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angela210793

  • 5 years ago

@amistre64 will u help me in smth pls?

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  1. amistre64
    • 5 years ago
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    depends on what I have to do with smth :)

  2. angela210793
    • 5 years ago
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    O.o..let me attach the problem pls...

  3. angela210793
    • 5 years ago
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    U have to find the area of the circle...

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  4. angela210793
    • 5 years ago
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    it is a rhomb ***

  5. amistre64
    • 5 years ago
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    i could see this as looking for the radius; which would indicate the a triangle in Q1

  6. amistre64
    • 5 years ago
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    its a 3-4-5 right triangle

  7. amistre64
    • 5 years ago
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    we then find the max area of a square that fits in it...maybe; or the max part of the circle to fit it

  8. angela210793
    • 5 years ago
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    w8..i think i got how to solve it....all we have to do is find the height at ODC triangle and that=r :)

  9. angela210793
    • 5 years ago
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    And then S=pi*r^2

  10. angela210793
    • 5 years ago
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    Right?

  11. amistre64
    • 5 years ago
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  12. angela210793
    • 5 years ago
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    this was what u were talking abt 2? Thnx a loot :)

  13. amistre64
    • 5 years ago
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    the equation of the line that we want is: (0,3) ; (4,0) ; the slope then being 3/4 or.... y = -(3/4)x + 3 right? that is the equation for the hyp ... that line of 5

  14. amistre64
    • 5 years ago
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    we want a perp line to this that goes thru the origin (0,0) if anything

  15. amistre64
    • 5 years ago
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    y = (4/3)x

  16. amistre64
    • 5 years ago
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    the distance R is the distance from the origin to the point where these 2 lines meet right?

  17. angela210793
    • 5 years ago
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    ehe...

  18. angela210793
    • 5 years ago
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    r=12/5?

  19. amistre64
    • 5 years ago
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    (4/3)x = (3/4)x +3 dunno yet still figuring this out the hard way lol

  20. watchmath
    • 5 years ago
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    I think it should be -(3/4)x=(4/3)x+4 Amistre...

  21. angela210793
    • 5 years ago
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    w8 is easy...S(odc)= B*H/2=3*4/2=6 S(odc) is also=OE*DC/2=6----> OE=12/DC=12/5

  22. amistre64
    • 5 years ago
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    the slope of the hyp I drew is -3/4 with a y int at (0,3) right?

  23. angela210793
    • 5 years ago
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    I didn't get ur way......:(:(:( why with coordinates? O.o :(

  24. amistre64
    • 5 years ago
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    my way is sound to me; i understand graphs and lines better than drawings

  25. watchmath
    • 5 years ago
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    Your solution is nice angela :)

  26. angela210793
    • 5 years ago
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    Wht abt my way? is it wrong?

  27. angela210793
    • 5 years ago
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    Oh thnx :) :)

  28. amistre64
    • 5 years ago
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    4 up and 3 over is the same region as 3 up and 4 over, ...

  29. amistre64
    • 5 years ago
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    i dont really know if your way is right or wrong yet ;)

  30. angela210793
    • 5 years ago
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    hmmmm.....

  31. watchmath
    • 5 years ago
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    Her solution based on computing the area of ODC in two ways. First: 1/2*3*4 Second* 1/2*CD*r = 1/2*5*r

  32. angela210793
    • 5 years ago
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    Ehe...

  33. amistre64
    • 5 years ago
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    (4/3)x = (-3/4)x + 3 [3] 4x = (-9/4)x +9 [4] 16x = -9x + 36 25x = 36 x = 36/25 did I get that right?

  34. angela210793
    • 5 years ago
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    I didn't get ur way so i can't tell it if it is right or not.... Thank u very much anyway :):):):)

  35. watchmath
    • 5 years ago
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    Amistre, the y-intercept is at (0,4)

  36. amistre64
    • 5 years ago
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    4 36 -- * --- = y 3 25 4 12 -- * --- = y = 48/25 1 25 R = sqrt(x^2 + y^2) R = sqrt((36/25)^2+(48/25)^2)

  37. amistre64
    • 5 years ago
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    watch; it can be; but im actually using the other side; even tho I inadvertantly flipped it; it doesnt matter if we have 4 over and 3; or 3 over and 4 up; its the same triangle just flipped on itself with the same amount of circle it it ...

  38. amistre64
    • 5 years ago
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    its the point of tangency that is important; in my figuring

  39. angela210793
    • 5 years ago
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    let me read it from the start once again...

