I'm just going to do minus a positive 3 times each of these.

We can multiply both sides of this equation by x plus two over one. x plus two.

I'm moving it 5 meters.

They work only during the day.

So in this case, n factorial was 7.

I'm gonna call you back.

I've actually given you some information here that we can't just assume when we were doing regular multiplication that, a times b is always equal to b times a.

So, 8 goes into, doesn't go into 1, it goes into 15 1 time 1 times 8 is 8

This is the same thing as 48 1 times 5 4.

So we can do it the same way.

So this term, in theory, should be that row times what?

I could either write that as x y, or I could just write that as xy.

Oh, actually, I made a mistake. y over x is equal to the natural log of x plus c.

You could say that the work is equal to 5 newtons times, I'll just write a dot for times.

3 times 3, plus 1 times 5, plus 2 times 5, right.

You can put them down, essentially out of the way.

And then to get to a cubed, what did we do?

And then to get to a to the fourth, what did we do?

To get that x plus one out of the denominator, we multiply both sides of this equation times x plus one over one times this side.

The time is 5 times 10 to the second seconds.

But of course, we're doing the chain rule.

Thank you.

It is time to go.

And those white markings are not natural markings.

And that makes sense because two to the three is two multiplying by itself three times, to the fifth is two multiplying by itself five times, and then we're multiplying the two, so we're going to multiply two eight times.

He's next. I'm sure of it.

I'll call back soon.

If you want to do it algebraically, you could cross multiply.

Now, a squared literally means a times a.

You can say x plus five over one.

Hey, look, I gotta go. Can I call you later?

They continued working.

Well, oxytocin doesn't only act on your brain.

Well, whenever we see an x in the denominator, the temptation is to multiply it by x.

So 7 minus 3 is 4, so 4 factorial.

Wherever we want to go, we could go arbitrarily we could scale a up by some arbitrary value.

You get a 1 and a 1.

And that sum I'm going to define as a times s.

If this was just 3 was so 7 times 6 times 5 times 4 times 3 times 2 times 1.

So it was 7 times 6 times 5 times 4 times 3 times 2 times 1.

Is it practical?

If it's false, then the flag hit will valuate to zero, 1 hit will be 1.

And then the denominator will have the b.

But we can't just multiply the left hand side by x.

So once you pass c, this thing becomes a 1, and you're just