Answer:
<em>The registers that are compared are instructions 3 and 4</em>
<em>Explanation:</em>
<em>From the question given,</em>
<em>Recall that we need to explain what the hazard detection unit is doing during the 5th cycle of execution and which registers are being compared.</em>
<em>Now,</em>
<em>The instructions on the 5th cycle, at the stage ID/EX and IF/ID:</em>
<em>The instruction values are in ID/EX : sub $t2, $t3, $t6 (instruction 3)</em>
<em>The instruction values are in IF/ID: sub $t3, $t1 $t5 (instruction 4)</em>
<em>The register $t3 is compared in the instructions 3 and 4</em>
<em>The hazard detection unit between instruction 4 and 5t o be compared, it need to find out the values of $t1</em>
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B. format because changing the brightness of the photo is changing its default format
Answer:
Lowest Level; Machine Language.
Explanation:
The lowest level of a computer is machine language, which are strings of 0's and 1's in bits, and it's possible to perform tasks at this level. It's however difficult to do and humans created <em>Assembly</em>; a type of low level programming language to be readable, and converts to machine language so that we don't have to work in binary.
Answer:
2^11
Explanation:
Physical Memory Size = 32 KB = 32 x 2^10 B
Virtual Address space = 216 B
Page size is always equal to frame size.
Page size = 16 B. Therefore, Frame size = 16 B
If there is a restriction, the number of bits is calculated like this:
number of page entries = 2^[log2(physical memory size) - log2(n bit machine)]
where
physical memory size = 32KB which is the restriction
n bit machine = frame size = 16
Hence, we have page entries = 2^[log2(32*2^10) - log2(16)] = 2ˆ[15 - 4 ] = 2ˆ11
Answer:
Adding extra horizontal scroll, Blocking mobile devices from viewing, Eliminating extra links, Resizing content to fit a screen.
Explanation: