USE. Russian language.

Task 13. How easy is it to do?

Task number 13- one of the most difficult. This is due to the fact that it is necessary to know a lot of rules for continuous, separate, hyphenated spelling of words. In addition, there are many words that you just need to remember. So there are difficulties.

I offer the easiest way to complete this task.

Algorithm for completing task No. 13

Continuous, separate, hyphenated spelling of words

Read the assignment carefully. It is necessary to find a sentence out of five proposed, in which the highlighted words are written together or apart. Even if the books you study mostly suggest finding conjoined writing words, an exam is an exam, you need to be prepared for anything. So it is with a careful reading of the task that its implementation begins.

In each sentence, eliminate words that are spelled with hyphen. Most often it is:

Words with suffixes SOMETHING, OR, SOMETHING and prefix CFU

Words anyway, exactly the same.

Adverbs with a prefix BY and suffixes OMU, HIM, SKI, LI:

in our opinion, in fox.

adjectives denoting colors, flavors(bright red, sweet and sour)

cardinal directions: southwest.

Words with roots floor: start at L(half a lemon) with a vowel(half an apple), capitalized(half of Europe).

Adjectives formed from homogeneous members, you can put a union between them AND(magazine-newspaper - that is, magazine and newspaper)

The first step has been taken. There will definitely be a word in a sentence that is written with a hyphen. Therefore, the number of proposals is reduced.

As if

In view of

Keep in mind

During

In continuation

Due to

Subsequently

Because

Whereas

That is

In order to

Despite

Regardless of

Immediately

As if

The third step is the most responsible. You need to clearly distinguish between words spelled together or apart.

To - what would

Same - the same

Also - the same

But - for that

Why - from what

Because - from that

Because - by that

And - at what

About (= o) - to the account (in the bank)

Remember: if a logical stress falls on a word, you highlight it with intonation, it is pronounced firmly, with some slowing down of intonation, and most importantly, you can imagine something concretely, then this word is spelled APART.

If none of the above is present, then this is an ordinary union, it is written ONE.

Compare.

TO should I give you a birthday present? (The emphasis falls on the word, we present the gift that we want to buy).

We met, TO discuss current affairs. (The word is pronounced quickly, as if casually, we cannot imagine anything, saying the word TO)

FOR THAT I received five assignments.

He prepared for a long time BUT passed the exam well.

Remember: if after SO SAME There is HOW AND, then it is always written separately. (The work was done AS QUALITY AS ALWAYS.)

Word SO it is written together, if this is an ordinary introductory word, something is summed up. ( SO, the work was completed before the holiday)

If we have an adverb and a union, then it is written separately, you can ask a question How?(So he spent all his free time (HOW did he spend it? - SO).

Remember that negative adverbs are always written together: nowhere, not at all, not at all, nowhere, nowhere etc.

These are the main cases to remember first.

All rules are on this site. Pay special attention to the tables with the spelling of adverbs, memorize the words.

EXAMPLE

Determine the sentence in which both underlined words are written ONE. Open the brackets and write out these two words.

Everything was (AS) THE SAME, (THAT) IS has not changed at all.

(WHAT) WOULD arrive on time (AT) THE MEETING, we left early in the morning.

(SOME) WHERE (IN) DALI could see the lights of the huts.

He disappeared (SO) AS suddenly as he appeared.

(And) SO let's start with the fact that I (IN) THE END met you.

EXPLANATION

We find sentences in which words are written with a hyphen. This is the first and third SOME WHERE, STILL. We exclude them. There are 3 offers left.

We find such words, in the separate spelling of which you have no doubt. This THAT IS(the first sentence, however, it has already been excluded)

There are 3 sentences left in which the words can be spelled correctly, thinking about their meaning.

2 sentence: where did we go? - TO MEET(for example, for a long-awaited meeting). That is, we clearly imagine the meeting that our heroes are going to. We write apart. Word TO here it is written together, since the lexical meaning in the word "What" No).

4 sentence - easy, it has AS WELL AS, so I write the word separately.

Remains number 5 is the correct answer: SO- introductory word FINALLY- adverb, when?

Do more tasks, and you will definitely succeed

Good luck!

Material prepared: Melnikova Vera Aleksandrovna

USE in mathematics profile level

The work consists of 19 tasks.
Part 1:
8 tasks with a short answer of the basic level of complexity.
Part 2:
4 tasks with a short answer
7 tasks with a detailed answer of a high level of complexity.

Run time - 3 hours 55 minutes.

Examples of USE assignments

Solving the task of the exam in mathematics.

Problem with solution:

In a regular triangular pyramid ABCS with a base ABC, the edges are known: AB \u003d 5 roots out of 3, SC \u003d 13.
Find the angle formed by the plane of the base and the straight line passing through the midpoint of the edges AS and BC.

Solution:

1. Since SABC is a regular pyramid, then ABC is an equilateral triangle, and the remaining faces are equal isosceles triangles.
That is, all sides of the base are 5 sqrt(3), and all side edges are 13.

2. Let D be the midpoint of BC, E the midpoint of AS, SH the height from point S to the base of the pyramid, EP the height from point E to the base of the pyramid.

3. Find AD from the right triangle CAD using the Pythagorean theorem. You get 15/2 = 7.5.

4. Since the pyramid is regular, point H is the intersection point of heights / medians / bisectors of triangle ABC, which means it divides AD in a ratio of 2: 1 (AH = 2 AD).

5. Find SH from right triangle ASH. AH = AD 2/3 = 5, AS = 13, by the Pythagorean theorem SH = sqrt(13 2 -5 2) = 12.

6. Triangles AEP and ASH are both right-angled and have a common angle A, hence similar. By assumption, AE = AS/2, hence both AP = AH/2 and EP = SH/2.

7. It remains to consider the right triangle EDP (we are just interested in the angle EDP).
EP = SH/2 = 6;
DP = AD 2/3 = 5;

Angle tangent EDP = EP/DP = 6/5,
Angle EDP = arctg(6/5)

Answer:

Do you know what?

Among all figures with the same perimeter, the circle will have the largest area. Conversely, among all figures with the same area, the circle will have the smallest perimeter.

Leonardo da Vinci derived the rule that the square of the diameter of a tree trunk is equal to the sum of the squares of the diameters of the branches, taken at a common fixed height. Later studies confirmed it with only one difference - the degree in the formula is not necessarily equal to 2, but lies in the range from 1.8 to 2.3. Traditionally it was believed that this pattern is due to the fact that a tree with such a structure has an optimal mechanism for supplying branches with nutrients. However, in 2010, the American physicist Christoph Elloy found a simpler mechanical explanation for the phenomenon: if we consider a tree as a fractal, then Leonardo's law minimizes the likelihood of breaking branches under the influence of wind.

Laboratory studies have shown that bees are able to choose the best route. After localizing the flowers placed in different places, the bee makes a flight and returns in such a way that the final path is the shortest. Thus, these insects effectively cope with the classic “traveling salesman problem” from computer science, which modern computers, depending on the number of points, can spend more than one day to solve.

If you multiply your age by 7, then multiply by 1443, the result is your age written three times in a row.

We consider negative numbers to be something natural, but this was far from always the case. For the first time negative numbers were legalized in China in the III century, but were used only for exceptional cases, as they were considered, in general, meaningless. A little later, negative numbers began to be used in India to denote debts, but they did not take root to the west - the famous Diophantus of Alexandria argued that the equation 4x + 20 = 0 is absurd.

American mathematician George Dantzig, being a graduate student at the university, one day was late for class and took the equations written on the blackboard for homework. It seemed to him more complicated than usual, but after a few days he was able to complete it. It turned out that he solved two "unsolvable" problems in statistics that many scientists struggled with.

In Russian mathematical literature, zero is not a natural number, but in Western literature, on the contrary, it belongs to the set of natural numbers.

The decimal number system we use arose due to the fact that a person has 10 fingers on his hands. The ability for abstract counting did not appear in people immediately, and it turned out to be most convenient to use fingers for counting. The Mayan civilization, and independently of them, the Chukchi historically used the decimal number system, using not only the fingers, but also the toes. The basis of the duodecimal and sexagesimal systems common in ancient Sumer and Babylon was also the use of hands: the phalanges of other fingers of the palm, the number of which is 12, were counted with the thumb.

One familiar lady asked Einstein to call her, but warned that her phone number is very difficult to remember: - 24-361. Remember? Repeat! Surprised Einstein answered: - Of course, I remember! Two dozen and 19 squared.

Stephen Hawking is one of the greatest theoretical physicists and popularizer of science. In a story about himself, Hawking mentioned that he became a professor of mathematics, having not received any mathematical education since high school. When Hawking began teaching mathematics at Oxford, he read his textbook two weeks ahead of his own students.

