Is there a way that I can declare the variable outside function?

Is there a way that I can declare a variable outside the function? $first and $second can't seem to read it outside function. and 1 more thing .remove() doesn't work.

function addition(){    let $first = parseInt($('#firstNum').val());    let $second = parseInt($('#secondNum').val());    $('h1').append($first + $second);  }  function subtraction(){    let $first = parseInt($('#firstNum').val());    let $second = parseInt($('#secondNum').val());    $('h1').append($first - $second);  }  function multiplication(){    let $first = parseInt($('#firstNum').val());    let $second = parseInt($('#secondNum').val());    $('h1').append($first * $second);  }  function division(){    let $first = parseInt($('#firstNum').val());    let $second = parseInt($('#secondNum').val());    $('h1').append($first / $second);  }  function clear(){    $('h1').remove();  }```  

https://stackoverflow.com/questions/65590041/is-there-a-way-that-i-can-declare-the-variable-outside-function January 06, 2021 at 12:04PM