MATH TIME!

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

  1. MarkFL

    MarkFL Guest

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    Okay, we have:

    A = (H/2)(b1+ b2)

    Multiply both sides by 2/H:

    2A/H = b1 + b2

    Subtract b2 from both sides:

    2A/H - b2 = b1 :)
     
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  2. *~BRAT~*

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    1(-2) - -3(1 - -2)^2
    -2 + 3 =4 = 5?????????
     
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  3. Mr.Lame

    Mr.Lame Veteran

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    @MarkFL

    Word problem woes

    The length of the rectangle is 9ft less than twice the width of the rectangle. The peimeter of the rectangle is 54 ft. Find the width and the length
     
  4. MarkFL

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  5. Mr.Lame

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  6. Mr.Lame

    Mr.Lame Veteran

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    Totally fluked the answer I'm shocked
     
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  7. MarkFL

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  8. Mr.Lame

    Mr.Lame Veteran

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    I'm still in shock about getting that answer right haha
     
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  9. Mr.Lame

    Mr.Lame Veteran

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    @MarkFL

    Hey Mark quick question.

    In 3x^3-5x^2=0

    I know we factor out the x^2 so it's x^2(3x-5)=0 But what do we do with the x^2? The rest is easy, just when I checked it magically disappears haha.
     
  10. MarkFL

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    You want to set each factor equal to zero and solve for x, to get all of the solutions...so solve these equations:

    x^2 = 0

    3x - 5 = 0

    What do you get?
     
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  11. Mr.Lame

    Mr.Lame Veteran

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    Root 0 is 0
    x=0

    3x=5
    X= 5/3

    So in zero solving, I can treat the x^2 as it's own x equation?
     
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  12. MarkFL

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    Yes, the "zero factor property" states that if the product of two or more factors is zero, then at least one of the factors must be zero. So, if we can find the values of x that make each factor zero, we find all of the values of x that make the original product equal to zero. :)
     
  13. Mr.Lame

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    Interesting, I guess my brain is trying to get over the "It's attached to the bracket" so treat it as part of the factor. Then I checked the next question as 2x^3-8x, pulled out 2x(x^2-4)=0 , 2x=0, x=0/2=0, x=0, (x+2)(x-2)=0 x=-2,2. Seems like the right steps?
     
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  14. MarkFL

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    You rocked it, bro! :gjob:

    Just a note: when you have gotten to:

    2x(x^2 - 4)=0

    Continue factoring:

    2x(x+2)(x - 2) = 0

    Now, at this point you can consider 2 and x to be separate factors...and setting 2 = 0 isn't going to yield anything (because it can never be true), so you can just set:

    x = 0

    x + 2 = 0

    x - 2 = 0

    Solving these gives you, as you also found:

    x = -2, 0, 2
     
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  15. Mr.Lame

    Mr.Lame Veteran

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    OH!!! Okay the brain hiccup has been resolved :D . Thanks Mark :)
     
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  16. MarkFL

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    Sometimes you may get factors that can't be further factored into linear factors (factors of the form ax + b). So, you then use other methods for determining the roots of those factors, such as the quadratic formula if the factor is quadratic. :)
     
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  17. Mr.Lame

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    This quiz I'm not gonna crash and burn on, I REFUSE :cussing:
     
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  18. Mr.Lame

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    My favorite formula :hearts:
     
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  19. MarkFL

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    There are also cubic and quartic formulas, but they are extremely cumbersome. For those that don't factor or have rational roots, I will use the Newton-Raphson method from differential calculus to determine approximations for the roots. :)
     
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  20. Mr.Lame

    Mr.Lame Veteran

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    I have yet to reach that level :rofl:
     
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