Senior 3 math review, ask for advice! Mathematics in the first year of high school, help to solve!!

The fastest time to improve the score of mathematics is to follow the teacher to do the topic, at that time must not be lazy, the teacher must try his best to finish, even if his own review plan must be placed behind the teacher's plan, follow the teacher one by one, the ability will gradually improve, and the forgetting will become less and less.........Be sure to listen carefully to the teacher's plans and lessons, you will benefit, the main learning method is not to give up, no matter how good or bad the test is, you can't give up, this is the most basic quality of high-scoring candidates! Keep in mind, keep ...... in mind

If you can't persist in regretting it, it's useless, encourage yourself often, and don't think too much about friendship or anything, the purer your heart, the more ideal your learning! I'm ...... from hereDon't do things that you will regret, think about your own burdens, responsibilities, relatives, classmates, ......Always be motivated, never give up, come on!

Hello, I just finished the college entrance examination this year, and I did well, although I am a candidate from Jiangsu, but I think most of the math studies are similar. First of all, we must first establish a goal, break through each during the review, and do more basic questions on weak knowledge points, although the time is very tight in the third year of high school, but these small questions must be practiced time. Secondly, I quite support the problem set, there can be a time interval but there should be a regularity, it is best to review and sort out once every 1 to 2 days.

The most important thing is to avoid the overly difficult finale questions, and you can take a look at some new thinking questions and add some ideas. Also, if you don't understand, you must ask the teacher.

Hope it will be useful to you and I wish you progress.

The college entrance examination has just ended, and the test is considered good 1. I think the biological clock of the college entrance examination is very important It is recommended that you practice during the time period of the mathematics college entrance examination.

2. Practising the basic questions can improve your self-confidence in mathematics.

3. Revision advice: Don't put too much time on difficult problems.

If you have the conditions, it is best to find a good tutor, after all, the college entrance examination is a big deal.

1. Look at the concept, read the book, do the example problems in the book, and then repeat it to summarize the method of doing the problem.

The premise of the concept is particularly important if it is to be memorized.

2 Do the questions. Do the questions sent by the teacher Don't look for the workbook The questions given by the school are done repeatedly, and the wrong ones are summarized in a variety of ways, and the questions that are missed are repeated regularly, for a month, or a week. I know it's particularly painful.

Because you will find that it is still the same as the wrong place ... But if you continue, it will change.

3 Calm ... You can't be in a hurry to do the questions... If you are in a hurry, you will not play at the level.

4 Science. Mathematics, physics and chemistry students read the book well, and physics and chemistry students read it without pulling a word, including the notes under the book. The physical theory section summarizes a timeline. And to summarize it from the perspective of the development of theory, the rest of physics is about the same as mathematics, and it is done over and over again.

Just listen to the lectures in chemistry.

Biological endorsements. Do Mind Mapping Mind mapping is a god horse, you can check it yourself. Very simple but very practical.

5 By the way, Chinese, English will follow the teacher. The language must follow the teacher. English depends on character.

1 All First of all, po is a flagpole, so perpendicular to ao and ob, apo is a right triangle, known oap=30°, po is h, then according to tan 30° = po ao, tan 30° = the root number of 3 points, it is easy to get ao = 3 h.

In the same way, if obp=45° is known, bo=h is easily obtained.

Knowing ab=20, aob=30°, the length of ao and bo, using the cosine theorem, ab = ao + bo 2·ao·bo·cos aob, bring in the data, get 400=h, h is the length, is a positive number, so h=20m.

Because oap=30° po=h

So ao=(3)h

Because obp=45°

So bo=po=h

According to the cosine theorem:

ab＾2＝ao＾2＋bo＾2－2＊ao＊bo＊cos（∠aob）ab=20m ∠aob=30°

Substitute the data.

So h 20m

Your answer will be studied and referenced by dozens or even tens of thousands of netizens, so please be responsible for your answer, and try to ensure that your answer is accurate, detailed and effective ab=20, aob=30°, the length of ao and bo, using the cosine theorem, ab = ao + bo 2·ao·bo·cos aob, bring in the data, and get 400=h, h is the length, which is a positive number, so h=20m.

Because pao=30°, ap=2h, ao=root number 3h

Because obp=45°, ob=po=h

Because, the po is perpendicular to the bottom, so ab = ao +on

Because OP ground, AOP and BOP are both right-angled triangles, and it is easy to obtain OA=ob=h Tan30°= 3H, in AOB, do OC AB, then AOC=15°, AC=10, OC=10*Tan75°=10(2+3), so by the Pythagorean theorem, .,3h = 100+100 (4 + 4 3 + 3), and the solution is h=20{ (2+ 3) 3}

Actually, it's very understandable.

Because , if a belongs to s, then 1 1-a belongs to s. So look at (1 1-a) as a whole, as "a". Now (1 1-a) belongs to s, then substituting in , there is:

1 [1-(1 1-a)] belongs to s, which is simplified to: 1-1 a

So there is: 1-1 a belongs to s

A and 1 1-a have equal status and can be substituted.

I chose A at first, and the answer said it was C, I don't understand, help me point out, I guess it will have something to do with A=B=? The title is incomplete, right? You're right a=b

Can it be seen as a function?

f(a)=1\1-a

f(1\1-a)=?

It may not be able to correspond, but you can refer to it.

Solution: Set the condition by the question, there are 0 y 6, y x 6. ∴0≤x≤π/6，0≤y≤x。

Original = (0, 6)dx (0,x)cosxdy x= (0, 6)cosxdx=sinx丨(x=0, 6)=1 2.

FYI.

hyperbola c: (x a)-y =1;right focal f( (a +1),0);

An asymptote: y=x a; Let the coordinates of the point p on the asymptote be (x,x a);

then po =x +(x a) =(1+1 a)x ; pf∣²=[x-√(a²+1)]²x/a)²;

From po = pf, po = pf ; i.e. (1+1 a)x =[x- (a+1)] x a) ;

Simplified: -2[ (A +1)]x+a +1=0;The abscissa of point p is x=(1 2) (a +1);

Area of OPF s = (1 2) of x A = (1 2)[ A +1)][1 2A) (A +1)] = (1 4A)(A +1).

a/4)+(1/4a)≧2√[(a/4)(1/4a)]=2×(1/4)=1/2；

i.e. the minimum value of the area of the opf = 1 2;Therefore, B should be chosen;

It's okay for Big Joe to send his teammates to the opposite spring.

If bp is selected on the mid-perpendicular line of of, you may wish to set p in the asymptotic equation y=x a where p passes one quadrant through the first quadrant

P(C2,C(2A)), C = Area of A+1δOPF.

s=(1/2)·c·(c/(2a)=c²/(4a)=(a²+1)/(4a)

1/4)(a+1/a)

1/4)·2√(a·(1/a))

i.e. s 1 2 and a=1"="

So choose B