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andriod09 Group Title

Algebra II question, a fraction on top of a fraction, please help me!

  • 2 years ago
  • 2 years ago

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  1. cpattison Group Title
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    a fraction on top of a fraction can be rewritten as say (2/3)/(3/4) then you can flip the second one because to divide you multiply by the reciprocal (2/3)*(4/3) then you multiply across the fractions (2*4)/(3*3).

    • 2 years ago
  2. andriod09 Group Title
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    \[\huge\frac{\frac{30x^2+27x+6}{9x^2-49}}{\frac{30x+15}{30x-70}}\] That is the fraction.

    • 2 years ago
  3. andriod09 Group Title
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    @cpattison

    • 2 years ago
  4. andriod09 Group Title
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    @ganeshie8 @Hero

    • 2 years ago
  5. cpattison Group Title
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    oh okay

    • 2 years ago
  6. cpattison Group Title
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    one second!

    • 2 years ago
  7. Hero Group Title
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    Put a divide sign between both of them, then put them next to each other. You'll know what to do then.

    • 2 years ago
  8. Hero Group Title
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    \[\huge\frac{30x^2+27x+6}{9x^2-49} \div \frac{30x+15}{30x-70} \]

    • 2 years ago
  9. cpattison Group Title
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    ((30x2+27x+6)/(9x2−49))/((30x+15)/(30x−70)) is where we start. then you can flip the bottom half and multiply ((30x2+27x+6)/(9x2−49))*((30x−70)/(30x+15)) then you just multiply though ((30x2+27x+6)*(30x−70))/((9x2−49)*(30x+15))

    • 2 years ago
  10. cpattison Group Title
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    yeah so flip that second fraction

    • 2 years ago
  11. cpattison Group Title
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    i really have to learn how to use the equation bar

    • 2 years ago
  12. andriod09 Group Title
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    @cpattison you should use the \\[LaTeX\] by using the \[\\] minus the \ in the middle.

    • 2 years ago
  13. cpattison Group Title
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    agreed

    • 2 years ago
  14. Hero Group Title
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    I just copied and pasted what @andriod09 wrote with a few minor changes :P

    • 2 years ago
  15. andriod09 Group Title
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    \[\huge\text i-know-how-to-use-LaTeX\]

    • 2 years ago
  16. Hero Group Title
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    Not as good as me :P

    • 2 years ago
  17. andriod09 Group Title
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    the answer isn't supposed to be like that. it is supposed to be on top of each other though. thats what it says. the answer it: \[\frac{2(5x+2)}{(x-7)}\]

    • 2 years ago
  18. Hero Group Title
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    You failed to execute the proper steps. That's why you didn't get it right

    • 2 years ago
  19. Hero Group Title
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    Did you flip the second fraction?

    • 2 years ago
  20. andriod09 Group Title
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    no. i don't know how to solve it like that, you have to factor it out is what the book says, but i don't get it.

    • 2 years ago
  21. Hero Group Title
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    If you had \[\frac{3}{6} \div \frac{3}{9}\] you would have no clue what to do?

    • 2 years ago
  22. andriod09 Group Title
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    thats simple: \[\frac{3}{6}\div\frac{3}{9}=\frac{1.5}{1}\]

    • 2 years ago
  23. andriod09 Group Title
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    \[:{\]

    • 2 years ago
  24. Hero Group Title
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    actually, you have it but you failed to execuate the reciprocal rule

    • 2 years ago
  25. andriod09 Group Title
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    \[:[\]

    • 2 years ago
  26. Hero Group Title
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    \[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6} \times \frac{9}{3} = \frac{3}{2}\]

    • 2 years ago
  27. Hero Group Title
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    But really, you're supposed to do the reciprocal then reduce in this manner: \[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6} \times \frac{9}{3} =\frac{1}{2} \times 3= \frac{3}{2}\]

    • 2 years ago
  28. Hero Group Title
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    Show me the full steps of how you got 1.5 from 3/6 div 3/9 @andriod09

    • 2 years ago
  29. Hero Group Title
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    You must have used a calculator bro, lol

    • 2 years ago
  30. Hero Group Title
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    Because you never divide by fractions to get a decimal. Only a calculator would do that.

    • 2 years ago
  31. andriod09 Group Title
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    \[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6}\times\frac{9}{3}=\frac{1}{2}\times3=\frac{3}{2}\]

    • 2 years ago
  32. andriod09 Group Title
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    btw, hero i literally typed all that out.

    • 2 years ago
  33. Hero Group Title
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    Good for you. Now do your original problem in the same manner.

