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anonymous

  • one year ago

Which of the following is a solution of x2 + 4x + 6? −2 + i times the square root of 2 2 + i times the square root of 2 −2 + three i times the square root of 2 −2 + three i times the square root of 2

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  1. anonymous
    • one year ago
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    you have a choice you can use the quadratic formula or you can complete the square you pick

  2. anonymous
    • one year ago
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    actually there is a third choice, namely to cheat

  3. anonymous
    • one year ago
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    how do i do that, any of it ?

  4. anonymous
    • one year ago
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    do you know the quadratic formula?

  5. anonymous
    • one year ago
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    i need a refresher.

  6. anonymous
    • one year ago
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    the solutions to \[ax^2+bx+c=0\] are \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] in your case \(a=1,b=4,c=6\) but also in your case completing the square is easier

  7. anonymous
    • one year ago
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    lets solve \[x^2+4x+6=0\] by completing the square ready?

  8. anonymous
    • one year ago
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    sorry i was working it out xD

  9. anonymous
    • one year ago
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    ok take your time

  10. anonymous
    • one year ago
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    so far I got |dw:1438546685018:dw|

  11. anonymous
    • one year ago
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    \[4^2-4\times 6=-8\]

  12. anonymous
    • one year ago
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    \[x=\frac{-4\pm\sqrt{-8}}{2}\]

  13. anonymous
    • one year ago
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    then \[\sqrt{-8}=\sqrt{4\times 2\times (-1)}=\sqrt{4}\sqrt{2}\sqrt{-1}=2\sqrt{2}i\]

  14. anonymous
    • one year ago
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    giving you \[\frac{-4\pm2\sqrt{2}i}{2}\] divide each term by \(2\) and get \[-2\pm\sqrt{2}i\]

  15. anonymous
    • one year ago
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    it would have been easier to complete the square there would have been no denominator

  16. anonymous
    • one year ago
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    is that supposed to be the answer ?

  17. anonymous
    • one year ago
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    lol yes it is "supposed to be" but in fact they wrote \[-2\pm i\sqrt{2}\]

  18. anonymous
    • one year ago
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    xD are you able to help me out with more ?

  19. anonymous
    • one year ago
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    sure

  20. anonymous
    • one year ago
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    Solve x2 + 4x + 1 = 0. x equals the quantity of 4 plus or minus square root 12 all over 2 x equals negative 4 plus or minus square root 3 x equals negative 2 plus or minus square root 3 x equals the quantity of 4 plus or minus square root 13 all over 2

  21. anonymous
    • one year ago
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    lets complete the square for this one, unless you want to use the formula again

  22. anonymous
    • one year ago
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    in fact i will just write down the steps for completing the square without skipping any so you can try one on your own

  23. anonymous
    • one year ago
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    \[x^2+4x=1=0\] subtract 1 from both sides get \[x^2+4x=-1\]

  24. anonymous
    • one year ago
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    alright

  25. anonymous
    • one year ago
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    then half of 4 is 2, and 2 squared is 4 go right to \[(x+2)^2=-1+4\] or \[(x+2)^2=3\]

  26. anonymous
    • one year ago
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    take the square root of both sides, don't forget the \(\pm\) get \[x+2=\pm\sqrt3\]

  27. anonymous
    • one year ago
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    subtract 2 from both sides to solve for x, get \[x=-2\pm\sqrt3\] done

  28. anonymous
    • one year ago
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    looks like option C in your choices

  29. anonymous
    • one year ago
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    xD I thought I wa supposed to do it lol

  30. anonymous
    • one year ago
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    Solve 6 over x minus 4 equals 4 over x for x and determine if the solution is extraneous or not. x = -8, extraneous x = -8, non-extraneous x = 8, extraneous x = 8, non-extraneous what about this one xD

  31. anonymous
    • one year ago
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    this is not a quadratic i guess

  32. anonymous
    • one year ago
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    \[\frac{x}{x-4}=\frac{4}{x}\] right?

  33. anonymous
    • one year ago
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    no that is not right \[\frac{6}{x-4}=\frac{4}{x}\] hows that?

  34. anonymous
    • one year ago
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    the x in the numerator is a 6

  35. anonymous
    • one year ago
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    as the math teacher say "cross multiply" and get \[6x=4(x-4)\] which is a linear equation can you solve that one?

  36. anonymous
    • one year ago
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    yea let me try xD

  37. anonymous
    • one year ago
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    i will wait

  38. anonymous
    • one year ago
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    x=-8?

  39. anonymous
    • one year ago
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    yes

  40. anonymous
    • one year ago
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    would it be extraneous?

  41. anonymous
    • one year ago
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    how do i tell if its extraneous or non-extraneous?

  42. anonymous
    • one year ago
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    if it does not make the denominator zero, then it is not extraneous or put another way, if it does make the denominator zero, then it is extraneous

  43. anonymous
    • one year ago
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    in this case it does not, so it is not extraneous

  44. anonymous
    • one year ago
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    oh now i get it

  45. anonymous
    • one year ago
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    \[\frac{6}{x-4}=\frac{4}{x}\] sustitute \(x=-8\) get \[\frac{6}{-8-4}=\frac{4}{-8}\] or \[-2=-2\] a perfectly good solution

  46. anonymous
    • one year ago
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    well actually \[-\frac{1}{2}=-\frac{1}{2}\] but you get the idea

  47. anonymous
    • one year ago
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    you up for some more xD? Or you tired out?

  48. anonymous
    • one year ago
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    go ahead

  49. anonymous
    • one year ago
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    What is the axis of symmetry for f(x) = 2x2 + 8x + 8? x = −2 x = −4 x = 4 x = 2

  50. anonymous
    • one year ago
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    oops axis of symmetry of \(y=ax^2+bx+c\) is \[-\frac{b}{2a}\]

  51. anonymous
    • one year ago
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    in your case \(a=2,b=8\)

  52. anonymous
    • one year ago
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    let me know when you get \[x=-2\]

  53. anonymous
    • one year ago
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    what is c?

  54. anonymous
    • one year ago
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    nothing?

  55. anonymous
    • one year ago
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    c in your case is 8 but it does not figure in to the axis of symmetry it is just \[x=-\frac{b}{2a}\]

  56. anonymous
    • one year ago
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    so 2?

  57. anonymous
    • one year ago
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    no there is a minus sign in the formula

  58. anonymous
    • one year ago
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    oh so x=-2?

  59. anonymous
    • one year ago
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    \[x=-\frac{8}{2\times 2}\\ x=-2\]

  60. anonymous
    • one year ago
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    What is the standard form equation of the line shown below? −3x + 2y = 5 3x − 2y = −5 y − 4 = three halves(x − 1) y = three halvesx + five halves

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  61. anonymous
    • one year ago
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    you need the slope first you know how to find it?

  62. anonymous
    • one year ago
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    y=mx+b?

  63. anonymous
    • one year ago
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    that is the "slope intercept" form not standard form but in any case you need \(m\) (the slope) first

  64. anonymous
    • one year ago
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    what is the standard form again ?

  65. anonymous
    • one year ago
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    standard form will be one of your first two choices \[ax+by=c \]

  66. anonymous
    • one year ago
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    im not understanding how i would use it

  67. anonymous
    • one year ago
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    in fact we don't have to do any work at all for this one since only choice B is in standard form just pick that one and move on

  68. anonymous
    • one year ago
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    Simplify open parentheses x to the 2 fifths power close parentheses to the 5 sixths power. x to the 37 over 30 power x to the 13 over 30 power x to the 10 elevenths power x to the 1 third power

  69. anonymous
    • one year ago
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    i really appreciate you helping me <3

  70. anonymous
    • one year ago
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    \[\huge (b^m)^n=b^{m\times n}\] is what you need for this one, i.e.multiply the exponents

  71. anonymous
    • one year ago
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    yw

  72. anonymous
    • one year ago
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    so what would the b be?

  73. anonymous
    • one year ago
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    multiply \[\frac{2}{5}\times \frac{5}{6}\]

  74. anonymous
    • one year ago
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    in this case \(b\) is \(2\)

  75. anonymous
    • one year ago
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    do i cross multiply?

  76. anonymous
    • one year ago
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    \[\huge (2^{\frac{2}{5}})^{\frac{5}{6}}\] \[\huge 2^{\frac{2}{5}\times \frac{5}{6}}\] hell no

  77. anonymous
    • one year ago
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    multiply just means multiply

  78. anonymous
    • one year ago
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    so\[\frac{ 10 }{ 30 }\]

  79. anonymous
    • one year ago
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    \[\frac{2}{5}\times \frac{5}{6}\] we can cancel stuff too

  80. anonymous
    • one year ago
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    the five

  81. anonymous
    • one year ago
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    \[\frac{2}{\cancel{5}}\times \frac{\cancel{5}}{6}=\frac{2}{6}\] is a start

  82. anonymous
    • one year ago
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    then you can cancel the 2 as well

  83. anonymous
    • one year ago
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    or you could cancel here \[\frac{1\cancel{0}}{3\cancel{0}}=\frac{1}{3}\]

  84. anonymous
    • one year ago
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    either way you get \(\frac{1}{3}\) and also you do NOT cross multiply

