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Inopeki Group Title

Algebra, Turingtest :D

  • 2 years ago
  • 2 years ago

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  1. TuringTest Group Title
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    FOIL Inopeki :D (a+b)(c+d)

    • 2 years ago
  2. Inopeki Group Title
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    ac+ad+bc+bd

    • 2 years ago
  3. TuringTest Group Title
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    nice (x+3)(x-4)

    • 2 years ago
  4. Inopeki Group Title
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    x^2-4x+3x-12

    • 2 years ago
  5. TuringTest Group Title
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    good, simplify

    • 2 years ago
  6. Inopeki Group Title
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    x^2-1x-12

    • 2 years ago
  7. TuringTest Group Title
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    Good, (no need to write the 1 though) (3p+t)(2p-4t)

    • 2 years ago
  8. Inopeki Group Title
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    Oh right 6p^2+3p*-4t+t*2p+5t?

    • 2 years ago
  9. TuringTest Group Title
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    not 5t... everything else is right though

    • 2 years ago
  10. Inopeki Group Title
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    -5t

    • 2 years ago
  11. TuringTest Group Title
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    no... look closely

    • 2 years ago
  12. Inopeki Group Title
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    -5t^2?

    • 2 years ago
  13. TuringTest Group Title
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    -4t^2 :P

    • 2 years ago
  14. Inopeki Group Title
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    Huh :O

    • 2 years ago
  15. TuringTest Group Title
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    You don't see it? (-4t)(t)=-4t^2 look at the last terms in the parentheses

    • 2 years ago
  16. Inopeki Group Title
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    Ohhh, now i see it

    • 2 years ago
  17. TuringTest Group Title
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    so you have what then, before simplifying?

    • 2 years ago
  18. Inopeki Group Title
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    6p^2+3p*-4t+t*2p+4t^2?

    • 2 years ago
  19. TuringTest Group Title
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    -4t^2 :P

    • 2 years ago
  20. Inopeki Group Title
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    Oh right xd

    • 2 years ago
  21. TuringTest Group Title
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    can you simplify it?

    • 2 years ago
  22. Inopeki Group Title
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    6p^2+3p*-4t+t*2p-4t^2 6p^2+6p^2-4t^2-4t^2?

    • 2 years ago
  23. TuringTest Group Title
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    no... how did you the p^2 in the middle? we had 6p^2+3p(-4t)+t(2p)-4t^2 parentheses look cleaner btw, you should use them here what is 3p(-4t) and t(2p) ???

    • 2 years ago
  24. Inopeki Group Title
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    -12pt? 2pt?

    • 2 years ago
  25. TuringTest Group Title
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    right, so then we have...?

    • 2 years ago
  26. Inopeki Group Title
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    6p^2-12pt+2pt-4t^2

    • 2 years ago
  27. TuringTest Group Title
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    yes :D and how does that simplify?

    • 2 years ago
  28. Inopeki Group Title
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    I dont know actually, maybe 6p^2-14pt+pt-4t^2?

    • 2 years ago
  29. Inopeki Group Title
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    No

    • 2 years ago
  30. TuringTest Group Title
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    they are like terms in the middle -12pt+2pt=-10pt this goes for any combination of variables a^2bc+3a^2bc=4a^2bc wr^4dt^3+3wr^2dt+6wr^4dt^3=7wr^4dt^3+3wr^2dt notice in this last example we couldn't combine all the terms because one had different exponents. We can only combine when they are the exact same in exponents and number of variables.

    • 2 years ago
  31. Inopeki Group Title
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    But dont i need to multiply them to get -10pt?

    • 2 years ago
  32. TuringTest Group Title
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    no, just add the coefficients factoring is one way to see it... factor out 'pt' from the middle two terms: \[6p^2-12pt+2pt-4t^2=6p^2+(12-2)pt-4t^2=6p^2-10pt-4t^2\]

    • 2 years ago
  33. Inopeki Group Title
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    Oh wait, i get it now, it would me multiplication if it was -10pt^2

    • 2 years ago
  34. TuringTest Group Title
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    not necessarily. You can't multiply things unless the equation tells you too, or you do a trick like multiply both sides if you have -10pt^2+7pt you cannot simplify that...

