Hi I legit hate coding idk y

Assume that the data on a worksheet consume a whole printed page and two columns on a second page. You can do all of the followi

UNO [17]

Answer:

Option (2) is the correct answer of this question.

Explanation:

We will increase the left and right margins of the data on a worksheet that will consume a whole printed page and columns on second page then the data will print all in one page.

Following are the steps to print the Data all in on page:-

Step 1. Choose the Ribbon Page Design button

Step2. select 1 page in the Scale to Fit category, and in the Width box, we will automatic choose the height box.

Step 3 .Finally we will choose 1 page in the Height box to print your worksheet to a single page.

"Other options are incorrect because they are not related to the given question.i.e, to print the data in on page in the worksheet".

4 0

3 years ago

Write an application that solicits and inputs three integers from the user and then displays the sum, average, product, smallest

Oksana_A [137]

Answer:

Please find below program

Explanation:

import java.util.Scanner;

public class Tester {

public static void main(String[] args) {

// create an object of scanner class

Scanner scanner = new Scanner(System.in);

//prompt user to enter first number

System.out.println("Enter first integer: ");

// read value from entered by user and keep it in num1

int num1 = scanner.nextInt();

//prompt user to enter second number

System.out.println("Enter second integer: ");

// read value from entered by user and keep it in num2

int num2 = scanner.nextInt();

//prompt user to enter third number

System.out.println("Enter third integer: ");

// read value from entered by user and keep it in num3

int num3 = scanner.nextInt();

//close scanner as we have done with user input

scanner.close();

//calculate sum

int sum = num1+num2+num3;

//calculate average

int average = sum/3;

//calculate product

int product = num1*num2*num3;

//find out largest number

int largest = 0;

if (num1>=num2){

if(num1>=num3)

largest = num1;

else

largest = num3;

}else {

if(num2>=num3)

largest = num2;

else

largest = num3;

}

// find out smallest number

int smallest = 0;

if (num1<=num2){

if(num1<=num3)

smallest = num1;

else

smallest = num3;

}else {

if(num2<=num3)

smallest = num2;

else

smallest = num3;

}

// print all the information

System.out.println("Sum of three numbers : "+sum);

System.out.println("Product of three numbers : "+product);

System.out.println("Average of three numbers : "+average);

System.out.println("Smallest number : "+smallest);

System.out.println("Largest number : "+largest);

}

3 0

4 years ago

A ____ is a software program that is designed to cause damage to a computer or perform some other malicious act.

Dafna11 [192]

Computer virus, a software program that is designed to cause damage to the computer system or perform some other
malicious act

5 0

2 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

To easily add an organizational chart to a document, users should select .

stepladder [879]

The answer is SmartArt. One can easily add an organizational Chart to a document by using Smart Art. It is a graphical tool to easily make an easy visual representation, to make organizational types of artwork such as Organizational Chart, SmartArt is a tool to turn ordinary text into something more visually appealing, drawing attention to important information or making information easier to interpret and understand.

3 0

3 years ago