So plus another 4 centimeters.

And in three minutes, we filled in another table in our

(Laughter)

There, we just set up a equation with one unknown and we can solve for x.

So you're going to get some combination of the i and j unit vectors.

Tons of studies should have been done it's a serious issue.

I'm going to multiply 8 times both sides of this equation.

And then we evaluate it.

Minus 2 times minus 1, that's plus 2, is equal to c.

Plus 1 s that's this right there times the Laplace transform of 1.

I got it from a friend.

Jay, you got them?

3 minus nothing is 3.

And there you go.

Now quickly, do me a favor and add all of those up.

And now I evaluate this at 1 and 0.

let's just divide by a again.

So this was c plud d i, so this is its conjugate.

So you get 10 billion for them.

No, lady, not here. LOUANNE

And of course, I returned with all sorts of bruises and things like that but nevertheless, it was such an unbelievable experience, because then, where on Earth are you going to experience that in a piece of music?

Where did you get these?

Do we have 20,000? Yes, we have 20,000!

And then minus 56 minus 10, so minus 66.

Let's see how this works.

I know there's a lot of fractions here, but if you just work through this, the fractions actually work out.

Add to the imaginary parts you have 5 i minus 6 i. You get negative i

The vessels looked better.

So we're in this scenario right over here.

Maybe Ben earned 50 and Abby earned 0.

I added 7x to both sides of that equation.

And so, if we just put this on the other side of the equation, what do we get?

These are a this is like a 15 million dollar line.

Ma We got the inspiration for our experiment from previous research on salmonella.

Transform of any arbitrary function very, very quickly.

Here we go the other way.

Prick a finger, you get some blood, about 50 microliters.

We can derive great pleasure from books.

When I multiplied a 2 by 3 vector times a 3 by 2 vector, what did I get?

Those are very valuable.

So even with our hand drawn graph, we were able to inspect it and see that, yes, we were able to come up with the point 3 comma 3, and that does satisfy both of these equations.

If you want, you can just multiply this guy out and you'll get that right there.

Theory and empirical evidence

That's actually shown right here in this graph.

From k is equal to 0, to n of a to the k.