I have a few questions I would like to ask you, the first year of junior high school mathematics sum

1.Solution: If the production of product A XKG, then the production of product B (50-X) kg

9x+4*(50-x)<=360

3x+10*(50-x)<=290

So 30<=x<=32, so 30kg for product A and 20kg for product B or 31kg for product A, product B.

19kg or 32kg for product A, 18kg2 for product BBecause (k+2)x-2=1-k(4-x), x=3 2-2k1)>0 gives k<3 4

2) <0 solution to k>3 4

3) <=2 solution to k>=-(1 4).

4) <=2 and k<1 are solved to -(1 4)<=k<13According to the title: (4a-3b)>0 and (2b-a) (4a-3b)=4 9, so b=5a6 and a<0

So the solution of the inequality ax b is x<5 6

4.Let the velocity of B be x, then the velocity of A is 12+x*1=12+x, 1+1 2)*x=(1 2)*(12+x) to get x=6, so choose a

1.Set up the production of X pieces of product A and Y pieces of product B.

then x+y=50, y=50-x

According to the title, there are 9x+3y 360

3x+10y≤290

i.e. 9x+3(50-x) 360

3x+10（50-x）≤290

Solution x 35

x 30 so it may be 30, 31, 32, 33, 34 or 35 pieces for product A and 35, 36, 37, 38, 39 or 40 pieces for product B.

So there are a total of 6 scenarios, which are:

30 pieces of product A, 40 pieces of product B, 31 pieces of product A, 39 pieces of product B, 32 pieces of product A, 48 pieces of product B, 33 pieces of product A, 37 pieces of product B, 34 pieces of product A, 46 pieces of product B, 35 pieces of product A, 35 pieces of product B, 2Let A produce x pieces then B produces 50-x pieces.

Then the cost of raw material A is 9x +4*(50-x) =200+5x

Consumption of B raw materials 3x+ 10*(50-x) =500-7x 200+5x<=360 x<=32 500-7x<=290 x>=30 x=30,31,32

2) y=x*700 +(50-x)*1200 =60000-500x, the bigger the x, the smaller the y, obviously when x=30, the profit is the largest 2 2(k+2)x-2=1-k(4-x)

k+2+k）x=3-4k

x=（3-4k）/（2k+2）

1）x>0

3-4k)/(2k+2)>0

3-4k)*(2k+2)>0

So k<-1 or k>3*4

2)x<0

The same goes for it: -1

You're so crazy, there's nothing you can solve, so many enthusiastic people tell you, be careful of your experience.

If x people are arranged, there will be (10-x) people planted B.

There are 10 people in total).

Because each person can plant 3 acres of A vegetables, x people can plant 3x acres, and in the same way (10-x), people can plant 2 (10-x) acres of B vegetables.

Yes, because a kind of vegetables can earn 10,000 yuan per mu, 3x mu of vegetables can earn 3x * 10,000 yuan, and 2 (10-x) mu B vegetables can earn 2 (10-x) * 10,000 yuan, because the total income should not be less than 10,000 yuan.

then x 4

So the original formula is decomposed in this way to get (x 5+x 2)*(x 5-x 2+10y) and then substitute x,y

Just substitute it. 3.（2a-b)^4÷(2a-b)^2=(2a-b)^24.（a^2-4）^2=(a+2)(a-2)(a+2)(a-2)

If the triangular ruler a coincides with the bottom edge of the quadrilateral in the figure, place another triangular ruler b above the first triangular ruler to make it at a right angle, and push b along a, if b coincides with the vertical edge of the quadrilateral, it proves that the bottom edge of the quadrilateral is perpendicular to the vertical edge, and the four sides of the quadrilateral are perpendicular to each other, then the opposite sides of the quadrilateral are parallel to each other

(1) The farthest place from home is 12:00

2) He started his first break at 10:30 and had a 30-minute break.

3) He stops advancing at 12:00 and rests for lunch.

4) He is 20 kilometers away from home at 10:30, 11:00 and 13:40.

The farthest is 12-13 hours.

The first break is for half an hour.

Stop for 12-13 hours and eat lunch.

The hour is 20 kilometers from home.

(1) At 12-13 o'clock (i.e., the time period of EF) (2) At time C, he starts his first break The time period of rest is CD If C is time, then the time of rest is hours (3) Normally, it should be around 12 o'clock for lunch Then the time of EF is the time for lunch The answer to this question is 12 o'clock (4) According to the image, the CD time period is when he leaves home at 9 o'clock Then it is -11 o'clock There is another one on the way home But it is not marked It doesn't matter Finally, summer vacation homework? Winter vacation, it looks like you're really in a hurry.

