How to solve task 8. Algorithm for completing the task

Their writing follows different patterns. Read the rules, look at the examples.

To complete the task correctly, you need to be able to determine which of these three cases the examples in your version of the CMM belong to.

The task requires knowledge of how words are written. To answer correctly, you need to be able to accurately identify the spelling. This action is similar to going through passport control at the border: the border guard looked at you, looked at the photo in your passport, and the way is clear. Before you are 5 words and the wording of the task. First understand what to look for.

If the task is on tested vowels, this is one thing, if on alternating vowels, then another.

Let's remember what is what.

Tested unstressed vowels in the root: goat´, goat - goats

In an unstressed position, the same vowel is written at the root of the word as under stress.

Why do we write the letter O, but not A in words: goats, goat? Because - co ´ s s!

How to check?

Choose the same word, but in a different form: goat´, goat - goats´(plural) or a related word, that is, a word with the same root: goat´, goat - goat´, goat´.

Do not confuse:

It is a mistake to select the following as test words:

• words with different meanings: frequency - pure, purity - often
• words with alternating vowels in the root: sunbathe - tan, dawn - dawn
• other types of verbs: be late - be late, assimilate - assimilate

Unverifiable unstressed vowels in the root: dog, cuttlefish, kamorka

This is the stupidest of all rules, because it is not a rule at all! Why? Because there is no action that would help decide which letter to write. The only thing you can do is find the right word in the dictionary. All textbooks advise this. But a dictionary may not be at hand. This means that such words need to be memorized. The more such words you know, the better. Gradually, you will develop intuition, and words with unchecked vowels will not create difficulties. So, despite the apparent stupidity, this rule is very useful!

If you cannot classify a word among words with alternating vowels in the roots and find test words (by changing the form of the word or looking for words with the same root), then you have a word with an untestable vowel.

Note:

After successfully passing the Unified State Exam, you can return to tip No. 1. It won't harm your life.

Alternating vowels in the root

The alternation of vowels in the roots of Russian words is a systemic phenomenon. There are many roots with alternating vowels, but in school practice they study a mandatory list. It includes 20 roots, which are presented below. You need to remember them and firmly know the entire list. This will help you avoid many annoying mistakes.
Sometimes they ask me in letters: “Why don’t you give the root -log-? After all, in fact, alternation is not - lag-//-lozh-, A -lag-//-log-//-false-?»
It's simple. I propose the option that is presented in most school textbooks. The unified exam is being prepared by FIPI, which promises that differences in programs will not affect the exam result. So far this setting has been followed. A - log- don’t let it worry you: it only comes with an accent! Forgery, pledge, pretext and others. And in the tasks you will only have words with unstressed roots. Therefore, I believe that you should not be distracted by this and other similar cases.

For those who want to know not a short, school list of 20 roots, but a complete list of all roots with alternating vowels, I recommend the most authoritative source: “Complete Academic Directory,” edited by V. Lopatin.

And here we will remember what you all studied at school.

1. Alternation of vowels in roots ber // bir, per//feast, measures//peace, burned //zhig, ter // shooting range, der//dir, h et//cheat, shine // shine, steel//steel: erase-erase

If in a word with roots: -ber-//-bir-, -per-//-feast-, -burned-//-jig-, -mer-//-world-, -ter-//-shooting range-, -der-//-dir-, -even-//-cheat-, - shiny-//-blist-, -steel-//-style-

there is a suffix A, write at the root And: at buy , shine ,
no suffix A, write e:takes, shine.

2. Alternation of vowels in roots kos//kas: touch - touch

If in a word

• there is a suffix A, write at the root A: touch, touching,
• no suffix A , write O:touch, at touching

3. Alternation of vowels in roots mok//poppy: soaked - to dip

If in words with roots -mok- //-poppy- meaning:

“to pass liquid, to absorb liquid,” then write o: you wet in the rain ,
“immerse in liquid”, write A: poppy .

4. Alternation of vowels in roots equal//equal: level - equalize

If in words with roots -even-//-equal- meaning:

“even, smooth”, write o: you smooth the road , behind level the holes ,
"equality", write A: at equal rights, equally third-party.

5. Alternation of vowels in roots mountains//gar, creation//creature, clone//clan: damn it's burnt

If the roots -gar- //-mountain- , -creature- //-creation- , -clan- //-clone-

no accent, write O: mountains and´t , cloned , creativity ,
under stress, write as you hear: behind ha R , cla ´n to be , yours R quality , twa ´ry .

Exception: at ´ soot

6. Alternation of vowels in roots dawn // dawn: dawn´

If the roots -zar- //-zor-

without stress, write a: dawn, lightning
under stress, write as you hear: dawn.

Exception: Zoreva ´t

7. Alternation of vowels in roots pilaf//swimming: float´ to- swimmers´(excl.)

If the roots are -float- //-float-

without accent, write a: float To,
under stress write as you hear: pla ´vat, pla ´ tion, swims V.

Exception: swimmers ´ , swimmer´ ts, swimmers´ ha, quicksand´

8. Alternation of vowels in roots grew // grow // grow, log//log, skak//skoch: the plant grew, term - addition, jump - upstart

If the root consonants are:

• With, write O: grew up,
  st or sch, That A: grow, grown,
• and, write O: offer ,
  G, That A: term,
• h, write O: upstart,
  To, That A: jump rope.

Exception: growth ´ to, height´ th, usurer´ to, height´ growth, teenage (along with the literary normsubro ´ stock ) and their derivatives: sprout, usurer and etc.

The eighth task of the Unified State Exam in Russian tests graduates' skills in the field of correct spelling of words. For correct execution you can receive one primary point. In the task you need to find a word in which a certain vowel is missing - either a checkable one, or an uncheckable one, or an alternating one. To do this, you need to have a good understanding of the spelling of roots with tested unstressed vowels, alternating vowels, as well as dictionary words, the correct spelling of which must be remembered. To make it easier to repeat this topic, we present a theory based on the materials of the eighth task of the Unified State Exam.

Theory for task No. 8 of the Unified State Exam in Russian

• tested unstressed vowel

This is the easiest option; to define it, you need to choose a form of the word in which the vowel will be stressed. For example, “reconcile”, “braggart”, “hardened” are checked by the words “peace”, “boast”, “callous”. Sometimes it is difficult to determine its meaning from a word without a vowel; for example, “to fade” can be understood both as “to see” and as “to fade.” This was taken into account when developing exam tasks: similar words are given in a contextual phrase.

There are not many alternating roots in the Russian language; you can simply memorize them. This table shows the alternating vowels in the root of a word and the rules for their use. However, you need to remember the exceptions that are missing from it: pretend, illuminate, combine, level, equally, peer, spasmodic.

• unchecked unstressed vowel

Here is a table of words from which are most often found in the exam.

Algorithm for completing the task

1. We carefully read the task, remember the rule (alternation of vowels in the root of a word, verifiable vowels in the root of a word, unchecked vowels in the root of a word).
2. We insert the missing vowels into each word given in the task, determine the rule on which the spelling of each word is based.
3. We find the right word, write it down, inserting the missing letter. We write down the answer.

Analysis of typical options for task No. 8 of the Unified State Examination in the Russian language

The eighth task of the demo version 2018

1. m..tsenat
2. look...see
3. mountainous (terrain)
4. grow up
5. comp..nent

Execution algorithm:

1. Maecenas come to terms with– the vowel being tested in the root of the word (smirny); mountainous terrain)– the vowel being tested in the root of the word (mountain); component
2. Nurture– a word in which the unstressed alternating vowel of the root is missing (roots rast - ros). We write a letter in place of the gap A, since it is followed by consonants ST.

Answer: nurture

First version of the task

Identify the word in which the unstressed vowel of the root being tested is missing. Write out this word by inserting the missing letter.

1. exp...dication
2. hydrogen...sli
3. k...lendary
4. k...shaky
5. post...pour

Execution algorithm:

1. An unstressed checkable vowel is a vowel that can be checked by changing the word and placing it under stress: MOUNTAIN – MOUNTAINS.
2. We insert the missing vowels into each word given in the task: expedition, calendar– you need to remember (unverifiable vowel at the root of the word); seaweed, lay– have an alternating vowel in the root.
3. Feline– a word in which the unstressed vowel being tested is missing. We select a test word where the vowel will be stressed: cat.

Answer: cat

Second version of the task

Identify the word in which the unstressed alternating vowel of the root is missing. Write out this word by inserting the missing letter.