The maximum number that can be written in Roman numerals without violating Schwartzman's rules (rules for writing Roman numerals) is 3999 (MMMCMXCIX) - you cannot write more than three digits in a row.

There are many parables about how one person offers another to pay him for some service as follows: he will put one grain of rice on the first cell of the chessboard, two on the second, and so on: each next cell is twice as much as the previous one. As a result, he who pays in this way is bound to be ruined. This is not surprising: it is estimated that the total weight of rice will be more than 460 billion tons.

In many sources, often with the aim of encouraging poorly performing students, there is an assertion that Einstein failed in mathematics at school or, moreover, studied badly in all subjects. In fact, this was not the case: Albert at an early age began to show talent in mathematics and knew it far beyond the school curriculum.

For the correct completion of the thirteenth task of the USE in the Russian language, graduates can receive one primary point. To do this, write out two words from the sentence that satisfy the condition. The task contains only such parts of speech as conjunctions, particles, prepositions, pronouns, adverbs and allied combinations. It is important to be able to distinguish between the meanings of homonymous words, which will help the theoretical material below.

Theory for assignment No. 13 USE in the Russian language

Spelling of different parts of speech

	Slitno	Apart	Through a hyphen
Unions	Because, because, because, because, because, but, why, also, too, moreover, moreover, as if, as if, so that, then so that, if	As if, that is, since, if, as if, or, due to the fact that, then ... then,
Allied combinations	Due to the fact that, due to the fact that,	not that... not that
	although
Pronouns	Nobody', nothing',	From someone, from something	Someone (-someone, -or, -someone, something)
	Not who, not what, not what	with no one, with no one,
		the same as; the same, the same, the same
Adverbs	From everywhere, from now on, in part, in general,	Tirelessly, side by side, eye to eye, in general, so far, abroad, abroad, everything is the same, exactly the same	Somehow (something, -something, something, something),
	soon, in an undertone, alone, in advance, at first, at first, at first, little by little, completely, why, then, immediately, immediately		Russian, I think
			hare, firstly,
			barely, barely, barely, today or tomorrow,
			de jure, de facto
Prepositions	As a result, due to	During, in continuation, in conclusion, in contrast, in order, in force, in measure, in the form, in the field,	Because of, from under, over
	towards, about	throughout, with respect to, except for,
	like, following	on account,
	over, instead of,	not to mention
	inside, in view
	like,
	despite,
	regardless of
Particles	Even, really, really	Whether, would,	Anyway, let's
		just what for

Derivative prepositions / nouns with a preposition

derived preposition	Noun with preposition
During	During
-during the winter	- in the course of the river = in the fast flow of the river
-in two hours
In continuation	In continuation
Answers the questions “How long? When?"	An adjective or participle can be placed between a preposition and a noun.
- for a whole week	-in the sequel of the movie = in the long-awaited sequel of the movie
- for two months
Due to	In consequence, in consequence
Can be replaced by the preposition "because of"; answers the question "Why?"	An adjective or participle can be placed between a preposition and a noun.
- due to illness	- in the investigation of the case = in the long investigation of the case
	-to the investigation of the case = to the sensational investigation of the case
Towards	For a meeting
Can be replaced by the preposition "to"; answers the question "Where?"
- towards him	- to meet a friend = to a long-awaited meeting with a friend
In view of	Keep in mind
Can be replaced by the prepositions "because of" and "because of"; answers the question "Why? From what?"	Set expression
due to bad weather
About	To account
Can be replaced with the preposition "o (about)"	An adjective can be placed between a preposition and a noun.
- make arrangements for a trip	-to the fund's account = to the fund's bank account
Like	Like
Can be replaced with the preposition "like"	I mean the geometric term
- like lunch
Finally	In custody
Can be replaced by the words "completing, in the end, in the end"	An adjective can be placed between a preposition and a noun.
	-in custody = in strict custody
Following	Track to track
It means "where"	Meaning "for"
-we watched the departing train	-Children follow each other

It is important to remember the spelling of the following words: in connection with, in contrast to, after (= behind), in the middle, near, instead of, throughout