    • 2 years ago
  35. Inopeki Group Title
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    Oh ok. But what i meant was that if it were to tell us to multiply it would look like this -10pt^2 and not this -10pt

    • 2 years ago
  36. TuringTest Group Title
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    if you multiplied you get \[(2pt)(-12pt)=-24p^2t^2\]

    • 2 years ago
  37. Inopeki Group Title
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    Oh right

    • 2 years ago
  38. Inopeki Group Title
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    Since you would be multiplying both p and t with itself

    • 2 years ago
  39. TuringTest Group Title
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    right

    • 2 years ago
  40. Inopeki Group Title
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    Whatever, were getting off track :)

    • 2 years ago
  41. TuringTest Group Title
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    not really, you need to get this down too. It's all related, and it's good to know where you're at. (2s^2+p)(s-3p^3)

    • 2 years ago
  42. TuringTest Group Title
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    hey I'm green! never been a mod for math before :)

    • 2 years ago
  43. Inopeki Group Title
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    Congratulations! You deserve it :)

    • 2 years ago
  44. TuringTest Group Title
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    thanks :) I can't delete things though, don't see the point oh well... (2s^2+p)(s-3p^3)

    • 2 years ago
  45. Inopeki Group Title
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    (2s^2+p)(s-3p^3) 2s^3+2s^3(-3p^3)+p(s)+3p^4

    • 2 years ago
  46. TuringTest Group Title
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    so far so good now try to simplify...

    • 2 years ago
  47. TuringTest Group Title
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    oh wait, small error

    • 2 years ago
  48. Inopeki Group Title
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    OH the -

    • 2 years ago
  49. TuringTest Group Title
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    2s^3+2s^3(-3p^3)+p(s)+3p^4 ^^ should be???

    • 2 years ago
  50. Inopeki Group Title
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    2s^3+2s^2(-3p^3)+p(s)-3p^4 ^^ plus that :)

    • 2 years ago
  51. TuringTest Group Title
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    good catch I missed the sign lol !!! yes excellent now can you simplify it

    • 2 years ago
  52. Inopeki Group Title
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    Thanks :D 2s^3+2s^2(-3p^3)+p(s)-3p^4 2s^3-6s^2p^3+sp-3p^4?

    • 2 years ago
  53. TuringTest Group Title
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    perfect last one: (a+b)(a-b)

    • 2 years ago
  54. Inopeki Group Title
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    a^2-ab+ab-b^2? a^2-2ab-b^2?

    • 2 years ago
  55. TuringTest Group Title
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    not quite, I don't think those terms in the middle add...

    • 2 years ago
  56. Inopeki Group Title
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    The simplification or the first one?

    • 2 years ago
  57. TuringTest Group Title
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    a^2-ab+ab-b^2 ^^^^^ what is -x+x ?

    • 2 years ago
  58. Inopeki Group Title
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    Oh right, 0

    • 2 years ago
  59. TuringTest Group Title
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    so (a+b)(a-b) is?

    • 2 years ago
  60. Inopeki Group Title
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    a^2-b^2?

    • 2 years ago
  61. pratu043 Group Title
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    Some kind of coaching class going on here?

    • 2 years ago
  62. TuringTest Group Title
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    And that is a very important formula @pratu Inopkie wants to learn Algebra, a good cause in my book :)

    • 2 years ago
  63. pratu043 Group Title
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    ok...

    • 2 years ago
  64. pratu043 Group Title
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    Shall I help?

    • 2 years ago
  65. TuringTest Group Title
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    Thanks but I think I have an idea what I want to show him, we've been working together for a while ;)

    • 2 years ago
  66. TuringTest Group Title
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    @Inopeki We just showed that\[(a+b)(a-b)=a^2-b^2\]That relation is called the 'Difference of Squares' formula. It is used all the time, as you will see, as a quick way to factor anything of the form\[a^2-b^2\]

    • 2 years ago
  67. Inopeki Group Title
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    so a^2-b^2=(a+b)(a-b)?

