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Inopeki

  • 2 years ago

Algebra, Turingtest :D

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  1. TuringTest
    • 2 years ago
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    FOIL Inopeki :D (a+b)(c+d)

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

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

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

  5. TuringTest
    • 2 years ago
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    good, simplify

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

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

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

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

  10. Inopeki
    • 2 years ago
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    -5t

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

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

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

  14. Inopeki
    • 2 years ago
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    Huh :O

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

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

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

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

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

  20. Inopeki
    • 2 years ago
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    Oh right xd

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

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

  23. TuringTest
    • 2 years ago
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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) ???

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

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

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

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

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

  29. Inopeki
    • 2 years ago
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    No

  30. TuringTest
    • 2 years ago
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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.

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

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

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

  34. TuringTest
    • 2 years ago
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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...

  35. Inopeki
    • 2 years ago
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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

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

  37. Inopeki
    • 2 years ago
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    Oh right

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

  39. TuringTest
    • 2 years ago
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    right

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

  41. TuringTest
    • 2 years ago
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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)

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

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

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

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

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

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

  48. Inopeki
    • 2 years ago
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    OH the -

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

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

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

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

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

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

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

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

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

  58. Inopeki
    • 2 years ago
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    Oh right, 0

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

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

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

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

  63. pratu043
    • 2 years ago
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    ok...

  64. pratu043
    • 2 years ago
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    Shall I help?

  65. TuringTest
    • 2 years ago
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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 ;)

  66. TuringTest
    • 2 years ago
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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\[a^2-b^2\]

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

  68. pratu043
    • 2 years ago
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    Its an identity.

  69. TuringTest
    • 2 years ago
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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

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

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

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

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

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

  75. TuringTest
    • 2 years ago
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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)

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

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

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

  79. TuringTest
    • 2 years ago
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    which part?

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

  81. TuringTest
    • 2 years ago
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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?

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

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

  84. Inopeki
    • 2 years ago
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    oooh

  85. Inopeki
    • 2 years ago
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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

  86. TuringTest
    • 2 years ago
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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.

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

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

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

  90. Inopeki
    • 2 years ago
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    Yeah

  91. TuringTest
    • 2 years ago
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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.

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

  93. TuringTest
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    which numbers on the RHS?

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

  95. TuringTest
    • 2 years ago
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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

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

  97. TuringTest
    • 2 years ago
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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.

  98. Inopeki
    • 2 years ago
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    Really?

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

  100. TuringTest
    • 2 years ago
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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.

  101. Inopeki
    • 2 years ago
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    Yes please :D

  102. TuringTest
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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

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

  104. Inopeki
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    Well not really

  105. Inopeki
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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?

  106. TuringTest
    • 2 years ago
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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

  107. TuringTest
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    scratch the *for instance :/

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

  109. TuringTest
    • 2 years ago
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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.

  110. TuringTest
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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.

  111. TuringTest
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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.

  112. TuringTest
    • 2 years ago
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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?

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

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

  115. TuringTest
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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?

  116. Inopeki
    • 2 years ago
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    Yeah

  117. TuringTest
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    That's right FFM does the curriculum meet your approval? I doubt it with your standards :P

  118. Inopeki
    • 2 years ago
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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

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

  120. Inopeki
    • 2 years ago
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    Ooooh

  121. TuringTest
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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?

  122. Inopeki
    • 2 years ago
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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.

  123. TuringTest
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    actually what is f(3) ?

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

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

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

  127. Inopeki
    • 2 years ago
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    YEah :D

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

  129. TuringTest
    • 2 years ago
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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...

  130. Inopeki
    • 2 years ago
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    Of course.

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

  132. Inopeki
    • 2 years ago
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    two*

  133. Inopeki
    • 2 years ago
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    :)

  134. TuringTest
    • 2 years ago
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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?

  135. Inopeki
    • 2 years ago
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    No way!

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

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

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

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

  140. TuringTest
    • 2 years ago
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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...

  141. Inopeki
    • 2 years ago
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    Yeah

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

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

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

  145. Inopeki
    • 2 years ago
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    NO

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

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

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

  149. TuringTest
    • 2 years ago
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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?

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

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

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

  153. Inopeki
    • 2 years ago
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    But yours is

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

  155. Inopeki
    • 2 years ago
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    No.

  156. TuringTest
    • 2 years ago
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    show

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

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

  159. Inopeki
    • 2 years ago
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    Yes

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

  161. Inopeki
    • 2 years ago
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    Yes

  162. Inopeki
    • 2 years ago
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    For all of them

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

  164. TuringTest
    • 2 years ago
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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?

  165. Inopeki
    • 2 years ago
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    No

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

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

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

  169. Inopeki
    • 2 years ago
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    Actually, no

  170. Inopeki
    • 2 years ago
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    It is a function

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

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

  173. Inopeki
    • 2 years ago
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    Yes! :D

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

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

  176. TuringTest
    • 2 years ago
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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.

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

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

  179. TuringTest
    • 2 years ago
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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?

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

  181. Inopeki
    • 2 years ago
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    Thats 0,0

  182. TuringTest
    • 2 years ago
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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

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

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

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

  186. Inopeki
    • 2 years ago
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    No

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

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

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

  190. Inopeki
    • 2 years ago
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    Yes, it is

  191. TuringTest
    • 2 years ago
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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)

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

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

  194. Inopeki
    • 2 years ago
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    Yes.

  195. TuringTest
    • 2 years ago
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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?

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

  197. TuringTest
    • 2 years ago
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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?

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

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

  200. TuringTest
    • 2 years ago
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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.

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

  202. TuringTest
    • 2 years ago
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    right :)

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

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

  205. Inopeki
    • 2 years ago
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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.

  206. TuringTest
    • 2 years ago
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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?

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

  208. TuringTest
    • 2 years ago
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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

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

  210. Inopeki
    • 2 years ago
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    exactly

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

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

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

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

  215. TuringTest
    • 2 years ago
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    I agree go ahead

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

  217. TuringTest
    • 2 years ago
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    exactly :)

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