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You need to be a little bit careful to make sure that you're doing float division.

Delete several items

We're multiplying it by 100 in the denominator.

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And we have 3 5 divided by 1 2.

It's not in the set of reals that's a complex number.

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You're gonna be left with 2 y's

Now the first thing you want to do when your divisor, the number that you're dividing into the other number, is a decimal, is to multiply it by 10 enough times so that it becomes a whole number so you can shift the decimal to the right.

Blah, blah, blah, blah, blah the part here at the bottom is the most important

And so in general you multiply or devide the significant figures in your product or the divisor, divident, the quotient , the quotient!

Delete the selected variable.

You want to keep everything in the correct places.

Destroyed millions of germs.

It don't know, there something wacky about that that the inverse of the number is really just the decimals

You remove the channel limit.

handle must be a valid handle to a printer.

Precedence and Associativity,

One thing you might do is you only include 1 variable, so I got rid of the s2 there.

And actually, let's just go in order.

When a circle becomes heavy because I'm locking it down,

We get 6 is equal to x, so the domain of this function is equal to the set of all real numbers except 6.

Now if you do that with the divisor, you also have to do that with the dividend, the number that you're dividing into.

But I'm going to rewrite it, just so it's a little bit neater.

Now for a quiz.

When cleaning the classroom, a few students carry the podium.

We need to divide 0.25 into 1.03075.

So once again, maybe we'll subtract 3 from both sides, just to get rid of this constant term.

Well, the problem is this division.

In this video, I want to tackle some inequalities that involve multiplying and dividing by positive and negative numbers, and you'll see that it's a little bit more tricky than just the adding and subtracting numbers that we saw in the last video.

I've been asked to make a video on algebraic division or algebraic long division.

In this example, Javier was our starting node and it had degree 3.

Now, the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that'll eliminate the fractions.

But I think you get what I'm saying.

And the easiest one I can think of is multiply both sides by 3.

It uses a method called Markov Chain Monte Carlo, or MCMC.

So you can't merge, when you're adding and subtracting, the real part to the imaginary part.

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