    • 2 years ago
  68. pratu043 Group Title
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    Its an identity.

    • 2 years ago
  69. TuringTest Group Title
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    Yes, it's always true for that form. This is handy as you will see because if you remember how factoring helped us solve other equations, it will do the same for us here: For instance\[x^2-4=0\]we can factor this..\[(x-2)(x+2)=0\]which as I mentioned befor by the 'zero factor property' means that we need to solve\[x-2=0\]and\[x+2=0\]to find the zeros of our polynomial

    • 2 years ago
  70. Inopeki Group Title
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    But isnt (x-2)(x+2) 0 because the 2s negate?

    • 2 years ago
  71. TuringTest Group Title
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    foil it out and tell me yourself

    • 2 years ago
  72. Inopeki Group Title
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    x^2+2x-2x+-2*2 Well pellet

    • 2 years ago
  73. TuringTest Group Title
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    right which simplified too what?

    • 2 years ago
  74. Inopeki Group Title
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    x^2-4

    • 2 years ago
  75. TuringTest Group Title
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    so that's not always zero, is it? if x=1 then f(x)=-3, not zero what the expression f(x) is equal to here depends on what we put for x. There are in fact only two values for x that will make f(x)=0 Which should make sense by what I said yesterday: The order of the equation is 2, so the total multiplicity of its roots is 2 (fundamental theorem of algebra)

    • 2 years ago
  76. Inopeki Group Title
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    isnt that f(1)=

    • 2 years ago
  77. TuringTest Group Title
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    yes, sorry, bad notation on my part f(1)=-3 f(0)-4 etc.

    • 2 years ago
  78. Inopeki Group Title
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    But how does that work?

    • 2 years ago
  79. TuringTest Group Title
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    which part?

    • 2 years ago
  80. Inopeki Group Title
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    Functions, maybe im not ready for that subject

    • 2 years ago
  81. TuringTest Group Title
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    No you are, perhaps I haven't been clear enough whenever I write x^2-4 If I want to know where this graph hits the x-axis it is useful to call this a function f(x) here we have f(x)=x^2-4 Think about a graph: When the graph is toughing the x-axis it is at y=0, correct?

    • 2 years ago
  82. TuringTest Group Title
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    I just 'called' it a function, and made it graph-able...

    • 2 years ago
  83. Inopeki Group Title
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    Yeah, when x and y intersect its 0,0

    • 2 years ago
  84. Inopeki Group Title
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    oooh

    • 2 years ago
  85. Inopeki Group Title
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    f(x)=y+2x-3 So if we want to figure out y we need to plug in something for x

    • 2 years ago
  86. TuringTest Group Title
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    well, wait now, if you have a y in the function then it is not f(x) here's one thing that I bet is confusing you: f(x)=y when we write a line equation y=mx+b we are saying something about the graph: that the y coordinate is proportional to the x coordinate in this way. However this is rather simplistic. It is really more that we have a function f(x)=mx+b and we have chosen to graph it by letting the y-coordinate of our graph be f(x) they are effectively the same here, but a function can be graphed in many different types of coordinates, so y would not be used unless we are in Cartesian coordinates as we are now.

    • 2 years ago
  87. TuringTest Group Title
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    hence it's better to think of graphs in terms of functions, represented in Cartesian coordinates by y

    • 2 years ago
  88. Inopeki Group Title
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    What are cartesian coordinates?

    • 2 years ago
  89. TuringTest Group Title
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    the regular ones you know x-axis and y-axis

    • 2 years ago
  90. Inopeki Group Title
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    Yeah

    • 2 years ago
  91. TuringTest Group Title
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    there are other coordinate systems like polar, spherical, and cylindrical coordinates If we tried to graph our function in polar coordinates, y would have no meaning. however the zeros of the function are independent of how we choose to graph it. The math will be the same. Hence when we want to know the zeros of some polynomial we call it f(x), and that allows us to put numbers into it and see what happens f(x)=x^2-4 f(0)=0-4=-4 f(1)=1-4=-3 f(2)=4-4=0 so we found a zero simply by calling this a function and putting numbers into it.

