Completed by: Shatny A.I.
Group RK5-42
Moscow 2004
Option 121c. The task:
Steel 40ХНМА (40ХН2МА) is used for the manufacture of crankshafts, connecting rods, gears, critical bolts and other loaded parts of complex configuration.
Indicate the optimal mode of heat treatment of the shaft d \u003d 40mm, made of steel 40ХНМА (40ХН2МА), build a graph t () for this steel.
Describe the structural transformations that occur during heat treatment.
Provide basic information about steel: GOST, chemical composition, properties, requirements for improved steels, advantages, disadvantages, the effect of alloying elements on the hardenability and toughness of steel.
Optimal shaft heat treatment d \u003d 40mm.
Hardening 850 С, oil. Vacation 620С, hardening with high frequency current.
Quenching is heat treatment, as a result of which a nonequilibrium structure is formed in the alloy. Structural and tool steels are hardened for hardening.
After quenching for martensite and high tempering, the properties of alloy steels are determined by the concentration of carbon in the martensite. The higher it is, the greater the hardness and strength, the lower the impact strength. Alloyed elements affect the mechanical properties indirectly by increasing or decreasing the carbon concentration in the martensite. Carbide-forming elements (Cr, Mo, W, V) increase the bond strength of carbon atoms with atoms of a solid solution, reduce the thermodynamic activity (mobility) of carbon atoms, and contribute to an increase in its concentration in martensite, i.e. hardening. Thus, the task of hardening is to obtain a martensite structure with a maximum percentage of carbon.
Consider hardening 40xnma (40xn2ma).
Critical temperatures for 40ХНМА (40ХН2МА):
A c3 \u003d 820C
A c1 \u003d 730C
When heated to a temperature of 730 ° C, the structure of the alloy remains constant - perlite.As soon as the point A c1 is passed, austenite begins to nucleate at the grain boundaries of pearlite. In our case, we have full hardening, because the temperature exceeds A c3, then all the pearlite passes into austenite. Thus, heating up to 820 ° C, we obtained a single-phase structure \u003d austenite, while the temperature rises above 800 ° C, the grain grows.
To obtain a martensitic structure, it is necessary to supercool austenite to the temperature of martensitic transformation; therefore, the cooling rate must exceed the critical one. Such cooling is most simply carried out by immersing the part to be hardened into a liquid medium (water or oil) with a temperature of 20-25 ° C. As a result of such processing, a heat-resistant martensite, with some residual austenite.
Vacation at 620С for 1.5 hours in water.
Tempering is heat treatment, as a result of which phase transformations occur in pre-hardened steels, bringing their structure closer to equilibrium.
40ХНМА (40ХН2МА)subject to vacation at t \u003d 620С - high vacation. It should be borne in mind that at tempering temperatures of more than 500 ° C, cooling is carried out in water.
At high heating in carbon steels, structural changes occur that are not associated with phase transformations: the shape, size carbidesand structure ferrite... Is happening coagulation: cementite crystals enlarge and approach a spherical shape. Changes in the structure of ferrite are detected starting from a temperature of 400 ° C: the density of dislocations decreases, the boundaries between lamellar crystals of ferrite are eliminated (their shape approaches equiaxial).
So, the phase hardening that has arisen during the martensitic transformation is removed. The ferrite-carbide mixture that forms after such tempering is called sorbitol release.
After that, hardening with a high frequency current (HFC) - surface hardening: at a high current frequency, the current density in the outer layers of the conductor is many times greater than in the core. As a result, almost all of the thermal energy is released on the surface and heats the surface layer to the hardening temperature. Cooling is carried out with water supplied through a sprayer.
In this case, the surface layers are hardened, significant compressive stresses arise in them.
The chart shows the distribution of copper smelting in the countries of the world (in thousand tons) for 2006. Among the countries represented, the first place in copper smelting was occupied by the USA, the tenth place - by Kazakhstan. Where did Indonesia rank?
DecisionTask 2. Option 255 Larina. USE 2019 in mathematics.
A parallelogram is shown on the coordinate plane. Find its area.
Decision Task 3. Option 255 Larina. USE 2019 in mathematics.
During the psychological test, the psychologist asks each of the two subjects A. and B. to choose one of three numbers: 1, 2 or 3. Assuming that all combinations are equally possible, find the probability that A. and B. have chosen different numbers. Round the result to the nearest hundredth
DecisionTask 4. Option 255 Larina. USE 2019 in mathematics.
Solve the equation
Decision
... If your equation has more than one root, write down the smallest root in your answer.Task 5. Option 255 Larina. USE 2019 in mathematics.
In the picture, angle 1 is 46 °; angle 2 is 30 °; angle 3 is 44 °. Find angle 4. Give your answer in degrees.
Decision Task 6. Option 255 Larina. USE 2019 in mathematics.
The figure shows the graph of the function f (x). The tangent to this graph, drawn at a point with an abscissa of −4, passes through the origin. Find f` (-4).
Decision Task 7. Option 255 Larina. USE 2019 in mathematics.
Find the square of the distance between the vertices D and C2 of the polyhedron shown in the figure. All dihedral angles of a polyhedron are straight.
Decision Task 8. Option 255 Larina. USE 2019 in mathematics.
Find the meaning of the expression
Decision Task 9. Option 255 Larina. USE 2019 in mathematics.
It is planned to use a cylindrical column to support the canopy. The pressure P (in pascals) exerted by the canopy and the column on the support is determined by the formula, where m \u003d 1200 kg is the total mass of the canopy and the column, D is the column diameter (in meters). Considering the acceleration of gravity g \u003d 10 m s /, and pi \u003d 3, determine the smallest possible diameter of the column if the pressure applied to the support should not exceed 400,000 Pa. Express your answer in meters
DecisionTask 10. Option 255 Larina. USE 2019 in mathematics.
Igor and Pasha can paint a fence in hours. Pasha and Volodya can paint the same fence in 12 hours, and Volodya and Igor in hours. How many hours will the boys take to paint the fence as the three of them work?
DecisionTask 11. Option 255 Larina. USE 2019 in mathematics.
Find the largest function value
Decision
on the segment [-9; -1]Task 12. Option 255 Larina. USE 2019 in mathematics.
a) Solve the equation
Decision
b) Indicate the roots of this equation belonging to the interval (-pi / 3; 2pi]Task 13. Option 255 Larina. USE 2019 in mathematics.
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Decision Task 14. Option 255 Larina. USE 2019 in mathematics.
Solve inequality
Decision Task 15. Option 255 Larina. USE 2019 in mathematics.
Given a triangle ABC, in which AB \u003d BC \u003d 5, the median. On the bisector CE, a point F is chosen such that CE \u003d 5CF. A straight line l is drawn through point F, parallel to BC. A) Find the distance from the center of the circle circumscribed about the triangle ABC to the straight line l B) Find in what ratio the straight line l divides the area of \u200b\u200bthe triangle ABC
DecisionTask 16. Option 255 Larina. USE 2019 in mathematics.
On January 15, it is planned to take out a bank loan for 9 months. The conditions for its return are as follows: - on the 1st day of each month, the debt increases by 4% compared to the end of the previous month; - from the 2nd to the 14th day of each month, it is necessary to pay part of the debt; - On the 15th day of each month, the debt must be the same amount less than the debt on the 15th day of the previous month. It is known that in the fifth month of crediting it is necessary to pay 44 thousand rubles. How much should be returned to the bank during the entire loan term?
DecisionTask 17. Option 255 Larina. USE 2019 in mathematics.
For what values \u200b\u200bof the parameter a the system
Decision
has the only solutionTask 18. Option 255 Larina. USE 2019 in mathematics.
In a sequence of natural numbers a1 \u003d 47, each next term is equal to the product of the sum of the digits of the previous term and a1 A) Find the fifth term of the sequence B) Find the 50th term of the sequence C) Calculate the sum of the first fifty terms of this sequence ..
The Novosibirsk-Krasnoyarsk train leaves at 15:20 and arrives at 4:20 the next day (Moscow time). How many hours does the train take?
DecisionTask 1. Option 255 Larina. USE 2019 in mathematics.
Aristarkh Lukov-Arbaletov takes a walk from point A along the paths of the park. At each fork, he randomly chooses the next path without going back. The track diagram is shown in the figure. Some of the routes lead to the village S, others to the field F or to the swamp M. Find the probability that Aristarchus wanders into the swamp. Round the result to the nearest hundredth.
Answer: 0.42.$$ \\ frac (1) (2) \\ cdot \\ frac (2) (4) + \\ frac (1) (2) \\ cdot \\ frac (1) (3) \u003d \\ frac (1) (4) + \\ Task 5. Training version of the exam number 221 Larina.
Solve the equation: $$ \\ sqrt (10-3x) \u003d x-2 $$
If your equation has more than one root, indicate the smaller one in your answer.
Answer: 3.
ODZ: $$ \\ left \\ (\\ begin (matrix) 10-3x \\ geq0 \\\\ x-2 \\ geq0 \\ end (matrix) \\ right. $$ $$ \\ Leftrightarrow $$$$ \\ left \\ (\\ begin (matrix) x \\ leq \\ frac (10) (3) \\\\ x \\ geq2 \\ end (matrix) \\ right. $$ \\ Leftrightarrow $$
$$ 10-3x \u003d x ^ (2) -4x + 4 $$
$$ \\ left \\ (\\ begin (matrix) x_ (1) + x_ (2) \u003d 1 \\\\ x_ (1) \\ cdot x_ (2) \u003d - 6 \\ end (matrix) \\ right. $$ $$ \\ $$ \\ left \\ (\\ begin (matrix) x_ (1) \u003d 3 \\\\ x_ (2) \u003d - 2 \\ end (matrix) \\ right. $$
$$ - 2 \\ notin $$ ODZ $$ \\ Rightarrow $$ 3 - root
Task 6. Training version of the exam number 221 Larina.
{!LANG-c7650bb106bc1b085d78be42bbcc98f3!}
{!LANG-8914eb80eebb6e097a920eaceb5eb77b!}
The quadrilateral ABCD is inscribed in a circle, with BC \u003d CD. The ADC is known to be 93 °. Find at what sharp angle the diagonals of this quadrilateral intersect. Give your answer in degrees.
1) $$ \\ bigtriangleup AOD \\ sim \\ bigtriangleup COB $$ $$ \\ Rightarrow $$
$$ \\ angle ADO \u003d \\ angle OCB \u003d \\ alpha $$
$$ \\ angle DAO \u003d \\ angle OBC \u003d \\ beta $$
2) $$ \\ bigtriangleup DOC \\ sim \\ bigtriangleup AOB $$ $$ \\ Rightarrow $$
$$ \\ bigtriangleup DCB $$ - isosceles
$$ \\ angle COB \u003d \\ angle DCB \u003d \\ beta $$ $$ \\ Rightarrow $$ $$ \\ alpha + \\ beta \u003d 93 ^ (\\ circ) $$
$$ \\ angle AOD \u003d 180 ^ (\\ circ) - \\ alpha- \\ beta \u003d 87 ^ (\\ circ) $$
Task 8. Training version of the exam number 221 Larina.
In a regular triangular prism $$ ABCA_ (1) B_ (1) C_ (1) $$, the sides of the bases of which are 2, the side edges are 1, cut through the vertices $$ ABC_ (1) $$. Find its area.
1) According to Pythagoras: $$ AC_ (1) \u003d \\ sqrt (AA_ (1) ^ (2) + A_ (1) C_ (1) ^ (2)) \u003d \\ sqrt (5) $$
$$ AC_ (1) \u003d BC_ (1) $$
2) Construct $$ C_ (1) H \\ perp AB $$, $$ C_ (1) H $$ - median, height $$ \\ Rightarrow $$
$$ C_ (1) H \u003d \\ sqrt (C_ (1) B ^ (2) -HB ^ (2)) \u003d \\ sqrt (5-1) \u003d 2 $$
3) $$ S_ (AC_ (1) B) \u003d \\ frac (1) (2) \\ cdot C_ (1) H \\ cdot AB \u003d \\ frac (1) (2) \\ cdot2 \\ cdot2 \u003d 2 $$
Task 9. Training version of the exam number 221 Larina.
