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soldi70 [24.7K]
3 years ago
5

Please help asap need it done. What is the measure of ∠CED and ∠ACD?

Mathematics
2 answers:
worty [1.4K]3 years ago
4 0

Answer:

m\angle CED= 64\°  

m\angle ACD=124\°  

Step-by-step explanation:

In the figure given:

∠ABC = 93°

∠BAC = 31°

∠CDE = 60°

To find ∠CED and ∠ACD.

Solution:

In triangle ABC, we are given two vertex angles. We can find the third angle as angle sum of triangle = 180°.

∠ABC = 93° , ∠BAC = 31°

∠BCA=  180\°-(93\°+31\°)

∠BCA = 56°

m\angle BCA+m\angle ACD=180\°    [Supplementary angles forming a linear pair]

m\angle ACD=180\°-56\°

m\angle ACD=124\°   (Answer)

In triangle CDE:

m\angle CDE+m\angle CED = m\angle ACD   [Exterior angle theorem :Exterior angle of a triangle is equal to sum of opposite interior angles ]

60\°+m\angle CED = 124\°

m\angle CED= 124\°-60\°

m\angle CED= 64\°      (Answer)

nikklg [1K]3 years ago
3 0

Answer:

m\angle CED= 64\°  

m\angle ACD=124\°  

Step-by-step explanation:

get an A!

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Please help it will help a lot i really need this!!
Lorico [155]

Answer:

A' = (0,0)

B'=(8,0)

C'=(8,2)

D'=(0,2)

It is not a rigid motion.

Step-by-step explanation:

To use the mapping rule, substitute the original x and y values in it.

The coordinates of A are (0,0).

Using the mapping rule, the x coordinate of A' = 2x0 = 0

Using the mapping rule, the y coordinate of A' = (1/2)x0=0

So A' will not change locations. The image will be at (0,0).

The coordinates of B are (4,0)

Using the mapping rule, the x-coordinate of B' = 2x4 = 8

Using the mapping rule, the y-coordinate of B' = (1/2)x0=0

Therefore the image of B' will be located at coorindate (8,0)

The coorindates of C are (4,4).

Using the mapping rule, the x-coordinate of C' = 2x4=8

Using the mapping rule, the y-coordinate of C' = (1/2)x4=2

Therefore the image of C' will be located at coordinate (8,2)

The coordinates of D are (0,4)

Using the mapping rule, the x-coordinate of D' = 2x0 = 0

Using the mapping rule, the y-coordinate of D' = (1/2)x4=2

Therefore the image of D' will be located at coordinate (0,2)

<em>Is the transformation a rigid motion?</em>

NO, this transformation is not a rigid motion because the relative distance between the points does not stay the same after they have been transformed. The transformation is not a translation , rotation, reflection, nor glide reflection.

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