    • 2 years ago
  92. Inopeki Group Title
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    One question, what do we gain from functions, what are the numbers on the RHS?

    • 2 years ago
  93. TuringTest Group Title
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    which numbers on the RHS?

    • 2 years ago
  94. Inopeki Group Title
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    Like f(0)=0-4=-4<--- ^ |

    • 2 years ago
  95. TuringTest Group Title
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    well that's the simplified form of f(0): f(0)=-4 check it by plugging in x=0 into the function

    • 2 years ago
  96. Inopeki Group Title
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    But if we know that f(0)=-4 why do we do it?

    • 2 years ago
  97. TuringTest Group Title
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    lol, why /don't/ we want to do that?! Almost all things in physics can be represented by function. If you want to know when the water pressure of a system will become too high, what trajectory to shoot your spaceship, or where a quantum particle will be at a certain time, you must analyse functions in this way.

    • 2 years ago
  98. Inopeki Group Title
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    Really?

    • 2 years ago
  99. Inopeki Group Title
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    Then teach me! This is important!

    • 2 years ago
  100. TuringTest Group Title
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    Really, though of course actual physics problems are usually (though not always) a bit trickier than just plugging in a number but yes, functions are instrumental to calculus, and therefor physics As I said earlier calculus is about analyzing functions... I guess we should start to talk about functions in a very detailed way.

    • 2 years ago
  101. Inopeki Group Title
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    Yes please :D

    • 2 years ago
  102. TuringTest Group Title
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    the definition of a function for you will be: if you have something like y=f(x) (where f(x) is some arbitrary polynomial with x's in it) and for every x you put in you get exactly one number for y back, then y=f(x) is a function, and y is a function of x try to soak that in, I'll tell you more about it momentarily

    • 2 years ago
  103. Inopeki Group Title
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    So what you are saying is that if f(x)=y x=y?

    • 2 years ago
  104. Inopeki Group Title
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    Well not really

    • 2 years ago
  105. Inopeki Group Title
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    What you are saying is that if x is one and y is 2 and you add 1 to x you add one to y aswell?

    • 2 years ago
  106. TuringTest Group Title
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    no, just that we get only one value of y for each x for instance f(x)=y I defined that, whether or not it's a function however if for every x we put in we get only one y back, then f(x) is a function, and the variable y is a function of the variable x

    • 2 years ago
  107. TuringTest Group Title
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    scratch the *for instance :/

    • 2 years ago
  108. Inopeki Group Title
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    Gotta go eat dinner, brb

    • 2 years ago
  109. TuringTest Group Title
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    look at a graph like a circle (I know you haven't studied them much yet this way)|dw:1326043068666:dw|this is not a function because there is more than one y for each value of x... see you in a bit, I'm gonna have breakfast.

    • 2 years ago
  110. TuringTest Group Title
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    drawing a line from x to its corresponding y value you should know we need a vertical line|dw:1326043155512:dw|see how one x-coordinate corresponds to two different y values y1 and y2? But what about y=2x+5 ?|dw:1326043286670:dw|this graph has exactly one value of y for each value of x, so it is a function.

    • 2 years ago
  111. TuringTest Group Title
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    the idea above is what's called the 'vertical line test': if you can draw a vertical line at any point on the graph and intersect it twice, then it is not a function.

    • 2 years ago
  112. TuringTest Group Title
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    when you return look at this say we have a table of values for some y=f(x) and we want to know if f(x) is a function.... look at the table x |y=f(x) --------- -1 | 2 3 | 7 2 | 0 5 | 2 Is this a function? why or why nmot?

    • 2 years ago
  113. Inopeki Group Title
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    Its not. EXEPT the one that says 2 | 0<---------------< cause that means that there is only one value of x ^

    • 2 years ago
  114. Inopeki Group Title
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    The others seem to have two values of x

    • 2 years ago
  115. TuringTest Group Title
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    actually look at the x's each x has one y, so it is a function we have that f(5)=f(-1)=5 so two different x's give the same y, but that is okay the only thing against the rules would be to have something like this x | y ------ 1 | 3 2 | 5 1 | 4 because this means that f(1) has two values: f(1)=3 and 4 so this is not a function... look at the x to see which gives more than one f(x)... (I am using the terms f(x) and y here interchangeably) x | y ----- 2 | 0 3 | 1 5 | 6 3 | 6 1 | 0 function or not ant why?

