Please help me 555555 with the math strongman in junior high school

There is salvation, and time is still in time. There is no failure in this world, only a temporary cessation of success. The past is not equal to the future.

I am a math teacher in junior high 3, and I can see that your foundation is okay and you can't get high scores, there are two reasons: one is that you don't do enough exercises, and the other is that the knowledge structure is not very clear. It is recommended that you first straighten out the knowledge points from the first to the third year of junior high school, and secondly, find some time to do more practice, practice makes perfect.

Third, we must make a good study plan.

Overcome excessive nervousness, maintain a moderate amount of nervousness, and take the exam calmly.

Stress Psychological Regulation:

1) Knowledge moderation method: First of all, it is necessary to take a correct view of the exam. An exam is nothing more than a special assignment or exercise, a test or "diagnosis" of learning effectiveness.

A high or low score does not equal success or failure. Failing an exam doesn't mean anything, let alone decide everything. Even if it is really "failure", there is nothing, "victory and defeat are a common thing in soldiers", not to mention, "failure is the mother of success".

If you don't know how to do the question because you're not usually ready, just calmly accept the fact and come back next time.

2) Positive suggestion: Give yourself positive hints to reduce stress and believe that you are capable of making a question. For example, silently say to yourself: As long as I am calm, I am confident that I will pass the level; As long as you relax, you can make it.

3) Deep breathing regulation: When you feel nervous or anxious, close your eyes, relax your body, calm your mind, concentrate, take slow deep breaths, and pay attention to experience the feeling of relaxation throughout your body. If you do this several times, you can get the effect of relieving tension and relaxing yourself.

Wow It's finally over, there's too much on the character book, it's killing me o( o haha

Oh o )zz

1. Be careful.

2. Don't be nervous.

3. Do more Olympiad questions.

Take deep breaths when you have time, so that you can set your mindset.

It's old-fashioned, but it's effective - the wrong book.

If you understand all the questions, you'll be pretty much there-

I just took the high school entrance exam this year - 112 out of 120 in mathematics

If you are not careful enough, just be careful.

You just have to put yourself into the problem as you do it. When you are nervous, force yourself to think about the question, and you will be able to naturally devote yourself to the question.

The main thing is to fight psychological warfare, don't think too much, and think, "If you don't do well in the exam, the question is difficult, it's none of my business, I've done my best".

Mathematics in junior high school is actually very easy to learn, as long as you listen carefully in class, carefully recall what you have talked about in class after class, do more basic questions after class, and then do writing improvement questions, so that's it.

Adjust your mentality well, and you will not be afraid of everything, read more books, read examples, and be saved.

You have to believe in yourself and don't be nervous.

It's okay usually, but it's not good in the big exam, which means that the comprehensive ability is not good, and you have to do more comprehensive questions

There is salvation, mathematics is to do problems, read more Olympiad problems, and practice. Don't be nervous when you're doing the questions, believe in yourself that you can do it!

Read more books and look at example questions to save you.

It seems that the full score is 150 now, right, you said that it is like this in a big test, which means that it is usually good, right, that is a big reason for your psychological factors, your strength is there, don't be nervous, just play hard.

（1）s=

Let s=at 2+bt

Substitute (4,0),(2,-2).

Get the answer. 2) t=12 or t=-6 rounding.

It can be reached by the end of December.

3) t=8, s=10,000 yuan).

x^2-4x+m-1=0

Because there are two equal real roots, discriminant 0

4^2-4(m-1)=0

The m=5 equation is x2-4x+4=0

x-2)^2=0x=2

m=5, the root of the equation is x=2, using the perfect sum of squares.

A quadratic equation has two equal real roots.

B-square-4ac=0

In a quadratic equation, the square of a=1, b=-4, c=m-1 2(-4) 1 (m-1 2)=0m=9 2 When m=9 2, the unary quadratic equation for x, x-square-4x+m-1 2=0, has two equal real roots.

Substituting m=9 2 into the original equation yields:

x-square-4x+4=0

x-2)(x-2)=0

x-2=0x1=x2=2

Solution: Two equal roots of real numbers =0 b2-4ac=0 4*4-4*(m-1)=0

m=5 The root of the original equation is 2

x2-4x+4-4+m-1=0

x-2)2 -4+m-1=0

Since the original equation is known to have two equal real roots, -4+m-1=0, so m=5 so the root of the equation is 2

x2-4x+m-1=0 has two equal real roots.

Then 4*4-4*(m-1)=0 i.e. m=5, so the equation is x2-4x+4=0 so his root is x=2

1) x indicates the number of people in a class, because 40%=2 5=.

2) A represents the area of the triangle, in fact, I want to say that the area of the triangle is unknown. (That's how I understand it).

3) b is ten times the number but two less. (That's how I understand it) note: This is my personal understanding, and this is the number of junior high schools, I'll first say that I'm in the first year of junior high school, but I don't know if it's reliable).

Listen to my junior high school absolutely.

1) A indicates the total number of people in the class.

2) b represents one-half of the bottom edge of the triangle.

3) b is a ten-digit number.

1) x represents the total number of students in a class of 7th grade.

2) A is the area of the triangle.

3) b is a ten-digit number.

x is the class size.

