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xxTIMURxx [149]
2 years ago
14

Explain what are Graphics Cards

Computers and Technology
1 answer:
julia-pushkina [17]2 years ago
6 0

Answer:

Graphic Cards are designated to make the pc have more FPS and actually power (It's not actually accurated, but you may search on the internet)

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If an Administrator performs a clean install of Windows Server 2012 R2 on a new server, and then moves critical domain services
gayaneshka [121]

Answer:

The correct answer to the following question will be "Server role migration".

Explanation:

  • A process of moving data from one data to the next. Security concerns are the causes behind the system relocation, the hardware is also being changed and several other influences.
  • With this, you are setting up a new server, either virtual or physical, running a new version of Windows (Windows Server) and then moving your positions and facilities to the newly constructed Windows Server Virtual Machine/Physical.

Therefore, it's the right answer.

3 0
3 years ago
A laptop gets replaced if there's a hardware issue. Which stage of the hardware lifecycle does this scenario belong to?
Ilya [14]

Answer: Maintenance. This is the stage where software is updated and hardware issues are fixed if and when they occur. Retirement. In this final stage, hardware becomes unusable or no longer needed and it needs to be properly removed from the fleet

Explanation:

3 0
3 years ago
When should you use the Reply All function when replying to an email
Dmitrij [34]
When the email was sent as a group email 

5 0
3 years ago
Read 2 more answers
Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3
Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

3 0
3 years ago
You created an interactive application named GreenvilleRevenue. The program prompts a user for the number of contestants entered
Korolek [52]

Answer:

Explanation:

The program in this code is written correctly and has the messages applied in the code. Therefore, the only thing that would need to be done is pass the correct integers in the code. If you pass the integer 100 contestants for last year and 300 for current year. Then these inputs will provide the following output as requested in the question.

The competition is more than twice as big this year!

This is because the current year would be greater than double last years (100 * 2 = 200)

4 0
3 years ago
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