Why is this code not working properly?

Question

https://code.sololearn.com/wDC1qJ76tzjo/?ref=app

Answer 1

line no 2, 3: add $ to "num6"

Answer 2

STOP DOWNVOTE EVERYONE'S ANSWER!!!!

Answer 3

add da dollar sign to both "num6"s. u can see it in the console...

Answer 4

you forget to place $ in "num6"

Answer 5

Insert '#x27; sign to 'num6' in line 2 & 3.

Answer 6

?php
function mult($num1, $num2 ,$num6) {
$res = $num1 * $num2 + $num6 ;
return $res;
}

echo mult(8, 3, 6);
?>

Code should work fine now

Answer 7

num6 is not declared, need $num6

Answer 8

php looks a lot like perl to me. leaving off the "quot; in front symbol is the problem. In perl "quot; is used to access the scalar value of the symbol.

Answer 9

https://code.sololearn.com/wTzaNleiDt32/?ref=app

This works and you can see what's wrong there.

Answer 10

$ in num6

Answer 11

in php you need to add $ in the front of the variable in line 2 and line 3.

Answer 12

This code should work fine now
https://code.sololearn.com/wwuAu2kofRpz/?ref=app

Answer 13

you need to add $ to num6

Answer 14

you need to add $ sign before num6

Answer 15

you missed the dollar sigh on num6

Answer 16

There isn't a $ sign before num6

Answer 17

to declare variables in php, you should use $ sign in front of the name of variables. ex: $num1 $data_variable $result etc...

Answer 18

guys i cant link php to html document pls help...

Answer 19

you wrote num6 instead of $num6