There is not much time left for ninth grade graduates before they need to take the main state exam. This is a very important stage in life, since many students will go to study at technical schools and colleges, and to enter the desired budget place you need to do well on tests. I will solve the OGE grade 9 - simply an irreplaceable site. It will help you prepare for testing much faster than studying on your own in order to pass it with the highest grade of “5”.

How to prepare for exams?

In order to prepare for exams, schoolchildren use different methods. This applies to studying additional literature, classes with a professional tutor, as well as additional lessons with a school teacher.

Still the most effective method Undoubtedly, the use of specialized sites such as “I will solve the OGE” is considered. It helps prepare both children from the fifth and 9th grade.

Website I will solve the OGE

Why is this service so popular? It gives you the opportunity to feel the same as in the case of the exam itself. For preparation, tests from previous years are given, because according to statistics, most of the “new” tasks will be very similar to those from previous years.

An important advantage is that you do not need to resolve tickets comprehensively each time if this is not necessary. You can complete separate assignments on a specific topic, which will be very convenient if you need to prepare for specific knowledge.

How to find the necessary information on the site?

What does any visitor see as soon as he enters the portal? At the very top of the page is the site header, and under it, in convenient icons, are the names of those subjects that you can choose for the exam. First of all, there are the following:

• mathematics;
• physics;
• chemistry;
• Russian language;
• Informatics.

Disciplines

This list is incomplete, because to find the necessary subject for which you need to prepare, you just need to go to the website. You can immediately select the desired discipline and then the portal will display all the information on this subject.

Below the list of items are fifteen popular tickets, selected by moderators as indicative ones.

Test options

If a student only goes through them, and then sorts out his mistakes together with the teacher, this will increase his chances several times. successful solution OGE for 9th grade.

Option No. 6561231

New User Registration

Such a desire to solve the OGE for 9th grade is natural for any student. This requires good preparation. To use the entire service with already solved tasks in full, you must go through the registration process. This will give you the opportunity not only to take as many tests as you want, but also to keep your own statistics.

Statistics in your personal account

It will allow you to understand which tasks need additional work in order to significantly raise your level of knowledge to the required level. You can also open access to this data to a teacher or tutor so that he can determine which topics are best for the student to pay attention to and what to work on additionally.

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To register for the site I will solve the OGE grade 9, it is important to indicate certain user data, including the following:

• address Email;
• password;
• teacher or student.

The most important thing in this case is to indicate your email. Because useful information for the user will begin to arrive at the registered address. Additionally, it is worth noting the possibility that if a student forgets his password, then using e-mail it will be possible to recover this information. This means that a new temporary code will be sent to the address, which can then be replaced.

Catalog of popular tasks

Catalog of tasks

After the user has successfully registered on the website I will solve the OGE grade 9, namely, the students of this class will be fully prepared for the exams. In the list on the left you can find a button labeled “Task Catalog” and then click on it.

There, all the tasks are already divided by topic, and you can safely go to the place with which information you need to further work on. For example, select "Actions with ordinary fractions" By clicking on this link, the student will get acquainted with the list of tasks that he may have on the exam.

Useful information for experts

School of Experts

This site is visited not only by students, but also by teachers, who will subsequently check assignments. Because each form must be checked in the same way as hundreds of thousands of others without bias towards the student.

To get acquainted with the information in more detail, it is important to go to the “Expert” tab. There are specific guidelines to check each task. Also, for training, you can check specifically selected tasks, and then receive comments on the grading: how to do it correctly, and how to avoid making mistakes next time.

The unique website “I will solve the OGE” will help you prepare more effectively for the main state exam. Each student will know exactly what to expect during testing, and all examiners will be familiar with the requirements for marking papers.

I WILL SOLVE OGE - this is an application for preparing for the state exam. exam. Thanks to it, you can practice completing tasks right on your Android device.

Important information

This solution does not offer a theoretical part. To study the theory, you should use other programs or resort to school textbooks.

Exam preparation

SOLVE OGE gives you the opportunity to prepare for the upcoming test online or offline - the student’s choice. Available subjects include Russian language, mathematics, computer science, geography, history, physics and chemistry. You can prepare in several modes, including a simulated exam mode, in which the user is severely limited in time.

