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Svet_ta [14]
1 year ago
8

Create a timeline of the evolution of computers and their impact on society

Computers and Technology
1 answer:
OleMash [197]1 year ago
3 0

While the conceptual idea after a computer was developed in the 19th century, the first electronic computer was created in the 1940s.

<h3>How does the evolution of computers impacts the society?</h3>

Computers have transformed the way people relate to one another and their living environment, as well as how humans manage their work, their communities, and their time. Society, in turn, has affected the development of computers via the needs people have for processing data.

<h3>What are the evolution of computers?</h3>

The first modern computer was built in the 1930s and was called the Z1, which was followed by large machinery that took up whole rooms. In the '60s, computers evolved from professional use to private use, as the first personal computer was presented to the public.

To learn more about computers, refer

brainly.com/question/24540334

#SPJ9

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Numerous engineering and scientific applications require finding solutions to a set of equations. Ex: 8x + 7y = 38 and 3x - 5y =
Gre4nikov [31]

The brute strength method determines if each x and y value satisfies both conditions. To do this, we must iterate through each result in the specified range and insert them into both models.

<h3>What are Elegant mathematical technique?</h3>

ALGORITHM :

1. Take the values of all coefficients and SET flag = FALSE

2. Run a FOR loop for x in range (-10, 11). 11 won't be included.

2a. Inside the first FOR, start the y FOR loop in range(-10,11). 11 won't be included

2b. Inside the y FOR loop, check IF for a particular value of x and y both the equations satisfied.

2c. If yes, set flag TRUE and print values of x and y.

2d. ELSE the flag stays FALSE.

END FOR Y

END FOR X

3. Check if flag = FALSE, then print NO SOLUTION

PYTHON CODE :

a=int(input()) #taking input for each coefficient

b=int(input())

c= int(input())

a1= int(input())

b1= int(input())

c1= int(input())

flag = False

for x in range(-10,11): #checking for all values of x in range -10 to 10

for y in range(-10,11): #checking for all values of x in range -10 to 10

if (a  x + b  y – c == 0) and (a1*x + b1*y - c1 == 0) : #checking if the x and y values satisfy the equation

flag = True #setting the flag if solution is found

print('Solution : x = {-10, 10}, y = {-10, 10}'.format(x,y)) #if they satisfy print x & y

if flag = False : #if flag stays false, that means there is no solution

print('No solution')

OUTPUT : The output is given below.

More about the Elegant mathematical technique link is given below.

brainly.com/question/27934739

#SPJ1

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How many nibbles are in 18 bytes??
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When you are printing handouts, which of these can you do? A. Save paper by placing more than one slide per page. B. Check the f
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Sound effects are located within the Microsoft® Office applications. In the Insert ribbon, what would you select to find pre-pro
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Sometimes clip art has sound files in them, it may have gotten updated though, that answer is the only one that could work.
7 0
3 years ago
1)
spayn [35]

Answer:

1)

for(i = 0; i < NUM_VALS; ++i) {

  if(userValues[i] == matchValue) {

     numMatches++;  }    }

2)  

for (i = 0; i < NUM_GUESSES; i++) {

      scanf("%d", &userGuesses[i]);   }

  for (i = 0; i < NUM_GUESSES; ++i) {

         printf("%d ", userGuesses[i]);    }

3)

sumExtra = 0;

for (i = 0; i < NUM_VALS; ++i){

     if (testGrades[i] > 100){  

        sumExtra = testGrades[i] - 100 + sumExtra;    }       }

4)

for (i = 0; i < NUM_VALS; ++i) {

    if (i<(NUM_VALS-1))  

   printf( "%d,", hourlyTemp[i]);

    else  

    printf("%d",hourlyTemp[i]); }      

Explanation:

1) This loop works as follows:

