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anonymous

  • one year ago

Does anyone know how to add and subtract rational expressions? help very last assignment D:

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  1. anonymous
    • one year ago
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    Its an assignment I have to do, but its all about adding and subtracting rational expression, only 5 problems

  2. Nurali
    • one year ago
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    Hope this will help

  3. anonymous
    • one year ago
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    Ok c: Thanks

  4. Nurali
    • one year ago
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    My Pleasure.

  5. anonymous
    • one year ago
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    is it ok if i do one problem and show you to make sure im not doing anything wrong?

  6. anonymous
    • one year ago
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    @Nurali

  7. UsukiDoll
    • one year ago
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    post an example

  8. UsukiDoll
    • one year ago
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    I think the other user is afk or something

  9. anonymous
    • one year ago
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    ill draw a picture of it

  10. anonymous
    • one year ago
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    |dw:1433843733453:dw|

  11. UsukiDoll
    • one year ago
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    hmmm... they share a common denominator so all we have to do is to combine like terms in the numerator

  12. anonymous
    • one year ago
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    How would we do that? would we add 2x to x and 1 to the other 1?

  13. UsukiDoll
    • one year ago
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    yeah

  14. anonymous
    • one year ago
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    would the numerator come out like this = 2x^2+2 or 3x+2

  15. UsukiDoll
    • one year ago
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    it would be 3x+2. because 2x+x = 3x 1 + 1 =2

  16. anonymous
    • one year ago
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    Ok! thanks, can you help with four more adding/subracting q's?

  17. UsukiDoll
    • one year ago
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    sure

  18. anonymous
    • one year ago
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    Thankyou so much! so the next question basically wants me to use the answer we just got (3x+2/2x) subtracted by x/x+3, but it also says to * identify any possible restrictions that exist with (or in) the resulting rational expression*

  19. UsukiDoll
    • one year ago
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    \[\frac{3x+2}{2x}-\frac{x}{x+3}\] it said any possible restrictions... so I"m not sure what that means 100% maybe what value makes the result undefined ?>! Not sure.. any way our lcd = (2x)(x+3) so we multiply x+3 on the first fraction and multiply 2x on the second fraction

  20. anonymous
    • one year ago
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    so like cross multiplying?

  21. UsukiDoll
    • one year ago
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    umm... similar but instead we need to find the least common denominator see example 3 on the cliffnotes site... noticed how they took the lcd of that problem?

  22. anonymous
    • one year ago
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    It looks kinda confusing sine the problem big :x

  23. anonymous
    • one year ago
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    on cliff notes

  24. UsukiDoll
    • one year ago
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    \[\frac{3x+2}{2x} \times \frac{x+3}{x+3}-\frac{x}{x+3} \times \frac{2x}{2x}\]

  25. anonymous
    • one year ago
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    ok that makes alot more since lol

  26. UsukiDoll
    • one year ago
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    example 3 on cliffnotes from Nurali was just x and y as the denominators. it was the later examples that I didn't choose because I knew it would be too much

  27. anonymous
    • one year ago
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    would we just times the bottom demonitaers togethor? like 2x(x+3)

  28. UsukiDoll
    • one year ago
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    yes...

  29. UsukiDoll
    • one year ago
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    \[\frac{(3x+2)(x+3)-2x^2}{2x(x+3)}\] we have to remember that whenever we have a - sign on the right side of the equation, we need to distribute that sign all over

  30. UsukiDoll
    • one year ago
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    \[-x(2x) \rightarrow -2x^2\]

  31. UsukiDoll
    • one year ago
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    and we need to use FOIL method to expand the left side

  32. anonymous
    • one year ago
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    I know how to do that cx ill show you after im done foiling it c:

  33. UsukiDoll
    • one year ago
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    ok

  34. anonymous
    • one year ago
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    would we do the distributive property to the denomater? or does that stay the same

  35. UsukiDoll
    • one year ago
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    umm I think it depends on the numerator... because we wouldn't want to combine the denominator and end up that we can cancel something out...taking a step backwards

  36. anonymous
    • one year ago
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    i foiled the top but didn't add up yet, is this correct so far? 4x+9x+2x+6

  37. UsukiDoll
    • one year ago
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    \[(3x+2)(x+3) -2x^2 = 3x^2+9x+2x+6 = 3x^2+11x+6 -2x^2 = x^2+11x+6\]

  38. UsukiDoll
    • one year ago
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    remember \[(x)(x) \rightarrow (x^1)(x^1) \rightarrow x^{1+1} \rightarrow x^2\]

  39. UsukiDoll
    • one year ago
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    so... \[(3x)(x) \rightarrow 3x^{1+1} \rightarrow 3x^2\]

  40. anonymous
    • one year ago
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    Ohhh ok, so now i have x^2+11x-6/2x(x+3)

  41. anonymous
    • one year ago
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    would we cancel our the 6 from 3?

  42. UsukiDoll
    • one year ago
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    no

  43. UsukiDoll
    • one year ago
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    I don't think this can be factored... the only pairs for 6 is 6 x 1 1 x 6 2 x 3 3 x2 that's 6+1 =7 3+2 = 5 and subtracting will make it worse

  44. anonymous
    • one year ago
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    maybe thats the restrictions they talk about?

  45. anonymous
    • one year ago
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    i think

  46. UsukiDoll
    • one year ago
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    \[\frac{x^2+11x-6}{2x(x+3)}\] maybe I wish it was more specific.. there is also what values of x makes this undefined too

  47. UsukiDoll
    • one year ago
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    2x(x+3) = 0 split into 2 problems 2x=0 x+3=0 solving x ... X = 0,-3

  48. anonymous
    • one year ago
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    those are probably the restrictions Cx , i think it does mean undifined

  49. anonymous
    • one year ago
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    ok so the next question wants me to do this |dw:1433845443996:dw|

  50. UsukiDoll
    • one year ago
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    \[\frac{x^2+11x-6}{2x(x+3)} \times \frac{x+3}{x+1}\]

  51. anonymous
    • one year ago
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    its x-1 cx

  52. UsukiDoll
    • one year ago
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    we can cancel out the x+3... it's in the numerator and denominator \[\frac{x^2+11x-6}{2x(x+3)} \times \frac{x+3}{x-1}\]

  53. anonymous
    • one year ago
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    oh true! so they would disapear?

  54. UsukiDoll
    • one year ago
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    YUP! :D

  55. anonymous
    • one year ago
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    yes! the easiest step on here lol cx so then the problem would be x^2+11-6/2 times x-1

  56. anonymous
    • one year ago
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    *divided by 2x not 2

  57. anonymous
    • one year ago
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    how would that work? since the second one isnt a denominater nor a numerator?

  58. UsukiDoll
    • one year ago
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    \[\frac{x^2+11x-6}{2x} \times \frac{1}{x-1} \rightarrow \frac{x^2+11x-6}{2x(x-1)}\]

  59. UsukiDoll
    • one year ago
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    huh?! ok wait I just latex the fractions without x+3 because the numerator had one and the denominator had one .

  60. anonymous
    • one year ago
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    ohhh ok i remebr learning that step now, so it would stay like that? :o

  61. anonymous
    • one year ago
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    what happened :o

  62. UsukiDoll
    • one year ago
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    what is the next question?

  63. anonymous
    • one year ago
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    you said they cancled eachother out cx, and now i have to divide that equation by 6/x-1, but it also says to (Discuss why the degree of the resulting denominator did not change from your expression’s degree)

  64. UsukiDoll
    • one year ago
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    \[\frac{x^2+11x-6}{2x(x-1)} \] divided by \[\frac{6}{x-1}\]

  65. anonymous
    • one year ago
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    yea

  66. UsukiDoll
    • one year ago
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    if you flip the second fraction... surprise x-1 gets canceled out too.

  67. UsukiDoll
    • one year ago
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    \[\frac{x^2+11x-6}{2x(x-1)} \times \frac{x-1}{6}\]

  68. anonymous
    • one year ago
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    ohh and it turns into a mutplication by flipping?

  69. UsukiDoll
    • one year ago
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    when dividing fractions... you flip the second fraction... in the end you're multiplying

  70. anonymous
    • one year ago
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    Oh ok, that makes it easier cx

  71. UsukiDoll
    • one year ago
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    \[\frac{x^2+11x-6}{2x} \times \frac{1}{6} \rightarrow \frac{x^2+11x-6}{12x}\]

  72. UsukiDoll
    • one year ago
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    I think it's safe now to combine the denominator...

  73. anonymous
    • one year ago
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    oh yeah i keep forget that the single number would be on bottom and the 1 on top, and yea it does xD

  74. anonymous
    • one year ago
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    so then i would have to explain *Discuss why the degree of the resulting denominator did not change from your expression’s degree*

  75. anonymous
    • one year ago
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    what the flip does it mean by that lol :x

  76. UsukiDoll
    • one year ago
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    what is that?!

  77. UsukiDoll
    • one year ago
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    12x that's first degree isn't it?! awww c*** I haven't dealt with these questions in a while. pure math doesn't have this

  78. anonymous
    • one year ago
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    well we got the retrictions Q , and i just wanna finish this one assignemnt thingy lol, you look like your doing it right! cx its just these wierd Q's

  79. UsukiDoll
    • one year ago
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    how did it not change from the expression's degree? could it be referring to the original question?!

  80. UsukiDoll
    • one year ago
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    the first question you had ... contained 2x our fourth question had the denominator 2x(6) -> 12x I think they are the same first degree. That's just a guess ..

  81. anonymous
    • one year ago
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    maybe they mean degree by like * x, x^2, x^3

  82. UsukiDoll
    • one year ago
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    yeah I'm thinking that we compare the last answer we have to the first question? hmmmmmmmm?! the degrees are the same if we look at it that way?!

  83. anonymous
    • one year ago
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    haha yea, ill just say, i shouldnt get that many points taken off since were doing the rest right

  84. anonymous
    • one year ago
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    say that*

  85. anonymous
    • one year ago
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    Ok, so thats our final answer for Q 4, now we just have one left!

  86. UsukiDoll
    • one year ago
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    o-o I thought that was the final one?!

  87. anonymous
    • one year ago
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    no this one is the final one lol,

  88. anonymous
    • one year ago
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    i had 5 Q's

  89. anonymous
    • one year ago
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    4* i mean

  90. UsukiDoll
    • one year ago
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    ????????????

  91. anonymous
    • one year ago
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    its the very last one, please helpp :3

  92. UsukiDoll
    • one year ago
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    ok

  93. anonymous
    • one year ago
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    |dw:1433847285555:dw|

  94. anonymous
    • one year ago
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    its that one^

  95. UsukiDoll
    • one year ago
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    ok first combine the second and third fraction

  96. anonymous
    • one year ago
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    i meant subract 6

  97. anonymous
    • one year ago
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    not + and ok cx

  98. anonymous
    • one year ago
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    2x/2x+2 ?

  99. UsukiDoll
    • one year ago
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    |dw:1433847473742:dw|

  100. UsukiDoll
    • one year ago
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    x + x = 2x .. the denominator stays the same..... OMG EVEN I DID +6 #FAIL

  101. anonymous
    • one year ago
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    I|dw:1433847507052:dw|

  102. anonymous
    • one year ago
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    lol its, its late at night lol xD

  103. UsukiDoll
    • one year ago
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    \[\frac{x^2+11x-6}{12x} + \frac{2x}{x+1}\]

  104. UsukiDoll
    • one year ago
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    now use the least common denominator which is 12x(x+1)

  105. anonymous
    • one year ago
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    so we would do that whole timezing the first by the equation and the second by the second one?

  106. UsukiDoll
    • one year ago
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    but only multiply what is misssing for the first fraction we have 12x.. x+1 is missing so multiply x+1/x+1

  107. UsukiDoll
    • one year ago
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    \[\frac{x^2+11x-6}{12x} \times \frac{x+1}{x+1} + \frac{2x}{x+1} \times\frac{2x}{2x}\]

  108. UsukiDoll
    • one year ago
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    OOPS

  109. anonymous
    • one year ago
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    wait, so why did we mutiply 2x instead of 12?

  110. UsukiDoll
    • one year ago
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    cuz I typoed

  111. UsukiDoll
    • one year ago
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    \[\frac{x^2+11x-6}{12x} \times \frac{x+1}{x+1} + \frac{2x}{x+1} \times\frac{12x}{12x}\]

  112. anonymous
    • one year ago
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    ok lol, and for example on the first bottom ones we would do (12x)(x+1) ?

  113. UsukiDoll
    • one year ago
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    yeah

  114. UsukiDoll
    • one year ago
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    numerator on the left is going to be a bit tough though

  115. anonymous
    • one year ago
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    ok :x ill do the denomnaters and show u

  116. anonymous
    • one year ago
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    |dw:1433847901459:dw|

  117. UsukiDoll
    • one year ago
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    |dw:1433847957533:dw|

  118. UsukiDoll
    • one year ago
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    \[(x^2+11x-6)(x+1) = x^3+11x^2-6x+x^2+11x-6 = x^3+12x^2+5x-6\]

  119. UsukiDoll
    • one year ago
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    \[x^3+12x^2+5x-6+24x^2 = x^3+36x^2+5x-6\]

  120. UsukiDoll
    • one year ago
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    \[\frac{x^3+36x^2+5x-6}{12x(x+1)}\]

  121. anonymous
    • one year ago
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    is that the end point? xO

  122. anonymous
    • one year ago
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    or do we do stuff with the denomatr to the numerator now?

  123. UsukiDoll
    • one year ago
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    hmmm... I think that's it?!

  124. UsukiDoll
    • one year ago
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    we can't use factor by grouping with the numerator

  125. UsukiDoll
    • one year ago
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    not a perfect cube either

  126. anonymous
    • one year ago
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    yea, it should be lol i cant belive 4 problems took an hour :X, and thank you soOO much! i've asked this Q many times on here and noone has helped me like you have! and yea it doesnt look like a perfect cube

  127. UsukiDoll
    • one year ago
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    I think it was due to my latex writing too.. that takes a bit of time and thinking xD

  128. anonymous
    • one year ago
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    i dont think so cx these types of Q's look like theyd take forever lol, and we were moving pretty fast cx

  129. anonymous
    • one year ago
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    again thank you so muccccch! very appreciated that you helped me cx

  130. UsukiDoll
    • one year ago
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    :D

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