    • 2 years ago
  116. Inopeki Group Title
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    Yeah

    • 2 years ago
  117. TuringTest Group Title
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    That's right FFM does the curriculum meet your approval? I doubt it with your standards :P

    • 2 years ago
  118. Inopeki Group Title
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    x | y ----- 2 | 0 =f(2)=2? im sorry, i dont really understand 3 | 1 5 | 6 3 | 6 1 | 0

    • 2 years ago
  119. TuringTest Group Title
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    f(2)=0 that means when x=2, y=0, yes?

    • 2 years ago
  120. Inopeki Group Title
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    Ooooh

    • 2 years ago
  121. TuringTest Group Title
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    that's all there is to that so is there any f(x) on the list that gives more than one number for a single x?

    • 2 years ago
  122. Inopeki Group Title
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    So, x | y ----- 2 | 0 = f(2)=0 3 | 1 = f(3)=1 5 | 6 = f(5)=6 3 | 6 = f(3)=6 1 | 0 = f(1)=0 So, no.

    • 2 years ago
  123. TuringTest Group Title
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    actually what is f(3) ?

    • 2 years ago
  124. Inopeki Group Title
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    Oh! 1 and 6!

    • 2 years ago
  125. TuringTest Group Title
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    exactly, so not a function :)

    • 2 years ago
  126. Inopeki Group Title
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    f(3)=1+6 so its not a function!

    • 2 years ago
  127. Inopeki Group Title
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    YEah :D

    • 2 years ago
  128. Inopeki Group Title
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    And that means that f(3) could be graphed in a circle?

    • 2 years ago
  129. TuringTest Group Title
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    right not 1+6=7 of course, it equal the set\[f(3)=\left\{ 1,6 \right\}\] It doesn't mean it would be a circle but it would fail the vertical line test, watch...

    • 2 years ago
  130. Inopeki Group Title
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    Of course.

    • 2 years ago
  131. Inopeki Group Title
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    Y cant have to places

    • 2 years ago
  132. Inopeki Group Title
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    two*

    • 2 years ago
  133. Inopeki Group Title
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    :)

    • 2 years ago
  134. TuringTest Group Title
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    thanks FFM, praise form Caeser :) |dw:1326046327322:dw|now look at the points of our possible function above. is there any place we can draw a vertical line that would intersect two points?

    • 2 years ago
  135. Inopeki Group Title
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    No way!

    • 2 years ago
  136. TuringTest Group Title
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    lol *Praise from Ceaser

    • 2 years ago
  137. TuringTest Group Title
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    no way to draw a vertical line and hit more than one point you say?

    • 2 years ago
  138. Inopeki Group Title
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    Not more than 3.

    • 2 years ago
  139. Inopeki Group Title
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    And there are 5 so..

    • 2 years ago
  140. TuringTest Group Title
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    more than 3? You only need to hit 2 points to show that this is NOT a function and here it is:|dw:1326046665544:dw|so again we show by this that f(3) has more than one value...

    • 2 years ago
  141. Inopeki Group Title
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    Yeah

    • 2 years ago
  142. TuringTest Group Title
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    are the following functions?|dw:1326046773020:dw|(I'll fill in different graph above)

    • 2 years ago
  143. Inopeki Group Title
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    f(x)=y?

    • 2 years ago
  144. TuringTest Group Title
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    yes f(x)=y because we have defined it that way now:|dw:1326046848239:dw|function?

    • 2 years ago
  145. Inopeki Group Title
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    NO

    • 2 years ago
  146. TuringTest Group Title
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    draw the vertical line that hits more than one point on the graph if that is true

    • 2 years ago
  147. Inopeki Group Title
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    No wait, there are no points

    • 2 years ago
  148. Inopeki Group Title
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    |dw:1326046981078:dw|

    • 2 years ago
  149. TuringTest Group Title
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    there are infinite points on a line.... a line intersects another line at a point|dw:1326047055394:dw|P is our intersect point can you draw a vertical line that hits the curve at more than one point?

    • 2 years ago
  150. Inopeki Group Title
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    Oh, so y is a vertical line?

    • 2 years ago
  151. Inopeki Group Title
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    Then this could be a function

    • 2 years ago
  152. Inopeki Group Title
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    |dw:1326047157974:dw| not function

    • 2 years ago
  153. Inopeki Group Title
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    But yours is

    • 2 years ago
  154. TuringTest Group Title
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    now you got the idea :D|dw:1326047282819:dw|function?

    • 2 years ago
  155. Inopeki Group Title
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    No.

    • 2 years ago
  156. TuringTest Group Title
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    show

    • 2 years ago
  157. Inopeki Group Title
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    |dw:1326047285288:dw|

    • 2 years ago
  158. TuringTest Group Title
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    just one line is sufficient, but yes :)||dw:1326047413725:dw|function?

    • 2 years ago
  159. Inopeki Group Title
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    Yes

    • 2 years ago
  160. TuringTest Group Title
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    good :) x | y ----- 1 | 3 2 | 3 3 | 3 4 | 3 function?

    • 2 years ago
  161. Inopeki Group Title
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    Yes

    • 2 years ago
  162. Inopeki Group Title
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    For all of them

    • 2 years ago
  163. Inopeki Group Title
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    But if it was 1|3 1|4 It would not

    • 2 years ago
  164. TuringTest Group Title
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    good x | y ----- 0 | 3 0 | 2 0 | 1 0 | 0 I think you already proved you know this, but function?

    • 2 years ago
  165. Inopeki Group Title
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    No

    • 2 years ago
  166. Inopeki Group Title
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    0 has 4 values of y

    • 2 years ago
  167. TuringTest Group Title
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    nice job, exactly slightly different: f(x)=x+3 function?

    • 2 years ago
  168. Inopeki Group Title
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    I guess that depends on what you plug in for x?

    • 2 years ago
  169. Inopeki Group Title
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    Actually, no

    • 2 years ago
  170. Inopeki Group Title
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    It is a function

    • 2 years ago
  171. Inopeki Group Title
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    It doesnt show multiple values of y, just x+3

    • 2 years ago
  172. TuringTest Group Title
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    Excellent!!!!!!!!

    • 2 years ago
  173. Inopeki Group Title
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    Yes! :D

    • 2 years ago
  174. TuringTest Group Title
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    perfect observation :) every x we put in gives us only one y. y=0 is y a function of x?

    • 2 years ago
  175. Inopeki Group Title
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    Only if we know that 0=x

    • 2 years ago
  176. TuringTest Group Title
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    not so, plug in various x into the function f(x)=0, we get the table x | y ---- 0 | 0 1 | 0 2 | 0 3 | 0 and so on... and I think you would agree that this chart constitutes a function.

    • 2 years ago
  177. TuringTest Group Title
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    tricky one, but now you know ;)

    • 2 years ago
  178. Inopeki Group Title
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    Yes, that chart constitutes a function. But if there were no chart?

    • 2 years ago
  179. TuringTest Group Title
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    They are all in agreement, the chart, the graph, everything. we don't need x=0 to get one value of y back; we always get one value of y. The same value of y, namely zero. do you know what the graph of y=0 looks like?

    • 2 years ago
  180. Inopeki Group Title
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    |dw:1326048301595:dw|

    • 2 years ago
  181. Inopeki Group Title
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    Thats 0,0

    • 2 years ago
  182. TuringTest Group Title
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    yes but I don't want the point (0,0) y=0 is a line: it's actually the x-axis the line x=0 is the y-axis think about that for a minute

    • 2 years ago
  183. Inopeki Group Title
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    Oh, right. |dw:1326048543616:dw|

    • 2 years ago
  184. Inopeki Group Title
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    Didnt really think of that

    • 2 years ago
  185. TuringTest Group Title
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    exactly, and does the graph of y=0 pass the vertical line test?

