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(C) For equation $I$: $\sqrt{x} = \frac{(18)^{15/2}}{x^2}$Multiplying both sides by $x^2$,we get $x^{1/2} \cdot x^2 = (18)^{15/2}$$x^{5/2} = (18)^{15/

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if $x < y$

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(C) For equation $I$: $\sqrt{x} = \frac{(18)^{15/2}}{x^2}$Multiplying both sides by $x^2$,we get $x^{1/2} \cdot x^2 = (18)^{15/2}$$x^{5/2} = (18)^{15/2}$Raising both sides to the power of $2/5$,we get $x = (18)^{(15/2) \cdot (2/5)} = 18^3 = 5832$.For equation $II$: $\sqrt{y} = \frac{(19)^{9/2}}{y}$Multiplying both sides by $y$,we get $y^{1/2} \cdot y = (19)^{9/2}$$y^{3/2} = (19)^{9/2}$Raising both sides to the power of $2/3$,we get $y = (19)^{(9/2) \cdot (2/3)} = 19^3 = 6859$.Comparing the values,$x = 5832$ and $y = 6859$.Since $5832 < 6859$,therefore $x < y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{x} - \frac{(18)^{15/2}}{x^2} = 0$
$II.$ $\sqrt{y} = \frac{(19)^{9/2}}{y}$

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Solve the given two equations and select the correct answer from the given options.$I.$ $\sqrt{x} - \frac{(18)^{15/2}}{x^2} = 0$$II.$ $\sqrt{y} = \frac{(19)^{9/2}}{y}$