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1. A completes one-fifth of the work in 3 days, then completes 1 15 in one day, and B completes two-fifths of the work in 4 days, then completes 1 10 in one day, and two people can complete this work in one day.
The two worked together for 6 days to complete the job
2. A travels for 4 hours, and B takes 6 hours.
It takes 10 hours to complete the journey of line A, and the complete journey of line B takes 10 hours
10 4 6 15 hours.
The second trip takes 15 hours.
It is planned to produce 360 parts.
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A completes 1 5 3 = 1 15 in one day
B completes 2 5 4 = 1 10 in one day
A and B cooperate in one day to complete 1 15 + 1 10 = 2 30 + 3 30 = 1 6
It takes 1 1 6 = 6 days to complete this work.
Analysis: B traveled for 6 hours and A for 4 hours.
So B travels A for 10 hours.
6/x =4/10
x=15 master does 1 9 an hour apprentice does 1 12 an hour master and apprentice does 8 hours (1 9 + 1 12) * 8 = 14 9 more parts 14 9-1 = 5 9
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1 5 * 1 3 = 1 15 1 5 5 * 1 4 = 1 10 2 5 * 1 4 = 1 10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Car A travels for 4 hours = Car B travels for 6 hours, Car A travels for 1 hour, and Car B takes 6 4 = 3 2 hours to travel.
Therefore, car A travels for 6 hours, and car B takes 9 hours to travel, so it takes 9 + 6 = 15 hours for car B to travel the whole way.
3.The first plan is to produce x, so the number of units per hour produced by the master is: x 9 The number of units produced per hour by the apprentice is: x 12
The number of units produced by two people in one hour of cooperation is: x 9 + x 12 = 7 x 36 The number of production produced by the master and apprentice in 8 hours is 200 more than planned, so: 7 x 36 * 8 = x + 200, and the solution is x = 360
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1. Let A complete x every day and B complete Y every day.
3x=1/5 x=1/15
4y=2/5 y=1/10
A and B cooperation is completed in one day 1 10 + 1 15 = 1 6 Two people cooperate in 6 days to complete.
2. Let the velocity of A be x and the velocity of B be y. The total distance is s.
6x+6y=s
6x+4x=s
y=1/15 s
So B takes 15 hours.
3. Set up x planned production.
x/9+x/12)*8=x+200
x=360
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1. A completes in one day: 1 15, and B completes in one day: 1 10
Two people cooperate in one day: 1 15 + 1 10 = 1 6 Cooperation: 15 + 10 = 25 (days) (pieces).
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2 2 3 3 3
The sum of these five "errors" is.
The process is a bit annoying, it's hard to write, but it's right, if you need it, ask me again, I don't know if you want the result or the process.
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I don't know if it's right, but it's a primary school question after all, right, then divide 13 by 5 = replace it with 5 numbers, then it will be 3 3 3 2 2 The sum is 13 and the maximum two numbers in the original question will be replaced by 3 The error is (
then the error is.
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The sum of the integers of the five numbers is 10, and the difference between 13 is three, so let the three decimal places count the big ones in, and the small ones are rounded, you can calculate them yourself.
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Is it so complicated or elementary school math?
What elementary school I'm curious!
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In elementary school, there is only one unknown.
There are x people in class A and 93-x people in class B.
Class A 1 5 is x 5
Class B 1 6e
93-x)/6
Then and to be 17 people is.
x/5+(93-x)/6=17
It turns out that there are 45 students in Class A.
There are 48 students in Class B.
The two places are 540 km apart
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Three even numbers x-2, x, x+2, the smallest and the middle number are twice as many as the sum of the maximum, 20, x-2+2x=x+2+20
2x=24x=12
So, these three numbers are 10, 12, 14.
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Shipping Issues:
The cost of shipping to A is: 600x+400 [15-(18-x)].
The cost of transportation to B is: 500 (17-x) + 800 [14-(17-x)].
So y=600x+400[15-(18-x)]+500(17-x)+800[14-(17-x)].
1300x+4900
As can be seen from the above formula, if you want to minimize the freight, then as long as X is the smallest, and since 18 units are needed in place A, and there are only 15 units in place B, then at least 3 units must be transported from place A to place A, that is, the minimum value of X is 3. When x=3, y=8800. That is, three units are transported from place A to place A.
The remaining 14 units were all shipped to B, and all 15 units from B were shipped to A.
Clothing Problem: Category A Clothing Profit: 45x. Cloth to use:
A, B. Then the remaining fabric is a::, and the number of clothes that can be made is :
and (, because it contains the variable x, then when (, x is approximately equal to because the number of clothing sets is an integer, i.e. x takes 1. And when x>1, (> the number of sets of clothing at this time is (, and vice versa.
