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mathmath333

  • one year ago

Prove

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Prove the area of the region enclosed by the graph }\hspace{.33em}\\~\\ & y=-|x\pm k|+5,y=0\hspace{.33em}\\~\\ & \normalsize \text{ is constant for} \ \ k,\ k\in \mathbb{R}\ \hspace{.33em}\\~\\ \end{align}}\)

  2. ganeshie8
    • one year ago
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    I'm finding it hard to visualize... could you show the area in graph ?

  3. mathmath333
    • one year ago
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    https://www.desmos.com/calculator/ylkcvfm5ss

  4. ganeshie8
    • one year ago
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    Ahh that should be easy

  5. ganeshie8
    • one year ago
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    First of all, we acknowledge that the the absolute value gives the distance between two points on number line: |dw:1434119749216:dw|

  6. perl
    • one year ago
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    You can prove it using calculus. The integral of that is constant.

  7. ganeshie8
    • one year ago
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    In light that of above thing, \(|x-k|\) represents the distance between \(x\) and \(k\).

  8. ganeshie8
    • one year ago
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    Set the given function equal to \(0\) and solve \(x\) intercepts : \[-|x-k|+5=0 \implies |x-k|=5\]

  9. mathmath333
    • one year ago
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    x=5+k.-5+k

  10. ganeshie8
    • one year ago
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    Yes, subtract them to get the base of triangle

  11. mathmath333
    • one year ago
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    base->10

  12. ganeshie8
    • one year ago
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    |dw:1434120100063:dw|

  13. mathmath333
    • one year ago
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    ok thnx

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