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ASAPT

  • one year ago

Given that D is equidistant to G and F, find m<GED A. 38 degrees B. 40 degrees C. 41 degrees D. 45 degrees

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  1. ASAPT
    • one year ago
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    @Michele_Laino @ganeshie8

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  2. Michele_Laino
    • one year ago
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    since the point D is equidistant from G and F, then the line ED is the bisector of the angle GEF, then we can write this: \[\Large 2x + 20 = 5x - 10\] please solve for x

  3. ASAPT
    • one year ago
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    @Michele_Laino im stuck

  4. Michele_Laino
    • one year ago
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    subtractin 2x from both sides, we get: \[\Large \begin{gathered} 2x + 20 - 2x = 5x - 10 - 2x \hfill \\ \hfill \\ 20 = 3x - 10 \hfill \\ \end{gathered} \] then adding 10 to both sides we get: \[\Large \begin{gathered} 20 + 10 = 3x + 10 - 10 \hfill \\ \hfill \\ 30 = 3x \hfill \\ \end{gathered} \] now divide both sides by 3, what do you get?

  5. Michele_Laino
    • one year ago
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    subtracting*

  6. ASAPT
    • one year ago
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    so 10?

  7. Michele_Laino
    • one year ago
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    ok! we have x=10 so the measure of the angle GED, is: (2*x+20)+(5*x-10)=(2*10+20)+(5*10-10)=...?

  8. ASAPT
    • one year ago
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    that's a lot of numbers I don't have much time I have like 17 left and a time limit!!

  9. Michele_Laino
    • one year ago
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    hint: \[\begin{gathered} \left( {2x + 20} \right) + \left( {5x - 10} \right) = \left( {2 \times 10 + 20} \right) + \left( {5 \times 10 - 10} \right) = ...? \hfill \\ = \left( {20 + 20} \right) + \left( {50 - 10} \right) = 40 + 40 = ...? \hfill \\ \end{gathered} \]

  10. Michele_Laino
    • one year ago
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    sorry, the angle GEF is 80 degrees, the measure of the angle GED is: \[2x + 20 = 2 \times 10 + 20 = ...?\]

  11. Michele_Laino
    • one year ago
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    \[2x + 20 = 2 \times 10 + 20 = 20 + 20 = ...?\]

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