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Alexxx [7]
3 years ago
6

WILL MARK BRAINLIEST!!!!!!!!!!!

Computers and Technology
2 answers:
gladu [14]3 years ago
5 0
I’m pretty sure it’s d if not then try C
Hope this helps
AysviL [449]3 years ago
3 0
Its either d or c!!!
You might be interested in
Write a date transformer program using an if/elif/else statement to transform a numeric date in month/day format to an expanded
boyakko [2]

I wrote it in Python because it was quick so i hope this is helpful

Answer:

date = input('Enter Date: ')

split_date = date.split('/')

month = split_date[0]

day = split_date[1]

if month == '1':

   english_month = 'January'

   spanish_month = 'Enero'

elif month == '2':

   english_month = 'February'

   spanish_month = 'Febrero'

elif month == '3':

   english_month = 'March'

   spanish_month = 'Marzo'

elif month == '4':

   english_month = 'April'

   spanish_month = 'Abril'

elif month == '5':

   english_month = 'May'

   spanish_month = 'Mayo'

elif month == '6':

   english_month = 'June'

   spanish_month = 'Junio'

elif month == '7':

   english_month = 'July'

   spanish_month = 'Julio'

elif month == '8':

   english_month = 'August'

   spanish_month = 'Agosto'

elif month == '9':

   english_month = 'September'

   spanish_month = 'Septiembre'

elif month == '10':

   english_month = 'October'

   spanish_month = 'Octubre'

elif month == '11':

   english_month = 'November'

   spanish_month = 'Noviembre'

elif month == '12':

   english_month = 'December'

   spanish_month = 'Diciembre'

US_English = f'{english_month} {day}'

International_Spanish = f'{day} {spanish_month}'

print(f'US English Form: {US_English}')

print(f'International Spanish Form: {International_Spanish}')

Input:

3/5

Output:

US English Form: March 5

International Spanish Form: 5 Marzo

Explanation:

You start by taking input from the user then splitting that at the '/' so that we have the date and the month in separate variables. Then we have an if statement checking to see what month is given and when the month is detected it sets a Spanish variable and an English variable then prints it to the screen.

Hope this helps.

7 0
3 years ago
CHALLENGE ACTIVITY |
fredd [130]

Answer:

maxSum = FindMax(numA, numB) + FindMax(numY, numZ);  

The maxSum is a double type variable which is assigned the maximum of the two variables numA numB PLUS the maximum of the two variables numY numZ using FindMax function. The FindMax() method is called twice in this statement one time to find the maximum of numA and numB and one time to find the maximum of numY numZ. When the FindMax() method is called by passing numA and numB as parameters to this method, then method finds if the value of numA is greater than that of numB or vice versa. When the FindMax() method is called by passing numY and numZ as parameters to this method, then method finds if the value of numY is greater than that of numZ or vice versa. The PLUS sign between the two method calls means that the resultant values returned by the FindMax() for both the calls are added and the result of addition is assigned to maxSum.

Explanation:

This is where the statement will fit in the program.

#include <iostream>

using namespace std;

double FindMax(double num1, double num2) {

  double maxVal = 0.0;    

  if (num1 > num2) { // if num1 is greater than num2,

     maxVal = num1;  // then num1 is the maxVal.    }

  else {          

     maxVal = num2;  // num2 is the maxVal.   }

  return maxVal; }  

int main() {

  double numA;

  double numB;

  double numY;

  double numZ;

  double maxSum = 0.0;  

maxSum = FindMax(numA, numB) + FindMax(numY, numZ);  

  cout << "maxSum is: " << maxSum << endl;   }

Lets take an example to explain this. Lets assign values to the variables.

numA = 6.0

numB = 3.0

numY = 4.0

numZ = 9.0

maxSum =0.0

maxSum = FindMax(numA, numB) + FindMax(numY, numZ);  

FindMax(numA,numB) method call checks if numA>numB or numB>numA and returns the maximum of the two. Here numA>numB because 6.0 is greater than 3.0. So the method returns numA i.e. 6.0

FindMax(numY, numZ) method call checks if numY>numZ or numZ>numY and returns the maximum of the two. Here numZ>numY because 9.0 is greater than 4.0. So the method returns numZ i.e. 9.0

FindMax(numA, numB) + FindMax(numY, numZ) this adds the two values returned by the method FindMax for each call.

FindMax returned maxVal from numA and numB values which is numA i.e. 6.0.

FindMax returned maxVal from numY and numZ values which is numZ i.e. 9.0.

Adding these two values 9.0 + 6.0 = 15

So the complete statement  FindMax(numA, numB) + FindMax(numY, numZ) gives 15.0  as a result.

This result is assigned to maxSum. Hence

maxSum = 15

To see the output on the screen you can use the print statement as:

cout<<maxSum;

This will display 15

6 0
3 years ago
I got enough points and enough brainliests how do I move on to my next rank?
emmasim [6.3K]

Answer: what do you mean exactly?

6 0
3 years ago
Read 2 more answers
While designing a website, a(n) ____ allows you to identify the specific amount of space used on a particular page before it con
Aleks [24]
The answer would be A: wireframe, im pretty sure...

Hope this helped!! :)

8 0
3 years ago
Read 2 more answers
When a device or service causes the system to hang during a normal boot, boot into __________ and disable the device or service
Olenka [21]

You should boot into <u>safe mode</u> and disable the device or service when it causes the computer system to hang during a normal boot.

<h3>What is safe mood?</h3>

Safe mood can be defined as a boot option in which the operating system (OS) of a computer system starts in <u>diagnostic mode</u> rather than in a normal operating mode, so as to enable the user correct any problems preventing the computer system to have a normal boot.

This ultimately implies that, safe mood is a diagnostic mode of the operating system (OS) of a computer system that is designed and developed to fix most problems within an operating system (OS) and for the removal of rogue software applications.

Read more on safe mood here: brainly.com/question/13026618

5 0
2 years ago
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