How to do general math problems, how to do math problems

Analysis: When the flood comes, the width of the water surface mn=32m is required to take emergency measures, only the length of the de, so only the radius r is required, and then the geometric algebraic solution is used to solve the r: no emergency measures need to be taken.

。。I don't know how the other one did it. My words are to take a look at the relevant example problems and apply the formula. It all feels the same, pretty much the same.

You have to look at what type, there are many ways to do it, but you are hungry to do more examples.

Original) This is a very general question. Although there are certain methods to find for test-taking math problems, there are generally certain methods or patterns to solve each type of problem. The methods are also different.

That's basically a couple of books. But as generic. Do a question.

It can be divided into three steps: reviewing, doing, and reflecting. It may seem abstract to put it this way. But every process counts.

First of all, review the question. 1.You must be calm, not in a hurry, see the question clearly, and understand the question accurately.

2.Combine the various factors, the known, the unknown, and the potential known conditions in the question. List it. It's about dissecting the problem.

3.Let's think about what knowledge points may be tested in this question, what formulas you can use to connect the factors decomposed in point 2, or what similar questions you have done in your mind that you can reduce to a method. You should understand the naturalization method.

I won't go into details here.

Then it's time to do the questions.

When you do the questions, it will test your usual hard work.

1。Do the questions and try to go through them all over again. Be accurate, reasonable and standardized in the process of calculation, reasoning and expression.

2.Strive to do it quickly. In order to speed up the speed of solving problems, one depends on thorough study of knowledge, and the other depends on hard training, which is the result of diligent study and hard training.

Rethink. After completing the homework, it is necessary to make a recall summary in time, review what basic knowledge has been learned, what skills and techniques have been mastered, what mathematical ability has been improved, what rules have been explored, and what new discoveries have been made, which can be reflected in the form of postscripts. Or you're stuck in ** train of thought, why can't you think of this train of thought and so on.

Or find out if there's an easier way.

Hope it helps you ...

Summary. How to do math problems.

The teacher is here.

Kiss, take a picture of the math problem and show it to the teacher.

2 or 3, kiss

Here's this. Two questions are required.

The last one is, right.

Right. Can you ask two questions?

Mmmm, yes, kiss.

The teacher will do the math for you.

Wait a minute, this is the calculation volume, it's very simple, let's say the height is y, and then 6 times 8 times y = 432y = 9

The height is 9cm

And then the third question is like this.

Seeing no kisses.

Ask about custom messages].

Okay, oh classmate.

Seeing that there are no classmates, the teacher's second height is 9cm, and the third question is high.

The kindergarten bought peaches for the children, and when distributing, if each child is divided into 5, there are 32 left, if 10 of the children are divided into 4, and the rest of the children are divided into 8, it is exactly finished. Then there are children in the kindergarten who bought a total of peaches.

Solution, obtain: Solution 1: Correspondence method].

Solution: In the first division, 1 person corresponds to 5 peaches, and there are 32 more peaches.

In the second division, 1 person corresponds to 8 peaches, and 10*(8-4) is missing = 40 peaches.

8-5 = 3, if one person corresponds to 3 peaches, the total number of peaches needed should be 32+40 = 72 (pcs).

Number of children = 72 3 = 24 (people), total number of peaches bought back = 24*5 + 32 = 152 (pcs).

Hopefully that's the solution you're referring to).

Solution: Set up x children with a total of Y peaches.

Then, the first division: 5x + 32 = y

The second division: 10*4 + x-10)*8 = y

Simultaneous solution, x = 24, y = 152

A: There are 24 children in the kindergarten and a total of 152 peaches were bought.

2.A piece of land, if ploughed with the same tractor. After 4 hours of ploughing, 8 hectares were left untilled, and after 6 hours of 3 tilling, 4 hectares were not ploughed. How many hectares is there in this area?

Solution: Set up each tractor and plough 1 per hour.

4 units for 4 hours, cultivated land 4 4 = 16 parts.

3 units for 6 hours, cultivated land 3 6 = 18 parts.

Difference: 18-16 = 2 parts.

These 2 servings are: 8-4 = 4 hectares.

So each serving is: 4 2 = 2 hectares.

The land has a total of:

16 2 + 8 = 40 hectares.

3.It costs 100 yuan to buy 2 chairs and a table, and 100 yuan more to buy 8 chairs than 2 tables.

Solution: According to "buying eight chairs is more than buying 2 tables", it can be known:

Buy 4 chairs and pay more than 1 table: 100 2 = 50 yuan.

Then "buying 2 chairs and a table" is equivalent to buying: 2 + 4 = 6 chairs, so.

The unit price of each chair is: (100+50) (2+4)=25 yuan.

The unit price of each table is: 100-25 2=50 yuan.

4.Master Wu processes a batch of parts, if he makes 50 parts a day, it will be completed 8 days later than the original plan, and if he does 60 parts a day, it can be completed 5 days in advance. How many parts are in this batch?

Solution: Solution: Set the original plan to be completed in x days.

50（8+x）=60（x-5）

400+50x=60x-300

10x=700

x=70 This batch of parts has:

60 (70-5) = 60 65 = 3900 (pcs).

You can also write this formula:

50 (70+8)=50 78=3900 (pcs).

A: There are 3900 parts in this batch.

Please type the question.

Question 22Find the derivative of the following function (5 points per question, 10 points in total: (1) y=(3x-5)10

Answer: 3=30(3x-5) 9 2(3x+2)'sin5x+sin5x'(3x+2)=3sin5x+5cos5x(3x+2) 3.

e^2x)'cos3x+e^2x(cos3x

Solution: This is just a staring and burying idea, and you can organize it yourself.

Let the AOD be high for H and the BOC high for H

then h h=od ob=ad bc

ad*h/2=9...1)

bc*h/2=16...2)

1)*(2) There is AD2BC2=9 16, so AAD BC=H=3, 4

S trapezoidal ABCD = (AD+BC)(H+H) 2=(7 3AD)*(7 3H) 2=49AD*H 18

ad*h 2=9

There is a substitute for the punch car on the type.

S trapezoidal ABCD = 49

1. Area of ACD = Area of BCD? With cd as the base, the high is equal (because ab cd).

2. Area of ACD = Area of AOD + Area of OCD Area of BCD = Area of missing BOD + Area of OCD.

3. So BOC area = area of AOD 6AOB area ob*H=9, area of AOD OD*H=6, so DO:ob = 6:9 with travel co:OA = 6:9

Solution: The corresponding angles of similar polygons are all equal, the corresponding sides are all proportional, the trapezoidal ABCD is an isosceles trapezoid, ab cd, ad=bc=3 2, ab=12, the difference is A=45°, and the potato is stupid with cd=ab-2*(ad*cosa)=6, bc b'c'=ab/a'b'=12 8=3 2, so b'c'=2 2, with the side of the inner corner of the virtual hand block complementary, so.

c= d=180°- a=135°, thank you!