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Romashka [77]
3 years ago
12

Triangle MNO has vertices M(6,5), N(2,2), and O(7,0). It’s image has vertices M’(-6,5), N’(-2,2), and O’(-7,0). Which line of re

flection maps triangle MNO onto its image?
A. x = 0
B. x = 1
C. y = 1
D. y = 0
Mathematics
2 answers:
Helga [31]3 years ago
8 0
Here, We can see the x-coordinates changing their sign in each instances, so the reflection would be over y-axis about the origin.

In short, Your Answer would be: Option D) y = 0

Hope this helps!
Natasha_Volkova [10]3 years ago
4 0

Answer:

D. y=0

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.
bezimeni [28]

Given:

In the circle, m(\overarc{VUX})=152^\circ and m(\angle MUV)=77^\circ.

To find:

The following measures:

(a) m\angle VUX

(b) m\angle UXW

Solution:

According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.

m(VUX)=2\times m\angle VWX

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\dfrac{152^\circ}{2}=m\angle VWX

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In a cyclic quadrilateral, the opposite angles are supplementary angles.

UVWX is a cyclic quadrilateral. So,

m\angle VUX+m\angle VWX=180^\circ          [Opposite angles of a cyclic quadrilateral]

m\angle VUX+76^\circ=180^\circ

m\angle VUX=180^\circ-76^\circ

m\angle VUX=104^\circ

Now,

m\angle UXW+m\angle UVW=180^\circ          [Opposite angles of a cyclic quadrilateral]

m\angle UXW+77^\circ =180^\circ

m\angle UXW=180^\circ-77^\circ

m\angle UXW=103^\circ

Therefore, m\angle VUX=104^\circ  and m\angle UXW=103^\circ .

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