Junior high school binary one time application questions! Thank you for the answer

Let the raw material of A need to be x, and the raw material of B needs to be y

x+ solution yields x=28 y=30

Later, you will go to the math know-it-all to ask, and there is a professional teacher who is answering, and it is very timely. I'm also a regular visitor there. Hey.

When X grams of raw materials for A and Y grams for raw materials for B are just enough to meet the needs of patients. According to the title, there are:

x+ biquint, y=30, x=28

That is, 30 grams of raw material A and 18 grams of raw material B are just enough to meet the needs of the patient.

Suppose raw material A needs to be x, and raw material B needs to be y

x+ solution yields x=28 y=30

Set A x grams, B y grams.

Rule. x+ solution yields x=28 y=30

That is, 28 grams for A and 30 grams for B.

Set X grams of raw materials for A and Y grams for raw materials of B. Get x+

solution, x=28, y=30

That is, 28 grams of raw material A and 30 grams of raw material B are required.

Solution: Set up x grams of raw material A and y grams of raw material B in each meal, according to the topic, get.

x+solutionx=28 y=30

Let A be x and B be y, and we get the system of equations x+

Finding the system of equations yields x=28 and y=30

Set a x gram, B y gram.

By the title:

x+ 2- gets: y=30

then x=28 so A needs 28 grams and B 30 grams.

Solution: Set the original price to X yuan.

x(1+solution, x=2500.)

Solution: Set a profit of x yuan.

solution, x=162

Hope it solves your problem, I'm also a freshman in junior high school.

1、Set the original price of each color TVxyuan,[x(1+

I didn't understand the question, so I directly thought that the "illegal income obtained" was the total amount of money that deceived customers.

Solution x = 450 yuan.

2. The total cost of books A and B is $ x and $ Y respectively.

x(1+,y(,x=1248,y=1500, so a total profit of x*yuan.

1.Set the original price to $x.

x (1 + solution 2500 yuan.)

2.Set a profit of $x.

x=1560-1560/(1+

solution, x=162

2 + root number 3 is the root of the equation so there is.

2 + root number 3) 2-4 (2 + root number 3) + c=0 gives c=1, so the equation is.

x^2-4x+1=0

So the other one is 4 (2-3).

- -I'm too lazy to count the little brother, I'll teach you, you can bring 2 + root number 3 into the following equation, and you can find the value of c. After finding the c value, then solve the one-dimensional two-dimensional equation, understand?

The answer is also given to you together, c = 1 + 4 root number 3 another value You can solve it yourself.

You yourself bring 2 + root number 3 into the equation. The value of c is found. Then bring C in and find the solution to the equation.

Since x is equal to 2 + root number 3, the square of (2 + root number 3) is -4 * (2 + root number 3) + c = 0, and -1 + c = 0

So, c=1

Substituting the value of c into the original equation yields that the square of x - 4x + 1 = the square of 0x - 4x + 4 - 4 + 1 = o

x squared - 2) squared = 3

x-2 = positive and negative root number 3

So x1 = 2 + positive root number 3 x2 = 2 - negative root number 3

Analysis: There is more than one design scheme that meets the requirements of this question, and there can be multiple design schemes. So it's an open-ended question.

Brief explanation: (1): The rectangle does not lean against the old wall.

If the width of the rectangular warehouse is x meters, the length is (50-x) meters.

From the solution: x=20 or x=30 of the question: x(50-x)=600 is found to be x=20 meets the requirements.

Since the square is also rectangular and has a larger area when the circumference is equal, a square warehouse with a side length of 25 meters is also eligible. At this time the area was 625 square meters.

2): The rectangular side is against the old wall.

Let the width of the rectangle be x meters, then x(100-2x)=600 solves: x= or , and the test knows that it meets the requirements.

At the same time, in order to make full use of the old walls, a rectangular warehouse designed to be 50 meters long and 25 meters wide also met the requirements. At this time, the area was 1250 square meters.

Let's start with the elementary answer: let the width be x and the length be y. The old wall was used to require a wide side of the rectangle, so as to save material and expand the area. The length and width meet the three conditions at the same time.

2。(50+x)+2y=100 "x is less than or equal to 0" or 50+2x+2y=100 "x is greater than 0".

1。When x is less than or equal to 0, the width is 50 (using the old wall to make another width), the length is 25, and the area is 1250

Set the length to x and the width to 100 2-x. (Unit: m)x(50-x) is greater than or equal to 600

The solution is 20 less than or equal to x less than or equal to 30

So the width is also 20-30 (m).

1.Solution: Set the distance of the flat road x kilometers, and the distance of the slope road is y kilometers, then according to the question.

y/12x/9

y 8x 9 solution: x

So the total distance is.

Kilometer. Attention to the uphill and downhill problems: 1) Use the known speed and set distance.

Indicates each period of time.

The production and sales cost of this Zen toy racing car is in yuan, but you have to ask if it can be cheaper, I can't be the master. It's not good to be in the spring.

3.Solution: First of all, the profit obtained by the company for a year of new products produced with these loans is:

10,000 yuan) and then according to the meaning of the topic, set the unknowns, find the equivalent relationship, and list the equation. Establish. These loans can be paid off after X years, according to the title, get:

40(1＋15%x)=26x

Solution:

x=2 has been tested to be in line with the topic.

A: You can pay off the loan after the 2nd year.

Question 1: If the time of going down the mountain is x, then the time of the flat land is: 55 60-x, then the distance of the mountain is: 12x, and the distance of the flat land is: (11 12-x)*9 and then the vertical beam is based on the speed and time of the return of the equation:

11 12-x)*9 8+12x 4=x=1 4, that is, the distance up the mountain is: 12*1 4=3 km, the distance on the flat land is: raid rent (11 12-1 4)*9=6 km, A and B total: 3+6=9 km.

1.If there are x people in the 2nd workshop, then the 1st workshop is.

x-10）*

x = 250 is 170 for the first workshop and 250 for the second workshop

2.Solution: The average speed from Beijing to Tianjin at the time of this test run is x kilometers per hour.

The average speed from Tianjin to Beijing is x+40 km/h.

List the equations according to the title:

x+40)*

Solution x=200

That is, the average speed from Beijing to Tianjin was 200 kilometers per hour during this test run.

1. Solution: If there are x people in the second workshop, then the first workshop has (people.

Solution x=250

Answer: 2. If the average speed from Beijing to Tianjin is x kilometers per hour during this test run, the average speed from Tianjin to Beijing is X + 40 kilometers per hour.

x+40)*

Solution x=200

A: The average speed from Beijing to Tianjin was 200 kilometers per hour.

！100% correct!!

1.Set the number of people in the first workshop x, then the number of people in the second workshop 4 5 * x - 30 column x+10 = (4 5 * x-30-10) * 3 42Set this test run from Tianjin to Beijing at a speed of x kilometers per hour.

Then Beijing to Tianjin speed X-40

30 60*x=(30+6) 60*(x-40), then the average velocity = 2*(30 60*x) [30+36) 60].

1. The second workshop x people first workshop 4 5x-30

3 4 (4 5x-30 + 10) = 4 5x-30-10x = 2501 workshop is 170, 2 workshop is 250

2. The distance between Tianjin and Beijing is x kilometers.

x (36 60) + 40 = x (30 60) x = 120 average speed = 2x [(36 + 30) 60] = km h).

1;Let the second workshop be m people, and the column equation is: 4 5 m-30 + 10 = (m-10) x3 4 m = 250 is the number of people in the second workshop, and the first workshop is 170 people.

1.Set the number of people in the second workshop to x, and the first workshop to y

4/5x-30=y

x-10=3/4x

Solution: x=40, y=2

So the first is 2 people, and the second is 40 people.