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Line UMK G.K. Muravina. Algebra and beginnings mathematical analysis(10-11) (deep)

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Mathematics

Preparation for the exam in mathematics ( profile level): tasks, solutions and explanations

We analyze tasks and solve examples with the teacher

The profile-level examination paper 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 tasks:

• part 1 contains 8 tasks (tasks 1-8) with a short answer in the form of an integer 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 (full record of the decision with the rationale for the actions performed).

Panova Svetlana Anatolievna, teacher of mathematics of the highest category of the school, work experience of 20 years:

“In order to receive a school certificate, a graduate must pass two compulsory exam in the form of the exam, one of which is mathematics. In accordance with the Concept for the Development of Mathematical Education in Russian Federation The USE in mathematics is divided into two levels: basic and specialized. Today we will consider options for the profile level.

Task number 1- checks with USE participants the ability to apply the skills acquired in the course of 5-9 classes in elementary mathematics in practical activities. The participant must have computer skills, be able to work with rational numbers, be able to round decimal fractions, be able to convert one unit of measure to another.

Example 1 In the apartment where Petr lives, a cold water meter (meter) was installed. On the first of May, the meter showed an 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 for May, if the price of 1 cu. 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 (cu m)

2) Find how much money will be paid for the spent water:

34.17 5 = 170.85 (rub)

Answer: 170,85.

Task number 2- is one of the simplest tasks of the exam. The majority of graduates successfully cope with it, which indicates the possession of the definition of the concept of function. Task type No. 2 according to the requirements codifier is a task for using 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 number 2 tests the ability to extract information presented in tables, diagrams, graphs. Graduates need to be able to determine the value of a function by the value of the argument with various ways of specifying the function and describe the behavior and properties of the function according to its graph. It is also necessary to be able to find the maximum or smallest value and build graphs of the studied functions. The mistakes made are of a random nature 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 purchased shares, and on April 13 he sold all the remaining ones. How much did the businessman lose as a result of these operations?

Solution:

2) 1000 3/4 = 750 (shares) - make up 3/4 of all purchased shares.

6) 247500 + 77500 = 325000 (rubles) - the businessman received after the sale of 1000 shares.

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

Answer: 15000.

Task number 3- is a task basic level the 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 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 ​​this figure, you can use the Peak formula:

To calculate the area of ​​this rectangle, we use the Peak formula:

S= B +	G
	2

where V = 10, G = 6, therefore

S = 18 +	6
	2

Answer: 20.

See also: Unified State Examination in Physics: solving vibration problems

Task number 4- the task 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 on the circle. Determine which polygons are larger: those with all red vertices, or those with one of the blue vertices. In your answer, indicate how many more of one than the other.

Solution: 1) We use the formula for the number of combinations from n elements by k:

all of whose vertices are red.

3) One pentagon with all red vertices.

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

whose vertices are red or with one blue vertex.

whose vertices are red or with one blue vertex.

8) One hexagon whose vertices are red with one blue vertex.

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

10) 42 - 16 = 26 polygons that use the blue dot.

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

Answer: 10.

Task number 5- the basic level of the first part tests the ability to solve the simplest 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 number 6 in planimetry for finding geometric quantities (lengths, angles, areas), modeling real situations in the language of geometry. The study of the 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- median line parallel to side AB. Find the area of ​​the trapezoid ABED.

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

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

Task number 7- checks the application of the derivative to the study of the function. For successful implementation, a meaningful, non-formal possession of the concept of a derivative is necessary.

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

Solution. 1) We use the equation of a straight line passing through two given points and find the equation of a straight line passing through the 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| · (-one)

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 slope of the tangent is the derivative of the function at the point of contact. Means, f′( x 0) = k 2 = –0,25.

Answer: –0,25.

Task number 8- checks the knowledge of elementary stereometry among the participants of the exam, 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 is the length of the edge of the cube), so

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 number 9- requires the graduate to transform and simplify algebraic expressions. Task number 9 advanced level Difficulty with short answers. Tasks from the section "Calculations and transformations" in the USE are divided into several types:

transformations of numerical rational expressions;

transformations of algebraic expressions and fractions;

transformations of numerical/letter irrational expressions;

actions with degrees;

transformation of logarithmic expressions;

1. conversion of numeric/letter trigonometric expressions.

Example 9 Calculate tgα 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

Hence, tan 2 α = ± 0.5.

3) By condition

3π	< α < π,
4

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

Answer: –0,5.

#ADVERTISING_INSERT# Task number 10- checks the ability of students to use the 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 are reduced 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 in the form of a whole number or a final 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 sin2α ≥ 50

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

We represent the solution of the inequality graphically:

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

Task number 11- is typical, but it turns out to be difficult for students. The main source of difficulties is the construction of a mathematical model (drawing up an equation). Task number 11 tests the ability to solve word problems.

Example 11. On the spring break 11-grader Vasya had to solve 560 training problems to prepare for the 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 on the last day of vacation.

Solution: Denote a 1 = 5 - the number of tasks that Vasya solved on March 18, d– daily number of tasks solved by Vasya, n= 16 - the number of days from March 18 to April 2 inclusive, S 16 = 560 - the total number of tasks, a 16 - the number of tasks that Vasya solved on April 2. Knowing that every day Vasya solved the same number of tasks more than the previous day, then you can use the 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 number 12- check students' ability to perform actions with functions, be able to apply the derivative to the study of the function.

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

Solution: 1) Find the domain 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; ∞). We define 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 work program in mathematics to the line of UMK G.K. Muravina, K.S. Muravina, O.V. Muravina 10-11 Download free algebra manuals

Task number 13- an increased level of complexity with a detailed answer, which tests 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,

	log3(2cos x) =	2	⇔		2cos x = 9	⇔		cos x =	4,5	⇔ because |cos x| ≤ 1,
	log3(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 .

It can be seen from the figure that the given segment has roots

11π	and	13π	.
6		6

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

Task number 14- advanced level refers to the tasks of the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes. The task contains two items. In the first paragraph, the task must be proved, and in the second paragraph, it must be 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 bases 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 the same 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 on a plane parallel to the bases of the cylinders is either 8 + 6 = 14, or 8 − 6 = 2.

Then the distance between 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 axis of the cylinder. 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's 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 the perpendicular bisector to this chord (it has a length of 8, as already noted) and from the center of the other base to another chord. They lie in the same plane β perpendicular to these chords. Let's call the midpoint of the smaller chord B, greater than 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 line of intersection of the base with the given plane.

So the required angle is

∠ABH = arctan	AH	= arctg	28	= arctg14.
	BH		8 – 6

Task number 15- an increased level of complexity with a detailed answer, checks the ability to solve inequalities, the most successfully solved among tasks with a detailed answer of an increased level of complexity.

Example 15 Solve the 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; +∞). In this case, this inequality can be rewritten in the form ( 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 a positive expression 3 x – x 2 , we get log 2 ( x + 1) ≤ 1, x + 1 ≤ 2, x≤ 1. Taking into account the area, we have x ∈ (0; 1].

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

Answer: (–1; –0.5] ∪ ∪ {3}.

Task number 16- advanced level refers to the tasks of 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 items. In the first paragraph, the task must be proved, and in the second paragraph, it must be calculated.

V isosceles triangle ABC with an angle of 120° at vertex A, a bisector BD is drawn. 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 ​​the rectangle DEFH if AB = 4.

Solution: a)

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

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

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

BD is the bisector of ∠B, so ∠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 number 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 - text task with economic content.

Example 17. The deposit in the amount 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 depositor annually replenishes the deposit by X million rubles, where X - whole number. Find highest value X, at which the bank will add less than 17 million rubles to the deposit in 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

(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 number 18- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection to universities with increased requirements for the mathematical preparation of applicants. Exercise high level complexity is not a task for applying one solution method, but for a combination of different methods. For the successful completion of task 18, in addition to solid mathematical knowledge, a high level of mathematical culture is also required.

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 as

	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 centered at the point (0, a). The set of solutions of the second inequality is the part of the plane that lies 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 of this system is the intersection of the solution sets of each of the inequalities.

Therefore, two solutions this system will have only in the case shown in Fig. one.

The points of contact between the circle and 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 the triangle PQR- rectangular isosceles. Dot Q has coordinates (0, a), and the point R– coordinates (0, – a). In addition, cuts PR and PQ are equal to the circle radius equal to 1. Hence,

QR= 2a = √2, a =	√2	.
	2

Answer: a =	√2	.
	2

Task number 19- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection to universities with increased requirements for the mathematical preparation of applicants. A task of a high level of complexity is not a task for applying one solution method, but for a combination of different methods. For the successful completion of task 19, it is necessary to be able to search for a solution, choosing various approaches from among the known ones, modifying the studied methods.

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

a) Give the formula P th member of this progression.

b) Find the smallest modulo sum S n.

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

Solution: a) Obviously, a n = S n – S n- one . 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) because S n = 2n 2 – 25n, then consider the function S(x) = | 2x 2 – 25x|. Her graph can be seen in the figure.

It is obvious that the smallest value is reached 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) It follows from the previous paragraph that sn positive since 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, with 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 full square is not achieved.

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

________________

*Since May 2017, the DROFA-VENTANA joint publishing group has been part of the Russian Textbook Corporation. The corporation also included the Astrel publishing house and the LECTA digital educational platform. CEO appointed Alexander Brychkin, graduate of the Financial Academy under the Government of the Russian Federation, candidate 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). Prior to joining the DROFA publishing house, he held the position of Vice President for strategic development and investments of the EKSMO-AST publishing holding. Today, the Russian Textbook Publishing Corporation has the largest portfolio of textbooks included in the Federal List - 485 titles (approximately 40%, excluding textbooks for remedial school). 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 to develop the country's production potential. The corporation's portfolio includes textbooks and study guides for elementary school awarded the Presidential Prize in Education. These are textbooks and manuals on subject areas that are necessary for the development of the scientific, technical and industrial potential of Russia.

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

Theory for task number 2

The rules for dealing with degrees can be represented as follows:

In addition, it should be recalled about operations with fractions:

Now we can move on to parsing. standard options! 🙂

Analysis of typical options for tasks No. 2 USE in mathematics of a basic level

The first version of the task

Find the value of an expression

Execution algorithm:

1. Express a negative number as a proper fraction.
2. Do the first multiplication.
3. Represent powers of numbers in the form prime numbers, replacing the powers by their 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, the multiplication of an integer by a proper fraction. To do this, multiply the numerator of the fraction by an integer, 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

The second version of the task

Find the value of an expression

Execution algorithm:

1. Express the first power of a number as an integer.
2. Express negative powers of numbers as proper fractions.
3. Perform integer multiplication.
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 present negative power numbers in the form common fraction, it is necessary to divide 1 by this number, but already in a positive degree.

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

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

Let's do integer multiplication.

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

The third version of the task

Find the value of an expression

Execution algorithm:

1. Express the powers of numbers as multiplication and calculate the value of the 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 as a 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:

The fourth option

Find the value of an 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 the common factor can be bracketed out.

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

Let's do the 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 option

Find the value of an expression

Execution algorithm:

1. Let us represent the power of a number in such a way that the common factor can be bracketed out.
2. Take the common factor out of the bracket.
3. Perform the action in parentheses.
4. Express 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 the common factor can be bracketed out.

Let's take the common factor out of the bracket

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

Let's do the parentheses.

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

Let's represent the power of a number as a multiplication. In order to represent the power of a number as a 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:

Variant of the second task from the USE 2017 (1)

Find the value 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 a standard form:

After that, you can divide 24 by 6, as a result we get 4.

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

Variant of the second task from the USE 2017 (2)

Find the value of the expression:

Solution:

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

After that, you can reduce the degrees of two: 6-5=1, for three: 8-5=3.

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

Variant of the second task of 2019 (1)

Execution algorithm

1. We apply to the numerator the holy degrees (a x) y = a xy. We get 3 -6.
2. We apply to the fraction of St. degrees a x /a y =a x-y.
3. Raise 3 to the power.

Solution:

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

Variant of the second task of 2019 (2)

Execution algorithm

1. We use for the degree in the numerator (14 9) St. (ab) x \u003d a x b x. We decompose 14 into the product of 2 and 7. We get the product of powers with bases 2 and 7.
2. Let's convert the expression into 2 fractions, each of which will contain powers with the same bases.
3. We apply to fractions of St-in degrees a x /a y =a x-y.
4. We find the resulting work.

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

Variant of the second task of 2019 (3)

Execution algorithm

1. We take out the common factor 5 2 =25.
2. We carry out the multiplication of numbers 2 and 5 in brackets. We get 10.
3. We perform the addition of 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

Variant of the second task of 2019 (4)

Execution algorithm

1. We square (-1). We get 1, because we are raising to an even power.
2. Raise (-1) to the 5th power. We get -1, because raised to an odd power.
3. Let's do the multiplication.
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

Variant of the second task of 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 characters.
3. We find the resulting sum.

Lexical means of communication:

1. Lexical repetition- repetition of the same word. Around the city on the low hills are forests, mighty, untouched. In the forests there were large meadows and deaf lakes with huge old pines along the banks.
2. Root words. Of course, such a master knew his own worth, felt the difference between himself and a not so talented person, but he knew perfectly well another difference - the difference between himself and a more gifted person. Respect for the more capable and experienced is the first sign of talent.
3. Synonyms. We saw an elk in the forest. Sukhaty walked along the edge of the forest and was not afraid of anyone.
4. Antonyms. Nature has many friends. She has fewer enemies.
5. Descriptive phrases. They built a highway. A noisy, swift river of life connected the region with the capital.

