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Annetta_Martin

  • one year ago

To solve |3x+2|<4, split the equation into 3x + 2 < 4 and 3x + 2 < -4 3x + 2 < 4 and 3x + 2 > - 4 3x - 2 < 4 and 3x - 2 > - 4 3x + 2 < - 4 and 3x + 2 > 4

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  1. Annetta_Martin
    • one year ago
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    Is it 3x + 2 < 4 and 3x + 2 < -4 ? (The first choice)

  2. snowman97
    • one year ago
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    need help

  3. Annetta_Martin
    • one year ago
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    Yes

  4. snowman97
    • one year ago
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    have answer choices

  5. Annetta_Martin
    • one year ago
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    There up there with the question

  6. Annetta_Martin
    • one year ago
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    They're

  7. Peter14
    • one year ago
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    it's not the first choice. see how the inequality signs point the same way? the inequality signs should point in opposite directions.

  8. Peter14
    • one year ago
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    do you see why?

  9. Annetta_Martin
    • one year ago
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    Yes, I do

  10. Annetta_Martin
    • one year ago
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    So then it'd be the second choice?

  11. Peter14
    • one year ago
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    yes

  12. Peter14
    • one year ago
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    an easy way to do this problem if you have a graphing calculator is to graph y=abs(3x-2)-4 (solve your equation for zero, then put y in instead of the zero)

  13. Annetta_Martin
    • one year ago
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    Ok, I'll try that!

  14. Annetta_Martin
    • one year ago
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    Could you help me with one more question?

  15. Peter14
    • one year ago
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    sure

  16. Peter14
    • one year ago
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    but I have a question for you afterward

  17. Annetta_Martin
    • one year ago
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    Ok!

  18. Annetta_Martin
    • one year ago
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    Which domain number causes this relation not to be a function? R = { (-1, -1), (2, 0), (2, 1), (3, 1), (4, 4) } -1 zero 1 2

  19. Peter14
    • one year ago
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    how much do you already know about what is and is not a function?

  20. Peter14
    • one year ago
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    a relation is not a function if it has two y-values for one x-value.

  21. Annetta_Martin
    • one year ago
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    I thought it was 1?

  22. Peter14
    • one year ago
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    domain is x-values which x-value appears twice and has two different y-values?

  23. Annetta_Martin
    • one year ago
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    2

  24. Peter14
    • one year ago
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    yes

  25. Annetta_Martin
    • one year ago
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    Ok, your question?

  26. Peter14
    • one year ago
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    I pm'd it to you

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