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mart [117]
3 years ago
14

A ____ resembles a circle of computers that communicate with each other.

Computers and Technology
1 answer:
Sidana [21]3 years ago
4 0

Answer:

ring network

Explanation:

In a ring network, the nodes are arranged in a circular pattern where each node is connected to two adjacent nodes. In this topology, any node can communicate with any other node via the intermediaries.

In comparison,

- in a star network every communication needs to pass through a central hub node

- in a bus, each node is connected to a shared linear communication bus.

- in a hierarchical network nodes are organized along a tree structure layout.

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You are developing a Windows forms application used by a government agency. You need to develop a distinct user interface elemen
-BARSIC- [3]

Answer:

The answer is "Developing the custom control for the user interface"

Explanation:

The major difference between customized control & user control would be that it inherit throughout the inheritance tree at different levels. Usually, a custom power comes from the system. Windows. UserControl is a system inheritance.

Using accuracy up is the major reason for designing a custom UI control. Developers must know how to use the API alone without reading the code for your component. All public methods and features of a custom control will be included in your API.

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3 years ago
All of the following are incentives proprietary schools use to attract a student except
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Idk what the answer choices are
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4 years ago
Read 2 more answers
In file hashing, a file is read by a special algorithm that uses the value of the bits in the file to compute a single number ca
Zielflug [23.3K]

Answer:

Option (B) i.e., Hash is the correct option.

Explanation:

A hash value is the number that has the fixed length and the data recognition is uniquely. It is also a type of file that is read by the special type algo, it uses the value of the bits in the following computer files which is used to manipulates the single number. That's why the following option is correct.

7 0
4 years ago
If you were going to construct a table that only included shape names with number prefixes, which shape would you include
NikAS [45]
The answer would be A. square
3 0
3 years ago
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
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