Find the value of the expression: $$ \\ frac (b ^ (3) \\ cdot \\ sqrt (b)) (\\ sqrt (b) \\ cdot \\ sqrt (b)) $$ for $$ b \u003d 4 $$
Answer: 64.$$ \\ frac (b ^ (3) \\ cdot \\ sqrt (b)) (\\ sqrt (b) \\ cdot \\ sqrt (b)) \u003d $$
$$ \u003d \\ frac (b ^ (3) \\ cdot b ^ (\\ frac (1) (12))) (b \\ frac (1) (21) \\ cdot b \\ frac (1) (28)) \u003d $ $
$$ \u003d b ^ (3+ \\ frac (1) (12) - \\ frac (1) (21) - \\ frac (1) (28)) \u003d $$
$$ \u003d b ^ (3) \u003d 4 ^ (3) \u003d 64 $$
Task 10. Training version of the exam number 221 Larina.
The stone throwing machine shoots stones at a certain acute angle to the horizon with a fixed initial speed. The flight path of a stone in the coordinate system associated with the machine is described by the formula $$ y \u003d ax ^ (2) + bx $$, $$ a \u003d - \\ frac (1) (25) $$, $$ b \u003d \\ frac ( 7) (5) $$ constant parameters, x (m) is the horizontal displacement of the stone, y (m) is the height of the stone above the ground. At what is the greatest distance (in meters) from the fortress wall with a height of 9 m should the machine be positioned so that the stones fly over the wall at a height of at least 1 meter?
Answer: 25.$$ - \\ frac (1) (25) x ^ (2) + \\ frac (7) (5) x \u003d 10 | \\ cdot25 $$
$$ 250 + x ^ (2) -35x \u003d 0 $$
$$ \\ left \\ (\\ begin (matrix) x_ (1) + x_ (2) \u003d 35 \\\\ x_ (1) \\ cdot x_ (2) \u003d 250 \\ end (matrix) \\ right. $$ $$ \\ Leftrightarrow $$
$$ \\ left \\ (\\ begin (matrix) x_ (1) \u003d 25 \\\\ x_ (2) \u003d 10 \\ end (matrix) \\ right. $$
Task 11. Training version of the exam number 221 Larina.
Two cars drove out of cities A and B towards each other at the same time at constant speeds. The speed of the first car was twice the speed of the second. The second car arrived at A 1 hour later than the first arrived at B. How many minutes earlier would the cars meet if the second car was traveling at the same speed as the first?
Answer: 10.Let $$ 2x-v_ (1) $$; $$ x-v_ (2) $$; $$ S_ (AB) \u003d 1 $$
$$ \\ frac (1) (x) - \\ frac (1) (2x) \u003d 1 $$ $$ \\ Leftrightarrow $$
$$ \\ frac (1) (2x) \u003d 1 $$ $$ \\ Leftrightarrow x \u003d 0.5 $$
Let $$ t_ (1) $$ be the meeting time in the first case:
$$ t_ (1) \u003d \\ frac (1) (0.5 + 2 \\ cdot0.5) \u003d \\ frac (1) (1.5) \u003d \\ frac (2) (3) $$
Let $$ t_ (2) $$ be in the second:
$$ t_ (2) \u003d \\ frac (1) (2 \\ cdot0.5 + 2 \\ cdot0.5) \u003d \\ frac (1) (2) $$
$$ t_ (1) -t_ (2) \u003d \\ frac (2) (3) - \\ frac (1) (2) \u003d \\ frac (1) (6) $$ (h) - difference
$$ \\ frac (1) (6) \\ cdot60 \u003d 10 $$ minutes
Task 12. Training version of the exam number 221 Larina.
Find the smallest value of the function $$ y \u003d \\ frac (x ^ (2) -6x + 36) (x) $$ on the segment $$$$
Answer: 6.$$ y "\u003d \\ frac ((2x-6) x-x ^ (2) + 6x-36) (x ^ (2)) \u003d $$
$$ \u003d \\ frac (2x ^ (2) -6x-x ^ (2) + 6x-36) (x ^ (2)) \u003d $$
$$ \u003d \\ frac (x ^ (2) -36) (x ^ (2)) $$
$$ f_ (min) \u003d f (6) \u003d \\ frac (6 ^ (2) -6 \\ cdot6 + 36) (6) \u003d 6 $$
Task 13. Training version of the exam number 221 Larina.
a) Solve the equation: $$ 7 \\ sin (2x- \\ frac (5 \\ pi) (2)) + 9 \\ cos x + 1 \u003d 0 $$
b) Indicate the roots of this equation that belong to the segment $$ [- \\ frac (3 \\ pi) (2); \\ frac (\\ pi) (3)] $$
Answer: a) $$ \\ pm \\ frac (2 \\ pi) (3) +2 \\ pi n, n \\ in Z $$ b) $$ - \\ frac (4 \\ pi) (3) $$; $$ - \\ frac (2 \\ pi) (3) $$.$$ 7 \\ sin (2x- \\ frac (5 \\ pi) (2)) + 9 \\ cos x + 1 \u003d 0 $$
$$ - 7 \\ sin (\\ frac (5 \\ pi-2x) (2)) + 9 \\ cos x + 1 \u003d 0 $$
$$ - 7 \\ cos2x + 9 \\ cos x + 1 \u003d 0 $$
$$ - 7 (2 \\ cos ^ (2) x-1) +9 \\ cos x + 1 \u003d 0 $$
$$ - 14 \\ cos ^ (2) x + 7 + 9 \\ cos x + 1 \u003d 0 $$
$$ 14 \\ cos ^ (2) x-9 \\ cos x-8 \u003d 0 $$
$$ D \u003d 81 + 448 \u003d 529 \u003d 23 ^ (2) $$
$$ \\ left \\ (\\ begin (matrix) \\ cos x \u003d \\ frac (9 + 23) (2 \\ cdot14) \u003d \\ frac (16) (14) \\\\\\ cos x \u003d \\ frac (9-23) ( 2 \\ cdot14) \u003d - \\ frac (1) (2) \\ end (matrix) \\ right. $$
$$ \\ Leftrightarrow $$ $$ \\ left \\ (\\ begin (matrix) \\ varnothing; | \\ cos x | \\ leq1 \\\\ x \u003d \\ pm \\ frac (2 \\ pi) (3) +2 \\ pi n, n \\ in Z \\ end (matrix) \\ right. $$
b) $$ - \\ pi- \\ frac (\\ pi) (3) \u003d - \\ frac (4 \\ pi) (3) $$
$$ - \\ pi + \\ frac (\\ pi) (3) \u003d - \\ frac (2 \\ pi) (3) $$
Task 14. Training version of the exam number 221 Larina.
The base of the pyramid DABC is a right-angled triangle ABC with right angle C. The height of the pyramid passes through the midpoint of the edge AC, and the side face of the ACD is an equilateral triangle.
a) Prove that the section of the pyramid by the plane passing through the edge BC and an arbitrary point M of the edge AD is a right-angled triangle.
b) Find the distance from the vertex D to this plane, if M is the midpoint of the edge AD, and the height of the pyramid is rank 6.
Answer: $$ 2 \\ sqrt (3) $$.
a) 1) Let $$ DH $$ be the height; $$ \\ Rightarrow DH \\ perp ABC $$
2) Let $$ MC \\ cap DH \u003d N \\ Rightarrow NH \\ perp AC $$
$$ \\ Rightarrow CH $$ - projection of $$ NC $$ on $$ (ABC) $$
3) since $$ AC \\ perp CB $$, then by the three perpendicular theorem $$ NC \\ perp CB $$
$$ \\ Rightarrow $$ $$ MC \\ perp CB $$
$$ \\ Rightarrow \\ bigtriangleup MCB $$ - rectangular
b) 1) since $$ AC \\ perp CB $$ and $$ CB \\ perp MC $$ $$ \\ Rightarrow CB \\ perp (ADC) $$
$$ \\ Rightarrow (BCM) \\ perp (ACD) $$
$$ \\ Rightarrow $$ distance from D to $$ (CBM) $$ - perpendicular $$ DL \\ in (ADC) $$
2) since $$ \\ bigtriangleup ACD $$ - equilateral and $$ AM-MD, then $$ CM \\ perp AD $$
$$ \\ Rightarrow DM $$ - required distance
3) $$ DC \u003d \\ frac (DH) (\\ sin C) \u003d \\ frac (6) (\\ sin60 ^ (\\ circ)) \u003d \\ frac (12) (\\ sqrt (3)) \u003d 4 \\ sqrt (3 ) $$
$$ \\ Rightarrow $$ $$ MD \u003d \\ frac (1) (2) AD \u003d \\ frac (1) (2) DC \u003d 2 \\ sqrt (3) $$
Task 15. Training version of the exam number 221 Larina.
Solve the inequality: $$ \\ frac (3 \\ log_ (0.5) x) (2- \\ log_ (0.5) x) \\ geq2 \\ log_ (0.5) x + 1 $$
Answer: $$ x \\ in (\\ frac (1) (4); \\ frac (1) (2)] \\ cup $$$$ \\ frac (10 + 2a + b) (3) \\ in N $$, while $$ 2a + b \\ in $$
$$ \\ Rightarrow $$ $$ 10 + 2a + b \\ in $$.
Select all multiples of 3 from this range: $$ 12; 15; 18; 21; 24; 27; 30; 33; 36 $$
1) $$ 10 + 2a + b \u003d 12 $$
$$ 2a + b \u003d 2 $$ $$ \\ Rightarrow $$ $$ a \u003d 1; b \u003d 0 $$ or $$ a \u003d 0; b \u003d 2 $$
2) $$ 10 + 2a + b \u003d 15 $$
$$ a \u003d \\ frac (5-b) (2) $$ $$ \\ Rightarrow $$ $$ a \u003d 0; b \u003d 5 $$ or $$ a \u003d 2; b \u003d 1 $$
or $$ a \u003d 2; b \u003d 1 $$
$$50505;52125;51315$$
3) $$ 10 + 2a + b \u003d 18 $$
$$ 2a + b \u003d 8 $$ $$ \\ Rightarrow $$ $$ a \u003d 4; b \u003d 0 $$
$$ a \u003d 3; b \u003d 2 $$ or $$ a \u003d 2; b \u003d 4 $$
$$ a \u003d 1; b \u003d 6 $$ or $$ a \u003d 0; b \u003d 0 $$
4) $$ 10 + 2a + b \u003d 21 $$
$$ 2a + b \u003d 11 $$ $$ \\ Rightarrow $$ $$ a \u003d 5; b \u003d 1 $$ or $$ a \u003d 4; b \u003d 3 $$
$$ a \u003d 3; b \u003d 5 $$ or $$ a \u003d 2; b \u003d 7 $$
5) $$ 10 + 2a + b \u003d 24 $$
$$ 2a + b \u003d 14 $$ $$ \\ Rightarrow $$
$$ a \u003d 7; b \u003d 0 $$ or $$ a \u003d 6; b \u003d 2 $$
$$ a \u003d 5; b \u003d 4 $$ or $$ a \u003d 4; b \u003d 6 $$
6) $$ 10 + 2a + b \u003d 27 $$
$$ 2a + b \u003d 17 $$ $$ \\ Rightarrow $$
$$ a \u003d 7; b \u003d 3 $$ or $$ a \u003d 6; b \u003d 5 $$
$$ a \u003d 5; b \u003d 7 $$ or $$ a \u003d 4; b \u003d 9 $$
7) $$ 10 + 2a + b \u003d 30 $$
$$ 2a + b \u003d 20 $$ $$ \\ Rightarrow $$
$$ a \u003d 9; b \u003d 2 $$ or $$ a \u003d 8; b \u003d 4 $$
$$ a \u003d 7; b \u003d 6 $$ or $$ a \u003d 6; b \u003d 8 $$
8) $$ 10 + 2a + b \u003d 33 $$
$$ 2a + b \u003d 23 $$ $$ \\ Rightarrow $$
$$ a \u003d 9; b \u003d 5 $$ or $$ a \u003d 8; b \u003d 7 $$
9) $$ 10 + 2a + b \u003d 36 $$
$$ 2a + b \u003d 26 $$ $$ \\ Rightarrow $$
Total: $$ 2 + 3 + 5 + 5 + 5 + 5 + 4 + 3 + 1 \u003d 33 $$ numbers
c) Taking into account point b) we get: 3 x digit numbers 3 pieces
4 x: $$ \\ frac (5aa5) (3) \u003d N $$
$$ \\ frac (10 + 2a) (3) \u003d N $$
$$ 2a \\ in $$ $$ \\ Rightarrow $$ $$ 10 + 2a \\ in $$
12: $$ 2a \u003d 2 $$ $$ \\ Rightarrow $$ $$ a \u003d 1 $$
15: $$ 2a \u003d 5 $$ $$ \\ Rightarrow $$ $$ \\ varnothing $$
18: $$ 2a \u003d 8 $$ $$ \\ Rightarrow $$ $$ a \u003d 4 $$
21: $$ 2a \u003d 11 $$ $$ \\ Rightarrow $$ $$ \\ varnothing $$
24: $$ 2a \u003d 14 $$ $$ \\ Rightarrow $$ $$ a \u003d 7 $$
27: $$ 2a \u003d 17 $$ $$ \\ Rightarrow $$ $$ \\ varnothing $$
Only 3 numbers.
That is, 3 x and 4 x digits in the amount of 6 pieces.
5 tees total 33 $$ \\ Rightarrow $$ together 39, we need 37, that is, the penultimate $$ \\ Rightarrow $$ 59295
The figure shows the profile of a diver diving to the bottom of the sea. The horizontal indicates the time in minutes, the vertical indicates the dive depth at a given time, in meters. Upon ascent, the diver stopped several times for decompression.
Decision
Determine from the figure how many times the diver has spent more than 5 minutes at the same depth.Task 2. Option 244 Larina. USE 2019 in mathematics.
The area of \u200b\u200bthe square is 10.
Find the area of \u200b\u200ba square whose vertices are the midpoints of the sides of this square.
Decision Task 3. Option 244 Larina. USE 2019 in mathematics.
In a ceramic tableware factory, 10% of the plates produced are defective. During product quality control, 80% of defective plates are detected. The rest of the plates go on sale.
Decision
Find the probability that the cymbal you randomly choose when you buy it is free from defects. Round your answer to the nearest ten thousandth.Task 4. Option 244 Larina. USE 2019 in mathematics.
Solve the equation.
Decision
In your answer, write down the largest negative root of the equation.Task 5. Option 244 Larina. USE 2019 in mathematics.
In triangle ABC, angle A is 48 °, angle C is 56 °. On the continuation of the AB side, the segment BD \u003d BC is postponed.
Find the angle D of triangle BCD.
Decision Task 6. Option 244 Larina. USE 2019 in mathematics.
The figure shows the graph of the derivative y \u003d f` (x) of the function f (x) defined on the interval (-4; 8).
At what point of the segment [-3; 1] does the function f (x) take the smallest value?
Decision Task 7. Option 244 Larina. USE 2019 in mathematics.
All edges of a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 are equal to 3
Find the area of \u200b\u200bthe side surface of the pyramid B A 1 B 1 C 1 D 1 E 1 F 1.
In the answer, indicate the received value multiplied by 18-3√7.
Decision Task 8. Option 244 Larina. USE 2019 in mathematics.
Find the meaning of the expression
Decision Task 9. Option 244 Larina. USE 2019 in mathematics.
The installation for demonstrating adiabatic compression is a vessel with a piston that sharply compresses the gas. In this case, the volume and pressure are related by the ratio pV 1.4 \u003d const, where p (atm) is the pressure in the gas, V is the volume of the gas in liters. Initially, the volume of gas is 24 liters, and its pressure is equal to one atmosphere.