    • 2 years ago
  186. Inopeki Group Title
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    No

    • 2 years ago
  187. TuringTest Group Title
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    are you saying that you can draw a vertical line that intersects it twice? show...

    • 2 years ago
  188. Inopeki Group Title
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    OH you mean that? Never mind lol

    • 2 years ago
  189. TuringTest Group Title
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    so is the graph indicative of a function or not?

    • 2 years ago
  190. Inopeki Group Title
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    Yes, it is

    • 2 years ago
  191. TuringTest Group Title
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    perfect, what about the line x=0? (standard cartesian notation here , don't get hung up on the names of variables, we could easily have them to t and p or whatever)

    • 2 years ago
  192. Inopeki Group Title
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    |dw:1326049009726:dw|

    • 2 years ago
  193. TuringTest Group Title
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    yes, so is it a function?

    • 2 years ago
  194. Inopeki Group Title
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    Yes.

    • 2 years ago
  195. TuringTest Group Title
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    be careful can we draw a vertical line that touches the graph at more than one point? what would the table of values look like?

    • 2 years ago
  196. Inopeki Group Title
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    No wait, this becomes f(x)=x right?

    • 2 years ago
  197. TuringTest Group Title
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    no, this would be f(x)|x=0 (read f of x such that x is zero) here f(x) takes on a range of values from negative to positive infinity: x | f(x) ------ 0 | 0 0 | 1 0 | 2 0 | -1 0 | -2 etc. not a function, right?

    • 2 years ago
  198. Inopeki Group Title
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    I think i get it, its because there is no y?

    • 2 years ago
  199. Inopeki Group Title
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    but f(x) has endless possibilities

    • 2 years ago
  200. TuringTest Group Title
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    no this chart is the same as x | y ------ 0 | 0 0 | 1 0 | 2 0 | -1 0 | -2 I told you that y and f(x) are interchangeable y is the name of the coordinate which we are using to represent f(x) so here they are the same the chart itself is of the same type as you saw before, we have one value of x corresponding to multiple (in this case infinite) values of y, so it is not a function.

    • 2 years ago
  201. Inopeki Group Title
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    Ohhh, but if you had y corresponding to multiple values of x it would be a function.

    • 2 years ago
  202. TuringTest Group Title
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    right :)

    • 2 years ago
  203. Inopeki Group Title
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    Ah, now i see!

    • 2 years ago
  204. TuringTest Group Title
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    let me try to clear a few things up... what is the dependent and what is the independent variable here?

    • 2 years ago
  205. Inopeki Group Title
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    The dependent variable is y because if its a function or not depends on y having multiple values corresponding to one value of x.

    • 2 years ago
  206. TuringTest Group Title
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    right, but I'm not sure about your reasoning, so what about f(y)=x=y^2+3y which is dependent and which is independent?

    • 2 years ago
  207. Inopeki Group Title
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    x is the dependent one i think

    • 2 years ago
  208. TuringTest Group Title
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    exactly, so here x is a function of y we could have therefor more than one y value for an x value in this case, but not more than one x value for each y, if it is to be a function. so really the rule is better stated this way: "a relation is a function if each value of the independent variable corresponds to exactly one value of the dependent variable" because we can change the names or swap positions of the variables

    • 2 years ago
  209. Inopeki Group Title
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    Because we can have more of y but not more of x

    • 2 years ago
  210. Inopeki Group Title
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    exactly

    • 2 years ago
  211. TuringTest Group Title
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    exactly, look at the new version of our rule above (in quotes)

    • 2 years ago
  212. Inopeki Group Title
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    Yeah, that sounds right.

    • 2 years ago
  213. TuringTest Group Title
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    so whenever we see f(t)=r for instance which is dependent and which ins independent?

    • 2 years ago
  214. Inopeki Group Title
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    We should move to another "thread" this one is lagging for me

    • 2 years ago
  215. TuringTest Group Title
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    I agree go ahead

    • 2 years ago
  216. Inopeki Group Title
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    r is the dependent and t is independent

    • 2 years ago
  217. TuringTest Group Title
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    exactly :)

    • 2 years ago
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