If you want an object to be at the bottom of a stack of objects, you would use send to back. true

The following function returns a string of length n whose characters are all 'x'. give the order of growth (as a function of n) of the running time. recall that concatenating two strings in java takes time proportional to the sum of their lengths. public static string f(int n) { if (n == 0) return ""; if (n == 1) return "x"; return f(n/2) + f(n - n/2); } options: a) constant b) logarithmic c) linear d) linearithmic e)quadratic f)cubic g) exponential