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3 years ago

What is MS-Word? Write some versions of MS-Word.

Computers and Technology

2 answers:

shtirl [24]3 years ago

8 0

Ms- word it is , hope it was helpful

tiny-mole [99]3 years ago

3 0

Answer:

MS- word is like Microsoft Word, Word 2010, Word 2013, Word 2016

the versions are 12.0. 14.0. 15.0. 16.0

Explanation:

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A web designer finds out after the launch that one of the external links did not have "http://' and therefore the link appeared

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Answer is oc design testing

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3 years ago

What should I change my fortnite name to: Barneys broken thumb or I dont have aim assist

Ket [755]

Answer: yes

Explanation:

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3 years ago

Write a program that asks the user to enter a number within the range of 1 through 10. Use a switch statement to display the Rom

Studentka2010 [4]

Answer:

// program in C++.

// headers

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

// Variable

int num;

cout << "Enter a number between 1 and 10: ";

// read input

cin >> num;

// validate input number

while(num<1||num>10)

{

// if input is wrong then print error message

cout<<"Wrong input!!Enter again:";

// read again

cin>>num;

}

// print output

cout << "The Roman numeral for "<<num<<" is:";

// switch

switch (num)

{

case 1:

// if input is 1

cout<<"I"<<endl;

// exit the switch

break;

case 2:

// if input is 2

cout<<"II"<<endl;

// exit the switch

break;

case 3:

// if input is 3

cout<<"III"<<endl;

// exit the switch

break;

case 4:

// if input is 4

cout<<"IV"<<endl;

// exit the switch

break;

case 5:

// if input is 5

cout<<"V"<<endl;

// exit the switch

break;

case 6:

// if input is 6

cout<<"VI"<<endl;

// exit the switch

break;

case 7:

// if input is 7

cout<<"VII"<<endl;

// exit the switch

break;

case 8:

// if input is 8

cout<<"VIII"<<endl;

// exit the switch

break;

case 9:

// if input is 9

cout<<"IX"<<endl;

// exit the switch

break;

case 10:

// if input is 10

cout<<"X"<<endl;

// exit the switch

break;

// default

default:

break;

}

return 0;

}

Explanation:

Read a number from usr and assign it to variable "num". If the input number is less than 1 or greater than 10 then ask again to enter a number until user enter a number between 1-10 only.Then with the help of switch() function print the equivalent Roman number.

Output:

Enter a number between 1 and 10: -5

Wrong input!!Enter again:12

Wrong input!!Enter again:6

The Roman numeral for 6 is:VI

7 0

3 years ago

Need help finding the totals and with the empty spots

Lunna [17]

Answer:

weff WEF ef ef aeF EFaef EFE fe efaef ea f

Explanation:

8 0

3 years ago

Input 3 positive integers from the terminal, determine if tlrey are valid sides of a triangle. If the sides do form a valid tria

butalik [34]

Answer:

In Python:

side1 = float(input("Side 1: "))

side2 = float(input("Side 2: "))

side3 = float(input("Side 3: "))

if side1>0 and side2>0 and side3>0:

if side1+side2>=side3 and side2+side3>=side1 and side3+side1>=side2:

if side1 == side2 == side3:

print("Equilateral Triangle")

elif side1 == side2 or side1 == side3 or side2 == side3:

print("Isosceles Triangle")

else:

print("Scalene Triangle")

else:

print("Invalid Triangle")

else:

print("Invalid Triangle")

Explanation:

The next three lines get the input of the sides of the triangle

side1 = float(input("Side 1: "))

side2 = float(input("Side 2: "))

side3 = float(input("Side 3: "))

If all sides are positive

if side1>0 and side2>0 and side3>0:

Check if the sides are valid using the following condition

if side1+side2>=side3 and side2+side3>=side1 and side3+side1>=side2:

Check if the triangle is equilateral

if side1 == side2 == side3:

print("Equilateral Triangle")

Check if the triangle is isosceles

elif side1 == side2 or side1 == side3 or side2 == side3:

print("Isosceles Triangle")

Otherwise, it is scalene

else:

print("Scalene Triangle")

Print invalid, if the sides do not make a valid triangle

else:

print("Invalid Triangle")

Print invalid, if the any of the sides are negative

else:

print("Invalid Triangle")

4 0

3 years ago