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anonymous

  • one year ago

Adding rational expressions. I keep getting these wrong. x/x^2+6x+9 + 7/x^2+8x+15

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  1. Nnesha
    • one year ago
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    \[\frac{ x }{ x^2+6x+7 } +\frac{ 7 }{ x^2+8x+15 }\] like this right ?

  2. anonymous
    • one year ago
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    that was dauntingly fast !

  3. anonymous
    • one year ago
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    x^2+6x+9 on the first one

  4. anonymous
    • one year ago
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    5(y+3)(y−1)/(y−3)2(y+1) my answer but I can't risk getting it wrong again.

  5. anonymous
    • one year ago
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    I've been doing something wrong with factoring on these. it's the correct answer but i'm assuming they want the full answer, not factored. but I have no clue

  6. Nnesha
    • one year ago
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    haha alright try to factor quadratic equations apply the AC (headphone method) find two number when you multiply them you should get product of AC and when you add or subtract them you should get middle term Ax^2+Bx+C=0 a=leading coefficient b=middle term c=constant ter

  7. Nnesha
    • one year ago
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    okay gimme a sec :)

  8. Nnesha
    • one year ago
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    so so what are the factors of x^2+6x+9 ??

  9. Nnesha
    • one year ago
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    so so what are the factors of x^2+8x+15??

  10. anonymous
    • one year ago
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    I have several of these to answer. will you help me with a few of them? I'm suppose to get 8 out of 10 correct and I was suppose to have all of this finished 22 minutes ago and I don't know if I submit this after i'm finish if it will even count. so i'm kinda panicked. haha I've done this test more than 10 times in the past week and still haven't figured out what i was doing wrong. decided to pay for owlbucks and have someone help me considering the deadline has already passed.

  11. Nnesha
    • one year ago
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    hmm okay good :)

  12. Nnesha
    • one year ago
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    do you know how to find factors quadratic equations ?

  13. Nnesha
    • one year ago
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    * of

  14. anonymous
    • one year ago
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    No, i don't

  15. Nnesha
    • one year ago
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    alright here is the easy method AC method (aka headphone method ) where you have to find two numbers when you multiply them you should get product of AC and when you add or subtract them you should get middle term |dw:1439411259059:dw| a= leading coefficient which is one b=middle term = 8 c=constant term = 15 so find two number when you multiply them you should get 15 but when you add or subtract them you should get 8

  16. anonymous
    • one year ago
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    I solved it with factoring and it was correct.

  17. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @21pmakesmehappy 5(y+3)(y−1)/(y−3)2(y+1) my answer but I can't risk getting it wrong again. \(\color{blue}{\text{End of Quote}}\) is it like this ? \[\huge\rm \frac{ 5(x+3)(x-1) }{ (x-3)^2(x+1) }\] variable is x not y

  18. anonymous
    • one year ago
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    I redid it and got something different

  19. anonymous
    • one year ago
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    I have another question now. Can we try this one?

  20. Nnesha
    • one year ago
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    alright! :) post

  21. Nnesha
    • one year ago
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    well answer for that one is incorrect we should try it again!

  22. anonymous
    • one year ago
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    2y/y^2 -2y+1 + 8/y^2+1y-2

  23. anonymous
    • one year ago
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    I got the previous question correct once i redid it.

  24. Nnesha
    • one year ago
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    ohh okay.

  25. Nnesha
    • one year ago
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    have you tried to solve the 2nd one ?

  26. anonymous
    • one year ago
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    yes. i'm getting 2(y^2+6y-4)/(y-1)^2(y+2)

  27. anonymous
    • one year ago
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    but I know it'll be wrong if I put it as my answer.

  28. anonymous
    • one year ago
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    is it not factored?

