In the first year of junior high school, you will not be able to solve math problems there are many

1. In a highway blasting, after the operator ignites the fuse, he should run to a place 400 meters away before the explosives.

In other words: run at least 400 meters.

It is assumed that the fuse must be at least x meters long to prove the safety of the operator.

The critical situation is that the operator has run exactly 400 meters when the fuse of x meters has been burned.

By the title: The time for the fuse to burn out is the same as the time for the operator to run 400 meters.

Column equation: x

Solution x = m.

Therefore, the fuse must be more than a meter to ensure the safety of the operator?

2, let 5-3x=0, and solve x=5 3

You can verify the result by bringing x=4 (greater than 5 3) into the equation and the result is negative, and x=1 (less than 5 3) into the equation and knowing that the result is positive. So here's the result.

Result: At x>5 3, the value of algebraic equations 5-3x is negative.

When x=5 3, the value of algebraic equation 5-3x is 0

x<5 3, the value of algebraic equations 5-3x is positive.

ok？Settle??

Because it is 12 centimeters per second, then the speed of the fuse can be obtained by conversion as meters per second.

The time taken by the person is 400 5 = 80 seconds.

So the fuse is meters long.

A: The fuse must be more than meters.

Because 5-3x is negative.

So 5-3x<0

x>5/3

Because 5-3x is 0

So 5-3x<0

x=5 3 because 5-3x is a positive number.

So 5-3x<0

x<5/3

1.Let the length of the fuse be set to x cm, then.

x/>=400/5

The solution is x>=96 (cm).

0 x<5/3

5-3x=0 x=5/3

5-3x<0 x>5/3

1。Solution: If the fuse is x cm long, then there is.

x gives x=96 (cm).

2。Solution: 1) 5-3x<0 x>5 3

2) 5-3x=0 x=5/3

3) 5-3x>0 x<5/3

You should think more about the problem and communicate with your classmates, and the collision of ideas can have a spark of inspiration It is recommended that you communicate more with your classmates!! Ask fewer questions to learn to make greater progress!! Thank you!!

Let the speed of car B be x, according to the title.

The absolute speed of car B is x-60

When car B leaves the front of car A, it travels 120 + 130 = 250m = column equation: (x-60) * 3 60 =

x = 65 kmh.

Let the speed of car B be x kilometers per hour, then.

When the two cars are driving in the same direction, the speed of car B relative to car A is (x-60), and car B can overtake and leave car A in 3 minutes, so (120+130) 1000(x-60)=3 60 solution: x = 65

B velocity is x

In 3 minutes, B travels 130+120=250m=3min=3 60h=1 20h, that is, 1 20x-60x1 20=

The solution is x=65km h

60000 3600=10 6 m min.

Solution: Let the speed of car B be x meters.

130+120）/（x-（10/6））=3x=85

If a type of deposit is $10,000, then a type B deposit (40-x) is $10,000.

Then by the title:

x*x=10 acres with a loss of 100,000 yuan, B a first deposit of 300,000 yuan.

If the land leaks and loses the deposit of A for $x, then the deposit for B is (400,000-x) dollars, yes.

Just solve it.

Solve the dust and forest a deposit to send the mu money x yuan to check the loss.

B deposit Y yuan.

x+y=400000

The solution is x=100000

y=300000

Trick: When they meet for the second time, A and B exercise Duan Feng's argument at the same time.

Let the distance of ab be x

According to the time = distance speed, when A meets for the second time, the distance traveled is one ab+(ab-34), then the time is (x+(x-34)) 70

In the second encounter, B travels a distance of ab+34 and the time is (x+34) 52

Column equation: x+(x-34)) 70

x+34)/52

The solution is x=122

A: ......

1. Suppose there are x students in the extracurricular book group, and the equation is as follows: 1 3x+4=1 2x

Gives x=24

2. Suppose that the unit price of B beverage is X yuan bottle, and the unit price of A beverage is (X-1) yuan bottle, according to the conditions, the equation is as follows.

2（x-1）+3x=13

Gives x=3

1. Assuming that there are x classmates in the extracurricular group, then the equation is as follows: 1 3*x+4=1 2*(x+4).

Launch x=12

2. Suppose the unit price of B beverage is X yuan bottle, then the unit price of A beverage is (X-1) yuan bottle, and the equation is as follows according to the conditions.

2*（x-1）+3x=13

Roll out x=3

1. If there are x people in the original group, then.

1/3x+4=1/2x

2. Let the unit price of beverage B be x, and the unit price of beverage A is (x-1), then 2 (x-1) + 3x = 13

5x=15x=3

One-half x - one-third x = 4 x total number of original people x = 24

a x element b x+1

2x+3（x+1）=13

x=2b=2+1=3 yuan.

1.Let the number of people in the group be x before, according to the title.

1/3 x＋4＝1/2（x＋4）

2.Let ** of a be x, then b be x 1

2x＋3（x＋1）＝13

1. Let the original x students have the column equation x (x+4) = (1 3) (1 2).

2. Let the unit price of B beverage be x yuan, and the column equation is 2 (x-1) + 3x = 13

Question 1: In the original extracurricular math group, there are students x 1 3 * x +4 = 1 2 * (x+4).

Question 2: Let the unit price of b be $x, 3*x+2*(x-1)=13, and the solution is $3.

From 1 to x> and round 3 8

Resolve 2 to x=a

Because 2 has only one integer solution, that is, an integer greater than 3 and 8, which is 1, as long as a is less than 2 and meets the first condition.

The range of a is a large collapse of 3 8 and less than 2

The landlord said that the X in the middle of the equation is numerator or the whole formula.

You are terrible at math, are you an inequality, how can you make an equation?

By the coincidence of 3 2 by a a 3 2 by a a 3 2 the solution of an integer should be the first integer 2 that is larger

From 1 we get x>3 8

2 is obtained by x, because 2 formula only has an integer solution, that is, an integer 1 greater than 3 8, so as long as a is less than 2 and greater than the infiltration pin pose 1, the socks meet the first condition.

Parallel relationships. Proof of the process –

Known: MCN= BCM=360- a- b- AMC, (CM is the bisector of BCD, and the sum of the inner angles of the quadrilateral is 360).

fnd=360-∠d-∠e-∠efn=360-∠a-∠b-∠efn=360-∠a-∠b-∠nfm,（∠a=∠d,∠b=∠e）

Old. MCN- FND=360- A- B- AMC-(360- A- B- NFM)= NFM- AMC= NFM-(180- FMC)= NFM+ FMC-180, (180 for the sum of angles and complements)=360- MCN- CNF-180 (360 for the sum of the inner angles of the quadrilateral)=180- MCN- CNF

So. 2∠mcn=180+∠fnd-∠cnf=180+∠fnd-(180-∠fnd)=2∠fnd

i.e. mcn= fnd

So, cm is parallel to fn.

The parallel lines that pass f to make mc intersect cd at point p have amc= mfp mcd= fpd

Because cm bisects bcd, bcm= mcd i.e. bcm= fpd

Because a= d, b= e

So efp+ fpd= amc+ bcm so efp= amc so efp= mfp so fp is afe and the bisector is known to coincide p with n so cm fn