Well, I had to stop that last presentation because

last video for any particular number.

last video you'll get your solution set.

In the last video, I defined what dimensionality is.

In the last video, we counted.

We saw that in the last video.

And I showed you that in the last video.

In the last video, we figured out that we could just use the

In the last video we started off with two linear transformations.

And this is what we got in the last video.

We've seen point slope form in the last video.

So in the last video I showed you that these guys are

Which we saw in the last video is equivalent to 6.4 times 3.2.

And we saw an example of that in the last video.

In the last presentation we learned the derivative of sin is what?

Based on the examples in the last video

This is exactly what we've been doing in the last video, right?

In the last video we saw a method of figuring out what the basis for column space is.

In the last video on even and odd functions,

And if you watched the last video, what's two times zero?

In the last video we learned what it meant to take the product of two matrices.

Because in the last video we figured out that two to the eighth was equal to two hundred and fifty six.

And you know, in the last video, we did this for a particular x.

In the last video we talked about a particular way of finding a area of a rhombus.

In the last presentation, I hopefully gave you a little bit of an intuition of what a derivative is.

And a good composite problem was that problem that I had introduced you to at the end of that last seminar.

Well, after the last video, hopefully, we're a little familiar with how you add matrices.

Let's tackle a slightly harder problem than what we saw in the last video.

And let me just be clear it's not just also any polynomial, and even in the last video, when we had x 3 1, this was neither even nor odd.

In the last video, I was confusing you with compound interest, and now I will continue to do so.

In the last video, we found the slope at a particular point of the curve y is equal to x squared.

And that was very fast compared to what we had to do at the end of the last video.

And now, we've gotten ourselves to a problem that's just like the problems we did in the last video.

So the slope, if you saw the last module that just introduces how we calculate the slope, that's just rise over run.

And like we saw in the last video we had to be very careful because we have an inequality here.

Like we did in the last video, let's solve for cosine of 2t and solve for sine of 2t.

Now, using what we've just learned and what we learned in the last video, let's do some more complicated problems.

In the last video we saw how a matrix and figuring out its inverse can be used to solve a system of equations.

In the last video I think I, just off the cuff, said that all your null space has to be empty.

In the last hyperbola video I didn't get a chance to do some concrete examples.

In the last video, I spoke somewhat theoretically about what a null space is and we showed that it is a valid subspace.

In the last video we did a couple of lattice multiplication problems and we saw it was pretty straightforward.

Well, I hinted in the last presentation that those coefficients were actually terms of a Pascal triangle.

In the last video, we saw that these can be pretty straightforward to solve if the left hand side is a perfect square.

In the last video, we showed that if you had a line, which we'll call a directrix we draw the directrix.