Why dose the SAR instruction always clear the overflow flag?

Based on the diagram below, what is the result for register AL after executing the mnemonic opcode/operand instructions? MOV AX, 0x0083 MOV BL 0x2 DIV BL AX BL AL АН 0083 02 01 Dividend Divisor Quotient Remainder The resulted AL is 43. The resulted AL is 41. Ob. The resulted AL is 30. O c. The resulted AL is 40. Od.

Programming Project: For Loops Write an Assembly Language Program that produces the Times Table for a specific integer value between 1 to 12. NOTE: You choose and hard code only one value (i.e. 7) and produce the times table for that value. Also note that since you are using registers and memory locations make sure that at each step you screen shot each result for each part of the Times Table. For example 7 x 12 = 84 but in Hexadecimal the value is 54. Sample code in C #include void main() ( int i, j; printf("Enter an integer: "); scanf("%d", &i); printf("Times %d Table\n", i). for(j = 0; j<= 12; j =j+1) { printf("%d x %d = %d\n", j, 1. j'i); } printf("\n");

Given the number Ox21100006, what is it when interpreted as a MIPS instruction? 1. The instruction format is type |. (R, I, or J?) 2. The opcode value is (specify as a hexadecimal value, e.g., OxfEE3, lower-case x, upper case hex values.)