\( V_{\text{rms}} \text{(mono)} > V_{\text{rms}} \text{(dia)} > V_{\text{rms}} \text{(poly)} \)
\( V_{\text{rms}} \text{(dia)} < V_{\text{rms}} \text{(poly)} < V_{\text{rms}} \text{(mono)} \)
\( V_{\text{rms}} \text{(mono)} < V_{\text{rms}} \text{(dia)} < V_{\text{rms}} \text{(poly)} \)
\( V_{\text{rms}} \text{(mono)} = V_{\text{rms}} \text{(dia)} = V_{\text{rms}} \text{(poly)} \)
The root mean square speed \( V_{\text{rms}} \) is given by: \[ V_{\text{rms}} = \sqrt{\frac{3RT}{m}} \] Since the gases are at the same temperature and pressure, the root mean square speed depends on the molar mass \( m \). For neon (monoatomic), chlorine (diatomic), and uranium hexafluoride (polyatomic), the molar mass increases in the order: \[ V_{\text{rms}} \text{(mono)} > V_{\text{rms}} \text{(dia)} > V_{\text{rms}} \text{(poly)} \] Thus, the correct answer is \( V_{\text{rms}} \text{(mono)} > V_{\text{rms}} \text{(dia)} > V_{\text{rms}} \text{(poly)} \).

\[ f(x) = \left\{ \begin{array}{ll} 1 - 2x & \text{if } x < -1 \\ \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2 \\ \frac{11}{18} (x-4)(x-5) & \text{if } x > 2 \end{array} \right. \]