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NatalieandKayla Group Title

solve the inequality. graph your solution. |x+3| > or equal to 15

  • one year ago
  • one year ago

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  1. jim_thompson5910 Group Title
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    Hint: |x| > k breaks into x < -k or x > k where k is some positive number

    • one year ago
  2. NatalieandKayla Group Title
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    is it still correct if the arrows go in the same direction when I graph it?

    • one year ago
  3. jim_thompson5910 Group Title
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    no the arrows should be going in opposite directions

    • one year ago
  4. NatalieandKayla Group Title
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    oh so you flip the sign when you make k a negative?

    • one year ago
  5. jim_thompson5910 Group Title
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    exactly

    • one year ago
  6. jim_thompson5910 Group Title
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    what did you get

    • one year ago
  7. NatalieandKayla Group Title
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    x> or equal to 12 and x < or equal to -18

    • one year ago
  8. jim_thompson5910 Group Title
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    good, you got it

    • one year ago
  9. NatalieandKayla Group Title
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    thanks :D

    • one year ago
  10. jim_thompson5910 Group Title
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    you're welcome

    • one year ago
  11. NatalieandKayla Group Title
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    find the slope of the line passing through the given points. then tell whether the line rises, falls, is horizontal, or is vertical. (-7,4) (5,-3)

    • one year ago
  12. jim_thompson5910 Group Title
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    use the slope formula m = (y2 - y1)/(x2 - x1)

    • one year ago
  13. jim_thompson5910 Group Title
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    in this case (x1,y1) = (-7,4) (x2,y2) = (5,-3)

    • one year ago
  14. NatalieandKayla Group Title
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    how do I find out whether the line rises, falls..etc ?

    • one year ago
  15. jim_thompson5910 Group Title
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    what did you get for the slope

    • one year ago
  16. NatalieandKayla Group Title
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    -7/12

    • one year ago
  17. NatalieandKayla Group Title
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    oh so the line decreases bc it has a negative slope?

    • one year ago
  18. jim_thompson5910 Group Title
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    you nailed it

    • one year ago
  19. NatalieandKayla Group Title
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    Do you mind if i ask you questions from time to time ? I'm trying to study for a trig final on monday.

    • one year ago
  20. jim_thompson5910 Group Title
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    sure I can help

    • one year ago
  21. NatalieandKayla Group Title
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    thanks! alright so this question has the same guidelines as the one we just did but with these coordinates (-5/4, 3); (2/3, 3) my friend had an error in her calculator trying to find the slope , but I got m=0 ...im confused

    • one year ago
  22. jim_thompson5910 Group Title
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    you're correct, it's 0 how do I know so quickly? notice that the y coordinates are both equal to 3 since the y coordinates are equal, the slope will always be 0 because you have a horizontal line through y = 3

    • one year ago
  23. jim_thompson5910 Group Title
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    all horizontal lines have a slope of 0

    • one year ago
  24. jim_thompson5910 Group Title
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    using m = (y2 - y1)/(x2 - x1) we get m = (y2 - y1)/(x2 - x1) m = (3 - 3)/(2/3 - (-5/4)) m = 0/(2/3 - (-5/4)) m = 0

    • one year ago
  25. NatalieandKayla Group Title
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    oh ok i understand

    • one year ago
  26. NatalieandKayla Group Title
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    what if the x coordinates are equal? would that make the line vertical?

    • one year ago
  27. jim_thompson5910 Group Title
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    exactly

    • one year ago
  28. jim_thompson5910 Group Title
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    and the slope would be undefined

    • one year ago
  29. NatalieandKayla Group Title
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    oh okay

    • one year ago
  30. NatalieandKayla Group Title
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    how would i graph y=|x-3|?

    • one year ago
  31. jim_thompson5910 Group Title
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    do you know how to graph y = | x |

    • one year ago
  32. NatalieandKayla Group Title
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    i think so? would there be 2 lines?

    • one year ago
  33. jim_thompson5910 Group Title
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    yes here is your blank xy axis |dw:1359244103163:dw|

    • one year ago
  34. jim_thompson5910 Group Title
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    the graph of y = | x | looks like this |dw:1359244135745:dw|

    • one year ago
  35. jim_thompson5910 Group Title
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    to graph y=|x-3|, you just shift over y = |x| three units to the right

    • one year ago
  36. NatalieandKayla Group Title
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    im not sure i get you there...

    • one year ago
  37. jim_thompson5910 Group Title
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    imagine you can pick up that v shaped graph and move it around well just move it over 3 to the right units like this |dw:1359244315265:dw|

    • one year ago
  38. NatalieandKayla Group Title
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    but how do i know the coordinates to create the v shape

    • one year ago
  39. NatalieandKayla Group Title
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    i understand the moving it to the right part

    • one year ago
  40. jim_thompson5910 Group Title
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    the sharp point on the V of y = |x| is at (0,0) move that point over 3 units to the right to get (3,0)

    • one year ago
  41. jim_thompson5910 Group Title
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    |dw:1359244469161:dw|

    • one year ago
  42. NatalieandKayla Group Title
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    yes i understand that but what are the other coordinates for that shape?

    • one year ago
  43. NatalieandKayla Group Title
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    for (3,0)..

