What is the solution to the inequality below?
12+x2 3(x-6)?

With alll of except for the 2 that i did already

Set $r$ is a set of rectangles such that (1) only the grid points shown here are used as vertices, (2) all sides are vertical or horizontal and (3) no two rectangles in the set are congruent. if $r$ contains the maximum possible number of rectangles given these conditions, what fraction of the rectangles in set $r$ are squares? express your answer as a common fraction.

The mean of 10 values is 19. if further 5 values areincluded the mean becomes 20. what is the meanthese five additional values? a) 10b) 15c) 11d) 22