    • 2 years ago
  34. cpattison Group Title
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    |dw:1349572930013:dw|

    • 2 years ago
  35. cpattison Group Title
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    |dw:1349573068242:dw|

    • 2 years ago
  36. Hero Group Title
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    Don't forget to reduce before executing

    • 2 years ago
  37. andriod09 Group Title
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    that doesn't help me any @cpattison

    • 2 years ago
  38. cpattison Group Title
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    |dw:1349573260282:dw|

    • 2 years ago
  39. cpattison Group Title
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    i'm trying to work this out too by the way

    • 2 years ago
  40. andriod09 Group Title
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    can you please use the \[LaTeX\]

    • 2 years ago
  41. andriod09 Group Title
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    just use the equation button if need be.

    • 2 years ago
  42. Hero Group Title
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    Try not to do it the hard way @cpattison

    • 2 years ago
  43. cpattison Group Title
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    \[\frac{ 30(10x^2+9x+2)(3x-7) }{ 15(9x^2-4)(2x+1) }\]

    • 2 years ago
  44. cpattison Group Title
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    i'm not sure i know the easy way. enlighten me?

    • 2 years ago
  45. Hero Group Title
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    First I'd have to make sure you're even doing this correctly

    • 2 years ago
  46. andriod09 Group Title
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    How is that the answer though... the answer, from the book, says that the answer is: \[\frac{2(5x+2)}{3x+7}\]

    • 2 years ago
  47. cpattison Group Title
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    i don't know how that happened unless we all have done something wrong. i get \[\frac{ 2(25x^2-27x-14) }{ 9x^2-4 }\]

    • 2 years ago
  48. cpattison Group Title
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    if i were you i would try and find someone in my class or around me to ask, and double check your equations

    • 2 years ago
  49. andriod09 Group Title
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    A)im homeschooled, B)this is me catching up

    • 2 years ago
  50. Hero Group Title
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    I was following what @cpattison did. I'm sure I would have gotten it if I had taken my own approach.

    • 2 years ago
  51. CliffSedge Group Title
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    Extremes-over-means, anyone? \[\large \frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}\]

    • 2 years ago
  52. Hero Group Title
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    Yep, @cpattison made a mistake because 9x^2 - 49 is in the denominator

    • 2 years ago
  53. andriod09 Group Title
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    cliff. dafaq/dahell does that even mean????

    • 2 years ago
  54. Hero Group Title
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    |dw:1349575081973:dw|

    • 2 years ago
  55. Hero Group Title
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    \[\frac{ 2(5x+2)(3x-7) }{(3x+7)(3x-7) }\]

    • 2 years ago
  56. cpattison Group Title
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    oh okay. sorry for that.

    • 2 years ago
  57. Hero Group Title
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    And now the 3x - 7 reduces to: \[\frac{ 2(5x+2) }{(3x+7) }\]

    • 2 years ago
  58. CliffSedge Group Title
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    @andriod09 "extremes-over-means" |dw:1349575148786:dw|

    • 2 years ago
  59. Hero Group Title
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    I already know about means over extremes. I was just trying to explain it in a way that @andriod09 should have already been familiar with.

    • 2 years ago
  60. CliffSedge Group Title
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    (It's the same thing . . )

    • 2 years ago
  61. andriod09 Group Title
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    I don't understand it.... \[:{\]

    • 2 years ago
  62. Hero Group Title
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    You ignored the part where I said something "he should have already been familiar with".

    • 2 years ago
  63. CliffSedge Group Title
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    Meditate on it.

    • 2 years ago
  64. CliffSedge Group Title
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    I ignored nothing.

    • 2 years ago
  65. Hero Group Title
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    Normal students other than @Cliffsedge learn how to change division to multiplication by flipping the second fraction.

    • 2 years ago
  66. CliffSedge Group Title
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    It's the same thing . . .

    • 2 years ago
  67. CliffSedge Group Title
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    |dw:1349575395748:dw|

    • 2 years ago
  68. Hero Group Title
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    I know that bro, but you should have registered that I was trying to show the poor kid something he should have already been familiar with. Ask the majority of students at a school what means over extremes mean and they'll give you a blank stare.

    • 2 years ago
  69. cpattison Group Title
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    http://www.khanacademy.org/math/arithmetic/fractions/v/dividing-fractions this could help. he's great

    • 2 years ago
  70. CliffSedge Group Title
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    Anyway, I was merely offering a method that might have been recognized. It wasn't, carry on.

    • 2 years ago
  71. CliffSedge Group Title
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    @Hero , but did you notice that I didn't just pop in and say, "extremes-over-means, yo!" and leave? I gave a formula that should be immediately recognized by the more modern phrasing of "keep-flip-change." I'll just leave now, because I don't want you to continue derailing the thread just to bark at me.

    • 2 years ago
  72. Hero Group Title
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    I'm not a dog, so unfortunately I don't know how to bark.

    • 2 years ago
  73. cpattison Group Title
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    none of that helps. @andriod09 is there anything else?

    • 2 years ago
  74. andriod09 Group Title
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    yea. i would like to understand how to work the stupid fraction problem. :{

    • 2 years ago
  75. cpattison Group Title
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    did khan help at all?

