1. Determine the output of each of the following circuits, assuming that the upper input is...

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1. Determine the output of each of the following circuits, assuming that the upper input is 1 and the lower input is 0. What would be the output when the upper input is 0 and the lower input is 1? a. b. C. 2.The following table represents the addresses and contents (using hexadecimal notation) of some cells in a machine's main memory. Starting with this memory arrangement, follow the sequence of instructions and record the final contents of each of these memory cells: Address 0x00 Contents OxAB 0x01 0x53 0x02 0x03 0xD6 0x02 Step 1 Move the contents of the cell whose address is 0x03 to the cell at address 0x00. Step 2 Move the value 0x01 into the cell at address 0x02. Step 3 Move the value stored at address 0x01 into the cell at address 0x03. 9. Perform each of the following additions assuming the bit strings represent values in two's complement notation. Identify each case in which the answer is incorrect because of overflow. a. 00101 + 01000 b. 11111 +00001 C. 01111 +00001 d. 10111 + 11010 e. 11111 + 11111 01100 f. 00111 1. Determine the output of each of the following circuits, assuming that the upper input is 1 and the lower input is 0. What would be the output when the upper input is 0 and the lower input is 1? a. b. C. 2.The following table represents the addresses and contents (using hexadecimal notation) of some cells in a machine's main memory. Starting with this memory arrangement, follow the sequence of instructions and record the final contents of each of these memory cells: Address 0x00 Contents OxAB 0x01 0x53 0x02 0x03 0xD6 0x02 Step 1 Move the contents of the cell whose address is 0x03 to the cell at address 0x00. Step 2 Move the value 0x01 into the cell at address 0x02. Step 3 Move the value stored at address 0x01 into the cell at address 0x03. 9. Perform each of the following additions assuming the bit strings represent values in two's complement notation. Identify each case in which the answer is incorrect because of overflow. a. 00101 + 01000 b. 11111 +00001 C. 01111 +00001 d. 10111 + 11010 e. 11111 + 11111 01100 f. 00111