Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
To make a horizontal line in mathematics you require some sort of constant function. Where no matter the input x the output y will always equal to a certain number n.
For eg. consider n = 2. Hence the graph of line has an equation of y = n making it always intercept in (x, n). As x increases the y is the same therefore the horizontal infinite line emerges.
So to plot or code this kind of program you need to have the same values of y therefore y is a constant function.
Hope this helps.
r3t40
Answer:
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Explanation:
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The answer to this question is, A. Cramming.
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Answer:
Find and replace header and footer
