If the positive square root of x is between $3$ and $11$ , then what inequality represents all possible values of x?

3 < x < 11

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9 < x < 11

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9 < x < 121

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x < 3

x < 11

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x < 9

x < 121

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Solution

The correct option is C $9 < x < 121$
Given, $3 < \sqrt{x} < 11$
Taking square on both sides we get $9 < x < 121$

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s i n^{- 1} (c o s {\frac{x^{2} + 3 | x | + 4}{x^{2} + 3 | x | + 5}}) = s i n (s i n^{- 1} {\frac{1 - 9 | x |}{9 | x |}}) + \frac{π}{2},

121 (x - a)^{4} (x - b)^{6} (x - c)^{12}

\frac{3}{11} + x = \frac{9}{11}

x

Find the value of x.

$3 x + \frac{9}{3} = 5 x + \frac{5}{9}$

Find the value of x in the equation $3 x + \frac{9}{3} = 5 x + \frac{5}{9} .$

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