Elementary school students please enter 20, friends of elementary school students please enter

QQ Dance, Crossfire, QQ Hall, QQ Speed, QQ Fantasy, Moore Manor and so on.

I love playing CrossFire.

It's a shootout game.

Much like CS. This game has a lot of feelings. It's very comfortable to play (except, of course, when you're unlucky.) ）

I'm in Beijing Netcom District 1, it's really hard to name O0 Corporal 6

I like to play Shanda Network's Superrun, and Shanda's other games are not bad either!

I love playing Warcraft, the undead, hahahaha.

QQ Hall, QQ Speed, Go-Kart.

Like Sasha the Angel of Love, I love to play Moore Manor.

At the same time, I also like to play with QQ pets.

I love playing Moore Manor.

Hope you enjoy it.

I like to play QQ Three Kingdoms and 4399 mini-games.

Neo pets, this game is not harmful to us. All-free browser game.

Go karting.

QQ Free Fantasy. QQ Three Kingdoms.

I'm level 84.

It's not a single purchase, it's a lot of roots in a small box.

It should be marked on the mechanical pencil.

If it's okay that night, just do it, and you can have a good time in the next few days. If you do it on the last day, then it's a kind of burden when you play, and it's not fun to play, right?

We did it that night, and we did it at school.

You may not want to do it that night, so you should do it all day.

You don't have to do it on any day, as long as you do it before you go to school, but I still recommend doing it the day before, because after you finish your homework, you won't have a burden in your heart when you go to play, and you will have a lot of fun!

Do it a few days before the start of school, decide on the amount of homework, plan it, and be able to complete it within the scheduled time

I like it, I usually buy clothes, accessories, and I buy whatever I see looking good!

Digital ones don't dare to buy clothes, pants and shoes are a little inconsistent in size The family bought clothes, pants and shoes, either big or small.

This sentence makes me feel: Dai girls are very graceful, like a group of beautiful fairies descending to earth. The swaying here makes me think they are very beautiful. This is a good word, and I love it!

It shows that the Dai girl is generous and decent, beautiful and useful. Very cleverly written.

1。The numerator of a fraction is 23 less than the denominator. If the numerator is added by 5, the denominator is reduced by 5, and the new fraction is about 2 3. Find the original score.

Let the numerator be x and the denominator be y.

The numerator is 23 less than the denominator: x+23=y

If the numerator is added by 5, the denominator decreases by 5, and the new fraction is approximately 2 3: x+5 y-5 = 2 3

Bring y=x+23 in, then x+5 x+18 = 2 3

x=21 y= 44

2。Class Six (1) has 62 students. If 1 5 and 8 girls of the boys go to the competition, then the remaining number of girls is as large as the number of boys. How many boys and girls are in this class?

There are x boys and y girls.

62 students; x+y = 62

If 1 5 of the boys and 8 girls go to the competition, then the remaining number of girls is as large as the number of boys: x-1 5x = y-8

Bringing x+y = 62 in, then = (62-x)-8

x = 30 y= 32

3。A, B, C, and D make a total of 360 parts, if A makes 10 more, B makes 10 less, the number of C makes is multiplied by 2, and the number of D is divided by 2, then the number of parts made by the four people is exactly equal. And how many parts were made?

AA, BB, CC, DD

a+b+c+d = 360

a+10 = b -10 = c*2 = d/2

The answer is your own!

4。a, b, c, d are known to have the following relationships:

a=b+3 c=a+a d=a+c+7 d=b+40

a=(15 ) b=(12 ) c=(30 ) d=（ 52）

c=2a, a=b+3 so c =2b+6

d=a+c+7=(b+3)+(2b+6)+7=3b+16=b+40

So 2b =24 b=12

a=15 c=30 d=52

5。Team A has (14+10a) people and team B has (10+2a) people. Now that the number of players in team A is to be tripled by the number of team B, how should the number of teams be allocated?

There are a total of 14+10a+10+2a=(12a+24) people.

B: (12a+24) 4=3a+6

A: (12a+24) 4*3= 9a+18

It's tiring to give points.