1. v..rsy
2. adventure
3. g..roar
4. accept..mother
5. adv..cat

Execution algorithm:

1. Roots with alternating vowels: ber - bir, kas - kos, lag - lodge, etc.
2. We insert the missing vowels into each word given in the task: nappy– test vowel – test word pile; adventure– unchecked vowel at the root of the word (you need to remember the spelling); grieve– the vowel being tested in the root of the word (gore). The alternation of “gor/gar” occurs in words such as “ tan, burnt, burn, scorch, cinder». Advocate– an untestable vowel at the root of the word (needs to be remembered).
3. Accept– a word in which the unstressed alternating vowel of the root is omitted (alternation nya/him): accept – accept.

Answer: accept

Third version of the task

Identify the word in which the unstressed unchecked vowel of the root is missing. Write out this word by inserting the missing letter.

1. op..sanie
2. l...lay
3. dry out
4. entertainment
5. calculate

Execution algorithm:

1. An untestable vowel at the root of a word is a vowel whose spelling needs to be remembered (for example: the vinaigrette).
2. We insert the missing vowels into each word given in the task: description– the vowel being tested in the root of the word (test word we write); get up to something– alternating vowel at the root of the word (creativity – utensils); entertainment– vowel being tested, test word have fun; calculate– the vowel being tested in the root of the word, test word numbers.
3. Cherish- a word with an unverifiable vowel at the root; its spelling must be remembered.

METHODS OF BIOLOGY – there was a question at the Unified State Exam 2018 about centrifugation.

How do we reason? You need to know all the terms by heart:cytology – the science of the cell, that is, all objects are very small.Centrifugation – a method of separating cell organelles, molecules, substances due to different densities of separated fractions. Centrifugal forces impart different accelerations and denser fractions settle first - for example. The cell nucleus, then less dense ones - mitochondria, chloroplasts, then the lightest ones, for example, ribosomes.Microscopy – examination under a microscope – yes, it is suitable for cytology.Chromatography – a method for separating mixtures of substances based on different speeds of movement of molecules that differ in structure, mass and size in the chromatograph. The method is more biochemical, but it is also suitable for cytology - separation of amino acids and chlorophylls.Heterosi s – this is in the so-called selection. "hybrid vigor", a phenomenon in which first-generation hybrids combine the best qualities of their parents. – not suitable for cytology.Monitoring – in ecology – long-term monitoring of the state of the environment. So the answer is: 4 and 545

ANSWER: 35

Evolutionary theory is the doctrine of the historical development of the organic world. The result of evolution is adaptation, new species. Idioadaptation is adaptation to a variety of existing environmental conditions without changing the level of organization. – yes, it fits evolution.

Divergence is an evolutionary path in which characters diverge

Diheterozygote – in genetics, this is AaBb – No

Aromorphosis is an evolutionary path in which the level of organization changes sharply (morphophysiological leap)

Hybridization is a method in selection.

Biosphere is the sphere of life.

Bioinert matter is soil.

Noosphere – the sphere of the mind. (Where is a person)

Gyres are what maintain the existence of the biosphere

Aromorphosis - in evolution

Hybridization - in selection

Answer: 45

Now decide for yourself.

Choose two correct answers out of five and write down the numbers under which they are indicated. Genetic engineering, unlike cellular engineering, includes research related to

1) cultivation of cells of higher organisms

2) hybridization of somatic cells

3) gene transplantation

4) transplantation of the nucleus from one cell to another

5) obtaining recombinant (modified) RNA and DNA molecules

Choose two correct answers out of five and write down the numbers under which they are indicated. The cellular level of organization coincides with the organismal level

1) bacteriophages

2) dysenteric amoeba

3) polio virus

4) wild rabbit

5) green euglena

Choose TWO correct answers out of five and write down the numbers under which they are indicated. The cytogenetic method makes it possible to study in humans

1) hereditary diseases associated with genomic mutations

2) development of symptoms in twins

3) the metabolic features of his body

4) its chromosome set

5) the pedigree of his family

Below is a list of terms. All of them, except two, are used to describe environmental patterns. Find two terms that “fall out” from the general series and write down the numbers under which they are indicated in the table.

1) parthenogenesis

2) symbiosis

3) succession

4) aromorphosis

5) consumer

Choose two correct answers out of five and write down the numbers under which they are indicated in the table. What scientific research methods are used to diagnose diabetes mellitus and identify the nature of its inheritance?

1) biochemical

2) cytogenetic

3) twin

4) genealogical

5) historical

Choose two correct answers out of five and write down the numbers under which they are indicated in the table. Biochemical research methods include:

1) microscopy

2) electrophoresis

3) inbreeding

4) X-ray spectrophotometry

5) hybridization

The lesson is devoted to the analysis of task 8 of the Unified State Exam in computer science

The 8th topic - “Programming algorithms with loops” - is characterized as tasks of a basic level of complexity, completion time - approximately 3 minutes, maximum score - 1

Algorithmic structures with loops

In task 8 of the Unified State Exam, algorithmic structures with cycles are used. Let's look at them using the Pascal language as an example.

• For introduction and repetition While loop, .
• For introduction and repetition For loop, .

Sum of arithmetic progression

Formula for calculation n th element of an arithmetic progression:

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

n terms of an arithmetic progression:

• a i
• d– step (difference) of the sequence.

Sum of geometric progression

Property of geometric progression:

b n 2 = b n+1 * q n-1

Formula for calculation denominator geometric progression:

\[ q = \frac (b_(n+1))(b_n) \]

Formula for calculation n th element of the geometric progression:

b n = b 1 * q n-1

Formula for calculation denominator geometric progression:

Formula for calculating the sum of the first n terms of geometric progression:

\[ S_(n) = \frac (b_1-b_(n)*q)(1-q) \]

\[ S_(n) = b_(1) * \frac (1-q^n)(1-q) \]

• b i– i-th element of the sequence,
• q is the denominator of the sequence.

Solving tasks 8 of the Unified State Exam in computer science

Unified State Examination in Informatics 2017, FIPI task option 15 (Krylov S.S., Churkina T.E.):

var k,s:integer; begin s:=512; k:=0; while s

• In a loop k increases by unit (k - counter). Respectively, k will be equal to the number of iterations (repetitions) of the loop. After the cycle is completed k is displayed on the screen, i.e. this is the result of the program.
• In a loop s increases by 64 . For simplicity of calculations, let’s take the initial s Not 512 , A 0 . Then the loop condition will change to s< 1536 (2048 — 512 = 1536):

• The loop will run until s<1536 , а s increases by 64 , it follows that the loop iterations (steps) will be:

• Respectively, k = 24.

Result: 24

For a more detailed analysis, we suggest watching a video of the solution to this 8th task of the Unified State Exam in computer science:

10 Training versions of exam papers to prepare for the Unified State Exam in Computer Science 2017, task 8, option 1 (Ushakov D.M.):

Determine what will be printed as a result of executing the following program fragment:

var k,s: integer; begin k:=1024; s:=50; while s>30 do begin s:=s-4; k:=k div 2; end; write(k)end.

Result: 32

For a detailed solution, watch the video:

Unified State Exam 8.3:

At what is the smallest integer number entered? d after the program is executed the number will be printed 192 ?

var k,s,d: integer; begin readln(d); s:=0; k:=0; while k< 200 do begin s:=s+64; k:=k+d; end; write(s); end.

Let's consider the program algorithm:

• The loop depends on a variable k, which increases by value each iteration of the loop d(input). The cycle will finish its work when k will be equal to 200 or exceed it ( k >= 200).
• The result of the program is the output of the variable value s. In a loop s increases by 64 .
• Since the assignment requires that the number be displayed 192 , then we determine the number of repetitions of the cycle as follows:

• Since in a cycle k increases by value d, and loop repetitions 3 (the loop ends when k>=200), let's create an equation:

• Since the number turned out to be non-integer, let’s check and 66 And 67 . If we take 66 , That:

those. the cycle will continue to work after three passes, which is not suitable for us.

• For 67 :

• This number 67 we are satisfied, it is the smallest possible, which is what is required by the assignment.

Result: 67

For an analysis of the task, watch the video:

Unified State Examination in computer science task 8.4 (source: option 3, K. Polyakov)

Determine what will be printed as a result of the following program fragment:

var k, s: integer; begin s:=3; k:=1; while k< 25 do begin s:=s+k; k:=k+2; end; write(s); end.

