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Inopeki

TuringTest, want to continue?

  • 2 years ago
  • 2 years ago

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  1. TuringTest
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    lol I like the attitude, I better eat though, give me a bit what time is it where you are?

    • 2 years ago
  2. Inopeki
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    9:45 pm, you? Then go eat :)

    • 2 years ago
  3. TuringTest
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    alright, just so you know it's easy to send messages via fan message now, so use that instead of a question to hail me

    • 2 years ago
  4. Inopeki
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    Oh, ill use that!

    • 2 years ago
  5. TuringTest
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    see you in a bit

    • 2 years ago
  6. Inopeki
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    Alright

    • 2 years ago
  7. No-data
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    How do you use those fan messages Turing?

    • 2 years ago
  8. Inopeki
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    Go to the profile and click "write a fan message"

    • 2 years ago
  9. TuringTest
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    factor 21x^2+14x

    • 2 years ago
  10. Inopeki
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    GCF of 21 and 14=7 GCF of x and x^2 is x 21x^2/7x+14x/7x=x(3x+2)?

    • 2 years ago
  11. TuringTest
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    almost but 1)be careful of notation 21x^2/7x+14x/7x is not equal x(3x+2) 2) remember to keep the whole GCF outside the parentheses

    • 2 years ago
  12. Inopeki
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    But i thought that if the 7 was on the LHS then it couldnt be on the RHS?

    • 2 years ago
  13. Inopeki
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    21x^2/7x+14x/7x=7x(3x+2)?

    • 2 years ago
  14. TuringTest
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    but they are not equal 21x^2/7x+14x/7x=3x+2 not 7x(3x+2) so you must write 21x^2+14x=7x(21x^2/7x+14x/7x)=?

    • 2 years ago
  15. Inopeki
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    I dont know...

    • 2 years ago
  16. Inopeki
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    7(3x^2+2)?

    • 2 years ago
  17. TuringTest
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    you almost have it but you need to put the GCF outside the parentheses what is the GCF?

    • 2 years ago
  18. Inopeki
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    7x

    • 2 years ago
  19. TuringTest
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    so that's what goes outside the parentheses

    • 2 years ago
  20. Inopeki
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    7x(3x^2+2) Cant be right.

    • 2 years ago
  21. TuringTest
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    what is 21x^2/7x=? rewrite it in a way more comfortable to you if it looks strange

    • 2 years ago
  22. Inopeki
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    21/7 * x^2/x?

    • 2 years ago
  23. TuringTest
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    yes...

    • 2 years ago
  24. Inopeki
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    Im pretty sure thats 3x man

    • 2 years ago
  25. TuringTest
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    it is, so why did you write 3x^2 above?

    • 2 years ago
  26. Inopeki
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    Cause you said the other one wasnt right :/ What is the correct way to write it?

    • 2 years ago
  27. TuringTest
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    here was your first answer x(3x+2) here was the next 21x^2/7x+14x/7x=7x(3x+2) the answer here is correct, I just pointed out that the two expressions above are not equal, I think that may have confused you. next you wrote 7x(3x^2+2) so I think we started to get off track... the answer is 21x^2+14x=7x(3x+2) I just made a point about how you show your work, and what it means.

    • 2 years ago
  28. Inopeki
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    Ohhhhh :)

    • 2 years ago
  29. TuringTest
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    write out the process as 21x^2+14x=7x(3x)+7x(2)=7x(3x+2) that is really the best way to show factoring

    • 2 years ago
  30. Inopeki
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    Oh ok :D Another one?

    • 2 years ago
  31. TuringTest
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    factor 3t^3+9t^2

    • 2 years ago
  32. TuringTest
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    sorry, so you know how to divide t^3/t^2 ???

    • 2 years ago
  33. Inopeki
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    x*x*x ----- right? It should jjust become x? x*x

    • 2 years ago
  34. Inopeki
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    t, sorry

    • 2 years ago
  35. TuringTest
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    you got it, so my question stands: factor 3t^3+9t^2

    • 2 years ago
  36. Inopeki
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    GCF of 3 and 9 is 3. GCF of t^2 and t^3 is t. 3t^3+9t^2=3t(3t^3/3t)+3t(9t^2/3t)=3t(t^3+3t^2)?