When solving the examples and problems offered by the program, you can enable viewing of the correct answers at the end of the entire test or check the answers after passing each task. Don't forget that to run a test without a network connection, you first need to download it to the Android device's memory.

Usage

The application is aimed at repeating previously studied material in preparation for passing the OGE. Forget about textbooks, which are hard to carry in your bag all the time! Forget that you can only prepare at home or in classroom, conveniently laying out notes and reference books on the desk. With the application I will SOLVE OGE you will be able to practice solving problems at any time and anywhere.

Key Features

• offers solutions to tasks for training before the main one state exam;
• does not contain a theoretical part;
• works without a network connection;
• can simulate a real exam (limit the user in time);
• has a simple and intuitive interface;
• works on all current versions of the Android operating system;
• distributed free of charge.

Secondary general education

Line UMK G. K. Muravin. Algebra and principles of mathematical analysis (10-11) (in-depth)

UMK Merzlyak line. Algebra and beginnings of analysis (10-11) (U)

Mathematics

Preparation for the Unified State Exam in mathematics ( profile level): tasks, solutions and explanations

We analyze tasks and solve examples with the teacher

Examination paper profile level lasts 3 hours 55 minutes (235 minutes).

Minimum threshold- 27 points.

The examination paper consists of two parts, which differ in content, complexity and number of tasks.

The defining feature of each part of the work is the form of the tasks:

• part 1 contains 8 tasks (tasks 1-8) with a short answer in the form of a whole number or a final decimal fraction;
• part 2 contains 4 tasks (tasks 9-12) with a short answer in the form of an integer or a final decimal fraction and 7 tasks (tasks 13–19) with a detailed answer (a complete record of the solution with justification for the actions taken).

Panova Svetlana Anatolevna, mathematic teacher highest category schools, work experience 20 years:

“In order to receive a school certificate, a graduate must pass two mandatory exams in the form of the Unified State Examination, one of which is mathematics. In accordance with the Concept of development of mathematics education in Russian Federation The Unified State Examination in mathematics is divided into two levels: basic and specialized. Today we will look at profile-level options.”

Task No. 1- tests the Unified State Exam participants’ ability to apply the skills acquired in the 5th to 9th grade course in elementary mathematics in practical activities. The participant must have computational skills, be able to work with rational numbers, be able to round decimals, be able to convert one unit of measurement to another.

Example 1. In the apartment where Peter lives, a cold water flow meter (meter) was installed. On May 1, the meter showed a consumption of 172 cubic meters. m of water, and on the first of June - 177 cubic meters. m. What amount should Peter pay for cold water in May, if the price is 1 cubic meter? m of cold water is 34 rubles 17 kopecks? Give your answer in rubles.

Solution:

1) Find the amount of water spent per month:

177 - 172 = 5 (cubic m)

2) Let’s find how much money they will pay for wasted water:

34.17 5 = 170.85 (rub)

Answer: 170,85.

Task No. 2- is one of the simplest exam tasks. The majority of graduates successfully cope with it, which indicates knowledge of the definition of the concept of function. Type of task No. 2 according to the requirements codifier is a task on the use of acquired knowledge and skills in practical activities and everyday life. Task No. 2 consists of describing, using functions, various real relationships between quantities and interpreting their graphs. Task No. 2 tests the ability to extract information presented in tables, diagrams, and graphs. Graduates need to be able to determine the value of a function from the value of the argument in various ways of specifying the function and describe the behavior and properties of the function based on its graph. You also need to be able to find the greatest or smallest value and build graphs of the studied functions. Errors made are random in reading the conditions of the problem, reading the diagram.

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Example 2. The figure shows the change in the exchange value of one share of a mining company in the first half of April 2017. On April 7, the businessman purchased 1,000 shares of this company. On April 10, he sold three-quarters of the shares he purchased, and on April 13, he sold all the remaining shares. How much did the businessman lose as a result of these operations?

Solution:

2) 1000 · 3/4 = 750 (shares) - constitute 3/4 of all shares purchased.

6) 247500 + 77500 = 325000 (rub) - the businessman received 1000 shares after selling.

7) 340,000 – 325,000 = 15,000 (rub) - the businessman lost as a result of all operations.

Answer: 15000.

