![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
How to do the applied problems of quadratic equation of one variable I won't be able to solve the application problems of one yuan twice at all. What can I do? What can I do to make me like to do the application problems? Maybe it's the problem of my understanding. What can I do? What book is better? Can I analyze the process of one yuan twice? Recommend it
The problem is a little long,
Xiao Li, Xiao Hong and Xiao Qiang participated in a social practice activity in a supermarket for their classmates. They were engaged in the sales of a certain kind of fruit. It is known that the purchase price of the fruit is 8 yuan / kg
Xiao Hong: if you sell it at 10 yuan / kg, you can sell 300 kg per day
Xiao Li: if you sell it at 13 yuan / kg, you can make a profit of 750 yuan per day
Xiaoqiang: I found that there is a functional relationship between the daily sales volume y yuan (kg) and the sales unit price X Yuan
1) Write down the sales quantity y at the price of 13 yuan / kg
2) 1. Find out the functional relationship between Y (kg) and X (yuan). (x > 0)
2. Let the daily profit of this kind of fruit in the supermarket be w yuan, and find out the functional relationship between W and X; and find out the maximum profit and the maximum profit when the sales unit price is w yuan [profit = sales × (sales unit price purchase price)]
(1) When the unit price is 13 yuan / kg, the sales volume is 750 / (13-8) = 150kg
Let the functional relation between Y and X be y = KX + B (K ≠ 0)
Substituting (10300) and (13150) respectively, we get: 10K + B = 300, 13K + B = 150
∴k=-50 b=800
The functional relationship between Y and X is y = - 50x + 800 (x > 0)
(2) ∵ profit = sales volume × (sales unit price purchase price)
∴W=（-50x+800）（x-8）
=-50x2+1200x-6400
=-50（x-12）2+800
When the unit price is 12 yuan, the daily profit is the biggest, and the biggest profit is 800 yuan
An applied problem of quadratic equation of one variable
A small ball starts to roll forward at the speed of 5 m / s, and decelerates at a constant speed. After 4 S, the ball stops rolling forward
1. How much deceleration does the ball roll per second?
2. It takes more time for the ball to roll to 5M (the result retains one decimal place)
Please list the quadratic equation of one variable and write the process and analysis process,
Could you please write it clearly? I've improved to 45
1. Suppose that the deceleration per second is x meters. Then: 4x = 5, x = 1.25. Because the ball decelerates uniformly, it takes 4 seconds from 5m / s to 0 (stop), so the deceleration per second is 5 △ 4 = 1.25m/s2. If the time is t, the ball speed in the T second is (5-1.25t) m / s
An applied problem of quadratic equation of one variable
Is it possible for a 56 meter fence to form two squares so that the sum of the two squares is 200 square meters? I have solved this equation. The answer is impossible. Why
Let a square side length be x, then the perimeter is 4x, and the other perimeter = 56-4x, side length = 14-x (14-x) &# 178; + X & # 178; = 2002x & # 178; - 28x + 196-200 = 02x & # 178; - 28x-4 = 0x & # 178; - 14x-2 = 0x & # 178; - 14x + 49 = 49 + 2 = 51 (X-7) &# 178; = 51x = (root 51) + 7, negative value is eliminated
(√ 1 / a - √ b) times √ AB, where a = 3 and B = 2
Original formula = 0 (1 / A * AB) - √ (b * AB)
=√b-b√a
=√2-2√3
If a is the root of the quadratic equation x & sup2; + BX + a = 0, and a ≠ 0, then the value of a + B is ()
A. - 1 B.1 C. - 1 / 2 D.1 / 2
If a is the root of the equation, then x = a, which is brought into the equation
a^2+ab+a=0
a(a+b+1)=0
Because a ≠ 0
Then a + B + 1 = 0
a+b=-1
The approximate solution of equation x ^ 2-5 = 0 on interval (2,3) can be obtained by dichotomy. After several dichotomy, the accuracy can reach 0.01
The interval length is 1, and the length is halved after each dichotomy
1/2^n 2^n>=100--> n>=7
That is 7 times
Let A1, A2,... As be an n-dimensional vector group. If s > N, then the vector groups A1, A2,... As are linear?
The rank of n-dimensional vector group is n at most. Vector groups A1, A2,... As are linearly correlated
Solving the equation of 5x + 5 = 3 (x + 5)
5x+5=3(x+5)
5x+5=3x+15
2x=10
x5
If the inequality m + X about X
x
RELATED INFORMATIONS
- 1. An applied problem related to quadratic equation of one variable The company has 120 taxis, and the daily rent of each taxi is 160 yuan. The taxi business is in short supply every day. In order to meet the market demand, with the approval of relevant departments, the company plans to increase the daily rent appropriately. According to the market survey, for every 10 yuan increase in the daily rent of a car, the number of taxis per day will correspondingly decrease by 6. If other factors are not taken into account, the number of taxis per day will increase by 10 yuan, When the company increases the daily rent of each car, the total daily rent income of the company will be 180 yuan more than before, and a certain number of vehicles will be maintained?
- 2. Application of quadratic equation of one variable In January 2010, the sales volume of Renming shopping mall was 1 million yuan, and the sales volume decreased by 10% in February. In March, the shopping mall took measures to reform the traffic strategy, so that the monthly sales volume increased significantly, and the sales volume in April reached 1.296 million yuan
- 3. The relation between the root and coefficient of quadratic equation of one variable, X & # 178; + X + 4N = 0 and 2x & # 178; + 7x + 3N = 0 have a common root, where n is not equal to 0. Try to find the value of their roots (not expressed in n)
- 4. On the relationship between root and coefficient It is known that the square of the quadratic equation x plus 3x minus 1 equals to 0 are X1 and x2. Please use the relationship between the root and the coefficient to find the absolute value of X1 minus x2 I want the process. Not just answers or theories.
- 5. Find the answer to the exercise paper about the relationship between the root and coefficient of quadratic equation of one variable
- 6. Elementary three quadratic equation of one variable, online, etc As shown in the figure, an isosceles trapezoid flowerbed should be designed. The upper bottom of the flowerbed is 100 meters long, the lower bottom is 180 meters long, and the distance between the upper and lower bottom is 80 meters. There is a transverse corridor at the middle line of the two waist, and there are two longitudinal corridors between the upper and lower bottom. The width of each corridor is equal. The area of the corridor is one sixth of the area of the ladder. How many meters should the width of the corridor be?
- 7. The solution of quadratic equation of one variable It is known that 4x ^ 2-12x + y ^ 2 + 2Y + 10 = 0 Find the value of 2x + Y / 2x-y It's not interesting at all
- 8. A problem of solving quadratic equation of one variable in the third grade of junior high school 2m & sup2; - 1 = 5m, 1 / N & sup2; + 5 / N = 2 and m ≠ n, find 1 / M + 1 / n
- 9. A problem of solving quadratic equation of one variable in the third day of junior high school X & sup2; - 2 | x | + 2 = m has exactly three real roots, What is the value of M? ∵ X & sup2; = | x | & sup2;; the original equation can be reduced to | x | & sup2; - 2 | x | & sup2; + (2-m) = 0, ∵ the equation has three real roots, | there must be one real root | x | = 0, (otherwise the equation has four real roots!); 2-m = 0, M = 2 What I don't understand is why | x | = 0?
- 10. The weight ratio of package a and package B is 4:1. If 13 grams are taken from package a and put into package B, the weight ratio of package a and package B becomes 7:5, then what are the weights of the two packages?
- 11. As shown in the figure, a warehouse with an area of 130m2 is to be built. One side of the warehouse is close to the wall (the length of the wall is 16m) and a 1m wide door is opened on the side parallel to the wall. At present, there are 32m long wooden boards which can be enclosed to calculate the length and width of the warehouse
- 12. Application of quadratic equation of one variable in grade two of junior high school The cost price of a commodity is 200 yuan, which is 50% higher than the cost price. Due to poor sales, the price after two consecutive price cuts is 192 yuan. If the two price cuts are the same, calculate the price reduction rate
- 13. A calculation problem (using the square difference formula) (1) The correct result is () A.(x-2)(2+x)=x^2-2 B.(3x-2)(x+2)=3x^2-4 C.(7x-y)(7x+y)=49x^2-y^2 D.(x^2-y^2)(x^2-y^2)=x^4-y^4 (x ^ 2 means the square of X, y ^ 2 means the square of Y, x ^ 4 means the fourth power of X, and Y ^ 4 means the fourth power of Y.) (2) (0.5x-1 / 2) (1 / 2x + 0.5) = (to write the calculation process) (1 / 2 means 1 / 2) After I check with the teacher, I'll see if you are right! I am the third level answer (only {he} her second question has a process)
- 14. The application of quadratic equation of one variable in the second grade mathematics of junior high school The same 8m long aluminum alloy window frame material is made into a rectangular window frame. What is the maximum lighting area of the window frame? What is the side length of the window frame when the daylighting area of the window frame is the largest
- 15. Junior two mathematics one yuan quadratic equation application problem. Seek detailed solution,. Thank you! A train goes on at constant speed. When it enters a 300 meter long tunnel, the knife completely passes through a kind of experience for 20 seconds. It also knows that a fixed equal vertical beam at the top of the tunnel irradiates the train for 10 seconds to find the length of the train
- 16. After two recent interest rate cuts, a bank has reduced the annual interest rate of one-year deposits from 2.25% to 1.98%. What is the average percentage of each interest rate cut?
- 17. In order to strengthen the flood control work, the municipal engineering team is preparing to reinforce a 2240 meter long embankment of Suzhou River. Due to the new reinforcement mode, the length of reinforcement planned every day is increased by 20 meters compared with the original plan. Therefore, the number of days required to complete the reinforcement project will be shortened by 2 days compared with the original plan. In order to further shorten the time of reinforcement project, if 224 meters are required to be reinforced every day, then On the basis of the present plan, how many meters will be added to the length of reinforcement every day?
- 18. There are 100 peach trees in an orchard, each of which bears 1000 peaches on average. Now we are preparing to increase the yield by planting more than one tree. Without planting more than one tree, the yield of each tree will be reduced by 2. However, the number of peach trees should not exceed 100. If we want to increase the yield by 15.2%, how many trees should be planted? My method is to set up multiple x trees and get (100 + x) (1000x-2x) = 100000 (1 + 15.2%) Look at how wrong I am. If it's right, help me figure it out. If it's wrong, help me change it! How do you solve the equation
- 19. In order to expand sales, increase profits, and reduce inventory, the mall decided to take appropriate price reduction measures. During the trial period, it was found that when the price of each commodity decreased by 1 yuan, the mall could sell 2 more per day. According to this rule, if the mall wants to make a profit of 1200 yuan per day, how much should the price of each commodity be reduced ？
- 20. System of linear inequalities with one variable A person's hair is about 100000-200000, and each hair grows about 0.32mm long every day. Xiaoying's hair is now 10cm long. How long does it take for her hair to grow to 16cm and 28cm,