    • 2 years ago
  37. TuringTest
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    check your answer and distribute to see if you get the original: \[3t(t^3+3t^2)=3t(t^3)+3t(3t^2)=3t^4+9t^3\neq3t^3+9t^2\]you didn't divide the terms in the middle right ...but I think more importantly the GCF of t^3 and t^2 is t^2, not t that is because each term can be evenly divided by t^2.

    • 2 years ago
  38. Inopeki
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    Hm

    • 2 years ago
  39. Inopeki
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    Oh right!

    • 2 years ago
  40. TuringTest
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    what is the GCF of t^3+t^5+t^7 ??

    • 2 years ago
  41. Inopeki
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    Something to the power of something doesnt follow the rules of usual numbers, forgot that. It should be t^5

    • 2 years ago
  42. TuringTest
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    no it's t^3 and it does follow the rules of regular numbers if you think about it, otherwise it wouldn't be true! look at 16,32,8 what is their GCF ?

    • 2 years ago
  43. Inopeki
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    8

    • 2 years ago
  44. TuringTest
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    now rewrite those three numbers as powers of 2

    • 2 years ago
  45. TuringTest
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    can you do that?

    • 2 years ago
  46. Inopeki
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    256,1024,64

    • 2 years ago
  47. Inopeki
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    ?

    • 2 years ago
  48. TuringTest
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    no I meant like 8=2^3 16=2^? 32=2^?

    • 2 years ago
  49. Inopeki
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    Oh! 8=2^3 16=2^4 32=2^5

    • 2 years ago
  50. TuringTest
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    ...and what is their GCF? 2^3 so the GCF of a set of variables is the highest common power to each term -in this case 3 so if we have y^4+y^7+y^9 the GCF is...?

    • 2 years ago
  51. Inopeki
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    y^4?

    • 2 years ago
  52. TuringTest
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    right :)

    • 2 years ago
  53. Inopeki
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    :D

    • 2 years ago
  54. TuringTest
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    good job :D so now back to our question... factor 3t^3+9t^2 what is the GCF?

    • 2 years ago
  55. Inopeki
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    3t^2?

    • 2 years ago
  56. TuringTest
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    right now can you factor it?

    • 2 years ago
  57. Inopeki
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    3t^3+9t^2=3t^2(3t^3/3t^2)+3t^2(9t^2/3t^2)=3t^2(t+3)?

    • 2 years ago
  58. TuringTest
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    very nice!!!!!!

    • 2 years ago
  59. Inopeki
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    I got it!! :D

    • 2 years ago
  60. TuringTest
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    you totally did :D that is very cool! so now lets see what's so great about factoring say we have f(t)=3t^3+9t^2 and we want to know the 'zeros' of the function that means when the function touches the x-axis, i.e. when f(t)=0 so to answer this we must solve 0=3t^3+9t^2 how can we solve that? by factoring... 0=3t^2(t+3) now you can solve it quickly, any idea which fundamental rule of algebra tells us how?

    • 2 years ago
  61. Inopeki
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    Nope, sorry

    • 2 years ago
  62. TuringTest
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    the rule is called the 'zero factor property' it states that if\[ab=0\]then either\[a=0\]or\[b=0\]or both does this rule make logical sense to you?

    • 2 years ago
  63. Inopeki
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    Yeah

    • 2 years ago
  64. TuringTest
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    so now look at our factored equation \[0=ab=3t^2(t+3)\]that means that either\[3t^2=0\]or \[t+3=0\]can you solve each of these equations?

    • 2 years ago
  65. TuringTest
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    *either or both I should say

    • 2 years ago
  66. Inopeki
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    OH so 3t^2 represents a and t+3 represents b!

    • 2 years ago
  67. TuringTest
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    exactly that's why i said to learn the rules on this page http://www.capitan.k12.nm.us/teachers/shearerk/basic_rules_of_algebra.htm almost all of algebra is in there, though sometimes it is hidden

    • 2 years ago
  68. Inopeki
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    3t^2=0 t=0-3t/t t=0-3 t=-3?

    • 2 years ago
  69. TuringTest
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    no, it's more simple than that, remember the same rule we just used: ab=0 then either a=0 or b=0 or both 3t^2=0 let 3=a t^2=b we know that a=3 cannot be zero, because 3 is never zero, it's a constant that leaves the possibility only of t^2=0 and the number number that times itself is zero is zero so\[3t^2=0\to t=0\]

    • 2 years ago
  70. Inopeki
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    Ohhhh.. Cause 3x0x0=0?

