Series “OGE. FIPI - school" was prepared by the developers of control measuring materials (CMM) of the main state exam.
The collection contains:
36 standard exam options compiled in accordance with the draft demo version of the KIM OGE in mathematics 2018;
instructions for performing work;
answers to all tasks;
solutions and criteria for evaluating tasks of part 2.
Completing the tasks of standard examination options provides students with the opportunity to independently prepare for the state final certification in the 9th grade, as well as to objectively assess the level of their preparation.
Teachers can use standard examination options to organize the control of the results of mastering by students educational programs main general education and intensive preparation of students for the OGE.

Examples.
Which of the following statements are correct?
1) Area of ​​a triangle less product two of its sides.
2) The angle inscribed in a circle is equal to the corresponding central angle based on the same arc.
3) Through a point not lying on a given line, one can draw a line perpendicular to this line.
In your answer, write down the numbers of the selected statements without spaces, commas, or other additional characters.

The first cyclist left the village on the highway at a speed of 21 km/h. An hour later, at a speed of 15 km/h, a second cyclist left the same village in the same direction, and an hour later, a third. Find the speed of the third cyclist if he first caught up with the second, and after 9 hours after that he caught up with the first.

Free download e-book in a convenient format, watch and read:
Download the book OGE, Mathematics, Typical exam options, 36 options, Yashchenko I.V., 2018 - fileskachat.com, fast and free download.

• OGE in mathematics from A to Z, Problems in geometry, Yashchenko I.V., Shestakov S.A., 2020
• OGE 2020, Mathematics, 50 options, Typical options for examination tasks, Vysotsky I.R., Yashchenko I.V., Khachaturyan A.V., Shestakov S.A.
• OGE 2020, Mathematics, 50 options, Typical options for examination tasks, Vysotsky I.R., Yashchenko I.V., Kuznetsova L.V.

The following tutorials and books.

When writing this work “OGE in Mathematics 2018. Option 1”, the manual “OGE 2018. Mathematics. 14 options. Typical test tasks from the developers of the OGE / I. R. Vysotsky, L. O. Roslova, L. V. Kuznetsova, V. A. Smirnov, A. V. Khachaturyan, S. A. Shestakov, R. K. Gordin, A S. Trepalin, A. V. Semenov, P. I. Zakharov; edited by I. V. Yashchenko. - M .: Publishing house "Exam", MTSNMO, 2018 ″.

Part 1

Module “Algebra”

Show Solution

To add two fractions, they must be reduced to common denominator. In this case, this is the number 100 :

Answer:

1. In several relay races that were held at the school, the teams showed the following results.

Team	I relay, points	II relay, points	III relay, points	IV relay, points
"Hit"	3	3	2	4
"Jerk"	1	4	4	2
"Takeoff"	4	2	1	3
"Spurt"	2	1	3	1

When summing up the scores of each team for all relay races are summed up. The team with the most points wins. Which team finished third?

1. "Hit"
2. "Jerk"
3. "Takeoff"
4. "Spurt"

Show Solution

First of all, we summarize the points scored by each team

“Strike” = 3 + 3 + 2 + 4 = 12
“Dash” = 1 + 4 + 4 + 2 = 11
“Takeoff” = 4 + 2 + 1 + 3 = 10
"Spurt" = 2 + 1 + 3 + 1 = 7

Judging by the result: the first place belongs to the “Strike” team, the second - to the “Jump” team, and the third - to the “Rise” team.

Answer:

The third place was taken by the team “Vlet”, number 3.

1. On the coordinate line, points A, B, C and D correspond to the numbers: -0.74; -0.047; 0.07; -0.407.

Which point corresponds to the number -0.047?

Show Solution

On a coordinate line, positive numbers are to the right of the origin, and negative numbers are to the left. So the only positive number 0.07 corresponds to point D. The largest negative number is -0.74, which means it corresponds to point A. Given that the remaining number is -0.047 more number-0.407, then they belong to points C and D, respectively. Let's show it on the drawing:

Answer:

The number -0.047 corresponds to point C, number 3.