Task No. 3- is a task basic level first part, tests the ability to perform actions with geometric shapes on the content of the course “Planimetry”. Task 3 tests the ability to calculate the area of ​​a figure on checkered paper, the ability to calculate degree measures angles, calculate perimeters, etc.

Example 3. Find the area of ​​a rectangle drawn on checkered paper with a cell size of 1 cm by 1 cm (see figure). Give your answer in square centimeters.

Solution: To calculate the area of ​​a given figure, you can use the Peak formula:

To calculate area given rectangle Let's use Peak's formula:

S= B +	G
	2

where B = 10, G = 6, therefore

S = 18 +	6
	2

Answer: 20.

Read also: Unified State Exam in Physics: solving problems about oscillations

Task No. 4- the objective of the course “Probability Theory and Statistics”. The ability to calculate the probability of an event in the simplest situation is tested.

Example 4. There are 5 red and 1 blue dots marked on the circle. Determine which polygons are larger: those with all the vertices red, or those with one of the vertices blue. In your answer, indicate how many there are more of some than others.

Solution: 1) Let's use the formula for the number of combinations of n elements by k:

whose vertices are all red.

3) One pentagon with all vertices red.

4) 10 + 5 + 1 = 16 polygons with all red vertices.

which have red tops or with one blue top.

which have red tops or with one blue top.

8) One hexagon with red vertices and one blue vertex.

9) 20 + 15 + 6 + 1 = 42 polygons with all red vertices or one blue vertex.

10) 42 – 16 = 26 polygons using the blue dot.

11) 26 – 16 = 10 polygons – how many more polygons in which one of the vertices is a blue dot are there than polygons in which all the vertices are only red.

Answer: 10.

Task No. 5- the basic level of the first part tests the ability to solve simple equations (irrational, exponential, trigonometric, logarithmic).

Example 5. Solve equation 2 3 + x= 0.4 5 3 + x .

Solution. Divide both sides of this equation by 5 3 + X≠ 0, we get

2 3 + x	= 0.4 or		2		3 + X	=	2	,
5 3 + X			5				5

whence it follows that 3 + x = 1, x = –2.

Answer: –2.

Task No. 6 in planimetry to find geometric quantities (lengths, angles, areas), modeling real situations in the language of geometry. Study of constructed models using geometric concepts and theorems. The source of difficulties is, as a rule, ignorance or incorrect application of the necessary theorems of planimetry.

Area of ​​a triangle ABC equals 129. DE– midline parallel to the side AB. Find the area of ​​the trapezoid ABED.

Solution. Triangle CDE similar to a triangle CAB at two angles, since the angle at the vertex C general, angle СDE equal to angle CAB as the corresponding angles at DE || AB secant A.C.. Because DE is the middle line of a triangle by condition, then by the property of the middle line | DE = (1/2)AB. This means that the similarity coefficient is 0.5. The areas of similar figures are related as the square of the similarity coefficient, therefore

Hence, S ABED = S Δ ABC – S Δ CDE = 129 – 32,25 = 96,75.

Task No. 7- checks the application of the derivative to the study of a function. Successful implementation requires meaningful, non-formal knowledge of the concept of derivative.

Example 7. To the graph of the function y = f(x) at the abscissa point x 0 a tangent is drawn that is perpendicular to the line passing through the points (4; 3) and (3; –1) of this graph. Find f′( x 0).

Solution. 1) Let’s use the equation of a line passing through two given points and find the equation of a line passing through points (4; 3) and (3; –1).

(y – y 1)(x 2 – x 1) = (x – x 1)(y 2 – y 1)

(y – 3)(3 – 4) = (x – 4)(–1 – 3)

(y – 3)(–1) = (x – 4)(–4)

–y + 3 = –4x+ 16| · (-1)

y – 3 = 4x – 16

y = 4x– 13, where k 1 = 4.

2) Find the slope of the tangent k 2, which is perpendicular to the line y = 4x– 13, where k 1 = 4, according to the formula:

3) The tangent angle is the derivative of the function at the point of tangency. Means, f′( x 0) = k 2 = –0,25.

Answer: –0,25.

Task No. 8- tests the exam participants’ knowledge of elementary stereometry, the ability to apply formulas for finding surface areas and volumes of figures, dihedral angles, compare the volumes of similar figures, be able to perform actions with geometric figures, coordinates and vectors, etc.