    • 2 years ago
  76. Hero Group Title
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    I could have sworn I just showed you. I suppose now you need help with factoring and reducing polynomials.

    • 2 years ago
  77. andriod09 Group Title
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    cp, i don't havethe time for khan, but yes, Sam is a really good tutor. i have been to that site nealy a year now. And Hero, YES PLEASE.

    • 2 years ago
  78. andriod09 Group Title
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    @Hero are you still here??

    • 2 years ago
  79. satellite73 Group Title
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    \[\frac{\frac{30x^2+27x+6}{9x^2-49}}{\frac{30x+15}{30x-70}}=\frac{30x^2+27x+6}{9x^2-49}\times \frac{30x-70}{30x+15}\]

    • 2 years ago
  80. satellite73 Group Title
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    factor and cancel if you can

    • 2 years ago
  81. jiteshmeghwal9 Group Title
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    A type of complex fraction\[\huge\frac{\frac{30x^2+27x+6}{9x^2-49}}{\frac{30x+15}{30x-70}}\]\[\LARGE{\frac{30x^2+27x+6}{9x^2-49} \div \frac{30x+15}{30x-70}}\]\[\LARGE{\frac{30x^2+27x+6}{9x^2-49} \times \frac{30x-70}{30x+15}}\]

    • 2 years ago
  82. satellite73 Group Title
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    for example \(9x^2-49=(3x+7)(3x-7)\)

    • 2 years ago
  83. andriod09 Group Title
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    ENGLISH PLEASE. K'THANKS

    • 2 years ago
  84. satellite73 Group Title
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    and \(30x-70=10(x-7)\) etc etc. there will be an orgy of cancellation

    • 2 years ago
  85. andriod09 Group Title
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    can you please type that in English for me, step by step?

    • 2 years ago
  86. andriod09 Group Title
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    @jiteshmeghwal9 2 things, 1) i luv your prof pic, 2) what do i do after all that> I dont get... :{

    • 2 years ago
  87. jiteshmeghwal9 Group Title
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    bro first factor the numerators & denominators of both fractions then tell me what do u gt ?

    • 2 years ago
  88. andriod09 Group Title
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    how do i do that?

    • 2 years ago
  89. jiteshmeghwal9 Group Title
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    what u haven't studied factorization yet

    • 2 years ago
  90. jiteshmeghwal9 Group Title
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    without studying factorization u can't do this question :(

    • 2 years ago
  91. andriod09 Group Title
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    not for fractions. i can factor \[x^2+10x+20\] easily. just not for fractions. plus, im homeschooled, so things aren't the same for me.

    • 2 years ago
  92. jiteshmeghwal9 Group Title
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    so factor \[30x^2+27x+6\]

    • 2 years ago
  93. jiteshmeghwal9 Group Title
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    @andriod09

    • 2 years ago
  94. andriod09 Group Title
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    \[30x^2+27x+6\] should i find a comon factor? because i don't think there is one. buy anyway. isn't it: \[2(5x+2)\]

    • 2 years ago
  95. jiteshmeghwal9 Group Title
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    \[ax^2+(a+b)x+b\]use this identity

    • 2 years ago
  96. andriod09 Group Title
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    thas makes no sense to me. :/

    • 2 years ago
  97. jiteshmeghwal9 Group Title
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    i mean break the middle term such that the sum becomes 27x & product becomes \(30x^2*6\)

    • 2 years ago
  98. andriod09 Group Title
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    but was the answer right? it should be right.

    • 2 years ago
  99. jiteshmeghwal9 Group Title
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    but after multiplying that u r not getting the right exression

    • 2 years ago
  100. jiteshmeghwal9 Group Title
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    \[30x^2+15x+12x+6\]\[15x(2x+1)+6(2x+1)\]\[(15x+6)(2x+1)\]

    • 2 years ago
  101. andriod09 Group Title
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    ohhh. okay. Thanks!

    • 2 years ago
  102. jiteshmeghwal9 Group Title
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    \[(3x)^2-(7)^2\]\[(3x+7)(3x-7)\]

    • 2 years ago
  103. jiteshmeghwal9 Group Title
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    \[30x-70\]\[10(3x-7)\]

    • 2 years ago
  104. jiteshmeghwal9 Group Title
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    \[30x+15\]\[15(2x+1)\]

    • 2 years ago
  105. andriod09 Group Title
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    TYSVM i get it now!

    • 2 years ago
  106. andriod09 Group Title
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    i g2g. Bye!

    • 2 years ago
  107. jiteshmeghwal9 Group Title
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    \[\frac{(15x+6)(2x+1)15(2x+1)}{(3x+7)(3x-7)10(3x-7)}\]

    • 2 years ago
  108. jiteshmeghwal9 Group Title
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    bye & thanx for liking my profile profile pic

    • 2 years ago
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