Take a look at the following code:

1    let x = 1;
2    function f1()
3    {
4        let x = 2;
5        console.log(x);
6    }
7    console.log(x);

Explain why line 4 and line 6 output different numbers.

Because the x that is inside the function is scope.

Take a look at the following code:

let x = 10

function f1()
{
    console.log(x)
    let y = 20
}

console.log(f1())
console.log(y)

What will be the output of this code. Explain your answer in 50 words or less.

The first console.log will run and log 10. The second one won't and will give a reference error. Because 'y' is scope not global

Take a look at the following code:

const x = 9;

function f1(val) {
  val = val + 1;
  return val;
}

f1(x);
console.log(x);

const y = { x: 9 };

function f2(val) {
  val.x = val.x + 1;
  return val;
}

f2(y);
console.log(y);

What will be the output of this code. Explain your answer in 50 words or less.

The first console.log will produce 9 output, The second function will write the object as {x:10} because it will mutate the contents of the object