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anonymous

  • one year ago

A student solved the following problem and made an error:

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    Line 1 segment AC equals 2, segment DF equals 2, segment AC is congruent to segment DF Line 2 ∠A ≅ ∠F Line 3 Length of segment AB. A (0, 2) B (2, 4) d equals square root of quantity x sub 2 minus x sub 1 squared plus quantity y sub 2 minus y sub 1 squared, then d equals square root of quantity 0 minus 2 all squared plus quantity 2 minus 4 all squared, then d equals square root of negative 2 squared plus negative 2 squared, then d equals square root of 4 plus 4, then d equals square root of 8 segment AB = 2.83 Line 4 Length of segment FD. D (2, 0) F (4, 2) d equals square root of quantity x sub 2 minus x sub 1 squared plus quantity y sub 2 minus y sub squared then d equals square root of quantity 2 minus 4 all squared plus quantity 0 minus 2 all squared, then d equals square root of negative 2 squared plus negative 2 squared, then d equals square root of 4 plus 4, then d equals square root of 8 segment FD = 2.83 Line 5 segment AC is congruent to segment FD Line 6 triangle ABC is congruent to triangle FDE by SAS

  3. anonymous
    • one year ago
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    In which line did the student make the first mistake? Line 3 Line 5 Line 1 Line 2

  4. anonymous
    • one year ago
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    Woah

  5. dix123
    • one year ago
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    I agree @sunleaf01

  6. dix123
    • one year ago
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    @SomeoneHelpPlz What is the student trying to prove?

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