1. Find the value of an expression

Show Solution

In this example, you need to be smart. If the root of 64 is 8, since 8 2 = 64, then the root of 6.4 is quite difficult to find in a simple way. However, after finding the root of the number 6.4, it must be immediately squared. So two steps: finding square root and squaring cancel each other out. Therefore we get:

Answer:

1. The graph shows the relationship atmospheric pressure from height above sea level. The horizontal axis shows altitude in kilometers, and the vertical axis shows pressure in millimeters of mercury. Determine from the graph at what height the atmospheric pressure is 140 millimeters of mercury. Give your answer in kilometers.

Show Solution

Let's find on the graph the line corresponding to 140 mmHg. Next, we determine the place of its intersection with the curve of dependence of atmospheric pressure on altitude above sea level. This intersection is clearly visible on the graph. Let's draw a straight line from the point of intersection down to the height scale. The desired value is 11 kilometers.

Answer:

Atmospheric pressure is 140 millimeters of mercury at an altitude of 11 kilometers.

1. Solve the Equation x 2 + 6 = 5X

If the equation has more than one root, write the smallest of the roots as your answer.

Show Solution

x 2 + 6 = 5X

Before us is the usual quadratic equation:

x 2 + 6 – 5X = 0

To solve it, you need to find the discriminant:

Answer:

The smallest root of this equation: 2

1. The mobile phone that went on sale in February cost 2,800 rubles. In September, it began to cost 2520 rubles. By what percent did the price of a mobile phone decrease between February and September?

Show Solution

So, 2800 rubles - 100%

2800 - 2520 \u003d 280 (p) - the amount by which the phone fell in price

280 / 2800 * 100 = 10 (%)

Answer:

The price of a mobile phone between February and September decreased by 10%

1. The diagram shows the seven largest countries in terms of area (in million km 2) of the world.

Which of the following statements are wrong?

1) Canada is the largest country in the world by area.
2) The territory of India is 3.3 million km 2.
3) The area of ​​the territory of China more area territory of Australia.
4) The area of ​​​​the territory of Canada is larger than the area of ​​\u200b\u200bthe United States by 1.5 million km 2.

In response, write down the numbers of the selected statements without spaces, commas, or other additional characters.

Show Solution

Based on the graph, Canada is inferior in area to Russia, which means the first statement wrong .

An area of ​​3.3 million km 2 is indicated above the histogram of India, which corresponds to the second statement.

The area of ​​China according to the graph is 9.6 million km 2, and the area of ​​Australia is 7.7 million km 2, which corresponds to the statement in the third paragraph.

The territory of Canada is equal to 10.0 million km 2, and the area of ​​the USA is 9.5 million km 2, i.e. almost equal. And that means statement 4 wrong .

Answer:

1. In every twenty-fifth package of juice, according to the terms of the promotion, there is a prize under the lid. Prizes are distributed randomly. Vera buys a package of juice. Find the probability that Vera does not find the prize in her bag.

Show Solution

The solution of this problem is based on the classical formula for determining the probability:

where, m is the number of favorable outcomes of the event, and n is the total number of outcomes

We get

So the chance of Vera not finding the prize is 24/25 or

Answer:

The probability that Vera will not find the prize is 0.96

1. Establish a correspondence between functions and their graphs.

In the table, under each letter, indicate the corresponding number.

Show Solution

1. The hyperbola shown in Figure 1 is located in the second and fourth quarters, therefore, function A can correspond to this graph. Let's check: a) at х = -6, y = -(12/-6) = 2; b) at x = -2, y = -(12/-2) = 6; c) at x = 2, y = -(12/2) = -6; d) at x = 6, y = -(12/6) = -2. Q.E.D.
2. The hyperbola shown in Figure 2 is located in the first and third quarters, therefore, function B can correspond to this graph. Perform the check yourself, by analogy with the first example.
3. The hyperbola shown in Figure 3 is located in the first and third quarters, therefore, function B can correspond to this graph. Let's check: a) at x = -6, y = (12/-6) = -2; b) at x = -2, y = (12/-2) = -6; c) at x = 2, y = (12/2) = 6; d) for x = 6, y = (12/6) = 2. As required.

Answer:

A - 1; B - 2; AT 3

1. The arithmetic progression (a n) is given by the conditions:

a 1 = -9, a n+1 = a n + 4.

Find the sum of the first six terms.