A indicates the bottom edge where 2cm is located.

b means (number in the top tens) minus 1 plus.

x represents the number of people.

a denotes the area of the triangle.

Question 3: I don't know what you're asking.

x represents the class size.

a represents the total area of the triangle.

b is the value of ten digits minus one-fifth.

1. x represents the number of people in the class.

2. A represents the area of the triangle.

3. b represents the number in the tens of digits of this number.

1.Total number of seventh grades.

2.The total area of the triangle.

3.Ten figures.

1. x represents the number of people in the class.

2. A represents the length of the three bottom sides. Because the area of the triangle = one-half of the base times the height, 1 2 base length 2 = base length.

3. b represents the number in the tens of digits of this number. Your question must be wrong, it should be 10b+5, otherwise it is impossible to express it as a two-digit number of 5. For example, 25 can be expressed as 10 2+5, so b represents the number in the tens of this number.

1. x indicates the number of people in the class.

2. A represents the area of the triangle.

3. b represents the number in the tens of digits of this number.

x represents the total class size in Year 7.

a represents the area of the triangle. b=

1) x indicates the number of people in a class.

2) I don't understand what you mean.

3) I don't understand what you mean.

ab=6-2√6

Answer: The perpendicular line DE of the BC side is made along the vertex D, the triangle DCE is the perpendicular line AF of the Isosceles Right Triangle along the vertex A as the DE side, and the triangle DAF is the isosceles right triangle AF=DF= 6, DE=CE=6-6, then AB=EF=6-2 6.

All that's left is to calculate the area of the triangle DAF, DCE, and the rectangular ABEF, which can be solved by summing them up.

Your idea is right, that is, the big triangle minus the small triangle, according to the angle relationship on the topic you can know (the inner angle of the quadrilateral and 360), both triangles are isosceles right triangles (you don't see this, causing you to walk into a dead end, no need to set unknown)! Then you draw od 2 = oc 2 2, oa = od-da, and the rest continues according to your train of thought!

It's annoying for you to do that, and it's as simple as this: according to 1::1:

2；ab=x, then oa is equal to 2x; Isn't ob equal to x, oc is not equal to x+6; Isn't od equal to 2x+2 3 and then according to the same ratio: od:od=( 2x+2 3):

x+6)=1：√2;Do the math:

2x+2√6=x+6;It's simple, x=6-2 6

1:：1：√2；ab=x, then oa is equal to 2x; Isn't ob equal to x, oc is not equal to x+6; od is not equal to 2x+2 3 and then according to the same ratio:

od：od=(√2x+2√3)：(x+6)=1：

2;Do the math:

2x+2√6=x+6;It's simple, x=6-2 6

First, it is proved that the triangle OAB and the triangle OCD are similar triangles. Then set OA or OB to X, and then similar to the side of the similar triangle, that is, OA ob=OC OD, and finally subtract the area of the small triangle OB from the area of the large triangle OCD, which is the area of the quadrilateral ABCD, think about it.

Hey, the child forgot the ratio of the sides of the isosceles right triangle. It's 1::1:

2。It is also possible to extend the ba, cd and make up the ** into a large isosceles right triangle with a point. Then it is also the area of the large triangle minus the area of the small triangle.

BC(6) is the right-angled side of the large triangle and AD is the right-angled side of the small triangle. This should be a simpler calculation, and there is no need to set x.

It really challenges my eyesight, my sister and I'm short-sighted.

You're doing the right thing, keep doing it!

You should have learned about similar triangles.

That's what you do.,Set ob to x oa od=ob oc and then you can solve x out.,It's just that the solution is weird.,You solve it.,I hope it's right.。

Original = 3 circles (9 root numbers, 4-4 root numbers, 9 root numbers) = 3 circles, (18-6) = 3 circles, 12 = 3 root numbers, 12 = 12 root numbers, 3 = 6 root numbers, 3 = 12 root numbers, 3 = 18 root numbers, 3 numbers

Let the duck x, the frog y, and the toad z be listed, then a ternary indefinite equation can be listed;

2x+4y+3z=13000

x+y+z=3600

The elimination element can obtain 2y+z=5800

The solutions to this indefinite equation will be many.

Duck x, frog y, toad z, 2x+4y+3z=13000, x+y+z=3600

Three unknowns, two equations.

There are six solutions:

Hope it helps.

x+y+z=3600

2x+4y+3z=13000

This is three unknowns and two equations, there should be no unique solution, right?

Take z=3600-x-y and bring it into the second equation.

2x+4y + 3(3600-x-y) = 13000-x + y = 2200

There are actually a lot of integer solutions, so they can't be solved.

Set x ducks, frogs y, and toads z.

then x+y+z=3600

2x+4y+3z=13000

3 unknowns, two equations, should not be easy to solve. I didn't try. There should be a lack of qualifications.

Set up ABC

2a+4b+3c=13000

a+b+c=3600

Get b = 2200 + a

c=1400-2a

Because abc is greater than or equal to 0 and is an integer.

Then linear programming.

Get a collection.

ABC belongs to the collection.

I'm at the elementary school level, and the answers can only be expressed in sets, and nothing else.

Set x ducks, frogs y, and toads z.

x+y+z=3600

2x+4y+3z=13000

Can't figure it out.

Correct the duck 100 legs.