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Nuetrik [128]
4 years ago
11

How are the properties of reflection used to transform a figure?

Mathematics
2 answers:
Lunna [17]4 years ago
8 0

Answer:When you reflect a point across the y-axis, the y-coordinate remains the same, but the x-coordinate is transformed into its opposite (its sign is changed). ... Remember that each point of a reflected image is the same distance from the line of reflection as the corresponding point of the original figure.

Step-by-step explanation:

Dmitriy789 [7]4 years ago
5 0

Answer:

The properties of reflection are used when we want to reflect a figure across a specific line, it can be across y-axis, x-axis, y = x or y = -x.

In either case, the result is a reflection, that is, those lines will work as a mirror, having the same shape and size but reflected across the line.

<em>It's important to say that a reflection is a rigid transformation, which means the shape or size of the figure won't chance.</em>

<em />

If we want to reflect across the y-axis, the transformation is:

(x,y) \implies (-x,y)

Across the x-axis: (x,y) \implies (x,-y)

Across the line y = x: (x,y) \implies (y,x)

Across the line y = -x: (x,y) \implies (-y,-x).

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Answer:

51 were the total number of the tickets sold at a school carnival were early-admission tickets.

Step-by-step explanation:

Total number of tickets sold by  = 100

Let x be the early-admission tickets sold by the school.

As 51% of the tickets sold at a school carnival were early-admission tickets.

so

x=\frac{51}{100}\times 100

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Therefore, 51 were the total number of the tickets sold at a school carnival were early-admission tickets.

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Identify the center and the radius of a circle that has a diameter with endpoints at (−5, 9) and (3, 5)
marusya05 [52]

Check the picture below, so the circle looks more or less like that one.

well, the center of it is simply the Midpoint of those two points, and its radius is simply half-the-distance between them.

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 -5}{2}~~~ ,~~~ \cfrac{ 5 + 9}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{14}{2} \right)\implies \stackrel{center}{(-1~~,~~7)} \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[3 - (-5)]^2 + [5 - 9]^2}\implies d=\sqrt{(3+5)^2+(-4)^2} \\\\\\ d=\sqrt{8^2+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}~\hfill \stackrel{\textit{half the diameter}}{\cfrac{4\sqrt{5}}{2}\implies \underset{radius}{2\sqrt{5}}}

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