So three four times.

Let me do that again.

You coil it around so that you get 1s and 2s and 3s on the outside, like

Three times three is nine.

If I do Rand three by three.

Let's do a quick check.

3 plus 3 is 0.

So we're integrating with respect to y first.

Well this is going to be 3 3 3.

Let's say I had this set of numbers.

They're literally saying, what are four threes?

So this could be a 3 and a 3, or it could be a 1 and a 9.

1... 2... 3...

March 3rd,

We know it's 1 3 cup.

So 10 3, so we might have 0, 1, 2, 3, 4.

We get I'll do it in a different color we get 3 2 c1 plus 3 2 c2 is equal to what's 3 2 times 2?

And in either case, what's it equal to?

Last one.

So let's divide them both by 3.

So the particular solution of this differential equation I will do in a different color, and it is y to the 3 2 minus x to the 3 2 is equal to 7.

Well, let's do that. Let's do the floor.

I'm going to repeat this process all the way through.

3 times 2 is 6.

So this g of 3 is equal to 6 plus f of what?

Minus 3 minus 4 is minus 7.

Well, 2 times 2 is 4.

And if we did that we get

And the answer is relatively easy.

This is equal to 645, and then we have our six zeroes.

One quarter times six is, what, three halves, about six over four is three halves, and one quarter times three is three quarters.

And how many of these star patches do I now have?

Negative 3 2 times negative 3 2 is positive 9 4.

And same thing for 3.

So x is two, so it's here.

It could also be rewritten as three two times.

But notice, I get the same answer.

How many apples would I have left?

Well, just like over here, we're left with just the question mark.

We could write this down as 4 3, which is the same thing as 3 3 3 3.

So, you know, one times three is three.

Zero times 3 is zero.

I have 1 3.

The next one is 1, 1 2 1. For the second expression, 2 3 1. 1 1 gives us a 1.

Well, it's going to be 3 of the 15 sections.