Math plz help

2. | y+4| <1 (1 point)

-5<y<-3
-3<y<5
-4<y<1
1<y<4

3. |2r| -5 =7. (1 point)

T=1 or -1
T=-6 or 6
T= 10 or -10
T= 12 or -12

4. |A| -3/4 = -5/8

1/8 or -1/8
7/8 or -7/8
1 3/8 or -1 3/8
No solution

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  1. I don't see your choices, so how can we check?

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    posted by Reiny
  2. My answers
    2.D
    3.also D
    4.A

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    posted by Gloony90
  3. clearly not 2D. Just plug in y=2
    |2+4| = |6| = 6 Bzzzt

    #3 is also wrong. You forgot to divide by 2.
    |2r| = 12, not |r| = 12

    #4 is correct

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    posted by Steve
  4. The last one is correct, the first two are wrong.

    Show me your steps so I can tell where you are going wrong.

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    posted by Reiny
  5. Just number 2 I'm not even understanding where to start

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    posted by Gloony90
  6. #2 is the main one i need help with

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    posted by Gloony90
  7. Why are you ignoring Reiny's advice?

  8. I said i don't even know where to start with it

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    posted by Gloony90
  9. #2
    | y+4| <1
    Use your definition:
    y+4 < 1 AND -y-4 < 1
    y < -3 AND -y < 5
    y < -3 AND y > -5

    or as they wrote it, -5 < y < -3

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    posted by Reiny
  10. #2. Remember that
    |x| = x if x >= 0
    |x| = -x if x < 0

    So, you have
    |y+4| < 1
    so, either y+4 >= 0, meaning y >= -4
    or y+4 < 0, meaning y < -4

    If y >= -4
    y+4 < 1
    y < -3

    If y < -4
    -(y+4) < 1
    -y-4 < 1
    -5 < y
    But also, we started with y < -4
    So the solution is -5 < y < -3

    Or, think of what |y+4| means. If y = -4, then y+4=0. So, |y+4| is the distance from y to -4. If that distance is less than 1, then you see the solution. Plot the value -4 on the number line. y must be less than 1 unit away from -4.

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    posted by Steve

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