  29. Nnesha
    • one year ago
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    \[\huge\rm \frac{ 2y }{ y^2-2y+1 } +\frac{ 8 }{ y^2+y -2 }\] find factor of quadratic equation y^2-2y+1 = find two number when we multiply them should get 1 and when we add them we should get -2 so -1 times -1= 1 -1-1 = -2 perfect (x-1)(x-1) are the factors of y^2-2y+1 equation

  30. Nnesha
    • one year ago
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    it's factorable

  31. anonymous
    • one year ago
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    i think i need to put in the unfactored answer

  32. anonymous
    • one year ago
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    because anytime i put it in factored, it's wrong

  33. Nnesha
    • one year ago
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    1st) factor quadratic equation 2nd) find common denominator

  34. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha \[\huge\rm \frac{ 2y }{ y^2-2y+1 } +\frac{ 8 }{ y^2+y -2 }\] find factor of quadratic equation y^2-2y+1 = find two number when we multiply them should get 1 and when we add them we should get -2 so -1 times -1= 1 -1-1 = -2 perfect (x-1)(x-1) are the factors of y^2-2y+1 equation \(\color{blue}{\text{End of Quote}}\) correction: (y-1)(y-1)

  35. Nnesha
    • one year ago
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    \[\huge\rm \frac{ 2y }{ y^2-2y+1 } +\frac{ 8 }{ y^2+y -2 }\] \[\rm \frac{ 2y }{ \color{reD}{(y-1)}(y-1) } +\frac{ 8 }{ (y+2)\color{ReD}{(y-1)} }\] (y+2) (y-1) are factors of y^2+y-2 now what is the common denominator ?

  36. anonymous
    • one year ago
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    (y-1)^2(y+2)

  37. anonymous
    • one year ago
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    i have no clue what im doing

  38. Nnesha
    • one year ago
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    yep right \[\huge\rm \frac{ 2y(y+2) +8(y-1)}{ (y-1)^2(y+2) }\] multiply the first fraction with the bottom of 2nd fraction multiply the 2nd fraction with the bottom of first one and f you distribute parentheses by outside terms you will get \[\frac{ 2y^2+12y-8 }{ (y-1)^2(y+2) }\] which is same as ur answer

  39. anonymous
    • one year ago
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    it says it was incorrect.

  40. Nnesha
    • one year ago
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    \[\rm \frac{ 2y^2+12y-8 }{ (y-1)^2(y+2) } \] =\[\frac{ 2(y^2+6y-4) }{ (y-1)^2(y+2)}\] take out the GCF(greatest common factor

  41. anonymous
    • one year ago
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    4x/x^2+2x+1 + 3/x^2+3x+2 is the new question

  42. anonymous
    • one year ago
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    4x^2 +11x+3/(x+1)^2(x+2 is the answer im getting

  43. anonymous
    • one year ago
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    and i did not. i submitted my answer before you said to take out the gcf so it was incorrect

  44. anonymous
    • one year ago
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    1 is the gcf, right?

  45. anonymous
    • one year ago
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    im confused

  46. anonymous
    • one year ago
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    so do I take 2 out of my final answer?

  47. Nnesha
    • one year ago
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    no.. GCF is greatest common factor 2 is common in 2y^2 +12y -8 \[\color{ReD}{2} \times 1 \] \[1 \times \color{Red}{2} \times 3 \times 6 \times 12\] factors of 12 \[1 \times \color{ReD}{2} \times 4 \times 8\] factors of 8 2 is common factor of all 3 terms so yes take out 2

  48. anonymous
    • one year ago
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    4x^2+11x+3/(x+1)^2x

  49. anonymous
    • one year ago
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    is that then?

  50. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha no.. GCF is greatest common factor 2 is common in 2y^2 +12y -8 \[\color{ReD}{2} \times 1 \] \[1 \times \color{Red}{2} \times 3 \times 6 \times 12\] factors of 12 \[1 \times \color{ReD}{2} \times 4 \times 8\] factors of 8 2 is common factor of all 3 terms so yes take out 2 \(\color{blue}{\text{End of Quote}}\) this is for 2nd question!

  51. anonymous
    • one year ago
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    i am completely lost

  52. Nnesha
    • one year ago
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    please type the number front of the new question

  53. Nnesha
    • one year ago
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    like 2nd ) question 3rd) question

  54. anonymous
    • one year ago
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    i have no clue what question im on

  55. anonymous
    • one year ago
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    im giving up.

  56. anonymous
    • one year ago
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    never give up, take a breath get a drink

  57. anonymous
    • one year ago
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    thanks for your help, though. i appreciate it a lot.

