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kozerog [31]
3 years ago
7

V=1/3πh^2 (3r-h) solve for r

Mathematics
1 answer:
stellarik [79]3 years ago
5 0
R=1v+h([pi]h^2) all over 3
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lina2011 [118]

\\ \sf\longmapsto \dfrac{1}{x-1}>4

  • Multiply x-1 on both sides

\\ \sf\longmapsto \dfrac{1}{x-1}\times (x-1)> 4(x-1)

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\\ \sf\longmapsto 1> 4x-4

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\\ \sf\longmapsto 1+4> 4x

\\ \sf\longmapsto 5> 4x

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\\ \sf\longmapsto \dfrac{5}{4}>x

\\ \sf\longmapsto x< \dfrac{5}{4}

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ATQ to the equation \bf \dfrac{1}{x-1} x should be greater than 1 .Because if it becomes 1 then the denominator will be 0 which is impossible.

Hence x>1

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\\ \sf{:}\!\implies 1

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\\ \therefore\boxed{\bf x\epsilon \left(1,\dfrac{5}{4}\right)}

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Answer:<BOC= 47

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