Fipi open job bank exam basic level. Ege in mathematics
Secondary general education
Line UMK Merzlyak. Algebra and the Beginnings of Analysis (10-11) (U)
Line UMK Merzlyak. Algebra and the Beginnings of Analysis (10-11) (B)
Line UMK G.K. Muravina. Algebra and the beginnings of mathematical analysis (10-11) (deep)
Line UMK G.K. Muravina, K.S. Muravina, O.V. Muravina. Algebra and the beginnings of mathematical analysis (10-11) (basic)
USE-2018 in mathematics, a basic level of: tasks 1-18
We bring to your attention an analysis of the tasks of the USE 2018 in mathematics. The article contains a detailed algorithm for solving 1-18 assignments and recommendations for relevant manuals for preparing for the Unified State Examination, as well as a selection of previously published materials on mathematics.The edition contains 30 training options examination papers to prepare for the exam. Each option is compiled in full accordance with the requirements of the Unified State Examination, includes tasks of the basic level. The structure of options is the same. At the end of the manual, answers to all tasks are given.
Exercise 1
The train left St. Petersburg at 23:50 (Moscow time) and arrived in Moscow at 07:50 the next day. How many hours did the train travel?
Solution
Given the fact that there are 24 hours in a day, and the day begins at 00:00 and ends at 24:00, the train is on the way for 10 minutes of the previous day and 7 hours 50 minutes of the next.
7 hours 50 minutes + 10 minutes = 8 hours
Answer: 8.
The points in the figure show the average air temperature in Sochi for each month in 1920. The numbers of the months are shown horizontally; vertically - temperature in degrees Celsius. For clarity, the points are connected by a line.
How many months has the average temperature been above 18 degrees Celsius?
Solution
Answer: 4.
A triangle is depicted on checkered paper with a cell size of 1 × 1. Find its area.
Solution
| S ∆ = | 1 | ha, |
| 2 |
where h- height, a- the side to which the height is drawn.
Answer: 6.
Task 4
There are a total of 25 tickets in the biology ticket book. Only in two tickets there is a question about mushrooms. At the exam, the student gets one randomly selected ticket from this collection. Find the probability that this ticket will contain a question about mushrooms.
Solution
Probability of an event BUT called the ratio of the number of favorable BUT outcomes to the number of all equally possible outcomes:
Answer: 0,08.
Handbook includes all topics school course and complies with modern educational standards and programs. The book consists of two parts: "Algebra and the beginning of analysis" and "Geometry". The main material of the school mathematics course is presented by the authors concisely and systematically: mathematical concepts, axioms, theorems, properties, etc. The book will be an indispensable assistant in studying and consolidating new material, repeating topics covered, as well as in preparing for final exams in the form of the Unified State Examination.
Task 5
Find the root of equation 3 x– 5 = 81.
Solution
3 x– 5 = 81
3 x– 5 = 3 4
x – 5 = 4
Answer: 9.
Task 6
Triangle ABC inscribed in a circle with center O. Corner BAC equals 32°. Find an angle BOC. Give your answer in degrees.
Solution
∠COB is the central angle, ∠ COB= arc CB
∠COB= 64°
Answer: 64.
The figure shows a graph of a differentiable function y = f(x). Nine points are marked on the x-axis: x 1 , x 2 , … , x 9 .
Find all marked points where the derivative of the function f(x) is negative. Enter the number of these points in your answer.
Solution
The derivative of a function is negative where the function is decreasing.
Points fall into these intervals x 3 , x 4 , x 5 , x 9 . Only 4 points.
Answer: 4.
The manual contains tables on all the most important sections of the school course in arithmetic, algebra, and the beginnings of analysis. The tables briefly outline the theory for each topic, provide basic formulas, graphs and examples of solving typical problems. There is a subject index at the end of the book. The manual will be useful for students in grades 7-11, applicants, students, teachers and parents.
Task 8
In the first cylindrical vessel, the liquid level reaches 16 cm. This liquid was poured into the second cylindrical vessel, the diameter of which is 2 times the diameter of the base of the first. At what height will the liquid level be in the second vessel? Express your answer in cm.
Solution
Formula for calculating the volume of a cylinder:
V = π R 2 H,
where R is the radius of the cylinder, H- his high.
Because the liquid level reaches 16 cm, so the height is 16.
V= π R 2 H = π R 2 16
The diameter of the second vessel is twice the diameter of the first.
Because d = 2R, then the radius of the second vessel is also twice the radius of the first, and is equal to 2 R.
h is the height of the liquid in the second vessel.
Find the volume of liquid in the second vessel:
V= π(2 R) 2 h= π4 R 2 h
When pouring a liquid into another vessel, its volume has not changed.
Equate the volumes of the liquid of the first and second vessels:
π R 2 16 = π4 R 2 h
4h = 16.
Answer: 4 cm
Task 9
Find sin2α if cosα = 0.6 and π< α < 2π.
Solution
sin2α = 2sinα cosα
(sinα) 2 + (cosα) 2 = 1
(sinα) 2 + (0.6) 2 = 1
(sinα) 2 = 1 - 0.36
(sinα) 2 = 0.64
(sinα) 2 = ±0.8
Because α ∈ 3 or 4 quarters, so
sin2α = 2 (–0.8) (0.6)
sin2α = -0.96
Answer: –0,96.
Task 10
The locator of the bathyscaphe, evenly immersing vertically downwards, emits ultrasonic signals with a frequency of 749 MHz. The receiver registers the frequency of the signal reflected from the ocean floor. The submersion rate of the bathyscaphe (in m/s) and frequencies are related by the relationship
| v = c | f – f 0 | , |
| f + f 0 |
where c\u003d 1500 m / s - the speed of sound in water, f 0 is the frequency of the emitted signal (in MHz), f is the frequency of the reflected signal (in MHz). Find the frequency of the reflected signal (in MHz) if the bathyscaphe is sinking at a speed of 2 m/s.
Solution
It follows from the condition that
v= 2 m/s
With= 1500 m/s
f 0 = 749 MHz
Substitute this data into the formula
| 2 | = | f – 749 |
| 1500 | f + 749 |
Answer: 751.
Task 11
In spring, the boat goes against the current of the river 1 2 / 3 times slower than downstream. In summer, the current becomes 1 km / h slower. Therefore, in summer, the boat goes against the current 1½ times slower than downstream. Find the speed of the current in spring (in km/h).
Solution
Boat own speed |
x(km/h) |
|
river speed |
y(km/h) |
y– 1 (km/h) |
With the flow |
x + y(km/h) |
x + y – 1 |
Against the stream |
x – y(km/h) |
x – y + 1 |
Because in the spring, the boat goes against the current more slowly than downstream, we compose the equation
Let's create a system:
| x + y | = | 5 | |
| x – y | 3 | ||
| x + y- 1 | = | 3 | |
| x – y+ 1 | 2 |
| 3(x + y) = 5(x – y) | |
| 2(x + y – 1) = 3(x – y + 1) |
| 2x = 8y | |
| x = 5y – 5 |
| x = 20 | |
| y = 5 |
5 km / h - the speed of the current in spring.
Answer: 5.
Mathematics: algebra and the beginnings of mathematical analysis, geometry. Algebra and beginning of mathematical analysis. Grade 11. A basic level of
The textbook is included in the teaching materials for mathematics for grades 10-11, studying the subject at a basic level. The theoretical material is divided into mandatory and optional, the system of tasks is differentiated by the level of complexity, each paragraph of the chapter ends control questions and assignments, and each chapter is a home control work. The textbook includes project topics and links to Internet resources.
Task 12
Find the maximum point of the function y=ln( x + 4) 2 + 2x + 7.
Solution
Considering that ln( x+ 4) 2 = 2ln │ x+ 4│ we have:
| y′ = | 2 | + 2, | x > –4 | |
| x + 4 | ||||
| 2 | + 2, | x < –4 | ||
| x + 4 |
Let us find the critical points of the function (the points at which the derivative is either equal to zero or does not exist), for this we equate y′ to 0.
y′ = 0.
| 2 | + 2 = 0 |
| x + 4 |
| x= –5 | |
| x ≠ –4 |
On the interval (–4; ∞) the derivative is positive, there are no critical points.
On the interval (–∞; –4) at the point –5, the derivative changes its sign from “+” to “–”, which means that the point X= -5 is the maximum point of the function.
Answer: –5.