1.Let the numerator be x and the denominator be y, then there is.

x=y-23

x+5）/（y-5)=2/3

The solution yields x=21 and y=44

2.There are x boys and y girls. Yes.

x+y=62

x-(1/5)x=y-8

The solution is x=30, y=32;

3.The idea of this question is as follows: four people are set up to make parts A, B, C, and D, and there are parts from the question.

a-10=b+10=2c=d/2；

(a-10)+(b-10)=2c+d 2;

i.e. a+b=2c+d 2;

and a+b+c+d=360;Substituting the above equation yields 2c+d 2+c+d=360, i.e. 3c+(3 2)d=360;

Re-synaccid 2c = d 2;

The solution yields c = 40, that is, C has made 40.

4.Let's talk about the idea of this question, and the right side of the above equation and equal sign can be represented by b to solve the value of b, and then go back to the generation.

Specifically, we get a=b+3;c=2b+6；d=a+c+7=3b+16；d=b+40；

There is d=3b+16=b+40;The solution is a=15, b=12, c=30, d=52;

5.From the question, it is known that people should be transferred from team B to team A.

If x people are transferred from team B to team A, it must be satisfied.

14+10a＋x＝3（10＋2a－x）

Simplifying the equation to 4(a+x)=16

a+x=4, so x=4-a;

That is to say, it is enough to transfer (4-a) from team B to team A.

The numerator is x and the denominator is y.

x+23=y

If the numerator is added by 5, the denominator decreases by 5, and the new fraction is approximately 2 3: x+5 y-5 = 2 3

Bring y=x+23 in, then x+5 x+18 = 2 3x=21 y= 44

2。There are x boys and y girls.

62 students; x+y = 62

If 1 5 of the boys and 8 girls go to the competition, then the remaining number of girls is as large as the number of boys: x-1 5x = y-8

Bringing x+y = 62 in, then = (62-x)-8x = 30 y = 32

3。AA, BB, CC, DD

a+b+c+d = 360

a+10 = b -10 = c*2 = d/24。a, b, c, d are known to have the following relationships:

a=b+3 c=a+a d=a+c+7 d=b+40a=(15 ) b=(12) c=(30) d=(52)c=2a,a=b+3 so c =2b+6d=a+c+7=(b+3)+(2b+6)+7=3b+16=b+40, so 2b =24 b=12

a=15 c=30 d=52

5。There are a total of 14+10a+10+2a=(12a+24) person B: (12a+24) 4=3a+6

A: (12a+24) 4*3= 9a+18 absolutely!

Trust me!

Hehe, indeed, college students have a headache when they see this problem, and they will only think of using equations and column solutions and get dizzy. I'm an elementary school graduate! Rattle!

A steamer sails downstream from port A to port B and then immediately returns. The reason why the first 4 hours are 60 km longer than the next 4 hours is because the downstream is faster than the reverse (the first three hours are divided into the downstream and the reverse section, the last three hours are all against the current, and the 60 km of the more sailing is faster in the downstream section).

So: sailing time against the water = 60 20 = 3 (hours) Then: sailing time against the water = 8-3 = 5 (hours).

Then: the speed of sailing against the water = 60 (5-3) = 30 (kmh) Then: the speed of sailing along the river is 30 + 20 = 50 (kmh) Then: the distance between the two ports of A and B = 5 30 = 50 3 = 150 (km) The equation is:

Let the speed against the water be x, and 3(x+20)=(8-3)x is obtained according to the same range of the forward and reverse water

It is launched when the speed of the solution against the water is 30 km.

The distance between the two ports = 5 30 = 50 3 = 150 (km).

I've heard that there are a lot of college students who can't figure it out, and only 3 out of 75 college students can answer correctly, so I would like to ask you what kind of college students you are looking for. Don't keep saying that college students are not, most of them are still good, ok.

60 20 = 3 hours, and the speed difference is 20, that is, there are 3 hours of sailing speed at high speed (downstream speed).

Let x be the speed of the water.

3x=(8-3)*(x-20)

3x=5x-100

2x=100

x = 50 km-h.

50 x 3 = 150 km.

I don't know where you found the 75 college students you were looking for, don't they just got drunk.

How happy it is for college students to have this question in their college exams.

150 km, calculated by column equation.

There are 20 to 80.

It's the same all over the country.