  40. amistre64
    • 5 years ago
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    like this

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  41. watchmath
    • 5 years ago
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    Yes, your computation gives the same r :)

  42. angela210793
    • 5 years ago
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    thank u both :)

  43. amistre64
    • 5 years ago
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    R = 12/5 yay!! lol

  44. angela210793
    • 5 years ago
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    :D ^_^

  45. amistre64
    • 5 years ago
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    so that verifies your answer angela :) I know my way is solid, which is why I went that route

  46. angela210793
    • 5 years ago
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    Will u see another 1 with inflexion points? (Is this how u call it?)

  47. amistre64
    • 5 years ago
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    inflection points are where the graph stays along a course in a general direction.. sure

  48. angela210793
    • 5 years ago
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    ^_^ ok w8 :)

  49. angela210793
    • 5 years ago
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    f(x)=x^4+c*x^3+x^2 what should b X so it will have 2 inflect. points? and then 1?

  50. amistre64
    • 5 years ago
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    'X' or 'c' ?

  51. amistre64
    • 5 years ago
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    f'= 4x^3 +3cx^2 +2x f'' = 12x^2 +6cx +2 = 0 for inflection points

  52. angela210793
    • 5 years ago
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    it says x...

  53. amistre64
    • 5 years ago
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    x, as far as I can tell, is an input and has no real effect on the graph of this thing; the constants describe the path it takes

  54. amistre64
    • 5 years ago
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    how are they defining inflection?

  55. angela210793
    • 5 years ago
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    they who? O.o

  56. amistre64
    • 5 years ago
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    they as in the people who designed the question, the authors of the material that you are using .... those guys

  57. amistre64
    • 5 years ago
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    I see the answers; but the question should be what value should 'c' be...

  58. angela210793
    • 5 years ago
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    C'mon...idk the words in english...............:S

  59. angela210793
    • 5 years ago
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    ok u solve it 4 c...

  60. amistre64
    • 5 years ago
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    f'' = 12x^2 +6cx +2 = 0 in order for this to have 2 real roots; we have to manipulate the discriminate of the quadratic formula....

  61. amistre64
    • 5 years ago
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    36c^2 -4(2)(12) > 0 for 2 inflections 36c^2 -4(2)(12) = 0 for 1 inflection

  62. amistre64
    • 5 years ago
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    36c^2 -96 = 0 36c^2 = 96 c^2 = 96/36 c = +- sqrt(96)/6 for one inflection

  63. amistre64
    • 5 years ago
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    and greater than that for 2 inflections

  64. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=sqrt%2896%2F36%29

  65. angela210793
    • 5 years ago
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    whr did tht 36c^2 come from?O.o Florida? :P

  66. amistre64
    • 5 years ago
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    it comes from taking the 1st and 2nd derivatives of the given equation

  67. amistre64
    • 5 years ago
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    youd have to understand a little calculus to get the idea of it; but basically I used proven methods to find when the 2nd derivative = 0

  68. amistre64
    • 5 years ago
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    y = x^4+c*x^3+x^2 y' = 4x^3 +3cx^2 +2x y'' = 12x^2 +6cx +2 ; when y'' = 0 we have a point of inflection

  69. amistre64
    • 5 years ago
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    y'' is a quadratic that can have its roots figured out by the quadratic formula; the discriminant of the quad formula tells us how many roots we will have

  70. amistre64
    • 5 years ago
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    b^2 -4ac (6c)^2 - 4(2)(12) 36c^2 -96 has to be greater than 0 for it to have 2 points for inflection 36c^2 -96 has to equal 0 for it to only have 1 point of inflection

  71. amistre64
    • 5 years ago
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    so you solve the equations of the discriminate for c..

  72. amistre64
    • 5 years ago
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    if we know what c is ro get the discriminate to = 0; then we should assume that larger values of c will make it greater then 0 for 2 roots

  73. angela210793
    • 5 years ago
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    ahaaa.....i got it....Thnx ^_^

  74. angela210793
    • 5 years ago
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    Will u help me in smth else?..... if u don't mind ofc...

  75. amistre64
    • 5 years ago
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    i gotta finish my international business homework; its due today..

  76. angela210793
    • 5 years ago
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    Oh..Ok..I'm sorry...didn't know tht.... Thnx a lot 4 helping me and good luck with ur hw :)

  77. amistre64
    • 5 years ago
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    thnx :) and good luck with yours :)

  78. angela210793
    • 5 years ago
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    ^_^

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