Respuesta :
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Answer:
C. segment EF is located at E (2, 0) and F (2, 6) and is twice the size of segment E prime F prime
Step-by-step explanation:
If the image is 1/2 the size of the original (scale factor = 1/2), the original is twice the size of the image. Each of the original coordinates is double the image coordinate value.
E(2, 0) = 2×E'(1, 0)
F(2, 6) = 2×F'(1, 3)
Answer:
segment EF is located at E (2, 0) and F (2, 6) and is twice the size of segment E prime F prime.
Step-by-step explanation:
just took test and got it correct