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Start your free trialAndres Vicente
1,665 PointsCreate a New Float Variable named $floatOne with a value of 1.5. Without changing the value of $integerOne or $floatOne,
I haven't find the solution
<?php
//Place your code below this comment
$integerOne = 1;
$integerTwo = 2 ;
$integerOne +=5;
$integerTwo -= 1;
$floatOne = 1.5
echo $integerOne * $floatOne;
?>
1 Answer
Daniel Stopka
13,520 PointsHi Andres, you are missing semicolon at the end of $floatOne variable;