    • 2 years ago
  71. TuringTest
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    right so t has to be zero...

    • 2 years ago
  72. Inopeki
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    Yeah

    • 2 years ago
  73. TuringTest
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    what about the other possibility t+3=0 ???

    • 2 years ago
  74. Inopeki
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    How can that be possible when t is 0?

    • 2 years ago
  75. TuringTest
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    there are two answers to every quadratic equation as you may recall me saying this is cubic so it has 3 actually, we say that zero ocurrs twice in 3t^2=0 because it leads to two 0's as you showed: 3x0x0 we call that a 'multiplicity of 2' so there will be multiple answers, the other is found by solving t+3=0 remember that either a=0 or b=0 or BOTH we don't know so we have to solve them all.

    • 2 years ago
  76. Inopeki
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    Oh right! This is a quadratic equation. t=-3 on this one so its (0,-3)?

    • 2 years ago
  77. TuringTest
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    like I said, cubic read what I wrote above please about multiplicity...

    • 2 years ago
  78. Inopeki
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    Three answers

    • 2 years ago
  79. Inopeki
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    So these are cubic coordinates?

    • 2 years ago
  80. Inopeki
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    (0,0,-3)?

    • 2 years ago
  81. TuringTest
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    not cubic coordinates (I don't know what that is exactly...), it's just that we say that 3t^3+9t^2 has zeros (0,3) - that means the graph of f(x)=3t^3+9t^2 hits zero there... where 0 here has a 'multiplicity' of 2 (that means it occurs twice) and 3 has a multiplicity of 1

    • 2 years ago
  82. Inopeki
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    So why isnt -3 involved?

    • 2 years ago
  83. TuringTest
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    sorry typo, meant (0,-3)* and -3 has multiplicity 1*

    • 2 years ago
  84. TuringTest
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    good catch

    • 2 years ago
  85. Inopeki
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    Thanks

    • 2 years ago
  86. TuringTest
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    so do you see what I mean? the zero's of\[f(t)=3t^2+9t^2\]are found by factoring and setting to zero\[0=3t^2(t+3)\]then solving each possibility\[3t^2=0\to t=0\]\[t+3=0\to t=-3\]where we say that for t=0 k=2 and for t=-3 k=1 where k is the multiplicity

    • 2 years ago
  87. Inopeki
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    Right, the k is cause t is to the power of 3 there.

    • 2 years ago
  88. TuringTest
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    for the part that had 3t^2=0 we had k=2 (because zero is the answer twice: 3x0x0) for t+3=0 we have k=1 because t is only to the first power and we only have one answer If what you mean is that you noticed that adding up all the k's gave youu 3, the order of the cubic, then you have noticed what is called the Fundamental Theorem of Algebra: "The sum of the multiplicities of the roots of a function is equal to the order of the polynomial" \[k_0+k_{−3}=2+1=3\]

    • 2 years ago
  89. TuringTest
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    - a very important theorem as the name implies...

    • 2 years ago
  90. Inopeki
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    I dont really get the theorem, please explain it.

    • 2 years ago
  91. TuringTest
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    Basically for whatever the highest power variable you have in a polynomial, that is how many answers you have. Note that they may not be all different answers, but the ones that occur more than once are counted as having a higher multiplicity, so if you add up the multiplicities (the k's) that's how many zeros the polynomial has. for example 7x^5+3x^3+2x^2+5x+3=0 must have 5 answers, because it is 5th order x^2+2x+2=0 must have 2 answers, because it is second order

    • 2 years ago
  92. Inopeki
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    So x^4+5x-4=0 must have 4 answers?

    • 2 years ago
  93. TuringTest
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    exactly, though it may not be 4 different answers for instance x^4=0 has only the answer x=0, but that zero has a multiplicity of k=4 because 0x0x0x0=0 is how it must be...

    • 2 years ago
  94. Inopeki
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    Oh, now i get it

    • 2 years ago
  95. TuringTest
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    good :) do you want to try some more factoring problems? or perhaps you should learn a little about exponents first? or perhaps you are ready top get some rest.... which is it?

    • 2 years ago
  96. Inopeki
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    I wanna learn something new :D

    • 2 years ago
  97. TuringTest
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    let's see if this is news to you: simplify\[{x^{14}\over x^{12}}\]

    • 2 years ago
  98. Inopeki
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    x^2 lol

    • 2 years ago
  99. TuringTest
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    good that saves a lot of time...