1st iteration:

i = 0

As i= 0 and NUM_VALS = 4 This means for condition i<NUM_VALS  is true so the body of loop executes

if(userValues[i] == matchValue) condition checks if element at i-th index position of userValues[] array is equal to the value of matchValue variable. As matchValue = 2 and i = 0 So the statement becomes:

userValues[0] == 2

2 == 2

As the value at 0th index (1st element) of userValues is 2 so the above condition is true and the value of numMatches is incremented to 1. So numMatches = 1

Now value of i is incremented to 1 so i=1

2nd iteration:

i = 1

As i= 1 and NUM_VALS = 4 This means for condition i<NUM_VALS  is true so the body of loop executes

if(userValues[i] == matchValue) condition checks if element at i-th index position of userValues[] array is equal to the value of matchValue variable. As matchValue = 2 and i = 1 So the statement becomes:

userValues[1] == 2

2 == 2

As the value at 1st index (2nd element) of userValues is 2 so the above condition is true and the value of numMatches is incremented to 1. So numMatches = 2

Now value of i is incremented to 1 so i=2

The same procedure continues at each iteration.

The last iteration is shown below:

5th iteration:

i = 4

As i= 4 and NUM_VALS = 4 This means for condition i<NUM_VALS  is false so the loop breaks

Next the statement: printf("matchValue: %d, numMatches: %d\n", matchValue, numMatches);  executes which displays the value of

numMatches = 3

2)

The first loop works as follows:

At first iteration:

i = 0

i<NUM_GUESSES is true as NUM_GUESSES = 3 and i= 0 so 0<3

So the body of loop executes which reads the element at ith index (0-th) index i.e. 1st element of userGuesses array. Then value of i is incremented to i and i = 1.

At each iteration each element at i-th index is read using scanf such as element at userGuesses[0], userGuesses[1], userGuesses[2]. The loop stops at i=4 as i<NUM_GUESSES evaluates to false.

The second loop works as follows:

At first iteration:

i = 0

i<NUM_GUESSES is true as NUM_GUESSES = 3 and i= 0 so 0<3

So the body of loop executes which prints the element at ith index (0-th) index i.e. 1st element of userGuesses array. Then value of i is incremented to i and i = 1.

At each iteration, each element at i-th index is printed on output screen using printf such as element at userGuesses[0], userGuesses[1], userGuesses[2] is displayed. The loop stops at i=4 as i<NUM_GUESSES evaluates to false.

So if user enters enters 9 5 2, then the output is 9 5 2

3)

The loop works as follows:

At first iteration:

i=0

i<NUM_VALS is true as NUM_VALS = 4 so 0<4. Hence the loop body executes.

if (testGrades[i] > 100 checks if the element at i-th index of testGrades array is greater than 100. As i=0 so this statement becomes:

if (testGrades[0] > 100

As testGrades[0] = 101 so this condition evaluates to true as 101>100

So the statement sumExtra = testGrades[i] - 100 + sumExtra; executes which becomes:

sumExtra = testGrades[0] - 100 + sumExtra

As sumExtra = 0

testGrades[0] = 101

So

sumExtra = 101 - 100 + 0

sumExtra = 1

The same procedure is done at each iteration until the loop breaks. The output is:

sumExtra = 8

4)

The loop works as follows:

At first iteration

i=0

i < NUM_VALS is true as  NUM_VALS = 4 so 0<4 Hence loop body executes.

if (i<(NUM_VALS-1))   checks if i is less than NUM_VALS-1 which is 4-1=3

It is also true as 0<3 Hence the statement in body of i executes

printf( "%d,", hourlyTemp[i]) statement prints the element at i-th index i.e. at 0-th index of hourlyTemp array with a comma (,) in the end. As hourlyTemp[0] = 90; So 90, is printed.

When the above IF condition evaluates to false i.e. when i = 3 then else part executes which prints the hourlyTemp[3] = 95 without comma.

Same procedure happens at each iteration unless value of i exceeds NUM_VAL.

The output is:

90, 92, 94, 95

The programs along with their output are attached.

4 0
4 years ago
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