Working With A Square

Discussion in 'Other Pre-University Math' started by nycmathguy, Jul 6, 2021.

  1. nycmathguy

    nycmathguy

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    20210706_023725.jpg See attachment.

    Let A = area of square.

    A = (side)^2

    Side = 2x + 8

    A(x) = (2x + 8)^2

    Domain = (-infinity, infinity).

    Yes?
     
    nycmathguy, Jul 6, 2021
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  2. nycmathguy

    MathLover1

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    A=8*8-2*x*x
    A=64-2x^2

    domain: -infinity, infinity
    range:-infinity,64
     
    MathLover1, Jul 6, 2021
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  3. nycmathguy

    nycmathguy

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    1. What is wrong with my effort?

    2. How did you determine the domain and range for this problem?

    3. Am I bombarding you with too many questions?

    I think less questions lead to a more detailed reply, which is what I need to really understand precalculus. Otherwise, I am just wasting my time and yours.
     
    nycmathguy, Jul 7, 2021
    #3
  4. nycmathguy

    MathLover1

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    you say
    A(x) = (2x + 8)^2 which is wrong
    the area of the square is 8^2=64
    from that you need to deduct the area x*x=x^2 from both sides, means minus 2x^2
    pay attention to the graph above: you see 4 little isosceles triangles, when you cut them off you can make 2 little squares whose side length is x
    these two little squares (2x^2) you need to deduct from total area (64)

    A(x) = 64-2x^2
     
    Last edited: Jul 7, 2021
    MathLover1, Jul 7, 2021
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  5. nycmathguy

    nycmathguy

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    Ok.
     
    nycmathguy, Jul 7, 2021
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