Grammar means of communication:

1. Personal pronouns. 1) And now I am listening to the voice of an ancient stream. He coos like a wild dove. 2) The call for the protection of forests should be addressed primarily to the youth. It is for her to live and manage on this earth, for her to 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 stars floated above the village. Such stars appear only in autumn. 2) Corncrakes screamed with a distant, sweet twitch. These corncrakes and sunsets are unforgettable; pure vision preserved them forever. - in the second text, means of communication - lexical repetition and 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. And so he did.
4. Unions(mostly writing) It was May 1945. Thundered spring. The people and the earth rejoiced. Moscow saluted the heroes. And joy soared into the sky with lights. With the same accent and laughter, the officers hurriedly began to gather; again put the samovar on the dirty water. But Rostov, without waiting for tea, went to the squadron.
5. Particles.
6. Introductory words and designs(in a word, so, firstly, etc.) Young people spoke about everything Russian with contempt or indifference and, jokingly, predicted the fate of the Confederation of the Rhine for Russia. In a word, the society was rather disgusting.
7. Unity of aspect tense forms of verbs- the use of the same forms of grammatical time, which indicate the simultaneity or sequence of situations. The imitation of the French tone of the time of Louis XV was in vogue. Love for the fatherland seemed pedantry. The wise men of the time praised Napoleon with fanatical obsequiousness and joked about our failures. All verbs are in the past tense.
8. Incomplete sentences and ellipsis, referring to the previous elements of the text: Gorkin cuts bread, distributes slices. He puts me too: huge, you will cover your whole face.
9. Syntax parallelism- the same construction of several adjacent sentences. Knowing how to speak is an art. Listening is culture.

Introductory word, union, particle, adverb	When is it used?
IN OTHER WORDS, IN OTHER WORDS	It is used when the author of the text wants to say the same thing, but more clearly.
MOREOVER	It is used when it is necessary to supplement what was said with some, in the opinion of the author, important thoughts or circumstances.
SO, SO, THEREFORE	Used when the author of the text sums up his reasoning.
FOR EXAMPLE, SO	They are used when the author wants to clarify what he was talking about before.
VICE VERSA	It is used when the author of the text contrasts one sentence with another.
FIRST, ON ONE SIDE	Indicates the order in which arguments are presented.
DESPITE IT, ALTHOUGH, DESPITE IT	The following meaning is introduced into the author's reasoning: "contrary to the circumstances indicated in the previous part of the text."
BECAUSE, AS, BECAUSE, THE THING IS THAT	The author uses when he indicates the cause of the phenomena described.
SO, SO, SO, FROM HERE	The author of the text uses when he wants to draw a conclusion from his reasoning.
THAT IS	Used to clarify what was said earlier.
HOWEVER, BUT, BUT	Used to contrast the meaning of one sentence with another.
EXACTLY, BECAUSE	They bring in the meaning of clarification and emphasize the importance of thought.
EVEN	Enter the gain value.
NOT BY CHANCE	It means "for this reason".
MEANS	The author wants to give an explanation to what was said before as a model, an illustration of his thought.

Semantic relations expressed by coordinating unions:

1. Connecting: and, yes(=and), and…and…, not only… but also, like… and, also, also
2. Dividers: or, either, then ... that, not that ... not that, or ... or, either ... or
3. Opposite: but, yes (= but), however, but
4. Gradation: not only, but also, not so much ... how much, not that ... but
5. Explanatory: that is, namely
6. Connecting: also, also, yes, and, moreover, moreover
7. also, yes and, that is, viz.

Semantic relations expressed by subordinating unions:

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

Evaluation

two parts, including 19 tasks. Part 1 Part 2

3 hours 55 minutes(235 minutes).

Answers

But you can make a compass Calculators on the exam not used.

the passport), pass and capillary or! Allowed to take with myself water(in a transparent bottle) and food

The examination paper consists of two parts, including 19 tasks. Part 1 contains 8 tasks of a basic level of complexity with a short answer. Part 2 contains 4 tasks of an increased level of complexity with a short answer and 7 tasks of a high level of complexity with a detailed answer.

For execution examination work in mathematics 3 hours 55 minutes(235 minutes).

Answers to tasks 1–12 are recorded as an integer or ending decimal. Write the numbers in the answer fields in the text of the work, and then transfer them to the answer sheet No. 1 issued during the exam!

When doing work, you can use the ones issued with the work. You can only use a ruler, but you can make a compass with your own hands. It is forbidden to use tools with reference materials. Calculators on the exam not used.

You must have an identity document with you for the exam. the passport), pass and capillary or gel pen with black ink! Allowed to take with myself water(in a transparent bottle) and food(fruit, chocolate, buns, sandwiches), but may be asked to leave in the hallway.