Solution ege profile. USE assignments profile mathematics - what to look for
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
Examination paper comprises two parts, including 19 tasks. Part 1 contains 8 tasks basic level Difficulty with short answers. Part 2 contains 4 tasks of increased difficulty with a short answer and 7 tasks high level Difficulties with extended answers.
To complete the examination work in mathematics is given 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 printed on them. 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.
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.
To complete the examination work in mathematics is given 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 printed on them. 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.
Average general education
Line UMK G.K. Muravina. Algebra and the beginnings of mathematical analysis (10-11) (deep)
Line UMK Merzlyak. Algebra and the Beginnings of Analysis (10-11) (U)
Maths
Preparation for the exam in mathematics ( profile level): tasks, solutions and explanations
We analyze tasks and solve examples with the teacherThe 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 obtain a school certificate, a graduate must pass two mandatory exams in USE form, 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 the ability of USE participants to apply the skills acquired in the course of grades 5-9 in elementary mathematics, in practical activities. The participant must have computational 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 An expense meter was installed in the apartment where Petr lives cold water(counter). 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 the 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 when various ways defining a function and describing 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 of the basic level of the first part, checks 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 |
S = 18 + |
6 | |
2 |
Read also: USE in physics: solving problems about vibrations
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. Let's split the two parts given equation for 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
Consequently, 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 exam participants, 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 No. 9 of an increased level of complexity with a short answer. Tasks from the section "Calculations and transformations" in the USE are divided into several types:
- conversion of numeric/letter trigonometric expressions.
transformations of numerical rational expressions;
transformations of algebraic expressions and fractions;
transformations of numerical/letter irrational expressions;
actions with degrees;
transformation of logarithmic 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. During 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 manualsTask 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 |
The circumference diameter 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.
In an isosceles triangle ABC with an angle of 120° at the 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 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 is a 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. 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 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.
Consequently, this system will have two solutions 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 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.
Graduating from high school is not easy these days. In order to say goodbye to the school desk, you need to pass several important exams, and not simple ones, but the Unified State Examination. Good certificate scores decide further fate graduate and give him a chance to enter a prestigious university. That is why the students are preparing for this test with all seriousness, and the conscious ones even begin to prepare for it from the very beginning. school year. What will be USE in mathematics 2017 and what changes await graduates in the delivery procedure, this article will tell.
It is worth noting that next year the number of compulsory subjects will not change. The guys, as before, must pass the Russian language and mathematics. The results are still evaluated on a 100-point scale, and in order to pass the exam, you must score at least the minimum number of points determined by the FIPI.
The mathematics exam will have a basic and profile direction.
Mathematics exam progress
No exact date yet conducting the exam in mathematics, but based on past years, it is easy to guess that it will take place around the beginning of June. In order to fully cope with the task, the student will be given as much as 3 hours. This time is enough to solve all the tests and practical tasks. Note that just before the exam, almost all personal belongings are taken away from graduates, leaving only a pen, a ruler and a calculator.
During the exam, it is prohibited:
- change;
- get up;
- to talk with neighbors;
- exchange materials;
- use audio devices to listen to information;
- go out without permission.
Do not forget that independent observers will be present in the classes at all times, so students must comply with all their requests regarding correct behavior during the exam!
Future changes
Every graduate who has ever taken the exam will tell you that the most difficult is mathematics. As a rule, only a few understand this subject, and far from many can solve all the test tasks. Unfortunately, no special indulgence in the content is planned, although some pleasant moments in passing the exam in mathematics in 2017 can still be noted. This applies to re-in case of defeat. Moreover, it will be possible to do it 2 times during the next academic year. In addition, if a student wishes to increase their scores, they can also apply for a retake.
The examination program will include not only assignments for grade 11, but topics from previous years. Recall that the basic level differs from the profile level in the knowledge assessment system: the basic one is based on a 20-point system, and the profile level is 100 points each. As statistics show, on average, only half of the students score 65 points at the profile level. Despite the fact that this is a rather low score, it is quite enough to enter an institute or university.
In 2017, they plan to increase the number of independent observers, as well as issue new forms for questions and answers. The test form will remain only in the mathematical exam, and then the specialists intend to add more practical tasks. This will avoid mere guessing and will help to soberly assess the knowledge of students.
Passing score of the basic level of the Unified State Examination in mathematics
The results of the exam can be viewed on the official portal, just by entering your passport data. To obtain a certificate, it is enough to earn only 7 points, which is equivalent to the usual “troika”. We suggest you familiarize yourself with the table for the basic level:
Passing score of the profile level of the Unified State Examination in mathematics
As mentioned above, to pass this exam, it is enough to score 65 points. This result guarantees the graduate a calm celebration of graduation and admission to the desired university in the country. In order to easily decipher the results of your knowledge, we suggest that you familiarize yourself with the points table for the profile level:
Exam Structure
Thanks to the demos that appear every year on the FIPI official website, the guys can go trial exam and see who's what. The exact structure of the exam, identical to the real one, has been developed in a special file. Note that the student will need to remember the program of all past years: trigonometry, logarithms, geometry, probability theory and much more. In 2017, the structure of the USE in mathematics is as follows:
All these tasks were compiled on the basis of the program studied during the school years. If the student studied diligently, performed all the work assigned by the teacher, it will not be difficult for him to pass the exam as “excellent”. In addition, going to tutors can increase the chances of a good grade.
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