So non negative multiples of 3.

Pay me back multiples.

So let's write all the multiples of 10.

It's equal to multiples of these two vectors.

The other ones are some multiples of that length.

Let me just mark only the multiples of 9.

Those are literally the nearest multiples of ten that

Can I split myself up into multiples?

We could just look at the multiples.

Our first coordinate is going to be multiples of v1.

The scalar multiples, obviously, are any real number.

We could pick one of these guys and keep taking their multiples, and then figuring out whether those multiples are both divisible by 5 and 15.

If they're collinear and kind of the scalar multiples the same.

And so anything that you were able to generate with V2 before, you can now generate with our U1, multiples of that, plus multiples of our new vector.

And I want to see what those multiples are.

And you'll see over here, these are multiples of the number 27.

This is v1 or this is v1, so it's multiples of v1.

And then I could do all the multiples of v1 like this.

All of the possible scalar multiples of v are going to be the same thing as all of the scalar multiples of our unit vector, u, which is just a scalar multiple of v.

But out of the ones we listed 40 and 80 are our common multiples.

Where you're starting at the origin right there, plus multiples of these two guys.

Let's see if we can identify what the common multiples are.

So I could draw another version of v2, just like that, just all the multiples of v2.

Multiples of 8 are 8, 16, 24, 32, 40, 48, 64, 72, 80 and so on.

European Expansion or Putin Expansion?

So let's write these in terms of multiples of 3, since we know that that's the increment level.

So you take all the multiples of v1, you're going to get every point along this line.

Let's say that, of your free variables, your free variables are going to be the scalar multiples.

Or they could be negative multiples, so they could also go in that direction.

It only has an i component, or all of its vectors are only multiples of the i unit vector.

Expansion of strategic partnerships

Expansion of financial services.

Expansion of productive employment

Expansion of truck lanes.

Because all of the derivatives of e are kind of multiples of the original function, and that's why we used it.

And it keeps going, because obviously, you could get negative multiples of my vector, and it also goes here.

(a) Expansion of employment opportunities

d. Expansion of the cooperatives

The expansion of AMIS continues.

(b) Expansion of marketing channels.

Expansion of access to treatment

(iv) Expansion of UNPROFOR headquarters

(c) Expansion of productive employment,

(c) Expansion of productive employment.

C. Expansion of productive employment