Show Solution

a 1 = -9, a n+1 = a n + 4.

a n + 1 = a n + 4 ⇒ d = 4

a n = a 1 + d(n-1)

a 6 \u003d a 1 + d (n-1) \u003d -9 + 4 (6 - 1) \u003d -9 + 20 \u003d 11

S 6 \u003d (a 1 + a 6) ∙ 6 / 2

S 6 \u003d (a 1 + a 6) ∙ 3

S 6 \u003d (–9 + 11) ∙ 3 \u003d 6

Answer:

1. Find the value of an expression

Show Solution

We open the brackets. Don't forget that the first parenthesis is the square of the sum.

Answer:

1. The area of ​​a quadrilateral can be calculated using the formula

where d 1 and d 2 are the lengths of the diagonals of the quadrilateral, a is the angle between the diagonals. Using this formula, find the length of the diagonal d 2 if

Show Solution

Remember the rule, if we have a three-story fraction, then the lower value is transferred to the top

Answer:

1. Specify the solution of the inequality

Show Solution

To solve this inequality, you need to do the following:

a) we move the term 3x to the left side of the inequality, and 6 to the right side, not forgetting to change the signs to the opposite. We get:

b) Multiply both sides of the inequality by a negative number -1 and change the inequality sign to the opposite.

c) find the value of x

d) the set of solutions to this inequality will be a numerical interval from 1.3 to +∞, which corresponds to the answer 3)

Answer:
3

Module “Geometry”

1. A 17 m long fire escape was attached to the window of the sixth floor of the house. The lower end of the ladder stands 8 m from the wall. How high is the window? Give your answer in meters.

Show Solution

In the figure, we see an ordinary right-angled triangle consisting of a hypotenuse (ladder) and two legs (the wall of the house and the ground. To find the length of the leg, we use the Pythagorean theorem:

AT right triangle the square of the hypotenuse is equal to the sum of the squares of the legs c 2 \u003d a 2 + b 2

So, the window is located at a height of 15 meters

Answer:

1. In the triangle ∆ ABC it is known that AB= 8, BC = 10, AC = 14. Find cos∠ABC

Show Solution

To solve this problem, you need to use the cosine theorem. The square of the side of a triangle is equal to the sum of the squares of the other 2 sides minus twice the product of these sides by the cosine of the angle between them:

a 2 = b 2 + c 2 – 2 bc cosα

AC² \u003d AB² + BC² - 2 AB BC cos∠ABC
14² = 8² + 10² - 2 8 10 cos∠ABC
196 = 64 + 100 – 160 cos∠ABC

160 cos∠ABC = 164 – 196
160 cos∠ABC = – 32
cos∠ABC = – 32 / 160 = -0.2

Answer:

cos∠ABC = -0.2

1. On a circle centered at a point O points are marked A and B so that ∠AOB = 15 o. Lesser arc length AB is 48. Find the length of the larger arc AB.

Show Solution

We know that a circle is 360 o. Based on this, 15 about is:

360 o / 15 o \u003d 24 - the number of segments in a circle of 15 o

So, 15 o make up 1/24 of the entire circle, which means the rest of the circle:

those. remaining 345 o (360 o - 15 o \u003d 345 o) make up the 23rd part of the entire circle

If the length of the smaller arc AB is 48, then the length of the larger arc AB will be:

Answer:

1. in a trapeze ABCD it is known that AB = CD, ∠BDA= 35 o and ∠ bdc= 58 o. Find the angle ∠ ABD. Give your answer in degrees.

Show Solution

According to the condition of the problem, we have an isosceles trapezoid. The angles at the base of an isosceles trapezoid (upper and lower) are equal.

∠ADC = 35 + 58 = 93°
∠DAB = ∠ADC = 93°

Now consider the triangle ∆ABD as a whole. We know that the sum of the angles of a triangle is 180°. From here:

∠ABD = 180 - ∠ADB - ∠DAB = 180 - 35 - 93 = 52°.

Answer:

1. A triangle is depicted on checkered paper with a cell size of 1x1. Find its area.

Show Solution

The area of ​​a triangle is equal to the product of half the base of the triangle (a) and its height (h):

a - the length of the base of the triangle

h is the height of the triangle.

From the figure, we see that the base of the triangle is 6 (cells), and the height is 3 (cells). Based on what we get:

Answer:

1. Which of the following statements is correct?

1. The area of ​​a rhombus is equal to the product of its two adjacent sides and the sine of the angle between them.
2. Each of the bisectors isosceles triangle is its median.
3. The sum of the angles of any triangle is 360 degrees.