Decision
To what volume must the gas be compressed in order for the pressure in the vessel to rise to 128 atmospheres? Express your answer in liters.Task 10. Option 244 Larina. USE 2019 in mathematics.
Ivan and Alexey agreed to meet in N-sk. They go to N-sku by different roads. Ivan calls Alexey and learns that he is 168 km from N-sk and is driving at a constant speed of 72 km / h. At the time of the call, Ivan is 165 km from N-ska and still has to make a 30-minute stop along the way.
Decision
How fast must Ivan go to arrive at N-sk at the same time as Alexei?Task 11. Option 244 Larina. USE 2019 in mathematics.
Find smallest function value
DecisionTask 12. Option 244 Larina. USE 2019 in mathematics.
a) Solve the equation
b) Indicate the roots of this equation belonging to the segment [-3π / 2; 0]
Decision Task 13. Option 244 Larina. USE 2019 in mathematics.
In a regular quadrangular pyramid SABCD with apex S AD \u003d 1/5 SD \u003d 1. A plane a is drawn through point B, intersecting the edge SC at point E and remote from points A and C by the same distance equal to 1/10. It is known that the plane a is not parallel to the line AC.
Decision
A) Prove that the plane a divides the edge SC in the ratio SE: EC \u003d 7: 1
B) Find the cross-sectional area of \u200b\u200bthe SABCD pyramid by plane a.Task 14. Option 244 Larina. USE 2019 in mathematics.
Solve inequality
Decision Task 15. Option 244 Larina. USE 2019 in mathematics.
The segment AD is the bisector of the right-angled triangle ABC (angle C \u003d 90 °).
Decision
A circle of radius √15 passes through points A, C, D and intersects side AB at point E so that AE: AB \u003d 3: 5. Segments CE and AD intersect at point O.
A) Prove that CO \u003d OE
B) Find the area of \u200b\u200bthe triangle ABC.Task 16. Option 244 Larina. USE 2019 in mathematics.
Oksana put a certain amount into a bank account for six months. Therefore, the deposit has a "floating" interest rate, that is, the number of accrued interest depends on the number of full months that the deposit has been on the account.
The table shows the conditions for calculating interest.Accrued interest is added to the deposit amount. At the end of each month, with the exception of the last one, Oksana, after accruing interest, adds such an amount that the contribution increases monthly by 5% of the original.
Decision
What percentage of the initial deposit is the amount accrued by the bank as interest?Task 17. Option 244 Larina. USE 2019 in mathematics.
Find all values \u200b\u200bof the parameter a, -π
has exactly three solutions.
DecisionTask 18. Option 244 Larina. USE 2019 in mathematics.
Is it possible to give an example of five different natural numbers, the product of which is 2800, and
Decision
a) five;
b) four;
at three o'clok
of them form a geometric progression?Task 19. Option 244 Larina. USE 2019 in mathematics.
The solution to option 244 of the exam in mathematics by Larin, as always, will not be simple and very interesting.
In general, many do not like Larin's options, because they are not standard, as many people think are more complex.
But in fact, Larin's options are the best teaching material and a very good example of
how one person can do the work of all institutes, ministries, etc. taken together absolutely free of charge,
Moreover, the work that the Ministry of Education does for a year, he does in a week without straining.
I strongly recommend everyone to use Larin's options to prepare for the exam in mathematics 2019.
Each option is unique and interesting in its own way, each task is aimed at making the student remember
and fixed this or that theorem.
Option 244 Larina will not be an exception, so I advise you to be ready on October 6 and
test your knowledge with option 244 of the exam in mathematics from Larin's website.
And we, in turn, will promptly provide a solution to Larin's version so that you can work on the errors.
The solution of option 244 of the exam Larina will be on our website on October 6, 2018 after publication on the website alexlarin.net
When paying for services through a payment terminal, a 9% commission is charged. The terminal accepts amounts in multiples of 10 rubles. The monthly Internet fee is 650 rubles.
What is the minimum amount to put into the terminal's receiving device so that the account of the company providing Internet services will have an amount not less than 650 rubles?
