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andriod09

  • 2 years ago

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

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

  3. andriod09
    • 2 years ago
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    @cpattison

  4. andriod09
    • 2 years ago
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    @ganeshie8 @Hero

  5. cpattison
    • 2 years ago
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    oh okay

  6. cpattison
    • 2 years ago
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    one second!

  7. Hero
    • 2 years ago
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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.

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

  9. cpattison
    • 2 years ago
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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))

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

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

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

  13. cpattison
    • 2 years ago
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    agreed

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

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

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

  17. andriod09
    • 2 years ago
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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: \[\frac{2(5x+2)}{(x-7)}\]

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

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

  20. andriod09
    • 2 years ago
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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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  48. cpattison
    • 2 years ago
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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

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

  50. Hero
    • 2 years ago
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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.

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

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

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

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

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

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

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

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

  59. Hero
    • 2 years ago
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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.

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

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

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

  63. CliffSedge
    • 2 years ago
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    Meditate on it.

  64. CliffSedge
    • 2 years ago
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    I ignored nothing.

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

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

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

  68. Hero
    • 2 years ago
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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.

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

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

  71. CliffSedge
    • 2 years ago
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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.

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

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

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

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

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

  77. andriod09
    • 2 years ago
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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.

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

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

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

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

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

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

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

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

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

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

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

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

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

  91. andriod09
    • 2 years ago
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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.

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

  93. jiteshmeghwal9
    • 2 years ago
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    @andriod09

  94. andriod09
    • 2 years ago
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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: \[2(5x+2)\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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