The volume of a cube circumscribed around a sphere is 216. Find the radius of the sphere.

Solution. 1) V cube = a 3 (where A– length of the edge of the cube), therefore

A 3 = 216

A = 3 √216

2) Since the sphere is inscribed in a cube, it means that the length of the diameter of the sphere is equal to the length of the edge of the cube, therefore d = a, d = 6, d = 2R, R = 6: 2 = 3.

Task No. 9- requires the graduate to have the skills of transformation and simplification algebraic expressions. Task No. 9 higher level Difficulty with a short answer. The tasks from the “Calculations and Transformations” section in the Unified State Exam are divided into several types:

transformation of numerical rational expressions;

converting algebraic expressions and fractions;

conversion of numeric/letter irrational expressions;

actions with degrees;

transformation logarithmic expressions;

1. converting numeric/letter trigonometric expressions.

Example 9. Calculate tanα if it is known that cos2α = 0.6 and

3π	< α < π.
4

Solution. 1) Let’s use the double argument formula: cos2α = 2 cos 2 α – 1 and find

tan 2 α =	1	– 1 =	1	– 1 =	10	– 1 =	5	– 1 = 1	1	– 1 =	1	= 0,25.
	cos 2 α		0,8		8		4		4		4

This means tan 2 α = ± 0.5.

3) By condition

3π	< α < π,
4

this means α is the angle of the second quarter and tgα< 0, поэтому tgα = –0,5.

Answer: –0,5.

#ADVERTISING_INSERT# Task No. 10- tests students’ ability to use acquired early knowledge and skills in practical activities and everyday life. We can say that these are problems in physics, and not in mathematics, but all the necessary formulas and quantities are given in the condition. The problems boil down to solving a linear or quadratic equation, or a linear or quadratic inequality. Therefore, it is necessary to be able to solve such equations and inequalities and determine the answer. The answer must be given as a whole number or a finite decimal fraction.

Two bodies of mass m= 2 kg each, moving at the same speed v= 10 m/s at an angle of 2α to each other. The energy (in joules) released during their absolutely inelastic collision is determined by the expression Q = mv 2 sin 2 α. At what smallest angle 2α (in degrees) must the bodies move so that at least 50 joules are released as a result of the collision?
Solution. To solve the problem, we need to solve the inequality Q ≥ 50, on the interval 2α ∈ (0°; 180°).

mv 2 sin 2 α ≥ 50

2 10 2 sin 2 α ≥ 50

200 sin 2 α ≥ 50

Since α ∈ (0°; 90°), we will only solve

Let us represent the solution to the inequality graphically:

Since by condition α ∈ (0°; 90°), it means 30° ≤ α< 90°. Получили, что наименьший угол α равен 30°, тогда наименьший угол 2α = 60°.

Task No. 11- is typical, but turns out to be difficult for students. The main source of difficulty is the construction of a mathematical model (drawing up an equation). Task No. 11 tests the ability to solve word problems.

Example 11. During spring break, 11th-grader Vasya had to solve 560 practice problems to prepare for the Unified State Exam. On March 18, on the last day of school, Vasya solved 5 problems. Then every day he solved the same number of problems more than the previous day. Determine how many problems Vasya solved on April 2, the last day of the holidays.

Solution: Let's denote a 1 = 5 – the number of problems that Vasya solved on March 18, d– daily number of tasks solved by Vasya, n= 16 – number of days from March 18 to April 2 inclusive, S 16 = 560 – total number of tasks, a 16 – the number of problems that Vasya solved on April 2. Knowing that every day Vasya solved the same number of problems more compared to the previous day, we can use formulas for finding the sum arithmetic progression:

560 = (5 + a 16) 8,

5 + a 16 = 560: 8,

5 + a 16 = 70,

a 16 = 70 – 5

a 16 = 65.

Answer: 65.

Task No. 12- they test students’ ability to perform operations with functions, and to be able to apply the derivative to the study of a function.

Find the maximum point of the function y= 10ln( x + 9) – 10x + 1.

Solution: 1) Find the domain of definition of the function: x + 9 > 0, x> –9, that is, x ∈ (–9; ∞).