Let's look at the program listing:

• The result of the program is the output of the value s.
• In a loop s changes, increasing by k, at the initial value s = 3.
• The cycle depends on k. The loop will end when k >= 25. Initial value k = 1.
• In a loop k is constantly increasing by 2 -> this means you can find the number of iterations of the loop.
• The number of loop iterations is:

(because k was originally equal to 1 , then in the last, 12th passage of the cycle, k = 25; loop condition is false)

• IN s the sum of an arithmetic progression accumulates, the sequence of elements of which is more convenient to start with 0 (not with 3 , as in the program). Therefore, imagine that at the beginning of the program s = 0. But let us not forget that in the end you will need to add 3 to the result!

• Then the arithmetic progression will look like:

• There is a formula for calculating the sum of an arithmetic progression:

s = ((2 * a1 + d * (n - 1)) / 2) * n

Where a1- the first term of the progression,
d- difference,
n— the number of terms of the progression (in our case — the number of loop iterations)

• Let's substitute the values ​​into the formula:

• Let's not forget that we must add to the result 3 :

• This is the meaning s, which is output as a result of the program.

Result: 147

Solution to this Unified State Exam assignment in computer science video:

Unified State Examination in computer science task 8.5 (source: option 36, K. Polyakov)

var s, n: integer; begin s:= 0; n:= 0; while 2*s*s< 123 do begin s:= s + 1; n:= n + 2 end; writeln(n) end.

Let's look at the program listing:

• Variable in the loop s constantly increasing per unit(works like a counter), and the variable n in a cycle increases by 2 .
• As a result of the program's operation, the value is displayed on the screen n.
• The cycle depends on s, and the cycle will end when 2 * s 2 >= 123.
• It is necessary to determine the number of cycle passages (cycle iterations): to do this, we determine the smallest possible s, to 2 * s 2 >= 123:

Or you would simply need to find the smallest possible even number >= 123 that, when divided by 2 would return the calculated root of the number:

S=124/2 = √62 - not suitable! s=126/2 = √63 - not suitable! s=128/2 = √64 = 8 - suitable!

• So the program will do 8 loop iterations.
• Let's define n, which increases each step of the cycle by 2 , Means:

Result: 16

The video of this Unified State Exam assignment is available here:

Unified State Examination in computer science task 8.6 (source: option 37, K. Polyakov with reference to O.V. Gasanov)

Write the smallest and largest value of the number separated by commas d, which must be entered so that after executing the program it will be printed 153 ?

var n, s, d: integer; begin readln(d); n:= 33; s:= 4; while s<= 1725 do begin s:= s + d; n:= n + 8 end; write(n) end.

Let's look at the program listing:

• The program loop depends on the value of a variable s, which in the loop constantly increases by the value d (d entered by the user at the beginning of the program).
• In addition, in the loop the variable n increases by 8 . Variable value n is displayed on the screen at the end of the program, i.e. on assignment n by the end of the program should n = 153.
• It is necessary to determine the number of loop iterations (passes). Since the initial value n = 33, and in the end it should become 153 , increasing in the cycle by 8 , then how many times 8 "will fit" in 120 (153 — 33)? :

• As we have determined, the cycle depends on s, which is at the beginning of the program s = 4. For simplicity of work, we assume that s = 0, then let’s change the loop condition: instead of s<= 1725 сделаем s <= 1721 (1725-1721)

• We'll find d. Since the loop is running 15 times, then you need to find an integer that, when multiplied by 15 would return a number greater than 1721:

• 115 - this is the least d at which n will become equal 153 (for 15 steps of the cycle).
• Let's find the greatest d. To do this, you need to find a number that corresponds to the inequalities:

• Let's find:

• Largest d= 122

Result: 115, 122

Watch the video of this 8th task of the Unified State Exam:

Task 8. Demo version of the Unified State Exam 2018 computer science:

Write down the number that will be printed as a result of the following program.

var s, n: integer; begin s:= 260; n:= 0; while s > 0 do begin s:= s - 15; n:= n + 2 end; writeln(n)end.