In response, write down the number of the selected statement.

Show Solution

This task is not a task. The questions listed here must be known by heart and be able to answer them.

1. This statement is absolutely right.
2. Wrong, because according to the properties of an isosceles triangle, it can have only one median - this is the bisector drawn to the base. It is also the height of the triangle.
3. Wrong because the sum of the angles of any triangle is 180°.

Answer:

Part 2

Module “Algebra”

1. Solve the Equation

Show Solution

Let's move the expression √6-x from the right side to the left

We reduce both expressions √6-x

Move 28 to the left side of the equation

Before us is the usual quadratic equation.

The range of acceptable values ​​in this case is: 6 – x ≥ 0 ⇒ x ≤ 6

To solve the equation, you need to find the discriminant:

D \u003d 9 + 112 \u003d 121 \u003d 11 2

x 1 \u003d (3 + 11) / 2 \u003d 14/2 \u003d 7 - is not a solution

x 2 \u003d (3 - 11) / 2 \u003d -8 / 2 \u003d -4

Answer:

1. The ship passes along the river to the destination for 210 km and after parking returns to the point of departure. Find the speed of the ship in still water, if the speed of the current is 4 km / h, the parking lasts 9 hours, and the ship returns to the point of departure 27 hours after leaving it.

Show Solution

x is the own speed of the ship, then

x + 4 - speed of the ship downstream

x - 4 - the speed of the ship against the current

27 - 9 = 18 (h) - the time of the ship's movement from the point of departure to the point of destination and back, excluding parking

210 * 2 \u003d 420 (km) - the total distance traveled by the ship

Based on the above, we get the equation:

reduce to a common denominator and solve:

To further solve the equation, you need to find the discriminant:

The figure above shows two graphs corresponding to the presented functions:

y = x 2 + 4x +4 (red line plot)

y = -45/x (graph depicted by the blue line)

Consider both functions:

1. y=x 2 +4x+4 on the interval [–5;+∞) is quadratic function, the graph is a parabola, and = 1 > 0 - the branches are directed upwards. If we reduce it according to the formula of the square of the sum of two numbers, we will get: y \u003d (x + 2) 2 - the graph shifts to the left by 2 units, which can be seen from the graph.
2. y \u003d -45 / x is inverse proportionality, the graph is a hyperbola, the branches are located in the 2nd and 4th quarters.

The graph clearly shows that the line y=m has one common point with the graph at m=0 and m > 9 and two common points at m=9, i.e. answer: m=0 and m≥9, check:
One common point at the top of the parabola y = x 2 + 4x +4

x 0 \u003d -b / 2a \u003d -4 / 2 \u003d -2

y 0 \u003d -2 2 + 4 (-2) + 4 \u003d 4 - 8 +4 \u003d 0 ⇒ c \u003d 0

Two common points at x \u003d - 5; y = 9 ⇒ c = 9

Answer:

1. Segments AB and CD are chords of the circle. Find the length of the chord CD, if AB = 24, and the distance from the center of the circle to the chords AB and CD are 16 and 12, respectively.

Show Solution

Triangles ∆AOB and ∆COD are isosceles.

AK=BK=AB/2=24/2=12

The segments OK and OM are heights and medians.

By the Pythagorean theorem: the square of the hypotenuse is equal to the sum of the squares of the legs, we have

OB 2 = OK 2 + BK 2

OB 2 = 16 2 + 12 2 = 256 + 144 = 400

Given that OB is the radius, we have:

OB=OA=OC=OD=20

From the triangle ∆COM, according to the Pythagorean theorem, we obtain:

CM 2 = OC 2 - OM 2

CM 2 = 20 2 - 12 2 = 400 - 144 = 256

CD=CM*2=16*2=32

The chord length CD is 32.

Answer:

1. in a trapeze ABCD with grounds AD and BC diagonals intersect at point O. Prove that the areas of triangles ∆ AOB and ∆ COD equal

Show Solution

Let AD be the bottom base of the trapezoid and BC the top, then AD>BC.