Task 13
Solution
Let's transform the left and right parts of the equation:
cos2 x= 1 – 2sin2 x(double angle formula for cosine)
1-2sin2 x= 1 - sin x
2sin 2 x– sin x= 0
sin x(2sin x– 1)= 0
| sin x = 0 | |
| 2sin x= 1 |
| x = π n, | n ∈ Ζ | |||
| x = | π | + 2π n, | n ∈ Ζ | |
| 6 | ||||
| x = | 5π | + 2π n, | n ∈ Ζ | |
| 6 | ||||
using a trigonometric circle.
| Answer: a) | π | + 2π n, n ϵ Ζ ; | 5π | + 2π n, n ϵ Ζ ; π n, n ϵ Ζ ; | ||
| 6 | 6 | |||||
| b) | –7π | ; | –11π | ; 2π. | ||
| 6 | 6 | |||||
Toolkit“Mathematics: algebra and the beginnings of mathematical analysis, geometry. Algebra and beginning of mathematical analysis. Grade 11. Basic level" is included in the "Vertical" system and complies with the Federal State Educational Standard.
All edges of a regular triangular prism ABCA 1 B 1 C 1 have length 6. Points M and N- the middle of the ribs AA 1 and A 1 C 1 respectively.
a) Prove that the lines BM and MN are perpendicular.
b) Find the angle between the planes BMN and ABB 1 .
Solution
1) Draw the height NB 1 in ∆ A 1 B 1 C 1 . BN 1 = √B1 C 1 2 – NC 1 2 = √6 2 – 3 2 = 3√3
2) ∆NB 1 B– rectangular with straight ∠ BB 1 N.
3) From ∆ NB 1 N according to the Pythagorean theorem: NB 2 = NB 1 2 + BB 1 2 = 6 2 + (3√3 ) 2 = 63.
4) From rectangular ∆ MAB according to the Pythagorean theorem: MB 2 = MA 2 + BA 2 = 6 2 + 3 2 = 45.
5) From rectangular ∆ MA 1 N according to the Pythagorean theorem: NM 2 = NA 1 2 + MA 1 2 = 3 2 + 3 2 = 18.
6) Consider ∆ MNB:
NB 2 = NM 2 + MB 2
Then, by the inverse Pythagorean theorem, we obtain that ∆ MNB rectangular, straight ∠ BMN. Means BM ⊥ MN. Ch.t.d.
B) 
1) Let's spend NK ⊥ A 1 B 1 .
2) NK ⊥ A 1 B 1 , NK ⊥ A 1 A, means NK ⊥ (A 1 B 1 B)
3) NK perpendicular to the plane NM- inclined KM– oblique projection NM to the plane ( A 1 B 1 B). According to the theorem inverse to the theorem on three perpendiculars, we have:
| sin ∠NMK= | 3√3 | ÷ 3√2 |
| 2 |
| sin ∠NMK= | √6 |
| 4 |
| ∠NMK= arc sin | √6 |
| 4 |
Solution
Let 3 x = y, 9x = y 2
| y 2 – 6y + 4 | + | 6y – 51 | ≤ y + 5 |
| y – 5 | y – 9 |
| y 2 – 6y + 5 – 1 | + | 6y – 54 + 3 | ≤ y + 5 |
| y – 5 | y – 9 |
| y 2 – 6y + 5 | – | 1 | + | 6y – 54 | + | 3 | ≤ y + 5 |
| y – 5 | y – 5 | y – 9 | y – 9 |
Expand the trinomial y 2 – 6y+ 5 for multipliers
y 2 – 6y + 5 = 0
D= (–6) 2 – 4 1 5 = 16
| y = | 6 + 4 | = 5 |
| 2 |
| y = | 6 – 4 | = 1 |
| 2 |
y 2 – 6y + 5 = (y – 5)(y – 1)
| (y – 5)(y – 1) | – | 1 | + | 6(y – 9) | + | 3 | ≤ y + 5 |
| y – 5 | y – 5 | y – 9 | y – 9 |
| y – 1 – | 1 | + 6 + | 3 | ≤ y + 5 |
| y – 5 | y – 9 |
| y – | 1 | + 5 + | 3 | ≤ y + 5 |
| y – 5 | y – 9 |
| – | 1 | + | 3 | ≤ 0 |
| y – 5 | y – 9 |
| –1(y – 9) + 3(y – 5) | ≤ 0 |
| (y – 5)(y – 9) |
| –y + 9 + 3y – 15 | ≤ 0 |
| (y – 5)(y – 9) |
| 2y – 6 | ≤ 0 |
| (y – 5)(y – 9) |
| 3 ≤ 3x | |
| 5 < 3x < 9 |
| 1 ≤ x | |
| log 3 5< x < 2 |
Answer:(–∞; 1] ∪ (log 3 5; 2)
The collection includes assignments for all sections and topics tested at the unified state exam of the basic level. Tasks of different difficulty levels are presented. At the end of the book answers are given that will help in monitoring and evaluating knowledge, skills and abilities. The materials of the manual can be used for the systematic repetition of the studied material and training in completing tasks. various types in preparation for the exam. They will help the teacher organize preparation for the unified state exam in mathematics at the basic level, and students will independently test their knowledge and readiness to pass the exam.
Two circles touch externally at a point K. Straight AB touches the first circle at a point A, and the second one is at the point B. Straight BK intersects the first circle at a point D, straight AK intersects the second circle at a point FROM.
a) prove that the lines AD and BC are parallel.
b) find the area of the triangle AKB, if it is known that the radii of the circles are 4 and 1.
Solution
- Draw a common tangent to the circles at the point K. She crosses AB at the point H.
- AH = HK, HK = HB(according to the property of tangents drawn from one point to a circle)
- In ∆ AKB, median KH equal to half a side AB, so it is rectangular, with ∠ AKB= 90°.
- ∠AKB = ∠AKD AKD based on diameter AD. Then AD ⊥ AB.
- ∠AKB = ∠CKB= 90° (as adjacent), so ∠ BKC based on diameter BC. Then BC ⊥ AB.
- consequently AD || BC.
b) Let R= 4 radius of the first circle centered O 1 , and r= 1 is the radius of the second circle with the center O 2 .
1) Consider ∆ CKB and ∆ AKD: vertex angles K straight lines, ∠ DAK = ∠ACB, as lying crosswise at AD || BC and secant AC. So ∆ CKB ~ ∆AKD at two corners.
| 2) | AK | = | KD | = | AD | = | 2R | = | 4 | = k |
| KC | BK | BC | 2r | 1 |
3) The ratio of the areas of similar triangles is k 2 (k- coefficient of similarity)
| S AKD | = 16, S AKD = 16S BKC |
| S BKC |
4) ∆AKB and ∆ AKD have a common height AK
| 5) | S AKD | = | DK | = | AD | = | 4 | , S BKA= | 1 | S AKD = | 1 | · 16S BKC = 4S BKC |
| S BKA | KB | BC | 1 | 4 | 4 |
6) ∆DCK and ∆ CKB have a common height CK, so their areas are related as the bases to which this height is drawn.
| S DKC | = | DK | = | 4 | , S DKC = 4S BKC |
| S BKC | KB | 1 |
7) Find the area of the trapezoid ABCD:
S ABCD = S DKA + S AKB + S CKB + S DCK
S ABCD = 16S BKC + 4S CKB + S CKB + 4S CKB
S ABCD = 25S CKB
8) Let's go to AD perpendicular O 2 S(trapezoid height)
9) From rectangular ∆ O 2 SO 1 by the Pythagorean theorem we find O 2 S:
O 2 S = √(O 2 O 1) 2 – (O 1 S) 2
O 2 S = √5 2 – 3 2 = 4
O2 S = AB = 4
S ABCD= 25S CKB
20= 25S CKB
S CKB= 0,8
S BKA= 4S BKC= 4 0.8 = 3.2.
Answer: 3,2.
The presented programs for the courses "Mathematics" for grades 5-6, "Algebra" and "Geometry" for grades 7-9, "Mathematics: Algebra and the beginning of mathematical analysis; Geometry" for grades 10-11 of educational institutions are created on the basis of a single concept of teaching mathematics in high school, developed by A.G. Merzlyak, V.B. Polonsky and M.S. Yakir.
Task 17
On January 15, it is planned to take a loan from a bank for six months in the amount of 1 million rubles. The conditions for its return are as follows:
- On the 1st of each month, the debt increases by r percent compared to the end of the previous month, where r is an integer;
- from the 2nd to the 14th of each month, part of the debt must be paid;
- On the 15th day of each month, the debt must amount to a certain amount in accordance with the following table.
Find highest value r, at which the total amount of payments will be less than 1.2 million rubles.
Solution
Debt to the bank (in million rubles) on the 15th day of each month should be reduced to zero according to the following scheme:
1; 0,6; 0,4; 0,3; 0,2; 0,1; 0.
Then the debt on the 1st of each month (together with interest) is equal to:
| one · | r | ; 0,6(1 + | r | ); 0,4(1 + | r | ); 0,3(1 + | r | ); 0,2(1 + | r | ); 0,1(1 + | r | ) |
| 100 | 100 | 100 | 100 | 100 | 100 |
Payments from the 2nd to the 14th of each month are:
The total payout is:
| 2,6(1 + | r | ) < 2,8 |
| 100 |
| 1 + | r | < | 28 |
| 100 | 26 |
Has only one solution.
Solution
Consider the first equation of the system:
1) When x≥ 0, we have the equation ( x – 5) 2 + (y d centered on a point G(5; 4) and radius 3.
2)When x≤ 0, we have the equation (– x – 5) 2 + (y – 4) 2 = 9, (x + 5) 2 + (y– 4) 2 = 9, this equation defines a circle With centered on a point F(–5; 4) and radius 3.
3) Equation ( x + 2) 2 + y 2 = a 2 defines a circle k centered at point H(–2; 0) and radius a, where a > 0.
Let us find at what values a circle k has a single common point with the circles d and c.
4) Draw from a point H Ray HG, it intersects the circle d at points O and P, dot O lies between the points H and G. The distance between the points is found by the formula | HG|= √(x G – x H) 2 + (y G – yH) 2
|HG|= √(5 + 2) 2 + (4 – 0) 2 = √65
HG = GO + Oh
Oh = HG – GO = √65 – 3
HP = √65 + 3
If a a< HO or a > HP circles d and k do not intersect.
If a HO< a < HP , then the circles d and k have two common points.
At a=HO or a = HP circles d and k
5) Draw a beam HF from a point H, it intersects the circle c at points M and N, wherein M lies between H and F. Find the distance between points HF,
|HF| = √(–5 + 2) 2 + 4 2 = 5
HM = 5 – 3 = 2
HN = 5 + 3 = 8
If a a< HM or a > HN circles c and k do not intersect.
If a HM< a < HN , then the circles c and k have two common points.
At a = HM or a = HN circles c and k touch each other at the same point.
The system has a unique solution if and only if the circle k touches exactly one of the two circles d and c and does not intersect with the other.
It is clear from the solution that HM < HO < HN < HP.
Then the condition of the problem is satisfied by the lengths of the segments HO= √65 + 3 and HM = 2.
Answer: √65 + 3; 2.
The publication contains 10 training options for examination papers to prepare for the exam. Each option is compiled in full accordance with the requirements of the Unified State Exam, includes tasks of a profile level. The structure of options is the same. At the end of the manual, answers to all tasks are given.
The USE in mathematics is the main discipline that all graduates take. The examination test is divided into two levels - basic and profile. The second is required only for those who plan to make mathematics the main subject of study in higher education. Everyone else passes the basic level. The purpose of this test is to check the level of skills and knowledge of graduate students for compliance with norms and standards. The division into core and basic levels was first used in 2017 so that students who do not need advanced mathematics for admission to a university do not waste time preparing for difficult assignments.
To get a certificate and apply to a university, you need to satisfactorily complete the tasks of the basic level. Preparation includes repetition school curriculum in algebra and geometry. Tasks in the USE of the basic level are available to students with different levels of knowledge. The basic level can be passed by students who were just attentive in the classroom.
The main recommendations for preparation are:
- Systematic preparation should be started in advance so that you do not have to be nervous, mastering all the tasks 1-2 months before the exam. The period required for high-quality training depends on the initial level of knowledge.
- If you are not sure that you will master the tasks on your own, ask a tutor for help - he will help you systematize your knowledge.
- Practice solving problems, examples, tasks, according to the program.
- Solve tasks online - "I will solve the exam" will help with regular training and preparation for the exam. With a tutor, you will be able to analyze mistakes, analyze tasks that cause particular difficulties.
In the process of preparation, solve as many tasks of varying complexity as possible, gradually switch to completing tasks for time. Get to know .