2) Find the derivative of the function:

4) The found point belongs to the interval (–9; ∞). Let's determine the signs of the derivative of the function and depict the behavior of the function in the figure:

The desired maximum point x = –8.

Download for free the working program in mathematics for the line of teaching materials G.K. Muravina, K.S. Muravina, O.V. Muravina 10-11 Download free teaching aids on algebra

Task No. 13-increased level of complexity with a detailed answer, testing the ability to solve equations, the most successfully solved among tasks with a detailed answer of an increased level of complexity.

a) Solve the equation 2log 3 2 (2cos x) – 5log 3 (2cos x) + 2 = 0

b) Find all the roots of this equation that belong to the segment.

Solution: a) Let log 3 (2cos x) = t, then 2 t 2 – 5t + 2 = 0,

	log 3(2cos x) =	2	⇔		2cos x = 9	⇔		cos x =	4,5	⇔ because |cos x| ≤ 1,
	log 3(2cos x) =	1			2cos x = √3			cos x =	√3
		2							2

then cos x =	√3
	2

	x =	π	+ 2π k
		6
	x = –	π	+ 2π k, k ∈ Z
		6

b) Find the roots lying on the segment .

The figure shows that the roots of the given segment belong to

11π	And	13π	.
6		6

Answer: A)	π	+ 2π k; –	π	+ 2π k, k ∈ Z; b)	11π	;	13π	.
	6		6		6		6

Task No. 14-advanced level refers to tasks in the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes. The task contains two points. In the first point, the task must be proven, and in the second point, calculated.

The diameter of the circle of the base of the cylinder is 20, the generatrix of the cylinder is 28. The plane intersects its base along chords of length 12 and 16. The distance between the chords is 2√197.

a) Prove that the centers of the bases of the cylinder lie on one side of this plane.

b) Find the angle between this plane and the plane of the base of the cylinder.

Solution: a) A chord of length 12 is at a distance = 8 from the center of the base circle, and a chord of length 16, similarly, is at a distance of 6. Therefore, the distance between their projections onto the plane is parallel to the bases cylinders is either 8 + 6 = 14 or 8 − 6 = 2.

Then the distance between the chords is either

= = √980 = = 2√245

= = √788 = = 2√197.

According to the condition, the second case was realized, in which the projections of the chords lie on one side of the cylinder axis. This means that the axis does not intersect this plane within the cylinder, that is, the bases lie on one side of it. What needed to be proven.

b) Let us denote the centers of the bases as O 1 and O 2. Let us draw from the center of the base with a chord of length 12 a perpendicular bisector to this chord (it has length 8, as already noted) and from the center of the other base to the other chord. They lie in the same plane β, perpendicular to these chords. Let's call the midpoint of the smaller chord B, the larger chord A and the projection of A onto the second base - H (H ∈ β). Then AB,AH ∈ β and therefore AB,AH are perpendicular to the chord, that is, the straight line of intersection of the base with the given plane.

This means that the required angle is equal to

∠ABH = arctan	A.H.	= arctan	28	= arctg14.
	B.H.		8 – 6

Task No. 15- increased level of complexity with a detailed answer, tests the ability to solve inequalities, which is most successfully solved among tasks with a detailed answer of an increased level of complexity.

Example 15. Solve inequality | x 2 – 3x| log 2 ( x + 1) ≤ 3x – x 2 .

Solution: The domain of definition of this inequality is the interval (–1; +∞). Consider three cases separately:

1) Let x 2 – 3x= 0, i.e. X= 0 or X= 3. In this case, this inequality becomes true, therefore, these values ​​are included in the solution.

2) Let now x 2 – 3x> 0, i.e. x∈ (–1; 0) ∪ (3; +∞). Moreover, this inequality can be rewritten as ( x 2 – 3x) log 2 ( x + 1) ≤ 3x – x 2 and divide by a positive expression x 2 – 3x. We get log 2 ( x + 1) ≤ –1, x + 1 ≤ 2 –1 , x≤ 0.5 –1 or x≤ –0.5. Taking into account the domain of definition, we have x ∈ (–1; –0,5].

