A switch’s port Gi0/1 has been correctly enabled with port security. The configuration sets the violation mode to restrict. A frame that violates the port security policy enters the interface, followed by a frame that does not. Which of the following answers correctly describe what happens in this scenario? (Choose two answers.)
A. The switch puts the interface into an err-disabled state when the first frame arrives.
B. The switch generates syslog messages about the violating traffic for the first frame.
C. The switch increments the violation counter for Gi0/1 by 1.
D. The switch discards both the first and second frame.

Write a pseudocode solution for each of these problems. 1. design a while loop that lets that user enter a number. the number should be multiplied by 10, and the result stored in a variable named product. the loop should iterate as long as product contains a value less than 100. 2. design a do-while loop that asks the user to enter two numbers. the numbers should be added and the sum displayed. the loop should ask the user whether he or she wishes to perform the operation again. if so, the loop should repeat; otherwise it should terminate. 3. design a for loop that displays the following set of numbers: 0, 10, 20, 30, 40, 50 100. 4. design a nested loop that displays 10 rows of # characters. there should be 15 # characters in each row. 5. convert this for loop to a while loop. declare integer count for count = 1 to 50 display count end for 6. find the error in the following pseudocode. declare boolean finished = false declare integer value, cube while not finished display “enter a value to be cubed.” input value; set cube = value ^ 3 display value, “ cubed is “, cube end while

Sarah is having a hard time finding a template for her advertising business that she may be able to use at a later date and also make it available to her colleagues

What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }