Preparation for the Unified State Exam in mathematics (profile level): assignments, solutions and explanations. Grammatical means of communication

Task 2 of the Unified State Exam in Society: how to solve

The difficulty of this task 2 of the Unified State Exam in social studies is that it requires you to find a generalizing word for a specified number of terms. A generalizing word is a generic term or concept that includes in its meaning the meanings of other concepts and terms. As in other Unified State Examination tasks on society, the topics of the tasks can be very different: social sphere, political, spiritual, etc.

Here, for example, is a task from a real Unified State Exam test in society:

It immediately becomes clear to intelligent boys and girls that the proposed words relate to the topic “Spiritual Sphere of Society”, namely to the topic of religion. If you find it difficult to answer right away, I recommend reading my previous post "" . Having read the terms for the most knowledgeable, it immediately becomes clear that there are only two options left for the answer: cult and religion. What will be more generalizing? A cult is the worship of something.

You can experiment by placing a broom in the corner of your room. And pray to him every day, talk to him... In a month this will be the most valuable item for you :). Create a cult of the broom. What is religion? This is a specific form of worldview, awareness of the world. It is clear that the concept of “religion” includes the concept of “cult,” since a worldview may include the worship of various deities. For example, paganism among the Eastern Slavs: some had the cult of Perun (the god of thunder and lightning), others had the cult of the god of swamps, etc.

Or, for example, Orthodox Christianity: there is the cult of Jesus Christ, there is the cult of the Holy Spirit, there is the cult of the Most Holy Theotokos... Got it?

OK. So the correct answer is: religion

Recommendation 2. You need to have a good knowledge of terms and concepts from various topics in social studies. Understand which terms are related to which ones, and which ones follow from them. For this purpose in my paid video course "Social studies: Unified State Exam 100 points " I have given the structure of terms for all topics of Social Science. I also highly recommend your article about.

Let's look at another task 2 of the Unified State Exam in social studies:

We immediately understand that task 2 of the Unified State Exam examines the topic Social Sphere. If you have forgotten the topic, download my free video course. If you don't do this, you will most likely make a mistake. Some people's logic is so crooked that it's simply brutal! Meanwhile, the correct answer: “agent of socialization” is a group or association that participates in an individual’s mastery of the rules and norms of society, as well as social roles. If you are not familiar with these terms, I again highly recommend downloading my free video course.

Recommendation 3. Be extremely careful! Solve tasks 2 of the Unified State Exam in social studies again and again to do this qualitatively on the machine. Here is an example of a similar task that is more difficult:

The theme “Science” from the spiritual sphere of society. By the way, I had a detailed article on this topic. People who are not very attentive will immediately make a mistake by indicating in the answer: classification basis, or theoretical validity. Between the correct answer: scientific knowledge , which includes different classifications and theoretical validity!

In the following posts we will definitely look at other difficult tasks on society, so !

I have attached a couple of tasks for Unified State Examination 2 in society for you to decide:

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): assignments, solutions and explanations

The profile level examination 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, mathematics teacher of the highest category of school, 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 for the Development of Mathematical Education in the 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, and 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 largest or smallest value from a function graph and build graphs of the studied functions. Errors made are random in reading the conditions of the problem, reading the diagram.

#ADVERTISING_INSERT#

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 basic level task of the first part, tests the ability to perform actions with geometric figures according to the content of the Planimetry course. Task 3 tests the ability to calculate the area of ​​a figure on checkered paper, the ability to calculate degree measures of 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 the area of ​​a given rectangle, we use Peak’s formula:

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

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 to transform and simplify algebraic expressions. Task No. 9 of an increased level of 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;

converting logarithmic expressions;

1. converting numeric/letter trigonometric expressions.

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

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

This means tan 2 α = ± 0.5.

3) By condition

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.

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.

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,

b) Find the roots lying on the segment .

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

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 a plane parallel to the bases of the 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

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.

√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.

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 the greatest 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

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

Answer: 24.

At what a system of inequalities

has exactly two solutions?

Solution: This system can be rewritten in the form

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

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 sets of textbooks in Russian schools in physics, drawing, biology, chemistry, technology, geography, astronomy - areas of knowledge that are needed for the development of the country's productive potential. The corporation's portfolio includes textbooks and teaching aids for primary schools, which were awarded the Presidential Award 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.

Lexical means of communication:

1. Lexical repetition- repetition of the same word. Around the city, forests spread across the low hills, mighty and untouched. In the forests there were large meadows and remote lakes with huge old pine trees along the banks.
2. Cognates. Of course, such a master knew his worth, felt the difference between himself and a less talented person, but he also knew perfectly well another difference - the difference between himself and a more talented person. Respect for the more capable and experienced is the first sign of talent.
3. Synonyms. We saw a moose in the forest. Sokhaty walked along the edge of the forest and was not afraid of anyone.
4. Antonyms. Nature has many friends. She has significantly fewer enemies.
5. Descriptive phrases. They built a highway. A noisy, fast-moving river of life connected the region with the capital.

Grammatical means of communication:

1. Personal pronouns. 1) And now I’m listening to the voice of an ancient stream. He coos like a wild dove. 2) The call for forest protection should be addressed primarily to young people. She should live and manage this land, she should decorate it. 3) He unexpectedly returned to his native village. His arrival delighted and frightened his mother.
2. Demonstrative pronouns(such, that, this) 1) A dark sky with bright, needle-like stars floated over the village. Such stars appear only in autumn. 2) The corncrakes screamed with distant, sweet twitching sounds. These corncrakes and sunsets are unforgettable; they were preserved forever by pure vision. – in the second text the means of communication are lexical repetition and the demonstrative pronoun “these”.
3. Pronominal adverbs(there, so, then, etc.) He [Nikolai Rostov] knew that this story contributed to the glorification of our weapons, and therefore it was necessary to pretend that you did not doubt it. That's what he did.
4. Unions(mostly composing) It was May 1945. Spring thundered. The people and the land rejoiced. Moscow saluted the heroes. And joy flew into the sky like lights. With the same chatter and laughter, the officers hastily began to get ready; again they put the samovar on dirty water. But Rostov, without waiting for tea, went to the squadron.”
5. Particles.
6. Introductory words and constructions(in one word, so, firstly, etc.) The young people spoke about everything Russian with contempt or indifference and, jokingly, predicted for Russia the fate of the Confederation of the Rhine. In short, the society was quite disgusting.
7. Unity of tense forms of verbs- the use of identical forms of grammatical tense, which indicate simultaneity or sequence of situations. Imitation of the French tone of the times of Louis XV was in vogue. Love for the fatherland seemed pedantry. The wise men of that time praised Napoleon with fanatical servility and joked about our failures. – all verbs are used in the past tense.
8. Incomplete sentences and ellipsis, referring to the previous elements of the text: Gorkin cuts the bread, distributes the slices. He puts it on me too: it’s huge, you’ll cover your whole face.
9. Syntactic parallelism– identical construction of several adjacent sentences. To be able to speak is an art. Listening is a culture.

Meaningful relationships expressed by coordinating conjunctions:

1. Connecting: and, yes (=and), and...and..., not only... but also, like... so and, also, too
2. Dividers: or, or, then...that, not that...not that, or...or, either...or
3. Nasty: a, but, yes (=but), however, but
4. Gradational: not only, but also, not so much... as, not really... but
5. Explanatory: that is, namely
6. Connecting: also, also, yes and, and moreover, and
7. too, yes and, that is, namely.

Meaningful relations expressed by subordinating conjunctions:

• Temporary: when, while, barely, only, while, just, barely, barely
• Causal: since, because, because, in view of the fact that, due to the fact that, due to the fact that, for (obsolete), due to the fact that
• Conditional: if (if only, if, if - obsolete), if, once, as soon
• Target: so that, in order to, in order to (obsolete), for the purpose of, in order to, then in order to
• Consequences: So
• Concessive: although, despite the fact that
• Comparative: as, as if, as if, exactly, than, as if, likewise, rather than (obsolete)
• Explanatory: what, how, to
• Conjunctions are not used at the beginning of a sentence: so, than, rather than, as well as explanatory conjunctions: what, how, so that.

In task No. 2 of the Unified State Examination in mathematics, it is necessary to demonstrate knowledge of working with power expressions.

Theory for task No. 2

The rules for handling degrees can be presented as follows:

In addition, you should remember about operations with fractions:

Now you can move on to analyzing typical options! 🙂

Analysis of typical options for tasks No. 2 of the Unified State Exam in basic level mathematics

First version of the task

Find the meaning of the expression

Execution algorithm:

1. Express a number with a negative exponent as a proper fraction.
2. Perform the first multiplication.
3. Represent powers of numbers as prime numbers, replacing powers by multiplication.
4. Perform multiplication.
5. Perform addition.

Solution:

That is: 10 -1 = 1/10 1 = 1/10

Let's perform the first multiplication, that is, multiplying a whole number by a proper fraction. To do this, multiply the numerator of the fraction by a whole number, and leave the denominator unchanged.

9 1/10 = (9 1)/10 = 9/10

The first power of a number is always the number itself.

The second power of a number is a number multiplied by itself.

10 2 = 10 10 = 100

Answer: 560.9

Second version of the task

Find the meaning of the expression

Execution algorithm:

1. Represent the first power of a number as an integer.
2. Represent negative powers of numbers as proper fractions.
3. Perform multiplication of integers.
4. Multiply whole numbers by proper fractions.
5. Perform addition.

Solution:

The first power of a number is always the number itself. (10 1 = 10)

To represent a negative power of a number as an ordinary fraction, you need to divide 1 by this number, but to a positive power.

10 -1 = 1/10 1 = 1/10

10 -2 = 1/10 2 = 1/(10 10) = 1/100

Let's multiply integers.

3 10 1 = 3 10 = 30

Let's multiply whole numbers by proper fractions.

4 10 -2 = 4 1/100 = (4 1)/100 = 4/100

2 10 -1 = 2 1/10 = (2 1)/10 = 2/10

Let us calculate the value of the expression, taking into account that

Answer: 30.24

Third version of the task

Find the meaning of the expression

Execution algorithm:

1. Represent powers of numbers in the form of multiplication and calculate the value of powers of numbers.
2. Perform multiplication.
3. Perform addition.

Solution:

Let's represent powers of numbers in the form of multiplication. In order to represent the power of a number in the form of multiplication, you need to multiply this number by itself as many times as it is contained in the exponent.

2 4 = 2 2 2 2 = 16

2 3 = 2 2 2 = 8

Let's do the multiplication:

4 2 4 = 4 16 = 64

3 2 3 = 3 8 = 24

Let's calculate the value of the expression:

Fourth version of the task

Find the meaning of the expression

Execution algorithm:

1. Perform the action in parentheses.
2. Perform multiplication.

Solution:

Let us represent the power of a number in such a way that we can take the common factor out of the bracket.

3 4 3 + 2 4 4 = 4 3 (3 + 2 4)

Let's perform the action in parentheses.

(3 + 2 4) = (3 + 8) = 11

4 3 = 4 4 4 = 64

Let us calculate the value of the expression, taking into account that

Fifth version of the task

Find the meaning of the expression

Execution algorithm:

1. Let us represent the power of a number in such a way that we can take the common factor out of the bracket.
2. Place the common factor out of brackets.
3. Perform the action in parentheses.
4. Represent the power of a number as a multiplication and calculate the value of the power of the number.
5. Perform multiplication.

Solution:

Let us represent the power of a number in such a way that we can take the common factor out of the bracket.

Let's take the common factor out of brackets

2 5 3 + 3 5 2 = 5 2 (2 5 + 3)

Let's perform the action in parentheses.

(2 5 + 3) = (10 + 3) = 13

Let's represent the power of a number in the form of multiplication. In order to represent the power of a number in the form of multiplication, you need to multiply this number by itself as many times as it is contained in the exponent.

5 2 = 5 5 = 25

Let us calculate the value of the expression, taking into account that

We perform multiplication in a column, we have:

Option for the second task from the Unified State Exam 2017 (1)

Find the meaning of the expression:

Solution:

In this task, it is more convenient to bring the values ​​to a more familiar form, namely, write the numbers in the numerator and denominator in standard form:

After this, you can divide 24 by 6, the result is 4.

Ten to the fourth power when divided by ten to the third power gives ten to the first, or simply ten, so we get:

Option for the second task from the Unified State Exam 2017 (2)

Find the meaning of the expression:

Solution:

In this case, we should note that the number 6 in the denominator is factored into factors 2 and 3 to the power of 5:

After this, you can perform reductions of degrees for two: 6-5 = 1, for three: 8-5 = 3.

Now we cube 3 and multiply by 2, getting 54.

Option for the second task of 2019 (1)

Execution algorithm

1. Apply to the numerator of holy powers (a x) y = a xy. We get 3 –6.
2. Apply to fractions of holy powers a x /a y =a x–y.
3. Raise 3 to the resulting power.

Solution:

(3 –3) 2 /3 –8 = 3 –6 /3 –8 = 3 –6–(–8)) = 3 –6+8 = 3 2 = 9

Option for the second task 2019 (2)

Execution algorithm

1. We use for the degree in the numerator (14 9) (ab) x =a x b x. Let us decompose 14 into the product of 2 and 7. We obtain the product of powers with bases 2 and 7.
2. Let's transform the expression into 2 fractions, each of which will contain powers with the same bases.
3. Apply to fractions of holy powers a x /a y =a x–y.
4. We find the resulting product.

Solution:

14 9 / 2 7 7 8 = (2 7) 9 / 2 7 7 8 = 2 9 7 9 / 2 7 7 8 = 2 9–7 7 9–8 = 2 2 7 1 = 4 ·7 = 28

Option for the second task 2019 (3)

Execution algorithm

1. We take the common factor 5 2 =25 out of brackets.
2. We multiply the numbers 2 and 5 in brackets. We get 10.
3. We add 10 and 3 in brackets. We get 13.
4. We multiply the common factor 25 and 13.

Solution:

2 5 3 +3 5 2 = 5 2 (2 5+3) = 25 (10+3) = 25 13 = 325

Option for the second task 2019 (4)

Execution algorithm

1. Square it (–1). We get 1, since it is raised to an even power.
2. Raise (–1) to the 5th power. We get –1, because raising to an odd power occurs.
3. We perform multiplication operations.
4. We get the difference of two numbers. We find her.

Solution:

6·(–1) 2 +4·(–1) 5 = 6·1+4·(–1) = 6+(–4) = 6–4 = 2

Option for the second task 2019 (5)

Execution algorithm

1. Let's convert the factors 10 3 and 10 2 into integers.
2. We find the products by moving the decimal point to the right by the appropriate number of decimal places.
3. Find the resulting amount.