Derived prepositions / gerunds

Conjunctions/pronouns with a particle

Union	Pronoun with particle
To	To
Can be replaced by unions "in order to, in order to"; "would" can not be removed from the phrase	Can be replaced by a noun; "would" can be rearranged
-I came to tell you = I came to tell you	-I asked what else I should read = I asked what book I should read = I asked what else I should read
Also, also	Same, same
You can replace the union "and", putting it at the beginning of the sentence; "same" can not be removed from the phrase	"yes" can be rearranged
- I also saw this movie = and I saw this movie	-I have also read this book = I have read the same book
But	For that
Can be replaced by the union "but"	"that" can be replaced by a noun, adjective or adverb
- we rarely see each other, but often call each other = we rarely see each other, but often call each other	- don't take on things you don't like = don't take on things you don't like
So	So
Can be replaced by the words "to sum up, therefore"	It means "very"
- so, we can say that = summing up, we can say that	We were tired and so hungry that we decided to stay at a hotel.
Because	From that
Meaning "because"	Can be replaced by a noun with a preposition
-it happened because = it happened because	-from what he will do = from the act that he will do
Moreover, and	At the same time, at what
Can be replaced by the words "in addition", "at the same time"	"volume" can be replaced by an adjective
-it works quickly and also qualitatively = it works quickly and also qualitatively	-that building has a beautiful garden = a tall building has a beautiful garden

Task execution algorithm

1. Read the assignment carefully.
2. We analyze each sentence, opening brackets in accordance with the spelling rules of the Russian language.
3. Write down the correct answer.

Analysis of typical options for task No. 13 USE in the Russian language

The thirteenth task of the 2018 demo

1. (FOR) BECAUSE L.N. Tolstoy, his relatives could guess (ON) HOW hard his brain is working now.
3. From the first pages, I experienced a strange feeling: AS if (WOULD) from a gloomy world I (THAT) HOUR was transferred to another world - sunny and bright.
4. (B) SUBSEQUENTly, researchers have repeatedly said that the apotheosis of Russian glory is the painting "Bogatyrs", in which V.M. Vasnetsov expressed his romantic and at the same time deeply civic understanding of Russia.
5. The physical properties of interstellar gas essentially depend (FROM) WHAT it is in relative proximity to hot stars or, (IN) REVERSE, is sufficiently remote from them.

Task execution algorithm:

1. Read the assignment carefully.
2. • (FOR) BECAUSE L.N. Tolstoy, his relatives could guess (ON) HOW hard his brain is working now.- FOR THAT we write separately, since this is a preposition with a demonstrative pronoun; HOW we write together, since it can be replaced by the pronoun HOW.
   • (C) CONSEQUENTLY, scientists have found that magnesium plays an important role in regulating potassium levels in the body, and ALSO regulates the functioning of the adrenal glands.- AFTER we write together, since this is an adverb that can be replaced by an adverb THEN; We ALSO write together, since it is impossible to omit the particle SAME without losing the meaning.
   • From the first pages, I experienced a strange feeling: AS if (WOULD) from a gloomy world I (THAT) HOUR was transferred to another world - sunny and bright .- AS if we are writing separately, since the particle can be omitted without loss of meaning; IMMEDIATELY we write together, as it can be replaced IN THE SAME INSTANT.
   • (B) SUBSEQUENTly, researchers have repeatedly said that the apotheosis of Russian glory is the painting "Bogatyrs", in which V.M. Vasnetsov expressed his romantic and at the same time deeply civic understanding of Russia.- AFTER we write together, since we can replace THEN; We write the SAME separately, since it is possible to omit the particle without losing the meaning.
   • The physical properties of interstellar gas essentially depend (FROM) THAT whether it is in relative proximity to hot stars or, (IN) REVERSE, is sufficiently remote from them.- FROM TOTO we write separately, since this is a preposition and a demonstrative pronoun that can be replaced by a noun; On the contrary, we write together, as it can be replaced with an adverb OPPOSITE.
3. Write down the correct answer: subsequently, also.

The first version of the assignment

Task execution algorithm:

1. Read the assignment carefully.
2. We analyze each sentence, opening brackets in accordance with the spelling rules of the Russian language:
   • We decided to go (IN) THIS alley, BECAUSE (BECAUSE) it is quiet: there is no traffic here at all.- both words are written separately; “because” is a union, and “for this” is written separately, since this is not an expression of a consequence and is the word “lane”.
   • (C) DEPENDING on the situation of communication, people behave (ON) DIFFERENTLY.- “depending on” is always written separately, and “differently” is hyphenated.
   • (NOT) ALWAYS we understand the meaning of toponyms, (FOR) OFTEN sounding strange to the ear of a modern person.- "not always" is never written together; “often” is written together, but the task says that both words in the sentence should be written together.
   • (THIS) HOUR, he strives to achieve his goal at WHAT (WHAT) IS POSSIBLE.- it is written together “now”, replaced by the combination “at the moment”; but "by all means" is always written separately.
   • Set an alarm clock, (WHAT) WOULD NOT oversleep and get up (FOR) EARLY.- “to” is written together, as it is replaced by the combination “in order to”. “Early” is an adverb that is always spelled together.
3. Write down the correct answer: to, early(Remember to write your answers without spaces or punctuation marks on the exam).