3) Finally, consider x 2 – 3x < 0, при этом x∈ (0; 3). In this case, the original inequality will be rewritten in the form (3 x – x 2) log 2 ( x + 1) ≤ 3x – x 2. After dividing by positive 3 x – x 2 , we get log 2 ( x + 1) ≤ 1, x + 1 ≤ 2, x≤ 1. Taking into account the region, we have x ∈ (0; 1].

Combining the solutions obtained, we obtain x ∈ (–1; –0.5] ∪ ∪ {3}.

Answer: (–1; –0.5] ∪ ∪ {3}.

Task No. 16- advanced level refers to tasks in the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes, coordinates and vectors. The task contains two points. In the first point, the task must be proven, and in the second point, calculated.

In an isosceles triangle ABC with an angle of 120°, the bisector BD is drawn at vertex A. Rectangle DEFH is inscribed in triangle ABC so that side FH lies on segment BC, and vertex E lies on segment AB. a) Prove that FH = 2DH. b) Find the area of ​​rectangle DEFH if AB = 4.

Solution: A)

1) ΔBEF – rectangular, EF⊥BC, ∠B = (180° – 120°): 2 = 30°, then EF = BE by the property of the leg lying opposite the angle of 30°.

2) Let EF = DH = x, then BE = 2 x, BF = x√3 according to the Pythagorean theorem.

3) Since ΔABC is isosceles, it means ∠B = ∠C = 30˚.

BD is the bisector of ∠B, which means ∠ABD = ∠DBC = 15˚.

4) Consider ΔDBH – rectangular, because DH⊥BC.

2x	=	4 – 2x
2x(√3 + 1)		4

1	=	2 – x
√3 + 1		2

√3 – 1 = 2 – x

x = 3 – √3

EF = 3 – √3

2) S DEFH = ED EF = (3 – √3 ) 2(3 – √3 )

S DEFH = 24 – 12√3.

Answer: 24 – 12√3.

Task No. 17- a task with a detailed answer, this task tests the application of knowledge and skills in practical activities and everyday life, the ability to build and explore mathematical models. This task - word problem with economic content.

Example 17. A deposit of 20 million rubles is planned to be opened for four years. At the end of each year, the bank increases the deposit by 10% compared to its size at the beginning of the year. In addition, at the beginning of the third and fourth years, the investor annually replenishes the deposit by X million rubles, where X - whole number. Find highest value X, in which the bank will accrue less than 17 million rubles to the deposit over four years.

Solution: At the end of the first year, the contribution will be 20 + 20 · 0.1 = 22 million rubles, and at the end of the second - 22 + 22 · 0.1 = 24.2 million rubles. At the beginning of the third year, the contribution (in million rubles) will be (24.2 + X), and at the end - (24.2 + X) + (24,2 + X)· 0.1 = (26.62 + 1.1 X). At the beginning of the fourth year the contribution will be (26.62 + 2.1 X), and at the end - (26.62 + 2.1 X) + (26,62 + 2,1X) · 0.1 = (29.282 + 2.31 X). By condition, you need to find the largest integer x for which the inequality holds

(29,282 + 2,31x) – 20 – 2x < 17

29,282 + 2,31x – 20 – 2x < 17

0,31x < 17 + 20 – 29,282

0,31x < 7,718

x <	7718
	310

x <	3859
	155

x < 24	139
	155

The largest integer solution to this inequality is the number 24.

Answer: 24.

Task No. 18- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection into universities with increased requirements for the mathematical preparation of applicants. Exercise high level complexity - this task is not about using one solution method, but about a combination of different methods. To successfully complete task 18, in addition to solid mathematical knowledge, you also need a high level of mathematical culture.

At what a system of inequalities

	x 2 + y 2 ≤ 2ay – a 2 + 1
	y + a ≤ |x| – a

has exactly two solutions?

Solution: This system can be rewritten in the form

	x 2 + (y– a) 2 ≤ 1
	y ≤ |x| – a

If we draw on the plane the set of solutions to the first inequality, we get the interior of a circle (with a boundary) of radius 1 with center at point (0, A). The set of solutions to the second inequality is the part of the plane lying under the graph of the function y = | x| – a, and the latter is the graph of the function
y = | x| , shifted down by A. The solution to this system is the intersection of the sets of solutions to each of the inequalities.

Consequently, this system will have two solutions only in the case shown in Fig. 1.