The vehicle drove out x kilometers with 60 people first, and the other 90 people walked.

x/36+(21-x)/4=(55+60)/60

x = 15 vehicles carrying 60 people drove 15 kilometers, dropped off these 60 people and walked to the train station, arriving at 7:55.

15 36 = 5 12 hours, 4 * 5 12 = 5 3 km.

At this time the back of 90 people walked for 5 12 hours, 5 3 kilometers.

15-5 3) (36+4) = 1 3 hours.

The vehicle returned, 1 to 3 hours with a meeting with 90 people behind.

4*1 3=4 3 km, 5 12 + 1 3 = 3 4 hours, 5 3 + 4 3 = 3 km.

The 90 people in the back walked for a total of 3 4 hours, 3 km, 21-3 = 18 km from the railway station.

Use the first method again:

y/36+(18-y)/4=(55+60)/60-3/4

y=15 Similarly: the vehicle carries 60 people and drives another 15 kilometers, drops off the 60 people and walks to the train station, arriving at 7:55.

15 36 = 5 12 hours, 4 * 5 12 = 5 3 km.

At this time, the 30 people in the back walked for another 5 12 hours, 5 3 kilometers, a total of 7 6 hours, 14 3 kilometers.

15-5 3) (36+4) = 1 3 hours.

The vehicle returned again, and 1 to 3 hours met 30 people behind.

4*1 3=4 3 km, 7 6 + 1 3 = 3 2 hours, 14 3 + 4 3 = 6 km.

The 30 people in the back walked for a total of 3 2 hours, 6 km, 21-6 = 15 km from the railway station.

15 36 = 5 12 hours, 5 12 + 3 2 = 23 12 hours.

Just arrived at all.

Another method: Solution: Divide 150 people into three groups of 50 people, walk at a speed of 4 kilometers, and the speed of a car is.

Solve x hours), that is, 90 minutes on foot and 25 minutes on a car ride per person. Three groups of people set off at the same time at 5 o'clock, the first batch of people.

Take a 25-minute bus to point A, get off and walk; The coach immediately returns from A and meets the second group of people on foot at point B, which is a 25-minute ride.

The second group got off the bus and walked, and the bus immediately returned, and at point C it met the third group on foot, and then straightened them up.

Pick-up and drop-off to the train station.

Thus arranged.

The first and second groups of people are no problem to get to the train station on time, but is the third group just a 25-minute train?

It must be calculated. The time of the second return is 20 minutes, and the time of the second return of the passenger car should also be 20 minutes, so when the passenger car is in phase with the third group of people.

At that time, the passenger car has been used 25 2 20 2 = 90 (points), and 115-90 = 25 (points), just enough to send the third batch of people on time.

Therefore, it can be arranged in the above way.

Hour 55 minutes = 23 12 o'clock, the car has to move non-stop during this time, so the distance traveled by the car during this time is:

36*23 12=69 km.

2. If there is no time limit, the car will send 150 people to the station in 3 times, and the distance taken should be:

21*5=105 km (5 is one way for two round-trips plus the third group).

3. In this way, the car actually travels 36 kilometers less, that is:

105-69 = 36 km.

These 36 kilometers require these 3 groups of people to walk on average, that is, each group of people walks 12 kilometers, but those who walk only need to go to the station and do not need to come back, so only 12 * 2 = 6 kilometers can be walked.

Here's how to do it:

1. The car sends the first group of people to a distance of 6 kilometers from the station, and allows them to walk to the station while giving way.

The second and third groups walked from the school to the station;

2,) The car turns back to pick up the second group to catch up with the first group and let the first.

Group 1 and 2 walk at the same time;

3) Then the car will turn back and pick up the third group and send it to the station, so that the three groups will arrive at the station at the same time.

Divide 150 people into three groups of 50 people, each group of people walks for 90 minutes and rides for 25 minutes. Three groups of people set off at the same time at 6 o'clock, and the first group of people took a 25-minute bus to point A, got off and walked; The bus immediately returned from A, and after 20 minutes of driving, it met the second and third groups of people on foot at point B, and the second group of people took a 25-minute bus to point C, got off and walked, and then returned immediately, and after another 20 minutes, it met the third group of people who came on foot at point D, obviously, the third group of people had walked for 90 minutes, and as long as they took another 25 minutes, they would arrive at the train station at 7:55.

4 unknown variables, 2 equations.

n*x+m*y=1575

x+2y=205

Even if it is a generation, it is a ternary equation, and there are many solutions.

At most, you can only give an interval range, and don't think about the correct solution.

In addition, the teacher who seriously despises this kind of topic for the first year of junior high school students is brain-dead and a waste of children's time.