The second version of the assignment

Task execution algorithm:

1. Read the assignment carefully.
2. We analyze each sentence, opening brackets in accordance with the spelling rules of the Russian language:
   • Nobody travels THIS path, BECAUSE the road here is in disrepair.-"On this path" - separately; can be replaced, for example, with the combination "the old way". "Because" - again, separately.
   • (TO) THE MEETING we were late, although (FROM) DUE TO the snowfall we left the house an hour earlier.-“For a meeting” - separately, since you can replace “for a business meeting”, or “for an important meeting”, and “because of” is written with a hyphen, as it indicates the reason.
   • AND(SO), (C)CLOSING let me thank you for your cooperation.- “So” is written together, but “in conclusion” - separately.
   • (B) DUE TO the unstable political situation, the trip to Egypt, which we (IN) HURRY planned, had to be postponed.- “In view of” is written together - it is replaced by the word “because of”; "hurriedly" - an adverb, written together.
   • (B) GAVE from civilization you AS (IF) are aware of all the imperfection of our modern world.- "away" is written together (replaced by "far"), but "as if" - always separately.
3. Write down the correct answer: I mean, hastily.

The third version of the task

Determine the sentence in which both underlined words are spelled ONE. Open the brackets and write out these two words.

Task execution algorithm:

1. Read the assignment carefully.
2. We analyze each sentence, opening brackets in accordance with the spelling rules of the Russian language:
   • WHAT (WHATEVER) they say about Russia abroad, but the country has long been (NOT) THE same as it was in the 90s.- “Whatever” and “not that” are written separately: in the first case, “would” can be rearranged - “whatever they say about Russia”, and in the second it is simply impossible to write together.
   • I tell you the SAME thing that Andrei, WHAT (WOULD) you have the same information.- “The same” is written separately, but “in order to” is written together, we replace it with “in order to”.
   • (APPOINTINGLY) Alexei did not hear what I answered him, (FOR) BECAUSE he repeated his question.- “Apparently” - an adverb that we write with a hyphen; “because” is merged as part of the union “because”.
   • Pushkin (THAT) HOUR became addicted to billiards, although he did not become a serious player (SO).- "Immediately" - an adverb, written together; "So and" are written separately.
   • My sister sang (B) HALF A VOICE, I THAT (SAME) began to quietly sing along.- The last sentence remains; "in an undertone" is always written together; We also write “too” together - we replace it with the union “and”.
3. Write down the correct answer: in an undertone, too.

Secondary 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)

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 get 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 Exam 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 5-9 grades 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 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 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 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 largest or smallest value from the function graph 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.

#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 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, it checks the ability to perform actions with geometric shapes according to 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) Let's use the equation of a straight line passing through two given points and find the equation of a straight 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 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:

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 tasks 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 of an 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 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 the 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. 1.

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- 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) 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. 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, LECTA digital educational platform) has been appointed General Director. 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 correctional schools). The corporation's publishing houses own the sets of textbooks in physics, drawing, biology, chemistry, technology, geography, astronomy, most in demand by Russian schools - the areas of knowledge that are needed to develop the country's production potential. The corporation's portfolio includes textbooks and teaching aids for elementary schools awarded the President's 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.

The lesson discusses the decision of the 13th task of the exam in computer science

Topic 13 - "Amount of information" - is characterized as tasks of an increased level of complexity, the execution time is about 3 minutes, the maximum score is 1

when working with text

• By using K bit can be encoded Q=2K various characters:
• Q- power of the alphabet
• K Q character options
• 2 - binary number system (data is stored in binary form)

N = 2i

• I, you need to multiply the number of characters N by the number of bits to store one character K:
• I
• N— message length (number of characters),
• K is the number of bits to store one character.
• These two formulas use the same variable:

Q=2K I=N*K

Consider an example using two formulas at the same time:

Example:
Message volume - 7.5 KB 7680 characters. What is the power of the alphabet?