Find the areas of triangles ∆ABD and ∆DCA:

S ∆ABD = 1/2 AD ∙ h1

S ∆DCA = 1/2 AD ∙ h2

Given that the size of the AD base and the height of both triangles are the same, we conclude that the areas of these triangles are equal:

S ∆ABD = S ∆DCA

Each of the triangles ∆ABD and ∆DCA consists of two other triangles:

S ∆ABO + S ∆AOD = S ∆ABD

S ∆DCO + S ∆AOD = S ∆DCA

If the areas of triangles S ∆ABD and S ∆DCA are equal, then the sum of the areas of their interior triangles is also equal. From here we get:

S ∆ABO + S ∆AOD = S ∆DCO + S ∆AOD

in this equality, the same triangle appears on both sides - S ∆AOD, which allows us to reduce it. We get the following equality:

S ∆ABO = S ∆DCO

Q.E.D.

Answer:

S ∆ABO = S ∆DCO

1. on the side BC acute triangle ABC how a semicircle is constructed on the diameter that intersects the height AD at the point M, AD = 9, MD=6, H- the point of intersection of the heights of the triangle ABC. Find AH.

Show Solution

To begin with, let's draw a triangle and a semicircle, as stated in the condition of the problem (Fig. 1).

We mark the point of intersection of the circle with the AC side with the letter F (Fig. 2)

BF is the height of the triangle ∆ABC, since for a circle ∠BFC is the inscribed angle that is supported by the 180° arc (BC is the diameter), therefore:

∠BFC=180°/2=90°

According to the “two secant” theorem, we have: AF * AC = AM * AK

Now consider the chord MK.

Segment BC is the perpendicular to segment MK passing through the center of the circle, so BC is the perpendicular bisector.

This means that BC bisects the chord MK, i.e. MD = KD = 6 (see problem statement)

Consider the triangles ∆AHF and ∆ACD.

The angle ∠DAC is common for both triangles.

And the angles ∠AFH and ∠ADC are equal, moreover, they are right angles.

Therefore, according to the first criterion for the similarity of triangles, these triangles are similar.

From here, by definition of similarity, we can write: AC / AH = AD / AF => AC * AF = AD * AH

Earlier, we considered the equality (by the two-secant theorem) AF * AC = AM * AK, from which we obtain

AM * AK = AD * AH

AH = (AM * AK) / AD

From the figure we find:

AM = AD - MD = 9 - 6 = 3

AK \u003d AD + KD \u003d 9 + 6 \u003d 15

AH = 3 * 15 / 9 = 45 / 9 = 5

Answer: AH = 5

All of us at one time went through the final examination tests, just like the future graduates of the 9th grade experienced and prepared for the exams in advance. Of course, modern youth, boys and girls of 2018 graduates, who after the ninth grade go to study at a technical school or college, have completely different technical and informational opportunities for preparing for future exams than their parents had.

Free access to all kinds of handy material in electronic form, which allows you to calmly prepare for final examinations, gives an undeniable advantage and great opportunities, well and systematically, and most importantly, in advance, to prepare for exams for future graduates, students of 9 classes of Russian schools.

Many of you will go to college or technical school in 2018 after you successfully pass the exam tests at the school called OGE (compulsory state exams). A considerable number of you expect to study in technical schools and colleges at the expense of state funding, in other words, for free, but not everyone is guaranteed such a place "under the sun", only the best of you can count on it.

We will talk about how to get state funding for education in colleges and technical schools later, further in the text, for now, let's get down to our main topic, which concerns final exams, namely mathematics. We will offer options for solving tasks in mathematics of the OGE in 2018 and 36 options with Yashchenko's answers for grade 9, you will learn about scores and exams in mandatory state tests, transfer them on a scale to grades, you can learn something else and perhaps understand ...

OGE scores in mathematics 2018 - translation into grades

Before proceeding to the presentation of 36 options for solving the OGE 2018 mathematics according to Yashchenko, we suggest that you familiarize yourself with the scale for translating examination scores into grades, including graduation mathematics, you can transfer from the score you received in the exam to the usual mark (grade).

0-7 points in OGE mathematics - grade "2"

8-14 points in OGE mathematics - grade "3"

15-21 points in OGE mathematics - grade "4"

22-32 points in OGE mathematics - grade "5"

Do not forget that passing the final exams in the 9th grade, you first of all confirm that you have mastered the program schooling up to the ninth grade, that is, you have successfully completed the basic general education and this is very important for you. This gives you the opportunity to calmly prepare for the entrance examinations to a technical school or college, so it is extremely important to pass the exams on the first try.