Then y=45x+50 [(when x>=1) or y=45x+50 [(when x<=1) is obtained.
y=6500-1330x (when x>=1) or y=when x<=1).
When x=1, the maximum value is 5170
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Question 1: If A is transported to X in A, it will be transported to B (17 x), B will be transported to A (18 x), and B will be transported to B [15-(18-x)] x 3
y=600x+500(17-x)+400(18-x)+800(x-3)
500x+13300
y is a primary function of x, and k 500 0
17 x 14, i.e. x 3
Y decreases as x decreases, so when x 3, y has a minimum value of 14800, so A is shipped to 3 units in place A and 14 units to place B. 15 units were shipped to place A and 0 units were transported to place B, at this time, the cost was the least, and the minimum cost was 14,800 yuan.
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1. An enterprise deposited a total of 200,000 yuan in deposits of two different natures, A and B. Assuming that the annual interest rate of deposit A is 9,500 yuan after deducting interest tax, the annual interest rate of deposit B is one year later. How many tens of thousands of yuan are the deposits of A and B?
Note: Interest tax = interest amount * 20%)
Solution: Let bank A deposit x yuan, bank B deposit y yuan, from the known conditions, get the following equation:
x+y=200000
x*[Solve the above equation to get.]
x= y=A: Deposit A is. 10,000 yuan, type B deposit is. 10,000 yuan
2 In order to make effective use of electricity resources, the electricity bureau of a city has implemented the implementation of residents since January"Bee Valley"Electricity pilot. If used"Bee Valley"Electricity, from 8:00 to 22:00 per kilowatt hour per day"Bee electricity"valence);
From 22:00 to 8:00 the next day, the electricity fee is RMB per kWh. One for home use"Bee Valley Electric", the electricity bill for a certain month is.
Yuan, after calculation, than not used"Bee Valley Electric"Saving yuan. Ask the family to use it for the month"Bee electricity"And.
Valley Electric"How many kWh each?
Let the peak current be x and the valley current be y, and the binary equation is listed.
x y=(Solve the equation to get x 140, y 60.)
The peak power is 140 kWh and the valley power is 60 kWh.
3 If 1 x+1 y=5 is known, then the value of the fraction 2x-3xy+2y x+2xy+y is (1).
1/x+1/y=5
x+y]/xy=5
x+y=5xy
2x-3xy+2y]/[x+2xy+y]
2*5xy-3xy]/[5xy+2xy]
7xy/7xy
14 In order to further alleviate traffic congestion, a city decided to build a light rail railway from the city center to the airport, and in order to complete the project three months ahead of schedule, it was necessary to increase the original work efficiency by 12%.Q: How many months did it take to complete the project?
The original plan was to complete the project in x days, and the daily workload was 1 x, and after the efficiency was increased by 12%, the daily workload was (1 12%) (1 x), that is, it would take 1 ( day, that is, x days, which was 3 days less than the original plan of x days, so the equation is.
x x and x 28 days.
5 When a is taken, is the solution of equation (x-1) (x-2)-(x-2) (x-1) = (2x+a) (x-2)(x-1) negative?
x-1)^2-(x-2)^2]/(x-2)(x-1)=(2x+a)/(x-2)(x-1)
x^2-2x+1-x^2+4x-4=2x+a
2x-3=2x+a
a=-3
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There are x number of employees in this workshop.
1/7x-6)+11(1/7x-6)=x
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Male workers are 1 + 11 = 12 times that of female workers.
Female workers account for 1 (1+12) = 1 13 of the total
According to the title, the total number of 1 13 is 6 less than the total number of 1 7 workshop employees: 6 (1 7-1 13) = 91 people.
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is f(2-x)+f(x-2)=2, because the condition given in the question is f(x)+f(-x)=2, and if 2-x is regarded as x by commutation, then -x=x-2. Therefore, the first way to write it is correct.
I also thought about this problem when I was in high school, first of all, the front multiple-choice questions should be done quickly, the method should be used flexibly, and it is not necessary to do the whole process, you can use a special method to bring in the method and a series of quick practices, and then fill in the blanks as much as possible, basically send points in front, there are two difficult points in the back, the first 2 questions of the big topic are very basic to ensure that they are all right, and the big questions behind should have the concept of step-by-step scoring, don't look at the type of question that you have not seen it and feel that it is difficult to have no confidence, the first few steps can still be scored, The next few steps are written to where it counts, and this is a score. In general, we should pay attention to the foundation, ensure that the basic score is not lost, the time should be allocated well, if the level of multiple-choice questions is good in 30 minutes, generally, about 40, fill-in-the-blank questions should have 30 minutes to do, and then there is about an hour, the first 2 big questions are 15 minutes, and the rest of the time try to do the rest of the questions! >>>More
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