✍ Solution:

• Let's use the formula:

I = N*K;
I- message size = 7.5 KB;
N- number of characters = 7680;
K- the number of bits per 1 character

• Let's find the number of bits required to store 1 character (first we convert the value to bits):

\[ K= \frac (7.5 * 2^(13))(7680) = \frac (7.5 * 2^(13))(15 * 2^9) = \frac (7.5 * 16 )(15) = 8 \]

those. K = 8 bits per 1 character

• Next, we use the formula:

Q=2K
K- the number of bits to store one character from Q character options (= 8)
Q is the cardinality of the alphabet, i.e. number of character options

• 8 bits per character allows you to encode:

2 8 = 256 different characters
256 characters is the power

Answer: 256

Measuring the amount of information
when working with various systems

• By using K bit can be encoded Q=2K different (numbers) objects of some system:
• Q- the total number of objects in some system, data about which is stored in a computer or transmitted in a message,
• K- the number of bits to store one object from the total Q,
• 2 - binary number system (data is stored in binary form).

* other designations are also accepted: N = 2i

• To find the information volume of a message I, you need to multiply the number of objects in the message - N- per number of bits K to store a single object:
• I- information volume of the message,
• N— the number of objects in the message
• K- the number of bits to store one object of the system.

Example:
The production has an automatic system for informing the warehouse about the need to deliver certain groups of consumables to the workshop. The system is designed so that through the communication channel to the warehouse the conditional number of consumables is transmitted(this uses the same, but the minimum possible number of bits in the binary representation of this number). It is known that a delivery request has been sent 9 groups materials from 19 used in production. Determine the amount of sent message (Give your answer in bits)

✍ Solution:

• Let's use the formula:

K- the number of bits for storing one material group number
Q- total number of numbers for different groups of consumables = 19

• to store the number of one group, a bit is required:

2 5 < 19 =>5 bit

Degree 4 we are not satisfied, because 2 4 = 16 , and groups 19 .

Next, we use the formula:

I = N*K;
I- message size = ? bit;
N— number of transmitted group numbers (= 9);
K- number of bits per 1 number (= 5)

Let's find the information volume of the message:

I = 9 * 5 = 45 bits

Answer: 45

Solving tasks 13 USE in computer science

USE in Informatics 2017 task 13 FIPI option 1 (Krylov S.S., Churkina T.E.):

7 33 -character alphabet. In the database for storing information about each user, the same and the smallest possible integer is allocated byte bit. In addition to its own password, additional information is stored in the system for each user, for which an integer number of bytes is allocated; this number is the same for all users.

To store information about 60 users needed 900 byte.

How many bytes are allocated to store additional information about one user?
In response, write down only an integer - the number of bytes.

✍ Solution:

• Let's decide on a password first. According to the formula Q = M N we get:

33 = 2N -> N = 6 bits per 1 character

The password consists of 7 characters:

-> 7*6 =42 bit just for a password

Since all user data is stored in bytes, we take the nearest number greater than 42 and multiple 8 :

48/8 = 6 42 bits ~ 6 bytes

Now let's find how many bytes are allocated to store information about one user:

900 bytes / 60 (users) = 15 bytes per user

Get the amount of memory to store additional information:

15 bytes (to store all information) - 6 bytes (to store the password) = 9 bytes for more information

Result: 9

A step-by-step solution to this 13 task of the exam in computer science is also available in the video lesson:

USE 2017 collection of D.M. Ushakov "10 training options ..." option 1:

The cable network holds a vote among viewers on which of the four films they would like to watch tonight. The cable network is used 2000 Human. Participated in the voting 1200 Human.
What is the amount of information ( in bytes) recorded by an automated voting system?

✍ Solution:

• Since the numbers of the four movies are stored in the computer system, we can find the number of bits needed to store the movie number:

Q = 2 k -> 4 = 2 k -> k = 2 bit

Since all 1200 people will vote for one of the films, respectively, the same amount of memory (i.e. 2 bits) must be allocated for each vote.

Find the number of bits needed to store all 1200 votes:

1200 * 2 = 2400 bits = 2400/8 bytes = 300 byte

Result: 300

USE 2017 collection of D.M. Ushakov "10 training options ..." option 6:

When registering in a computer system, each user is given a password consisting of 15 characters and containing only characters from 12 -character set A, B, C, D, E, F, G, H, I, K, L, M, N. In the database for storing information about each user, the same and the smallest possible integer is allocated byte. In this case, character-by-character coding of passwords is used, all characters are encoded in the same and the minimum possible number. bit. In addition to the password itself, additional information is stored in the system for each user, for which 12 bytes per user.