Yashchenko mathematics OGE 2018 - 36 solutions

Below the link, you can familiarize yourself with the tests, solutions and answers of tasks in mathematics of the OGE 2018 from Russian mathematician Yaschenko, test your knowledge online. This program for solving mathematics in the compulsory state exams is an excellent tool to test yourself and your knowledge, to test the level of your preparation in this examination school subject.

OGE in mathematics according to YASHCHENKO 2018 below ...

self-edu.ru/oge2018_36.php

The world, as they say, does not stand still, everything around is changing, so there have been changes in the examination tests of Russian schoolchildren, graduates of grade 9, which we will now talk about. They mostly already happened in the last school season, but we will remind you of them, refresh the memory, so to speak, of those who have forgotten about them.

Changes in the OGE in mathematics 2018

To date, there is no new information about the innovations of 2018 in the OGE or innovations, all those changes that occurred earlier, including in the last academic season, remain relevant today. We will tell you about the mandatory state exams, the changes that took place last year:

The change in the OGE of 2018 affected the subject of literature, where the criteria for assessing tasks that provide for detailed answers have been changed; in future exams they will be similar to the assessment criteria for the USE, due to which primary score(maximum) increased from 23 to 29, we remind you that this applies to the subject "Literature";

In the next academic season, both compulsory final exams and those that the student chooses at his own discretion will be taken into account in aggregate, as a result, the following will turn out - in order to receive a final certificate, you need to pass all four subjects for at least minimum grades (points);

In the next academic season, it will be possible to make three attempts in order to eventually pass the OGE;

Unlike the final exam for students in grade 11, the exam scores for the exam, as you know, are translated into grades. Those of you who are graduating from grade 9 in 2018 will have to take three modules in the subject of mathematics at once. In total, the graduate will be able to "earn" 36 points for completing the examination paper. Including for the module in the subject "Algebra" - 17 points, the subject "Geometry" - 11 points, and for the subject "Real Mathematics" - 8 points.

The recommended minimum threshold for mathematics at final examinations will be a minimum of 8 points, but on condition that for each of the modules ("Algebra", Geometry" and "Real Mathematics"), the student will score at least 2 points.

The main thing for each of you is to score a minimum of 2 points, to reach this threshold. In the event that the grade for the mathematics exam is below the annual grade, then only the annual grade is taken into account when it is set. If the examination grade for mathematics is higher than the annual grade, then both of them are taken into account when placing the final grade in the certificate.

OGE Mathematics Exams 2018

After you finish grade 9 and successfully pass the final exams in all school subjects, not only in compulsory Russian and mathematics, but also in two others of your choice, you will, of course, enter some kind of technical school or college, at least a significant proportion of those who graduate from school in 2018 and will not be transferred to the 10th grade.

Many of you have decided where you will go to study, which technical school or college, of which there are a huge number in Russia and they are available in almost all even the smallest towns of our vast country, and therefore go somewhere far from home to get a special education not particularly necessary.

Of course, many of you are counting on free education in the new academic season of 2018, receiving special education in a college or technical school at the expense of state funding, and of course some of you will achieve such an opportunity, get a coveted place, but not all of this is for sure.

To be among the elect, those who receive budget place, you need not only to have high graduation scores in exam tests, but also to pass the entrance exams better than your competitors in the college or technical school where you intend to go to study, get your future profession in your chosen specialty.

Do not forget that among the first applicants for such places, who will surely get a budget place in a college or technical school, these will be school medalists, as well as prize-winners and winners school olympiads in various subjects, of the most varied caliber, from a national scale to republican, regional and regional ones.

Therefore, it is simply extremely important for you to get high passing scores for the OGE in 2018 in order to have a chance to take such a place, including successfully passing entrance tests. Thanks to modern opportunities, you can prepare for the upcoming exams in advance and without leaving their home, as they say, it would be only your desire, but there are plenty of opportunities.

Where to go to study? Choose below...

Where to go to study after grade 9 in 2018?

The choice of where to go to study is faced by many future graduates of Russian schools, and an important question is which profession to choose, which specialty to enter a college or technical school in the next academic season of 2018, which one to give preference to.

Of course, most have already decided who they will study, many have a cherished dream, for example, to become an elementary school teacher, a teacher of physical education, music or history, someone has long dreamed of learning the profession of a paramedic or pharmacist, and someone sees himself as an agronomist or a veterinarian, and someone likes technical, architectural or construction professions, such as an auto mechanic, civil engineer, designer, and so on.

