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EastWind [94]
2 years ago
10

Use the drop-down menus to complete the proof. by the unique line postulate, you can draw only one segment,

Mathematics
1 answer:
Doss [256]2 years ago
6 0

The reflection of BC over I is shown below.

<h3>What is reflection?</h3>
  • A reflection is a mapping from a Euclidean space to itself that is an isometry with a hyperplane as a set of fixed points; this set is known as the reflection's axis (in dimension 2) or plane (in dimension 3).
  • A figure's mirror image in the axis or plane of reflection is its image by reflection.

See the attached figure for a better explanation:

1. By the unique line postulate, you can draw only one line segment: BC

  • Since only one line can be drawn between two distinct points.

2. Using the definition of reflection, reflect BC over l.

  • To find the line segment which reflects BC over l, we will use the definition of reflection.

3. By the definition of reflection, C is the image of itself and A is the image of B.

  • Definition of reflection says the figure about a line is transformed to form the mirror image.
  • Now, the CD is the perpendicular bisector of AB so A and B are equidistant from D forming a mirror image of each other.

4. Since reflections preserve length, AC = BC

  • In Reflection the figure is transformed to form a mirror image.
  • Hence the length will be preserved in case of reflection.

Therefore, the reflection of BC over I is shown.

Know more about reflection here:

brainly.com/question/1908648

#SPJ4

The question you are looking for is here:

C is a point on the perpendicular bisector, l, of AB. Prove: AC = BC Use the drop-down menus to complete the proof. By the unique line postulate, you can draw only one segment, Using the definition of, reflect BC over l. By the definition of reflection, C is the image of itself and is the image of B. Since reflections preserve , AC = BC.

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Find the values of y = c(x) = for x = 0, 0.008, 0.027, 0.064, 0.125, 0.216, 0.343, 0.512, , 0.729, and 1.
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c(x) is a function, then we will solve the problem for two different functions.

1. If c(x)=\frac{x}{0.5}+x

Then

y=c(x)

y=\frac{x}{0.5}+x

Using each of the values for x we have.

When x=0 the result is:   y=\frac{0}{0.5}+0=0

When x=0.008 the result is: y=\frac{0.008}{0.5}+0.008=0.016+0.008=0.024

When x=0.027 the result is: y=\frac{0.027}{0.5}+0.027=0.054+0.027=0.081

When x=0.064 the result is: y=\frac{0.064}{0.5}+0.064=0.128+0.064=0.192

When x=0.125 the result is: y=\frac{0.125}{0.5}+0.125=0.25+0.125=0.375

When x=0.216 the result is: y=\frac{0.216}{0.5}+0.216=0.432+0.216=0.648

When x=0.343 the result is: y=\frac{0.343}{0.5}+0.343=0.686+0.343=1.029

When x=0.512 the result is: y=\frac{0.512}{0.5}+0.512=1.024+0.512=1.536

When x=0.729 the result is: y=\frac{0.729}{0.5}+0.729=1.458+0.729=2.187

When x=1 the result is: y=\frac{1}{0.5}+1=2+1=3

2. If c(x)=sin(x)+2x

Then

y=c(x)

y=sin(x)+2x

Using each of the values for x we have.

When x=0 the result is:   y=sin(0)+2(0)=0

When x=0.008 the result is: y=sin(0.008)+2(0.008)=0.000139+0.016=0.016139

When x=0.027 the result is: y=sin(0.027)+2(0.027)=0.000471+0.054=0.054471

When x=0.064 the result is: y=sin(0.064)+2(0.064)=0.001117+0.128=0.129117

When x=0.125 the result is: y=sin(0.125)+2(0.125)=0.002182+0.25=0.252182

When x=0.216 the result is: y=sin(0.216)+2(0.216)=0.003769+0.432=0.435769

When x=0.343 the result is: y=sin(0.343)+2(0.343)=0.005986+0.686=0.691986

When x=0.512 the result is: y=sin(0.512)+2(0.512)=0.008936+1.024=1.032936

When x=0.729 the result is: y=sin(0.729)+2(0.729)=0.012723+1.458=1.470723

When x=1 the result is: y=sin(1)+2(1)=0.017452+2=2.017452

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