MATH TIME!

Discussion in 'User Topics' started by Mr.Lame, Jan 9, 2018.

  1. MarkFL

    MarkFL Guest

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    I forgot to mention the formulas of Euler and de Moivre, but my main point is there exist lots of choices for determining roots of polynomials.
     
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  2. Mr.Lame

    Mr.Lame Veteran

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    @MarkFL

    I'm not sure how to start out this equation.

    The perimeter of a rectangle is 40 inches and its area is 96 square inches. Find the length and the width of the rectangle.
     
  3. MarkFL

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    Okay, let's let x by the width and y be the height (both in inches). If the perimeter is 40 inches, then the semi-perimeter is 20 inches, and so we have:

    x + y = 20

    If the area is 96 inches squared, we have:

    xy = 96

    Now, solving the first equation for y, we have:

    y = 20 - x

    Substituting for y into the second equation, we have:

    x(20 - x) = 96

    Can you solve this quadratic equation?
     
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  4. Mr.Lame

    Mr.Lame Veteran

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    8 and 12?
     
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  5. Mr.Lame

    Mr.Lame Veteran

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    And it doesn't matter which is which right?
     
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  6. MarkFL

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    Yes:

    x(20 - x) = 96

    20x - x^2 = 96

    x^2 - 20x + 96 = 0

    (x - 12)(x - 8) = 0

    (x,y) = (8,12), (12,8)

    Can you explain how we can have two solutions?
     
  7. MarkFL

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    Exactly...so you answered my question about having two solutions. Excellent! :gjob:
     
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  8. Mr.Lame

    Mr.Lame Veteran

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    SCORE! :D
     
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  9. Mr.Lame

    Mr.Lame Veteran

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    Going to try this in an hour or so to see if I retained how to solve it o.o
     
  10. MarkFL

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    Suppose the semi-perimeter is S and and the area is A...then what are x and y?
     
  11. Mr.Lame

    Mr.Lame Veteran

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    Isn't the semi only half of the rectangle then?
     
  12. MarkFL

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    The semi-perimeter is half the perimeter. Instead of being given numbers for the semi-perimeter and area, I am giving you parameters, so that you can solve a more general problem. That's what I would do when I was a student, so that I could then look at how the solutions to the question would change as the parameters changed (especially as they approached the boundaries of the problem). It gave me a deeper insight to the behavior of the solutions. :)
     
  13. Mr.Lame

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    So in other words, I'm looking at P1/2=2L+2W ?
     
  14. MarkFL

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    lame021.png
     
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  15. Mr.Lame

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    @MarkFL

    A calculator and a battery together sell for 21. The price of the calculator is 20 more than the price of the battery. Find the price of the calculator and the price of the battery.

    So P= 21-(C+20)?
     
  16. MarkFL

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    I would let B be the price of the battery and C be the price of the calculator. From the given information, we may state:

    B + C = 21

    C - B = 20

    If we add the two equations, we get:

    2C = 41

    C = 41/2 = 20.5

    And so B = 0.5

    And so, we find the price of the battery is $0.50 and the price of the calculator is $20.50. :)
     
  17. Mr.Lame

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    Wait how does it become 2C without 2B?
     
  18. MarkFL

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    The B's cancel out to zero, because one is positive and the other is negative...this is a type of elimination...by adding the two equations, we eliminate B and wind up with an equation with just one variable. ;)
     
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  19. Mr.Lame

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    ohhhhhh I see it now!
     
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  20. JC.Martin

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    99 in calculus go to skool
     

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