Ask a question and also know where the classification is1

Lori control Lori must be lace, but Lori control sister is not necessarily lace.

Don't blame me for being so absolute, even if it wasn't at the time, it will be slowly.

It can't be worship or anything, oh, except for the brain-dead, I'm talking about real comic fans.

Lori controls Lori's? Probably not! If you're too young, don't call it Lori.,Lori--it's not that age but it's cute!

Lolita control Lori is also called Loli control.

In this case, it can be regarded as GL.

You just ask. I'll just forget it.

Not necessarily, maybe 15 or 6 years old. It's no more than 16 anyway.

Who knows? But that's a good thing.

It is usually divided into one.

Generally, (1) is not used as a title.

I feel like it's on the ** online shopping bar! There are also a lot of genuine goods***, foreign trade clothing, **very moderate, I have friends who often buy things on ** abroad, and the shipping is not expensive, compared to what you buy abroad.

Go to DFO, it's cheaper.

There are many kinds of jade spring auspicious styles, such as glass species, Fengsan water species, ice species, if you want to see **, you can go to the specific introduction to the jewelry of the Mianmian Fight.

to find out.

First of all, you need to divide 4 subnets, and you need to borrow 2 bits from the host bit.

The first subnet ID is the range that is the broadcast address).

The second subnet ID is the range is the broadcast address).

The third subnet ID is the range that is the broadcast address).

The fourth subnet ID is the range that is the broadcast address).

There are 14 addresses in the first subnet! A total of 56 addresses are available!

There are a total of 28 bits for the network address and 4 bits for the host address. i.e. the mask is 28 host ranges of 16 addresses in each subnet, of which 1 network number, one broadcast address, 14.

Sorry, you can't see all the problems of other users.

You can find it on the Category Questions page.

Hello friend.

Look at it.

When it is easy to find a=, i.e., a<1 3, take the left set.

Together, a>1 3, take the right collection. Because version A 2+1 is always greater than 2a, the same.

weight can be found b=, and a<1 3 and a>1 3 are analyzed respectively. Since b is contained in a, for a<1 3, there is 2a>=3a+1, and a 2+1<=2, the solution is a=-1, and for a>1 3, there is 2a>=2, and a 2+1<=3a+1, the solution is 1

Classified by whether 3a+1 is greater than 2.

When it is greater than or equal to 2.

3a+1<2a<2

When 3a+1 is less than 2.

2<2a<3a+1

2Note, the above formula can be taken equal sign, typing is inconvenient to simplify.

The equation b is less than or equal to a, and then the inequality is solved.