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3 years ago

How are reflections represented as a function? what is the relationship between a reflection and a rigid motion?

2 answers:

hjlf3 years ago

7 0

Reflections are a type of transformation that move an entire curve such that its mirror image lies on the other side of the x or y axis.

A vertical reflection is given by the equation y=−f(x) and results in the curve being "reflected" across the x-axis. A horizontal reflection is given by the equation y=f(−x) and results in the curve being "reflected" across the y-axis.

Reflection is one of the four basic rigid motions.

Four basic rigid motions:

1.Reflection

2.Glide Reflection

3.Rotation

4.Translation

Cloud [144]3 years ago

6 0

Answer with explanation:

→→→Reflection is done through a line.When you reflect an Object through a line the Pre-image before reflection and Image Obtained after reflection are congruent to one another.

Reflection is done through a line.

The equation of line can be

1. x=k

2. y=k

3.x +y=k

where, k is any real number.

→→→When you talk about rigid motion , then there are four kind of transformation that can take place in the object called Pre-Image which are

1.Translation

2.Dilation

3. Rotation

4. Reflection

Apart from dilation , in all other cases Pre-image and Image are congruent while in all cases the two are similar.

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The expression below is the factorization of what trinomial?

Lapatulllka [165]

The answer is: [A]: 45x² + 81x + 36 .
_________________________

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then, combine the "like terms", to simplify.
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(5x+4)(9x+9) = 45x² + 45x + 36x + 35 = 45x² + 81x + 36 .
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2 years ago

Consider a computer that uses 6 bits to represent integers: 1 bit for the sign and 5 bits for the actual number. What's the larg

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3 years ago

Given the circle with the equation (x + 1)2 + y2 = 36, determine the location of each point with respect to the graph of the cir

kati45 [8]

$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad 
radius=\stackrel{}{ r}\\\\
-------------------------------\\\\
(x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~
\begin{cases}
\stackrel{center}{(-1,0)}\\
\stackrel{radius}{6}
\end{cases}$

so, that's the equation of the circle, and that's its center, any point "ON" the circle, namely on its circumference, will have a distance to the center of 6 units, since that's the radius.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2}
\\\\\\
d=\sqrt{0+1}\implies d=1$

well, the distance from the center to A is 1, namely is "inside the circle".

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{6})\\\\\\
\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(6-0)^2}\implies d=\sqrt{(-1+1)^2+6^2}
\\\\\\
d=\sqrt{0+36}\implies d=6$

notice, the distance to B is exactly 6, and you know what that means.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad 
C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-8})
\\\\\\
\stackrel{distance}{d}=\sqrt{[4-(-1)]^2+[-8-0]^2}\implies d=\sqrt{(4+1)^2+(-8)^2}
\\\\\\
d=\sqrt{25+64}\implies d=\sqrt{89}\implies d\approx 9.43398$

notice, C is farther than the radius 6, meaning is outside the circle, hiking about on the plane.

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3 years ago

Read 2 more answers