Let's consider the algorithm:
• The loop depends on the value of a variable s, which is initially equal 260 . Variable in the loop s constantly changes its value, decreasing at 15.
• The cycle will complete its work when s<= 0 . So, you need to count how many numbers 15 "will be included" in the number 260 , in other words:

• This figure must correspond to the number of steps (iterations) of the cycle. Since the cycle condition is strict - s > 0, we increase the resulting number by one:

• Let's check it with a simpler example. Let's say initially s=32. Two passes through the loop will give us s = 32/15 = 2.133... Number 2 more 0 , accordingly, the cycle will run a third time.
• As a result of the work, the program prints the value of the variable n(the desired result). Variable in the loop n, initially equal 0 , increases by 2 . Since the cycle includes 18 iterations, we have:

Result: 36

For a detailed solution to this 8th task from the demo version of the Unified State Exam 2018, watch the video:

Solution 8 of the Unified State Examination task in computer science (control version No. 2 of the 2018 exam paper, S.S. Krylov, D.M. Ushakov):

Determine what will be printed as a result of executing the program:

var s, i: integer; begin i:= 1; s:= 105; while s > 5 do begin s:= s - 2; i:= i + 1 end; writeln(i)end.

• Let's consider the algorithm. The loop depends on a variable s, which decreases every iteration of the loop on 2.
• The loop also contains a counter - a variable i, which will increase per unit exactly as many times as there are iterations (passes) of the loop. Those. As a result of executing the program, a value equal to the number of iterations of the loop will be printed.
• Since the loop condition depends on s, we need to count how many times can s decrease by 2 in a cycle. For ease of calculation, let's change the loop condition to while s > 0 ; since we s decreased by 5 , accordingly, change the 4th line to s:=100 (105-5):

• In order to count how many times the loop will be executed, you need 100 divide by 2 , because s each loop step is reduced by 2: 100 / 2 = 50 -> number of loop iterations
• In the 3rd line we see that the initial value i is 1 , i.e. in the first iteration of the loop i=2. This means that we need to add to the result (50) 1 .
• 50 + 1 = 51
• This value will be displayed on the screen.

Result: 51

Solution 8 of the Unified State Examination task in computer science 2018 (diagnostic version of the 2018 exam paper, S.S. Krylov, D.M. Ushakov, Unified State Examination simulator):

Determine the value of a variable c after executing the next program fragment. Write your answer as an integer.

a:=-5; c:=1024; while a<>0 do begin c:=c div 2; a:=a+1 end;1000 do begin s : = s + n; n := n * 2 end ; write (s) end .

var n, s: integer; begin n:= 1; s:= 0; while n<= 1000 do begin s:= s + n; n:= n * 2 end; write(s) end.

Let's consider the algorithm:

• The loop condition depends on the variable n, which changes in a loop according to obtaining powers of two:

This means that on the 24th iteration of the loop the variables s And n received such values ​​after which the condition still remains true: 2 > 0. At the 25th step this condition is satisfied:

Result: 22

We invite you to watch a video of the solution to the task:

Task No. 8 in the profile level of the Unified State Examination in mathematics tests basic knowledge of stereometry. The tasks in this section are simple, on basic formulas - usually on the volumes of simple standard figures - cylinder, cube, pyramid, cone.

Theory for task No. 8

I will give formulas for the volume of figures, as this material is quite common.

Analysis of typical options for tasks No. 8 of the Unified State Exam in mathematics at the profile level

First version of the task (demo version 2018)

In a cylindrical vessel, the liquid level reaches 16 cm. At what height will the liquid level be if it is poured into a second vessel, the diameter of which is 2 times larger than the first? Express your answer in cm.

Solution algorithm:

1. Determine the area of ​​the base of the first and second vessels.
2. We write down volume formulas and equate them.
3. We remove the same values. We draw a conclusion.
4. We write down the answer.

Solution:

1. The area of ​​the base of the first vessel is determined by the formula

The diameter of the second vessel is 2 times larger. So the area of ​​its base is equal to

that is, 4 times more:

2 Write down the formula for the volume of liquid in each container.

Since the volume of liquid remains constant, we obtain the equation:

We remove the same values. From here

Second version of the task (from Yashchenko, No. 1)

The cylinder and cone have a common base and height. The volume of the cone is 28. Find the volume of the cylinder.

Solution algorithm:

1. We write down the formula for the volume of a cone.
2. We write down the formula for the volume of a cylinder.
3. We compare the formulas and draw a conclusion.
4. Calculate the volume of the cylinder.
5. We write down the answer.

Solution:

1. The volume of the cone is determined by the formula

where H is the height of the cone; R is the radius of the base of the cone.

2. The volume of the cylinder is determined by the formula

3. Compare both formulas. It is easy to see that the volume of the cylinder is 3 times the volume of the cone.

4. Calculate the volume of the cylinder:.