Preparation Methods
- Studying the subject at school;
- Self-education - solving problems by example;
- Lessons with a tutor;
- Training in courses;
- Online preparation.
There are 20 tasks (the number may change every year) for which short answers are required. This is enough for a student who plans to enter higher educational establishments for the humanities.
The subject is given 3 hours to complete the tasks. Before starting work, you must carefully read the instructions and act in accordance with its provisions. Accompanying the examination book are reference materials that are necessary for passing the examination test. For the successful completion of all tasks, 5 points are given, the minimum threshold score is 3.
Evaluation
3 hours(180 minutes).
20 short answer tasks and practical skills.
Answer
But you can make a compass Calculators on the exam not used.
the passport), pass and capillary or! Allowed to take with myself water(in a transparent bottle) and food
Allotted for work 3 hours(180 minutes).
The examination paper consists of one part, including 20 short answer tasks base level of difficulty. All assignments are for checking the development of basic skills and practical skills application of mathematical knowledge in everyday situations.
Answer for each of tasks 1–20 is an integer or a final decimal, or a sequence of digits. A task with a short answer is considered completed if the correct answer is recorded in the answer sheet No. 1 in the form provided for in the instructions for completing the task.
When doing work, you can use the ones containing the basic formulas of the mathematics course, issued along with the work. You can only use a ruler, but you can make a compass with your own hands. It is forbidden to use tools with reference materials printed on them. Calculators on the exam not used.
You must have an identity document with you for the exam. the passport), pass and capillary or gel pen with black ink! Allowed to take with myself water(in a transparent bottle) and food(fruit, chocolate, buns, sandwiches), but may be asked to leave in the hallway.
The USE tests in mathematics at the basic level are not difficult. All tasks are aimed at testing the development of basic skills and practical skills in applying mathematical knowledge in everyday situations.
Distribution of CMM tasks
The structure of the variant consists of one part, including 20 tasks with a short answer.
- Algebra - 10 tasks;
- Equations and inequalities - 3 tasks
- Functions - 1 task
- Beginnings of mathematical analysis - 1 task
- Geometry - 4 tasks
- Elements of combinatorics, statistics and probability theory - 1 task
The duration of the USE test in mathematics base
The examination time is 3 hours (180 minutes).
Each number needs to devote about 9 minutes.
Additional materials and equipment
The list of additional devices and materials, the use of which is allowed for the Unified State Examination, is approved by the order of Rosobrnadzor. The necessary reference materials are issued along with the text of the examination paper. When performing tasks, it is allowed to use a ruler.
How points will be transferred
The correct solution of each of the tasks 1-20 is estimated at 1 point. The task is considered completed correctly if the examinee gave the correct answer in the form of an integer or a final decimal fraction, or a sequence of numbers.
The maximum primary score for the entire work is 20.
Primary scores are not converted into test scores, only into grades:
- 0-6 = 2 (failed);
- 7-11 = 3 (satisfactory);
- 12-16 = 4 (good);
- 17-20 = 5 (excellent).
Based on the demo version of 2019, 10 basic level math options with solutions and answers have been developed;
After registering on the site, you can monitor the level of your knowledge.
Get ready and get the maximum score!
In this section, we are preparing for the USE in mathematics as a basic, specialized level - we present analysis of problems, tests, a description of the exam and useful recommendations. Using our resource, you will at least understand how to solve problems and be able to successfully pass the exam in mathematics in 2019. Begin!
The USE in mathematics is a mandatory exam for any student in the 11th grade, so the information presented in this section is relevant for everyone. Mathematics exam is divided into two types - basic and profile. In this section, I provide an analysis of each type of task with a detailed explanation for the two options. The tasks of the exam are strictly thematic, therefore, for each number, you can give precise recommendations and give the theory necessary specifically for solving this type of task. Below you will find links to tasks, by clicking on which you can study the theory and analyze examples. Examples are constantly updated and updated.
The structure of the basic level of the exam in mathematics
The Foundation Level Mathematics Exam consists of one part , including 20 tasks with a short answer. All tasks are aimed at testing the development of basic skills and practical skills in applying mathematical knowledge in everyday situations.
The answer to each of tasks 1–20 is integer, final decimal , or digit sequence .
A task with a short answer is considered completed if the correct answer is recorded in the answer sheet No. 1 in the form provided for in the instructions for completing the task.
