There are two math problems to be solved in detail

1.Let the water inlet pipe per minute be x, the outlet pipe will be y per minute, and the water outlet pipe will be opened t minutes later than the water inlet pipe; From the meaning of the title:

xt+8x-2*8y=0，……1） xt+5x-3*5y=0，……2）

Homogenize the above two formulas into the form x:y, and merge them to give 16 (8+t)=x:y=15 (5+t);

It can be solved: t=40;

2.The total production of cars is x; Under normal circumstances: the production volume of small displacement cars is 30%x, and the production volume of large and medium displacement cars is 70%x;

After the adjustment of the production structure, the production volume of large and medium displacement cars is 90%*(70%x)=63%x, and the total production volume is (1+, so the output of small displacement cars is;

The percentage increase in the production volume of small displacement cars compared with the normal situation is: (

In order to make it easier to understand, the second question is written in more detail).

1.Since the three outlets are the same, opening three outlets at the same time is equivalent to draining the water from the inlet pipe for 3 minutes than opening two outlets at the same time. It can be seen that the water discharged in 5 minutes from an outlet pipe is equivalent to the water discharged from an inlet pipe within 3 minutes.

Therefore, if you open 2 outlet pipes at the same time, the water discharged in 8 minutes is equivalent to the water inlet in the inlet pipe in minutes.

That is, the outlet pipe opens a minute later than the inlet pipe.

2.If the original total production is 1, then the current total production is the production of small displacement cars under normal conditions, and the production of large and medium displacement vehicles is.

Now the structure is adjusted, and the production volume of large and medium-displacement vehicles is.

The production volume of cars with small displacement is.

The production of small displacement cars has increased compared to normal (

1. Set the large bucket x and the small bucket y

then x+y=20

4x-2y=8；

Simultaneous two can be very disarmedx=8;y=12

2. A, B and C each have teeth x, y and z.

then 5x=7y=2z;

Obviously, xyz is an integer, and 2 can be chosen

The least common multiple of 57 is 70=5x=7y=2z, so x=14,y=10,z=35;is the minimum solution.

Fractions are equal to fractions of 25 equals 5 divided by 6 equals what division by 12 equals 36 fractions equal to 35 divided by what.

18=25 ()=5 6=() 12=() 36=35 ()Let (a) 18=25 (b)=5 6=(c) 12=(d) 36=35 (e).

then a=18x5 6=15, b=25 (5 6)=30

c=12x5/6=10

d=36x5/6=30

e=35÷（5/6）=42\

Then (15) 18=25 (30)=5 6=(10) 12=(30) 36=35 (42)2,

A divided by b is equal to c, the greatest common factor of a and b is (, and the least common multiple is ( ) then b a=c

b = ac, so a and b

The greatest common factor is .

a, the least common multiple is b

1.Solution: Connecting EG, since the four constituent lines are bisector lines, the angles are all 45 degrees, and the quadrilateral EFGH is rectangular.

A triangle AFD is an equilateral right-angled triangle.

So af=fd;

The triangle AEB is an equilateral right-angled triangle.

So ae=eb;

Extend CH, AF, and AD respectively

bc in p, q; then the quadrilateral apcq is a parallelogram;

So ap=cq; So bq=dp;

So the triangle beq is congruent with the triangle dgp;

So ae=be=dg;

and af=fd;

So ef=fg; Proven.

2.Solution: Let the unit velocity be x

km/min;

The speed of Xiaohua is 3x, and the speed of the creeps is 4x

The column equation has. 6/3x=10/4x－10

Solution x=20

km/min

Then Xiaohua's speed is.

60km/min

Creep speed is 80km min.

1. First of all, the four corners of EFGH are right angles, and secondly, due to symmetry, EH=HG, so the quadrilateral EFGH is a square.

2. Let the speed of Xiaohua and the creep be 3x, 4x, then 6 (3x)+10=10 (4x), and the solution is x=, so the speed of Xiaohua's creep is, kilometers per minute.

Because it is 4 corner bisectors.

So there is. afd

aebbch

CGDs are isosceles right triangles.

Therefore, EFGH is rectangular and AFD and BHC congruence.

AEB and CGD congruence.

So hg=ch-cg=df-dg=gf

Therefore, EFGH is a square, and the speed of Xiaohua is 3V

km h has.

6/3v+1/6=10/4v

v=1 3, so Xiaohua's velocity is 1km h

Creep speed is 3 4 km h

1.Assuming that the total sales in the first quarter are 100, the sales of Type A are 56 and the sales of B + C are 54

Q2: Type A 56*(1+23%) B+C is 54*(1-a%) Total: 100*(1+12%)

then 56*(1+23%)+54*(1-a%)=100*(1+12%). The answer seems to be 2

Of course, you can also set the total sales of the first quarter to x, but that's a bit of a hassle!

2.When the length of the rivet into the equipment part is sufficient, the first time is 2cm, the second time is 1cm, and the third time is, the requirement of the problem is that the effect of two times can not enter the equipment completely, and the effect of three times must all enter the equipment, so the total length of the rivet is between the length required for the first two and the first three times. and 1+2+>=a>1+2

The landlord is rewarded!

1.(for the total amount of the first quarter) you understand.

2. 3

Find an expert, I passed by the soy sauce maker.

One, the greatest common factor of 125 and 50 is 25

Second, the greatest common factor is 40, which can be divided into 40 portions, each serving of toffee 8, soy milk 3 apples 4

The first problem is to plant 12 trees, 5 trees on the long side 5x2=10, 3 trees on the wide side 2x3=6, 10+6=16, and then subtract the four trees with repeated corners.

The second question is 40 gifts, each with 8 toffee, 3 bottles of soy milk, and 4 apples.

1.A rectangular piece of land, 125 meters long and 50 meters wide, is now to be planted in and around his four corners, so that the distance between each adjacent two trees is equal. How many trees should be planted?

Plant a tree on a closed graph with the same number of trees and segments. The greatest common divisor of 125 and 50 is 25, and the perimeter of the figure is (125+50) 2=350 meters, at least 350 25=14 lessons.

1 Carry at least 66 trees.

2 Up to 40 points, 8 toffee per serving, 3 bottles of soy milk, 4 apples.