Answer:
<u>Figure A</u>
Step-by-step explanation:
See the attached figure which represents the given options
We are to select the correct pair of triangles that can be mapped to each other using a translation and a rotation about point A.
As shown: point A will map to point L, point R will map to point P and point Q will map to point K.
we will check the options:
<u>Figure A</u>: the triangle ARQ and LPK can be mapped to each other using a translation and a rotation about point A.
<u>Figure B: </u> the triangle ARQ and LPK can be mapped to each other using a translation and a reflection about the line RA.
<u>Figure C:</u> the triangle ARQ and LPK can be mapped to each other using a translation and a reflection about the line QA.
<u>Figure D:</u> the triangle ARQ and LPK can be mapped to each other using a rotation about point A.
So, the answer is figure A
<u>The triangle pairs of figure A can be mapped to each other using a translation and a rotation about point A.</u>
Answer: A) 23
Step-by-step explanation:
3^2 + ( 6 - 2 ) ⋅ 4 -6/3
3^2 + 4 ⋅ 4 -6/3
9 + 4 ⋅ 4 -6/3
9 + 16 -6/3
25 - 6/3
75/3 -6/3 = 69/3
69/3 = 23
Answer:
(1,2)
Step-by-step explanation:
x+4y = 9
2x -4y= -6
Add the equations together
x+4y = 9
2x -4y= -6
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3x +0y = 3
3x=3
Divide by 3
3x/3 = 3/3
x=1
Now find y
x+4y = 9
1 +4y =9
Subtract 1 from each side
4y = 8
Divide by 4
4y/4 = 8/4
y =2
<h3>
Answer: 80 degrees</h3>
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Explanation:
I'm assuming that segments AD and CD are tangents to the circle.
We'll need to add a point E at the center of the circle. Inscribed angle ABC subtends the minor arc AC, and this minor arc has the central angle AEC.
By the inscribed angle theorem, inscribed angle ABC = 50 doubles to 2*50 = 100 which is the measure of arc AC and also central angle AEC.
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Focus on quadrilateral DAEC. In other words, ignore point B and any segments connected to this point.
Since AD and CD are tangents, this makes the radii EA and EC to be perpendicular to the tangent segments. So angles A and C are 90 degrees each for quadrilateral DAEC.
We just found angle AEC = 100 at the conclusion of the last section. So this is angle E of quadrilateral DAEC.
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Here's what we have so far for quadrilateral DAEC
- angle A = 90
- angle E = 100
- angle C = 90
- angle D = unknown
Now we'll use the idea that all four angles of any quadrilateral always add to 360 degrees
A+E+C+D = 360
90+100+90+D = 360
D+280 = 360
D = 360-280
D = 80
Or a shortcut you can take is to realize that angles E and D are supplementary
E+D = 180
100+D = 180
D = 180-100
D = 80
This only works if AD and CD are tangents.
Side note: you can use the hypotenuse leg (HL) theorem to prove that triangle EAD is congruent to triangle ECD; consequently it means that AD = CD.