Well, here we have two numbers that add up let's say that a is equal to fj.

So this is equal to s a times the Laplace transform of sine of a t.

The second one is always true.

I could draw it like that.

First there's a bunch of logic to say that it should be the case that say the a red, a green, and a yellow variables, one of them should be true.

Plus this a times that a.

We return b.

So that means that f of x has to be less than positive a or greater than negative a.

Once again, two numbers when we add them they equal whats the coefficient here, it's negative one.

Once x reaches 0, z has to equal ab. Keep in mind that's exactly what we return the value of z.

Not A or B is true as long as B is true, and also true if not A is true, which means A is false.

So I don't want my_sort() to return a copy of the list. I want, after I call my_sort() on a , if I then refer to a

Let's call our function gcd. It takes 2 inputst that are integers a and b.

They could have just asked how many minutes after the meeting will it take Betty to reach point A.

Now, from there we look at A, and we see that there are two pages that A links to.

There are four combinations in total.

So the general or the standard form for an ellipse centered at the origin is x squared over a squared plus y squared over b squared is equal to 1.

So this is not a good bat, so either I'm super evil or this last one is the winner and it turns out that the last one is the winner.

It's x minus the x coordinate of the focus.

And what do I mean by perfect square?

Return sign of a 1 if a is positive and 1 if it 's negative.

The idea is as follows

In the case where it's odd, it's a little bit more complicated. Pull one of the b's at.

If test(a) is true, then we return the second parameter b.

And then 3 times 1 is 7.

You switch the top left and the bottom right.

So that's 32 16, which equals to 2 seconds.

One is the sum function.

The value of b is 123. That was what we passed in at the second input.

I just have to add the exponents cause I have the same base and I'm taking the product, that's going to be a to the eighth power, and I still have this three out front.

It works as follows

So this term now if we substitute r3 with that, this will just turn into a this will just turn into a plus 1.

For example, we can say that the alarm is true whenever the earthquake or burglary is true with this sentence.

But if I fix the bug, then I get 5, 1, 8, 1.

And then you could multiply both sides by a squared.

That means that the absolute value of f of x, or f of x has to be less than a away from 0.

Well, once again, it's nice if we can retain this pattern, where every time we decrease the exponent we're dividing by a.

left with an a in the numerator, and a divided by a is just 1.

The other possibility is 20 plus a is equal to 30.

Because that's the coefficient on the x squared term.

And now, let me play a trick on you.

So this will only happen when a is greater than b .

16 divided by 8 is 2. a squared divided by a is a.

Then x gets counted down, so the number of times the body of this loop is executed is exactly whatever a is, and there's two statements in the loop.

The one on the left is exactly twice the size, 0.0555 which is two thirds times 12, it's 1 18th.