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meriva
3 years ago
6

What is this line of code doing? scanf("%f", &height);

Computers and Technology
1 answer:
Andreas93 [3]3 years ago
8 0
It's saving a float value in your variable "height".

for example:

float height = 0.0;

printf("Insert your height: ");
scanf("%f", &height);
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Write a program that prompts the user to enter an equation in the form of 10 5, or 10-5, or 1*5, or 13/4, or 13%4. The program s
grandymaker [24]

Answer:

  1. equation = input("Enter an equation: ")
  2. if("+" in equation):
  3.    operands = equation.split("+")
  4.    result = int(operands [0]) + int(operands[1])
  5.    print(operands[0] + "+" + operands[1] + "=" + str(result))
  6. elif("-" in equation):
  7.    operands = equation.split("-")
  8.    result= int(operands [0]) - int(operands[1])
  9.    print(operands[0] + "-" + operands[1] + "=" + str(result))  
  10. elif("*" in equation):
  11.    operands = equation.split("*")
  12.    result = int(operands [0]) * int(operands[1])
  13.    print(operands[0] + "*" + operands[1] + "=" + str(result))  
  14. elif("/" in equation):
  15.    operands = equation.split("/")
  16.    result = int(operands [0]) / int(operands[1])
  17.    print(operands[0] + "/" + operands[1] + "=" + str(result))  
  18. elif("%" in equation):
  19.    operands = equation.split("%")
  20.    result = int(operands [0]) % int(operands[1])
  21.    print(operands[0] + "%" + operands[1] + "=" + str(result))

Explanation:

The solution code is written in Python 3.

Firstly prompt user to enter an equation using input function (Line 1).

Create if-else if statements to check if operator "+", "-", "*", "/" and "%" exist in the input equation. If "+" is found (Line 3), use split method to get the individual operands from the equation by using "+" as separator (Line 5). Output the equation as required by the question using string concatenation method (Line 6).  The similar process is repeated for the rest of else if blocks (Line 7 - 22).  

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Answer:

Word: A word processor

           File extension is .doc

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Answer:

Contiguous

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