    • one year ago
  44. jim_thompson5910 Group Title
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    (1,1) is on the original graph y = |x| move that 3 units to the right to get (4,1)

    • one year ago
  45. jim_thompson5910 Group Title
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    |dw:1359244638800:dw|

    • one year ago
  46. jim_thompson5910 Group Title
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    (-1,1) is on the original graph y = |x| move that 3 units to the right to get (2,1)

    • one year ago
  47. jim_thompson5910 Group Title
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    |dw:1359244675721:dw|

    • one year ago
  48. NatalieandKayla Group Title
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    ohhhh so the slope for it is rise 1 over 1?

    • one year ago
  49. jim_thompson5910 Group Title
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    for the right half of it, yes

    • one year ago
  50. jim_thompson5910 Group Title
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    |dw:1359244723365:dw|

    • one year ago
  51. jim_thompson5910 Group Title
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    left half, slope is -1 |dw:1359244745927:dw|

    • one year ago
  52. NatalieandKayla Group Title
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    ohhh ok so on the left side for any problem i will always decrease 1 and rise 1?

    • one year ago
  53. NatalieandKayla Group Title
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    im trying to understand the concept so i can graph any absolute value problem

    • one year ago
  54. jim_thompson5910 Group Title
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    on the left side, you go down 1 and over to the right 1 on the right side, you go up 1 and over to the right 1

    • one year ago
  55. NatalieandKayla Group Title
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    the problem they gave me was y=|x-3| would you have to move to graph from (0,0) to (0,-3) ? you moved it up 3

    • one year ago
  56. jim_thompson5910 Group Title
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    if it was |x| - 3, then you would move (0,0) to (0,-3)

    • one year ago
  57. jim_thompson5910 Group Title
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    but it's actually |x-3|

    • one year ago
  58. NatalieandKayla Group Title
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    oh

    • one year ago
  59. NatalieandKayla Group Title
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    would i be able to just plug this in to my calculator ?

    • one year ago
  60. jim_thompson5910 Group Title
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    yes you should be able to

    • one year ago
  61. jim_thompson5910 Group Title
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    do you know how to graph absolute value equations?

    • one year ago
  62. NatalieandKayla Group Title
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    I guess I forgot :/

    • one year ago
  63. jim_thompson5910 Group Title
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    what kind of calculator do you have

    • one year ago
  64. NatalieandKayla Group Title
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    i just found the absolute value button on my TI-83

    • one year ago
  65. NatalieandKayla Group Title
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    and i put it into the y= and i got what you showed me earlier

    • one year ago
  66. jim_thompson5910 Group Title
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    ok great, glad you did

    • one year ago
  67. NatalieandKayla Group Title
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    Solve the equation. \[4x ^{2}-6x+9=0\]

    • one year ago
  68. jim_thompson5910 Group Title
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    you can either factor or use the quadratic formula

    • one year ago
  69. jim_thompson5910 Group Title
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    oh wait, one sec

    • one year ago
  70. jim_thompson5910 Group Title
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    no you can't factor, so you'll have to use the quadratic formula

    • one year ago
  71. NatalieandKayla Group Title
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    how do you know you cant factor this problem?

    • one year ago
  72. jim_thompson5910 Group Title
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    D = b^2 - 4ac D = (-6)^2 - 4(4)(9) D = 36 - 4(4)(9) D = 36 - 144 D = -108 because the discriminant is negative, there are no real solutions so that's why you can't factor

    • one year ago
  73. NatalieandKayla Group Title
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    oh okay, and i can use the discriminant as a test on any problem that looks like this one?

    • one year ago
  74. jim_thompson5910 Group Title
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    yes if the discriminant is a) positive or zero and b) a perfect square then the quadratic can be factored

    • one year ago
  75. NatalieandKayla Group Title
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    ohhhhh okay

    • one year ago
  76. NatalieandKayla Group Title
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    solve the system of linear equations using any method. 2x+3y=47 7x-8y=-2 2x-y+3z=-19

    • one year ago
  77. jim_thompson5910 Group Title
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    solve the first two equations for x and y once you have x and y, use them to find z

    • one year ago
  78. NatalieandKayla Group Title
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    im not sure how i ended up with 111 for y... the answers say its wrong

    • one year ago
  79. jim_thompson5910 Group Title
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    no y should be 9, try it again

    • one year ago
  80. NatalieandKayla Group Title
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    i did elimination with the first two equations

    • one year ago
  81. NatalieandKayla Group Title
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    to get them to equal to 14 and -14 so they can cancel out but i keep getting 111

    • one year ago
  82. jim_thompson5910 Group Title
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    2x+3y=47 7x-8y=-2 8(2x+3y)=8*47 3(7x-8y)=3(-2) 16x+24y=376 21x-24y=-6 ------------- 72x+0y = 370 37x = 370 x = 370/37 x = 10 2x+3y=47 2(10)+3y=47 20+3y = 47 3y = 47-20 3y = 27 y = 27/3 y = 9 So x = 10 and y = 9 Use them to find z

    • one year ago
  83. NatalieandKayla Group Title
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    Oh! I understand now, thank you so much :)

    • one year ago
  84. jim_thompson5910 Group Title
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    you're welcome

    • one year ago
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