The points of contact of the circle with the lines will be the two solutions of the system. Each of the straight lines is inclined to the axes at an angle of 45°. So it's a triangle PQR– rectangular isosceles. Dot Q has coordinates (0, A), and the point R– coordinates (0, – A). In addition, the segments PR And PQ equal to the radius of the circle equal to 1. This means

Qr= 2a = √2, a =	√2	.
	2

Answer: a =	√2	.
	2

Task No. 19- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection into universities with increased requirements for the mathematical preparation of applicants. A task of a high level of complexity is a task not on the use of one solution method, but on a combination of various methods. To successfully complete task 19, you must be able to search for a solution, choosing different approaches from among the known ones, and modifying the studied methods.

Let Sn sum P terms of an arithmetic progression ( a p). It is known that S n + 1 = 2n 2 – 21n – 23.

a) Provide the formula P th term of this progression.

b) Find the smallest absolute sum S n.

c) Find the smallest P, at which S n will be the square of an integer.

Solution: a) It is obvious that a n = S n – S n- 1 . Using this formula, we get:

S n = S (n – 1) + 1 = 2(n – 1) 2 – 21(n – 1) – 23 = 2n 2 – 25n,

S n – 1 = S (n – 2) + 1 = 2(n – 1) 2 – 21(n – 2) – 23 = 2n 2 – 25n+ 27

Means, a n = 2n 2 – 25n – (2n 2 – 29n + 27) = 4n – 27.

B) Since S n = 2n 2 – 25n, then consider the function S(x) = | 2x 2 – 25x|. Its graph can be seen in the figure.

Obviously, the smallest value is achieved at the integer points located closest to the zeros of the function. Obviously these are points X= 1, X= 12 and X= 13. Since, S(1) = |S 1 | = |2 – 25| = 23, S(12) = |S 12 | = |2 · 144 – 25 · 12| = 12, S(13) = |S 13 | = |2 · 169 – 25 · 13| = 13, then the smallest value is 12.

c) From the previous paragraph it follows that Sn positive, starting from n= 13. Since S n = 2n 2 – 25n = n(2n– 25), then the obvious case, when this expression is a perfect square, is realized when n = 2n– 25, that is, at P= 25.

It remains to check the values ​​from 13 to 25:

S 13 = 13 1, S 14 = 14 3, S 15 = 15 5, S 16 = 16 7, S 17 = 17 9, S 18 = 18 11, S 19 = 19 13, S 20 = 20 13, S 21 = 21 17, S 22 = 22 19, S 23 = 23 21, S 24 = 24 23.

It turns out that for smaller values P a complete square is not achieved.

Answer: A) a n = 4n– 27; b) 12; c) 25.

________________

*Since May 2017, the united publishing group "DROFA-VENTANA" has been part of the Russian Textbook corporation. The corporation also includes the Astrel publishing house and the LECTA digital educational platform. Alexander Brychkin, a graduate of the Financial Academy under the Government of the Russian Federation, Candidate of Economic Sciences, head of innovative projects of the DROFA publishing house in the field of digital education (electronic forms of textbooks, Russian Electronic School, digital educational platform LECTA) was appointed General Director. Before joining the DROFA publishing house, he held the position of vice president for strategic development and investments of the publishing holding EKSMO-AST. Today, the publishing corporation "Russian Textbook" has the largest portfolio of textbooks included in the Federal List - 485 titles (approximately 40%, excluding textbooks for special schools). The corporation's publishing houses own the most popular Russian schools sets of textbooks on physics, drawing, biology, chemistry, technology, geography, astronomy - areas of knowledge that are needed for the development of the country's production potential. The corporation's portfolio includes textbooks and teaching aids For primary school, awarded the Presidential Prize in the field of education. These are textbooks and manuals in subject areas that are necessary for the development of the scientific, technical and production potential of Russia.

I’ll solve the GIA 2017 9th grade, like this search query is of interest to those who have started preparing for the State Examination online based on our textbooks, workbooks and State Examinations. Our website is an archive and collection of all GIA options for last years and contains all the tasks that you will meet at the final certification. Decide OGE options you will have to devote 2 to 3 hours every day to this activity throughout the year. This is necessary to do in order to consolidate the material and accurately monitor progress at the stage of preparation for the State Examination. When we say “I’ll solve the GIA in mathematics,” we usually mean the options that were developed and implemented by Dmitry Gushchin. It was he who created the website where he collected all the OGE tests in mathematics with answers, where you can conveniently solve the OGE options and test your skills in solving equations and problems.

