Conv Celsius to Fahrenheit

Conv Celsius to Fahrenheit
0

#1

Tell us what’s happening:

PerryChalmers, lukaszsobek, I appreciate your help for Somme reason I am at a total loss. Perry when I entered the formula I am doing it so that the var Fahrenheit is defined. Whenever I enter the formula not as part of the definition I get null or an error saying that it was expecting a function call but got an expression. so I think its how I’m entering it.
Your code so far

function convertToF(celsius) {
  var fahrenheit;
  // Only change code below this line
  fahrenheit = (convertToF * 9/5 + 32);
  
  // Only change code above this line
  return fahrenheit;
}

// Change the inputs below to test your code
convertToF(-10);

Your browser information:

Your Browser User Agent is: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/603.3.8 (KHTML, like Gecko) Version/10.1.2 Safari/603.3.8.

Link to the challenge:


#2

You are supposed to use the passed argument celsius to do the calculations but you are using the function convertToF instead.


#3

thank you I appreciate such a quick response. if you are talking about like this Fahrenheit = (celcius * (9/5)) + 32, I have tried that and it keeps returning an error saying that it can’t find the var celcius.


#4

The argument name is celsius not celcius :slight_smile: