Tell whether the given value is a solution of the Inequality, \( \begin{array}{ll}\text { 6. } \frac{x}{6} \leq 1 ; x=5 & \text { 7. } 1.4 v>16 ; v=11 \\ \text { 8. } t+9.8 \geq 21.01 ; t=11.1 & \text { 9. } 2 \frac{1}{2}<\frac{m}{30} ; m=90 \\ \text { 10. } \frac{1}{2}>3 p ; p=\frac{1}{6} & \text { 11. } 2.16 \geq 3 z-0.5 ; z=0.6\end{array} \)

Let’s evaluate each inequality with its corresponding value! 6. For \( \frac{x}{6} \leq 1 \) and \( x=5 \): - Plugging in gives \( \frac{5}{6} \leq 1 \), which is true since \( 0.8333... \leq 1 \). So, 5 is a solution! 7. For \( 1.4v > 16 \) and \( v=11 \): - Plugging in gives \( 1.4 \times 11 > 16 \) or \( 15.4 > 16 \), which is false. Thus, 11 is not a solution! 8. For \( t+9.8 \geq 21.01 \) and \( t=11.1 \): - Plugging in gives \( 11.1 + 9.8 \geq 21.01 \), which simplifies to \( 20.9 \geq 21.01 \). This is false, so 11.1 is not a solution! 9. For \( 2 \frac{1}{2} < \frac{m}{30} \) and \( m=90 \): - Converting \( 2 \frac{1}{2} \) to an improper fraction gives \( 2.5 < \frac{90}{30} \) or \( 2.5 < 3 \), which is true! Therefore, 90 is a solution! 10. For \( \frac{1}{2} > 3p \) and \( p=\frac{1}{6} \): - Plugging in gives \( \frac{1}{2} > 3 \times \frac{1}{6} \) or \( \frac{1}{2} > \frac{1}{2} \), which is false. So, \( \frac{1}{6} \) is not a solution! 11. For \( 2.16 \geq 3z - 0.5 \) and \( z=0.6 \): - Plugging in gives \( 2.16 \geq 3 \times 0.6 - 0.5 \) or \( 2.16 \geq 1.8 - 0.5 \), which simplifies to \( 2.16 \geq 1.3 \). This is true! Thus, 0.6 is a solution!