Determine the amount of memory ( in bytes) needed to store information about 30 users.
In the answer, write down only an integer - the number of bytes.

✍ Solution:

Result: 600

An example of solving this USE task is available in the video tutorial:

USE 2017 collection of D.M. Ushakov "10 training options ..." option 10:

Rehearsal exam at school 105 Human. Each of them is given a special number that identifies him in the automatic system for checking answers. When registering a participant to record his number, the system uses the minimum possible number of bit, the same for each participant.

What is the amount of information in bits, recorded by the device after registration 60 participants?

✍ Solution:

Result: 420

An example of solving this USE task is available in the video tutorial:

13 task. Demo version of the exam 2018 informatics:

10 characters. Capital letters of the Latin alphabet are used as symbols, i.e. 26 various characters. In the database, each password is stored with the same and the smallest possible integer byte. In this case, character-by-character coding of passwords is used, all characters are encoded in the same and the minimum possible number. bit.

Determine the amount of memory ( in bytes) needed to store data about 50 users.
In the answer, write down only an integer - the number of bytes.

✍ Solution:

• The main formula for solving this problem is:

Where Q is the number of character variants that can be encoded using N bit.

• To find the number of bits required to store one password, first you need to find the number of bits required to store 1 character in the password. According to the formula we get:

26 = 2N -> N ~ 5 bits

The password consists of 10 characters. This means that a bit must be allocated to the password:

10 * 5 = 50 bits total per password

Since password information is stored in bytes, we translate:

50 bits / 8 ~ 7 bytes (we take the nearest number greater than 50 and a multiple of 8: 57/8 = 7)

Now let's find how many bytes are allocated to store information about 50 users:

7 bytes * 50 (users) = 350 byte

Result: 350

For a detailed solution to task 13 of the USE demo version of 2018, see the video:

Solution 13 of the USE task in computer science (diagnostic version of the examination paper, USE simulator 2018, S.S. Krylov, D.M. Ushakov):

In some country, the license plate consists of 7 characters. Each character can be one of 18 different letters or decimal figure.

Each such number in a computer program is written in the minimum possible and the same integer number. byte, while character-by-character coding is used and each character is encoded by the same and the minimum possible number bit.

Determine the amount of memory in bytes, assigned by this program for recording 50 numbers.

✍ Solution:

• Since the number can contain either one letter from 18 , or one digit from 10 , then only one of the 28 characters:

18 + 10 = 28

Let's determine how many bits are needed to store one character in a number, for this we use the formula N = 2i:

28 = 2 i => i = 5

Since the total number of characters in the number is 7 , then we get the required number of bits for storing one number:

I = 7 * 5 = 35 bits

Since the number is allocated the same amount of storage byte, then convert to bytes:

35 / 8 ~ 5 bytes

The problem asks how much memory is required to store 50 numbers. We find:

I=50*5= 250 bytes to store 50 numbers

Result: 250

Video analysis:

Solution 13 of the USE task in informatics (control version No. 1 of the examination paper, Simulator 2018, S.S. Krylov, D.M. Ushakov):

Rehearsal exam passed 9 flows by 100 person in everyone. Each of them is allocated a special code consisting of a stream number and a number in the stream. In coding these participant numbers, the checking system uses the minimum possible number of bit, the same for each participant, separately for the number of the stream and the number in the stream. In this case, to write the code, the minimum possible and equally integer number is used. bytes.
What is the amount of information in bytes written by the device after registration 80 participants?
Give your answer only as a number.

✍ Solution:

• The code consists of two components: 1. stream number (in bits) and 2. sequence number (in bits). Find the number of bits needed to store them:

1. N = 2 i -> 9 = 2 i -> i = 4 bits (2 3 100 = 2 i -> i = 7 bits (2 6

Total we get 4 + 7 = 11 bits for one code. But by condition, an integer number of bytes is allocated for storing the code. So let's translate the resulting result into bytes:

11/ 8 ~ 2 bytes (one byte is not enough, 8

Since we need to get the amount of information after registration 80 participants, we calculate:

2 * 80 = 160 byte

Result: 160

Video analysis of the task:

Solution 13 of the USE task in informatics (K. Polyakov, v. 4):

Message volume - 7.5 KB. This message is known to contain 7680 characters. What is the power of the alphabet?

