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mikaa_toxica13

16 < 3x + 1 < 4 Part 1: Solve the inequality above. Part 2: Describe the graph

  • one year ago
  • one year ago

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  1. fernando12
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    16 < 3x + 1 < 4 ==> This is not possible, if it's not a typo then: 15 < 3x < 3 5 < x < 1 ==> This is an impossible outcome, there is no solution. But if it is: 4 < 3x + 1 < 16 3 < 3x < 15 1 < x < 5 interval notation: (1 , 5) Edit: Well what do you know! I never thought I see the day that in a group of nine 55% not only actually believe that 4>16 and 1>5 but they also proved it and wrote it down for those 45% in the group and the rest of the world to see that 16 & 5 are no longer greater than 4 & 1. lol

    • one year ago
  2. satellite73
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    another county heard from

    • one year ago
  3. mikaa_toxica13
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    @satellite73 what ??

    • one year ago
  4. satellite73
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    unless you have a typo there is no solution to your question, because it would mean that \(16<4\) which is false

    • one year ago
  5. mikaa_toxica13
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    well then how about -7y - 17 > 11 Part 1: Solve the inequality above. Part 2: Describe the graph of the solution. i gotta do one or the other

    • one year ago
  6. mikaa_toxica13
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    @satellite73

    • one year ago
  7. lyzzy7
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    for -7y-17 > 11 add 17 on each side -7y > 28 then divide each side by -7 y > -4

    • one year ago
  8. lyzzy7
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    for the graph it is an open circle on -4 and the numbers to the right of four are greater so you shade in the line to the right.

    • one year ago
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