Pl F -10--8 6--4-2 2 4 6 8-10 H S R. G -8- 10- Evelyn transforms EQFRGSHP so that the image of E is at (2, 0) and the image of R is at (6, – 7). Which transformation could Evelyn have used to show EQFRGSHP and its image are congruent? @ EQFRGSHP was reflected over the line y = x + 2. ® EQFRGSHP was translated right 7 units and down 4 units. © EQFRGSHP was rotated 135 degrees clockwise about the point Q. O EQFRGSHP was rotated 90 degrees clockwise about the point (-3,– 1).
Pl F -10--8 6--4-2 2 4 6 8-10 H S R. G -8- 10- Evelyn transforms EQFRGSHP so that the image of E is at (2, 0) and the image of R is at (6, – 7). Which transformation could Evelyn have used to show EQFRGSHP and its image are congruent? @ EQFRGSHP was reflected over the line y = x + 2. ® EQFRGSHP was translated right 7 units and down 4 units. © EQFRGSHP was rotated 135 degrees clockwise about the point Q. O EQFRGSHP was rotated 90 degrees clockwise about the point (-3,– 1).
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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GUEST, GUEST (Username: GUEST) GUEST SESSION
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GUEST, GUEST
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-10--8
--4
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G
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Evelyn transforms EQFRGSHP so that
image of E is at (2, 0) and the image of
at (6, – 7).
-
Which transformation could Evelyn have used to show EQFRGSHP and its image are congruent?
@ EQFRGSHP was reflected over the line y = x + 2.
® EQFRGSHP was translated right 7 units and down 4 units.
© EQFRGSHP was rotated 135 degrees clockwise about the point Q.
O EQFRGSHP was rotated 90 degrees clockwise about the point (-3, – 1).
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