    • 2 years ago
  100. TuringTest
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    simplify\[\sqrt[3]{x^{21}}\]

    • 2 years ago
  101. Inopeki
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    x^7?

    • 2 years ago
  102. TuringTest
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    good :) more time saved...

    • 2 years ago
  103. Inopeki
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    :D actually i didnt know that

    • 2 years ago
  104. TuringTest
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    the genereal rule is\[\sqrt[b]{x^{a}}=x^{a/b}\]any radical can be wriitten as a fractional exponent, for instance\[\sqrt x=x^{1/2}\]so... let's try some of that.

    • 2 years ago
  105. Inopeki
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    Yeah, i assumed that :)

    • 2 years ago
  106. TuringTest
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    good guess, try to really rationalize it if you can... since you seem to know the rules let's try a trickier one simplify \[\frac{\sqrt[3]{x^2}\sqrt[5]{x^3}}{\sqrt x}\]

    • 2 years ago
  107. Inopeki
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    x^2/3 * x^3/5 ----------- Its basically this, right? If so, i just need to make the variables "suitable" to x^1/2 be merged, right?

    • 2 years ago
  108. TuringTest
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    right, just remember that\[x^ax^b=x^{a+b}\]and that\[\frac{x^a}{x^b}=x^{a-b}\]so you're gonna have and subtract to add those fractions, so they all need a common denominator.

    • 2 years ago
  109. TuringTest
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    so you're gonna have to add and subtract those fractions*

    • 2 years ago
  110. Inopeki
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    Yeah, thats what i meant by making them suitable to merge

    • 2 years ago
  111. TuringTest
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    I figured, but that's not a known term to me, just making sure

    • 2 years ago
  112. Inopeki
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    x^2/3 * x^3/5 x^10/15 * x^9/15 x^19/15 ----------- = =-------------- x^1/2 x^1/2 Now i need to get 19/15 divisible by 2

    • 2 years ago
  113. TuringTest
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    good so far :)

    • 2 years ago
  114. Inopeki
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    19/15 ------------38/30

    • 2 years ago
  115. Inopeki
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    x^38/30 -------- = x^1/2 Actually, dont i need to get them divisible by 15 so i can get 2 as the denominator?

    • 2 years ago
  116. TuringTest
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    what can you multiply the bottom by to get 1/2 over 30 ?

    • 2 years ago
  117. Inopeki
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    Oh! x^38/30 -------- = 23/30? No, that cant be it.. x^15/30

    • 2 years ago
  118. TuringTest
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    yes it can and is but you have to leave the x of course... x^(23/30)

    • 2 years ago
  119. Inopeki
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    Oh right lol

    • 2 years ago
  120. TuringTest
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    so that was good, I won't test you on that do you know how to FOIL ?

    • 2 years ago
  121. Inopeki
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    What?

    • 2 years ago
  122. TuringTest
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    i.e. simplify (a+b)(c+d)

    • 2 years ago
  123. Inopeki
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    abcd?

    • 2 years ago
  124. TuringTest
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    FOIL= First Outer Inner Last|dw:1325979609902:dw|watch the arrows they multiply the first in the brackets, the outer terms, the inner, and the last seperateyl

    • 2 years ago
  125. TuringTest
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    it's just like distribution, but you have to do a and b seperate

    • 2 years ago
  126. Inopeki
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    I think i get it. give me another one and ill solve it fast

    • 2 years ago
  127. TuringTest
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    It's okay if you don't solve it fast actually remember that solving is not simplifying, here we have no = sign so we are simplifying: (x+3)(x+2)

    • 2 years ago
  128. Inopeki
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    x^2+2x+3x+6?

    • 2 years ago
  129. TuringTest
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    nice! you can simplify the middle terms

    • 2 years ago
  130. Inopeki
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    Oh right,x^2+5x+6

    • 2 years ago
  131. TuringTest
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    nicely done! especially for never having FOILed before. simplify (2x+5)(3x-1)

    • 2 years ago
  132. Inopeki
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    5x+2x+15x-5?

    • 2 years ago
  133. Inopeki
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    6x*

    • 2 years ago
  134. TuringTest
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    careful... what's the first? what's the outer? the others are right.

    • 2 years ago
  135. TuringTest
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    yes the first is 6x the outer though...