For those who have not decided on the choice of profession, are in thought, which profession to give preference to, which specialty to choose, we will facilitate their task. Next, we will introduce you to almost all the technical and humanitarian specialties, as well as the subjects studied by the college and technical schools of 2018, in which you can enter and get the profession of your choice.

In conclusion, I would like to say this - do not forget that the main thing in teaching is not knowledge, but whether you will be able to put them into practice, that is, put them into action, the practical plane of application. Do not forget that you need to successfully pass mathematics in the OGE in 2018, which will help you with tasks with solutions and answers by mathematician Yashchenko, and also remember that the guarantee of free education in colleges and technical schools is high passing scores and successful entrance exams. tests.

Prepare in advance, as for the OGE in mathematics and other exam subjects in 2018, in order to get high passing scores and become one of the contenders for a budget place in a college or technical school. Good luck to all Russian schoolchildren, 9th grade graduates on exams in mathematics, Russian, physics, chemistry, history and other subjects - let your desires turn into reality, thanks to your own efforts!

Not all MATHEMATICS , remember this and it's not a problem!

What else I would like to say is that the mathematical mindset is far from all of us, someone is more developed logical thinking, creative or artistic direction, therefore, probably those of you who are not inclined to such science should not choose a profession related to the exact sciences, calculations, numbers, etc.

Before choosing a profession for yourself, a specialty for which to enter a college or technical school in 2018, you should understand yourself, pass some tests in order to understand your true capabilities, what you are more inclined to, because having chosen a specialty today and lowering learn from it tomorrow, making a mistake can spoil a lot in your life.

Do not forget that your favorite work is just a vital necessity for a person, when you go to it with pleasure and do it with the same feeling, which means you have an incentive to grow further, to improve. It is extremely important for any person to be able to do in their professional life what the soul lies in, remember this, think about it and make the right decision on choosing future profession- peace to you and success in life!

OGE matematika 9 class 2018 ekzameni Yaschenko 36 variantov

OGE mathematics 9 class 2018, exams Yashchenko 36 options

Series “USE. FIPI - School" was prepared by the developers of control measuring materials (KIM) of the unified state exam. The collection contains:
36 standard exam options compiled in accordance with the draft demo version of the KIM USE in mathematics of the profile level in 2018;
instructions for performing the examination work;
answers to all tasks;
solutions and criteria for assessing tasks 13-19.
Completing the tasks of standard examination options provides students with the opportunity to independently prepare for the state final certification, as well as to objectively assess the level of their preparation.
Teachers can use standard examination options to organize the control of the results of mastering the educational programs of secondary general education by schoolchildren and intensive preparation of students for the Unified State Examination.

Examples.
The distance between piers A and B is 77 km. From A to B, a raft set off along the river, and after 1 hour a motor boat set off after it, which, having arrived at point B, immediately turned back and returned to A. By this time, the raft had sailed 40 km. Find the speed of the motorboat in still water if the speed of the river is 4 km/h. Give your answer in km/h.

There are 35 different things written on the board. natural numbers, each of which is either even or its decimal notation ends with the number 3. The sum of the written numbers is 1062.
a) Can there be exactly 27 even numbers on the board?
b) Can exactly two numbers on the board end in 3?
c) What is the smallest number of numbers ending in 3 that can be on the board?

Free download e-book in a convenient format, watch and read:
Download the book USE, Mathematics, Profile level, Typical exam options, 36 options, Yashchenko I.V., 2018 - fileskachat.com, fast and free download.

• I will pass the Unified State Examination, Mathematics, Self-study course, Problem solving technology, Profile level, Part 3, Geometry, Yashchenko I.V., Shestakov S.A., 2018
• I will pass the Unified State Examination, Mathematics, Self-study course, Problem solving technology, Profile level, Part 2, Algebra and the beginning of mathematical analysis, Yashchenko I.V., Shestakov S.A., 2018
• I will pass the Unified State Examination, Mathematics, Self-study course, Problem solving technology, Basic level, Part 3, Geometry, Yashchenko I.V., Shestakov S.A., 2018
• I will pass the exam, Mathematics, Profile level, Part 3, Geometry, Yashchenko I.V., Shestakov S.A., 2018

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