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RoseRoque

  • 3 years ago

Find the value of c such that the point P(a, b) lies on the graph of the function f. f(x) = x*sqrt(36 − x^2) + c; P(5, 10)

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  1. primeralph
    • 3 years ago
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    |dw:1367696013938:dw|

  2. surjithayer
    • 3 years ago
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    for f(x) to be defined36-x^2>=0,36>=x^2, or x^2<=36 \[\left| x \right|\le6,-6 \le x \le6\] c can have any real value \[but x \in \left( -6,6 \right)\]

  3. RoseRoque
    • 2 years ago
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    @jim_thompson5910

  4. RoseRoque
    • 2 years ago
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    @Callisto

  5. RoseRoque
    • 2 years ago
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    Can ANYONE help...both people who have posted are wrong and I need help preparing for my Final Exam

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