I will pass the State Examination Test online, express preparation for the final certification

I will solve the GIA 2017 online - this is a proven collection of exercises where you can dwell on the topic in more detail, help a friend and repeat the material covered. The importance of the Russian language in human life is difficult to overestimate. We all communicate, pass information to each other, read newspapers and tabloids, where all news articles are written in Russian. This is why we use educational process tests and assignments to prepare for the State Academic Examination 2017 in the Russian language. For a deeper perception of information, we recommend solvers and workbooks, which can also be downloaded on our website. I will decide GIA Gushchin 9th grade is effective way test yourself and the quality of teaching with secondary special education educational institutions Russian Federation.

State Examination in the Russian language, tests and assignments to prepare for the State Examination 2017

Our Russian language teacher Alevtina Ivanovna Sinichko forced us to solve the GIA options online in 2017, when we were young and playful and ran through the cool puddles in the courtyard of our school No. 346 in Moscow. She was right, it was necessary to encourage students to study more, solve theorems, prove equations with three unknowns and be proud that you studied at a Moscow school. In those young years, we poorly understood the importance of the GIA in mathematics and hoped that we would get away with all the pranks, but time passed and the day of judgment came when the savior came and said “I will solve the GIA Gushchin” and all the students fell prostrate before him.

Now that decades have passed, we remember with a smile our state final certification and think about how we were afraid, how we hoped for a high GPA how we dreamed of entering Lomonosov Moscow State University. Now our outlook on life has changed and we see from the height of the years that we have lived that it was enough to say “I will solve the GIA 2017 9th grade in Gushchin” and all doors would be wide open for such a brave applicant. You say “Heresy!”, I will answer “Yes!”. However, let me remind you that you should study your answers, otherwise you will not pass the exam.

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The educational portal for preparing for exams according to Dmitry Gushchin’s system was originally created as a means of self-control and a way for students to self-prepare for the state final certification. However, the creators of this most useful resource built into it mechanisms for interaction with teachers.

In these notes, I share my experience of using resource materials offline - without using computers connected to the Internet. For convenience, the material is presented in the form of a text document and in PDF format.

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I WILL TAKE THE GIA: I WILL SOLVE THE OGE AND THE USE educational portal Dmitry Gushchin. Experience of use

The tools offered to the user of the portal I will SOLVE the OGE and the Unified State Exam are very wide, and allow you to pass testing both on individual topics and on a complex task. The function of randomly selecting tasks from the catalog is extremely useful. It is implemented with buttonsHomework template And Sample test work .

When the so-called Homework then the advantages of a randomized choice of task are offset by the fact that each task contains its number. Students open the catalog on another tab and, having found the answer there using this identifier, enter it into the text of the work.

When assigning a test, a number is not entered, but in some types of problems you can very quickly find the correct answer from the text. Thus, the resource can now only be used as a means of preparing students for testing in the classroom.

To obtain objective results of assessing the quality of training, the following application options are possible: testing on the Gushchin website PASS GIA: I WILL SOLVE the OGE and the Unified State Exam, testing using other systems or paper media. Experience shows that the only way to conduct objective testing on the described resource is to disable access to the Internet after students receive the assignment, and turn it on to send the results. If such technical feasibility no, then tasks have to be copied to another testing system. I use MyTestX from A. Bashlakova (http://mytest.klyaksa.net ).

I’ll move on to the algorithm for using the resource I’ll pass the State Exam: I’ll solve the OGE and the Unified State Exam to create a multi-choice test work. The OGE is a standardized exam, that is, each type of task corresponds to a specific topic, and all you need to do is learn the theory and solve, solve, solve tasks sequentially on each topic, gradually improving your hand until it becomes automatic. Any unexpected or unrelated school curriculum There can be no assignments on the exam. Experience shows that good method preparation is the solution large quantity tasks of the same type, this is exactly the test we will create.

Go to the “Teacher” tab

Let's set the test parameters , available from this tabprint version.