Ask the seniors who are good at math to help!!

Don't be in a hurry, mathematics should actually pay attention to the basics, master the basic knowledge on the basis of appropriate do some questions, on the one hand, consolidate knowledge, on the other hand, expand the ideas and some common skills, improve the accuracy, and ensure that you will do resolutely not lose points OK! Come on!

It is necessary to do the questions, but you can't do them blindly. Generally, the exam only tests so many types of questions (you should be preparing for the exam) first carefully do a set of questions, summarize the types of questions and the distribution of knowledge points. Start with the key (i.e., high-value) question types, and be sure to understand them.

Remember, do the questions first and then read the book

The mind is calm and naturally cool. Nothing can be rushed. Have you mastered the questions you have done, will you do the same problem next time, what is the reason why you did it wrong, sloppy or you really can't, or you didn't think of it, why.

The foundation is critical. And then there's the learning method. Focus on efficiency.

Summarizing more is the most important thing, and the improvement of mathematics scores definitely requires a certain amount of questions, but not to do more questions, but to do fine questions, and to summarize after doing it. If you do the test question correctly, you should think about doing it again in a different way, and if you don't do it right, you should think about what is wrong?

The main thing is to study the question type and find a way to solve this kind of problem, rather than doing it rigidly. The most important thing is the method, which will be easier to learn. Mathematics does not rely on doing problems, but really mastering the method, ask the teacher in time if you don't understand something, ask your classmates, and understand the solution ideas of the problem.

∵|pf1|>pf2|, p is on the x positive semi-axis, and by the way, the focal points are: f1 (- 5,0), f2 ( 5,0).

1) p is not a right-angled vertex and |pf1|>|pf2|There is a |pf1|⊥|f1f2|, pass f2 as a perpendicular line to obtain the intersection point with the ellipse is p: (5, 4 3) or (5, -4 3).

From the Pythagorean theorem: |pf1| / |pf2|= 7/2 ；

2) p is a right-angled vertex, then there is |pf1|⊥|pf2|, you can set the p coordinate as: (3cos, 2sin), pf1|>|pf2|，∴cosα>0

Then: vector |pf1|：(5 - 3cosα，-2sinα)

Vector |pf2|：(5 - 3cosα，-2sinα)

pf1|⊥|pf2|, the product of the two vectors is 0

0 = 9(cosα)^2 - 5 + 4(sinα)^2 = 5(cosα)^2 - 1

Combine cos >0 to give cos = 1 5, sin = 2 5 or -2 5

By symmetry, take sin = 2 5.

As pa of2 in a, it is easy to obtain a(3 5 ,0), f2a = 2 5, and pa = 4 5

pf1f2∽△apf2 ，∴pf1|/|pf2|=pa/f2a = 2

Did you write the equation wrong? is x a + y b = 1 take a special point on the ellipse (0, b), focus (c, 0), (c, 0), then the root number (b + c) + root number (b + -c) = 2 3, the solution b + c = 3

a²=b²+c²∴a²=3

The elliptic alignment x=-a c=-3 gives c=1, b =a -c =2 The equation for the ellipse m: x 3 + y 2 = 1

Let the top angle be x and the bottom angle be y

So x+y=140, x+2y=180 (sum of triangle inner angles=180), xy can be solved.

The sum of the three inner angles of the triangle is 180°, and the sum of one of its top angles and one bottom angle is 140°, then the other base angle is 180°-140°=40°, and the top angle is 140°-40°=100° or 180°-2 40°=100°

If the number of top angles is x, then the number of ground angles is 140-x.

Column equation: x+140-x+140-x=180x=100

So: top corner: 100, bottom corner: 40

It should be rectangular: 6cm on the long side and 4cm on the short side

There can be 2 different types of cylinders:

1: Take the short side as the axis, that is, rotate around the central axis with 4cm The volume is: the base area multiplied by the height of the volume.

2: Take the long side as the axis, that is, rotate around the central axis with 6cm The volume is: the base area multiplied by the height of the volume.

I'm not a great god and can't solve this problem.

x-y=80①

1+20%)y-(1-25%)x=30 pairs of finishing:

y=200 generationsY=200 into

x=80+200=280

That is, the solution of the system of equations is x=280;y=200

The posterior is the preceding x= 280, and the x= 280 is substituted into the preceding tense.

280-y=80 y=200 so x= 280 y=200

1001=7*11*13

Because 1001 can be written as the sum of 10 numbers: 91, 91, 91, 91, 91, 91, 91, 91, 91, 182

So the maximum value of d is 91

Isn't it a different number of 10?

The minimum factor greater than 55 of 1+2+3+4+5+6+7+8+9+10=551001 is 77

So d=13

1001=7x11x13

So d max is 13x7=91

Because there can be 1001 = 91x(11) - as long as there is any number that adds up to 11 left, it is true.

For example: 1001 = 91 + 91 + 91 + 91 + 91 + 91 + 91 + 91 + 91 + 182

And 11 in () is not 7 or 13 -- because.

To be the maximum, 2And the number in () should be greater than 10, so that 1001 can be broken down into ten natural numbers and added.

The greatest common divisor of these 10 natural numbers will be d,:a1=d*a1,a2=d*a2....a10=d*a10

then there is d*(a1+a2+a3+..a10)=1001=7*11*13 Because a1, a2, and a3 are all natural numbers.

So a1 + a2 + a3 + .a10>=10, the maximum fixed d is 7*13=91, that is, a1, a2, a3....The greatest common divisor of a10 is d, and the maximum can only be 91

Let a1=d*a1, a2=d*a2,...a10=d*a10。Therefore 1001=d*(a1+a2+...+a10）。And 1001 = 7 * 11 * 13, and a1 + a2 + ...+a10) is greater than or equal to (0+1+....+9) = 45, so d = 1001 m (m>45), m should be as small as possible.

After testing, the minimum m is 77, and d is 13 at this time

First of all, 1001 can be split into 7*11*13. Next, let's analyze that since d is the greatest common factor, we can simply get the formula, nd=1001, and n is a natural number. Because d should be the largest, so n should be the smallest, and because n should be greater than the number of natural numbers 10, so when n takes 11, d can take the maximum value of 13*7=91.