  85. anonymous
    • one year ago
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    ok gotcha cx makes more sense now

  86. anonymous
    • one year ago
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    whew final answer \[\huge 2^{\frac{1}{3}}\]

  87. anonymous
    • one year ago
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    What is the simplified form of the quantity 12 z-squared minus 7z minus 12 over the quantity 3 z-squared plus 2z minus 8 ? 4z plus 3 over z plus 2 4z minus 3 over z plus 2 4z plus 3 over z minus 2 4z minus 3 over z minus 2

  88. anonymous
    • one year ago
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    lol

  89. anonymous
    • one year ago
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    wait wouldnt it be x1/3

  90. anonymous
    • one year ago
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    \[\frac{12z^2-7z-12}{3z^2+2z-8}\] factor and cancel

  91. anonymous
    • one year ago
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    hold on let me go back and look

  92. anonymous
    • one year ago
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    yeah sorry \[\huge x^{\frac{1}{3}}\] the base was x, not 2

  93. anonymous
    • one year ago
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    ahh gotcha

  94. anonymous
    • one year ago
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    for the last one you have to factor and cancel i suck at factoring, but i am good at cheating http://www.wolframalpha.com/input/?i= \frac{12z^2-7z-12}{3z^2%2B2z-8}

  95. anonymous
    • one year ago
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    i dont understand how to use that xD

  96. anonymous
    • one year ago
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    this will work better http://www.wolframalpha.com/input/?i=%2812z^2-7z-12%29%2F%283z^2%2B2z-8%29 apparently it is \[\frac{4z+3}{z+2}\]

  97. anonymous
    • one year ago
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    A?

  98. anonymous
    • one year ago
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    click on the link to the second one and see how i wrote it in wolfram

  99. anonymous
    • one year ago
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    yeah A

  100. anonymous
    • one year ago
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    questions are all over the map must be an on line course ...

  101. anonymous
    • one year ago
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    What is the simplified form of the seventh root of x times the seventh root of x times the seventh root of x times the seventh root of x? this one would be \[\sqrt[7]{4x}\] right?

  102. anonymous
    • one year ago
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    no

  103. anonymous
    • one year ago
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    yea it is XD

  104. anonymous
    • one year ago
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    and oh i thought it was

  105. anonymous
    • one year ago
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    one answer is \[x^{\frac{4}{7}}\]

  106. anonymous
    • one year ago
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    another is \[\sqrt[7]{x^4}\]

  107. anonymous
    • one year ago
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    the power is four, not the coefficient

  108. anonymous
    • one year ago
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    \[x\times x\times x\times x=x^4\]not \(4x\)

  109. anonymous
    • one year ago
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    4 times the seventh root of x x to the four-sevenths power the seventh root of 4x x to the seven fourths power

  110. anonymous
    • one year ago
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    thes are the answers

  111. anonymous
    • one year ago
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    \[\huge \sqrt[7]{x^4}=x^{\frac{4}{7}}\]

  112. anonymous
    • one year ago
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    make that a B

  113. anonymous
    • one year ago
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    Which of the following represents the zeros of f(x) = 6x3 + 25x2 - 24x + 5? -5, one third, one half 5, -one third, one half 5, one third, - one half 5, one third, one half

  114. anonymous
    • one year ago
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    cheat like crazy for this one

  115. anonymous
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    http://www.wolframalpha.com/input/?i=+6x3+%2B+25x2+-+24x+%2B+5%3D0

  116. anonymous
    • one year ago
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    \[\{-5,\frac{1}{2},\frac{1}{3}\}\]

  117. anonymous
    • one year ago
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    so A?

  118. anonymous
    • one year ago
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    otherwise you have to figure out that it factors as \[(3 x-1) (2 x-1) (x+5)\]

  119. anonymous
    • one year ago
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    yeah A

  120. anonymous
    • one year ago
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    Use the graph below for this question: What is the average rate of change from x = −1 to x = −2?

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  121. anonymous
    • one year ago
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    \[5-7=-2\]

  122. anonymous
    • one year ago
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    although the question is confusing since it says "from -1 to -2" which is backwards

  123. anonymous
    • one year ago
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    maybe they want you to say 2, i am not clear on that the question is ill posed

  124. anonymous
    • one year ago
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    not uncommon for an on line class btw

  125. anonymous
    • one year ago
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    sorry i was talking to my brother

  126. anonymous
    • one year ago
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    i gotta run now good luck with the rest keep posting in a new thread, i am sure you will get lots of answers

  127. anonymous
    • one year ago
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    thank you so much!!!

  128. anonymous
    • one year ago
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    yw any time!

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