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romanna [79]
3 years ago
9

Which of the following helps in developing a microservice quickly? API Gateway Service registry Chassis Service Deployment

Computers and Technology
1 answer:
Arturiano [62]3 years ago
5 0

Answer:

Micro Service is a technique used for software development. In micro services structure, services are excellent and arrange as a collection of loosely coupled. API Gateway helps in developing micro services quickly.

Explanation:

The API gateway is the core of API management. It is a single way that allows multiple APIs to process reliably.

Working:

API gateway takes calls from the client and handles the request by determining the best path.

Benefits:

Insulated the application and partitioned into micro services.

Determine the location of the service instances.

Identify the problems of services.

Provide several requests.

Usage:

API stands for application program interface. It is a protocol and tool used for building software applications. Identify the component interaction and used the Graphical Interface component for communication.

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Turns on her laptop and gets the error message "OS NOT FOUND." She checks the hard disk for damage or loose cable, but that is n
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It needs an Operating System like a cable or something that will help it operate look for more and double check
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3 years ago
Omega Software Inc. is currently running a training program for employees to upgrade their software skills so that they are able
MArishka [77]

Answer:

Omega Software is attempting to make a change in the area of people.

Explanation:

Types of Organizational Change:

Most organizations have to go through a change to keep up with the changing market dynamics, trends and technologies. There are four major types of organizational change:

  1. Structural
  2. Strategic
  3. People
  4. Process

People change:

One of the organizational change is people change where organizations strive towards the improvement of their employee's skills set and productivity. This is usually achieved by launching various specialized and general training programs for employees to enhance their knowledge and widen their skills set so that they can work more efficiently for the organization.

Therefore, it can be concluded that Omega Software is attempting to make a change in the area of People.

3 0
3 years ago
Aniyah is setting up a group of computers in her office that will share several devices. Which type of operating system should s
jenyasd209 [6]

Answer:

B

Explanation:

the software needs to share thus should be networked

8 0
2 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
Which one of the following is malicious software that denies you access to your files or your organization's files unless you pa
gladu [14]

Answer:

Ransomware

Explanation:

Just as its name implies, you have to pay a 'ransom' to get a locked file that  belongs to you

4 0
1 year ago
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