  58. Nnesha
    • one year ago
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    relax! you are doing a great job!

  59. Nnesha
    • one year ago
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    alright i didn't give u a GREAT explanation so yeah i'll tag other users

  60. Nnesha
    • one year ago
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    let's see..

  61. anonymous
    • one year ago
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    @triciaal 's great

  62. Nnesha
    • one year ago
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    @Hero @triciaal @nincompoop <!!!

  63. anonymous
    • one year ago
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    is the problem factoring ?

  64. anonymous
    • one year ago
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    the only reason I have these answers is because I've done them each seriously 10 times today. I know how to solve these problems. I don't need an explanation as to how to solve these problems. i want to know why we match answers and when i put it into the system, it's wrong. I've gotten help from several people over the past week and i can't figure out why my answers are wrong. I appreciate the help but there is honestly no point in attempting this anymore. I've taken this test so many times and the deadline was over an hour ago. I could get all 10 correct and submit it and have it not even count. Your explanation was great and I genuinely appreciate the help. The problem is i get to my final answer, i enter it, it's wrong. I assume it has something to do with the factoring. someone suggested that I give "the full answer" "unfactored" but i have not one clue what that means.

  65. Nnesha
    • one year ago
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    i would like to see the screenshot of *multiple choice or answer box what is the statement ??

  66. anonymous
    • one year ago
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    okay. how can i enter a photo?

  67. Nnesha
    • one year ago
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    Press PrntScr to take a screenshot and do you see to blue *attach file* button to attach file ?

  68. anonymous
    • one year ago
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    I don't see an attach file button

  69. anonymous
    • one year ago
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    1 Attachment
  70. Nnesha
    • one year ago
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    http://prntscr.com/83xios

  71. Nnesha
    • one year ago
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    answer is not correct

  72. Nnesha
    • one year ago
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    show your work so i can loook over and find out some mistakes that would be great

  73. anonymous
    • one year ago
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    This is the last question and i've already gotten 3 wrong. i can go back and take the test after i finish (if im not locked out) so hold on.

  74. triciaal
    • one year ago
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    do you still need help?

  75. anonymous
    • one year ago
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    I got locked out. My account is no longer active.

  76. anonymous
    • one year ago
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    besides factoring the nominator i don't see anything wrong (i'm might be doing it wrong too lol) what exactly is wrong @Nnesha

  77. Nnesha
    • one year ago
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    okay like you said you already know how to solve so it's better to post the work then no one would waste your time

  78. Nnesha
    • one year ago
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    ik numerator isn't correct so that's mean answer is not correct

  79. anonymous
    • one year ago
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    well the numerator is correct (again i might be wrong ) just not factored

  80. triciaal
    • one year ago
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    |dw:1439414757668:dw|

  81. Nnesha
    • one year ago
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    alright let's see \[ \frac{ 6a }{ a^2+2a+1 } +\frac{ 7 }{ a^2+3a+2 }\]

  82. Nnesha
    • one year ago
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    ohhhhhh i see *facepalm* it's 2 not 7

  83. triciaal
    • one year ago
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    so your denominator is wrong

  84. Nnesha
    • one year ago
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    was looking at the wrong question.

  85. anonymous
    • one year ago
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    im pretty sure LCD is (a+2)(a+1)^2

  86. Nnesha
    • one year ago
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    ye answer is correct i'm sorry pma i was lookin at the wrong question.so i apologize

  87. anonymous
    • one year ago
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    how are you getting \[(a+1)(a+2)\] for LCD @trwatkins1 ?

  88. triciaal
    • one year ago
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    when one is a factor of the other you do not include

  89. triciaal
    • one year ago
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    |dw:1439415089426:dw|

  90. triciaal
    • one year ago
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    oops so sorry I forgot the square when I factored

  91. triciaal
    • one year ago
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    sorry yes the LCD = (a +1)^2*(a+2)

  92. Nnesha
    • one year ago
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    don't factor out the GCF *find the sum* try to submit your answer

  93. triciaal
    • one year ago
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    |dw:1439415305928:dw|

  94. triciaal
    • one year ago
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    |dw:1439415500982:dw|

  95. anonymous
    • one year ago
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    this is a horrible way of learning to factor

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