    • 2 years ago
  136. Inopeki
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    -1?

    • 2 years ago
  137. TuringTest
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    times what?

    • 2 years ago
  138. Inopeki
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    2x, maybe thats -2x..

    • 2 years ago
  139. TuringTest
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    there ya go ;-)

    • 2 years ago
  140. TuringTest
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    so it should be...?

    • 2 years ago
  141. Inopeki
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    6x+-2x+15x-5

    • 2 years ago
  142. TuringTest
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    scratch the extra plus sign, but yes, now simplify...

    • 2 years ago
  143. Inopeki
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    29x-5?

    • 2 years ago
  144. TuringTest
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    oh my mistake, you forgot that the first term would be squared First=2x(3x)=6x^2 now what do you get?

    • 2 years ago
  145. Inopeki
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    6x^2-2x+15x-5?

    • 2 years ago
  146. TuringTest
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    yes, and now simplify

    • 2 years ago
  147. Inopeki
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    4x^2+15x-5?

    • 2 years ago
  148. TuringTest
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    like terms, x with x.... not x with x^2...

    • 2 years ago
  149. Inopeki
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    so 6x^2+13x-5?

    • 2 years ago
  150. TuringTest
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    yes. good job. now simplify (2p-3)(5p-1)

    • 2 years ago
  151. Inopeki
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    10p^2-2p-15p-3? Simplified:10p^2-17p-3

    • 2 years ago
  152. TuringTest
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    good, but watch the last term negative times negative is...?

    • 2 years ago
  153. Inopeki
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    Oh right, positive. So 10p^2-17p+3?

    • 2 years ago
  154. Inopeki
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    2r^2+2r*3t+t*r+3t^2 Simplified: 2r^2+2r^2+3t^2+3t^2

    • 2 years ago
  155. TuringTest
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    stop before your simplification (which is wrong, sorry) look at the first step: 2r^2+2r*3t+t*r+3t^2 ^^^ should everything be positive?

    • 2 years ago
  156. Inopeki
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    Oh right.. 2r^2+2r*-3t+t*r

    • 2 years ago
  157. TuringTest
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    yes but you forgot the last term you should also put parentheses around the negative terms

    • 2 years ago
  158. Inopeki
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    2r^2+2r*(-3t)+t*(-r)?

    • 2 years ago
  159. TuringTest
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    lol it got deleted, what was the original problem please?

    • 2 years ago
  160. Inopeki
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    I cant remember :(

    • 2 years ago
  161. TuringTest
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    ok another, because we started losing terms in the last one (2y+x)(4y-3x)

    • 2 years ago
  162. Inopeki
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    8y+2y*3x+x*4y(-3x^2)

    • 2 years ago
  163. TuringTest
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    there should be four terms what is your first? write out the middle step

    • 2 years ago
  164. TuringTest
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    it's close in many ways, but there are quite a few mistakes

    • 2 years ago
  165. Inopeki
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    Ill put them in prenthesis (8y)+(2y*3x)+(x*4y)-(3x^2)

    • 2 years ago
  166. TuringTest
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    better, but the first is 2y(4y) no? and the outer is a positive times a negative as well

    • 2 years ago
  167. Inopeki
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    Oh, ill solve this one and then sleep :) (8y^2)+(2y*-3x)+(x*4y)-(3x^2)

    • 2 years ago
  168. TuringTest
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    nice, so to simplify this first multiply the coefficients...

    • 2 years ago
  169. Inopeki
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    (8y^2)+(8y^2*-3x^2)-(3x^2)

    • 2 years ago
  170. TuringTest
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    not I think you're getting tired (8y^2)+(2y*-3x)+(x*4y)-(3x^2)=8y^2-6xy+4xy-3x^2=?

    • 2 years ago
  171. TuringTest
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    no*

    • 2 years ago
  172. TuringTest
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    can you simplify the last step?

    • 2 years ago
  173. Inopeki
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    Yeah, i need to sleep.

    • 2 years ago
  174. TuringTest
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    for future reference:\[(8y^2)+(2y*-3x)+(x*4y)-(3x^2)\]\[=8y^2-6xy+4xy-3x^2=8y^2-2xy-3y^2\]goodnight!

    • 2 years ago
  175. Inopeki
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    Good night! Thanks for all youve taught me :D See you tomorrow?

    • 2 years ago
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