✍ Solution:

• Let's use the formula:

I - message size N - number of characters K - number of bits per 1 character

In our case N=7680 characters for which I = 7.5 KB of memory. Let's find the number of bits required to store one character (first converting Kbytes to bits):

I = 7.5 KB = 7.5 * 2 13 bits

\[ K = \frac (7.5 * 2^(13))(7680) = \frac (7.5 * 2^(13))(15 * 2^9) = \frac (7.5 * 16 )(15) = 8 \]

8 bits per character allow you to encode:

2 8 = 256 various symbols
(according to the formula Q = 2 N)

256 characters is the power

Result: 256

Video analysis of the task is presented after the next task.

Message encoding (text):

Solution 13 of the USE task in informatics (K. Polyakov, v. 6):

The power of the alphabet is 256 . How many KBytes of memory is required to save 160 pages of text containing on average 192 characters on every page?

✍ Solution:

• Let's find the total number of characters on all pages (for convenience, we will use powers of two):

160 * 192 = 15 * 2 11

According to the formula Q = 2n find the number of bits required to store one character (in our case Q=256):

256 = 2 n -> n = 8 bits per character

Let's use the formula I=N*K and find the required volume:

\[ I = (15 * 2^(11)) * 2^3 bits = \frac (15 * 2^(14))(2^(13)) KB = 30 KB \]

I= 30 KB

Result: 30

See a detailed analysis of text encoding tasks: from 1 to 2100), month number (number 1 to 12) and the number of the day in the month (number from 1 to 31). Each field is written separately from other fields using the minimum possible number of bits.
Determine the minimum number of bits required to encode one record.

✍ Solution:

• Formula needed Q = 2n.
• Let's calculate the required number of bits to store each item of the entire record:

1. 2100 choices: 2100 ~ 2 12 -> n = 12 bits 2. 12 choices: 12 ~ 2 4 -> n = 4 bits 3. 31 choices: 31 ~ 2 5 -> n = 5 bits

Let's find the total number of bits for the entire record:

12 + 4 + 5 = 21

Solution 13 of the USE task in informatics (K. Polyakov, v. 33):

The car number consists of several letters (the number of letters is the same in all numbers), followed by three digits. At the same time, they use 10 digits but only 5 letters: N, O, M, E And R. Must have at least 100 000 various numbers.
What is the minimum number of letters that should be in a car number?

✍ Solution:

• Formula needed Q = m n.

Q - number of options m - cardinality of the alphabet n - length

Let's compose the right side of the formula, based on the data of the task conditions (an unknown number of letters (out of five options) and three numbers (out of 10 options)):

5 ... 5 10 10 10 = 5 x * 103

This result must be at least 100000 . Substitute the rest of the data in the formula:

100000

From here we find the smallest suitable x:

x= 3 : 5 3 * 1000 = 125000 (125000 > 100000)

Result: 3

We offer you to watch the video analysis of the task:

Solution 13 of the USE task in informatics (K. Polyakov, v. 58):

When registering in a computer system, each user is given a password consisting of 9 characters. Used as symbols uppercase and lowercase letters of the Latin alphabet (in it 26 characters), and decimal digits. The database stores information about each user with the same and the smallest possible integer number of bytes. In this case, character-by-character coding of passwords is used, all characters are encoded with the same and the minimum possible number of bits. In addition to the password itself, additional information is stored in the system for each user, for which 18 bytes per user. Allocated in the computer system 1 Kb to store information about users.

What is the maximum number of users that information can be stored in the system? In your answer, write down only an integer - the number of users.

✍ Solution:

• Since both uppercase and lowercase letters are used, we get the total number of character options for encoding:

26 + 26 + 10 = 62

From the formula Q = 2 n we get the number of bits required to encode 1 character of the password:

Q = 2n -> 62 = 2n -> n = 6

Since the password has 9 characters, we get the number of bits to store 1 password:

6 * 9 = 54

Let's translate into bytes (since, by condition, passwords are stored in bytes):

54 / 8 = 7 bytes

18 bytes are allocated for storing additional information. Let's get the number of bytes to store all the information for one user:

18 + 7 = 25 bytes

According to the condition, 1 Kb is allocated for storing information about all users. Let's convert this value to bytes:

1 KB = 1024 bytes

Get the possible number of users:

1024 / 25 = 40,96

Let's drop the fractional part: 40

Result: 40

Watch the video with the solution of the task: