circle centered at (4, 0) with radius 5.

Step-by-step explanation:

The distance between two points (x₁, y₁) and (x₂, y₂) on the x-y plane can be calculated with:

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Since 7.28 > 5, the point (-3, 2) lies outside of the circle centered at (4, 0) with radius 5.

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