Spelling the roots of words is, at first glance, a simple topic. Moreover, it is studied in Russian language lessons already in elementary school. However, it is in the roots that students very often make mistakes.
Reasons for misspelling word roots:
- Ignorance of the rules for writing vowels and consonants at the root.
- Inability to correctly select the word to be tested, which can easily be used to check both vowels and consonants.
- Errors in identifying roots with alternating vowels. Checking such vowels with stress is a grave mistake. Alternating vowels should be written only according to the rule.
- There are frequent cases when, among words with missing spellings, those in which the letter is missing are offered in the console!!! Be careful not to confuse the prefix with the root (for example: d...reliable, the O in the prefix is missing here)
As we can see, the main reason is ignorance of the rules. You need to learn the rules of the Russian language, guys. Only then will you be able to write words correctly.
On the Unified State Exam in Russian, in task No. 8 you need to find the word from a list of words with the one being checked unstressed vowel in the root and write this word on the answer form. Thus, the task, compared to previous years, has become significantly more difficult. Now you need not only to find this word, but also to know very well how to spell it. A word spelled incorrectly but found correctly will be an incorrect answer.
Learn to choose correctly test words. In them, the vowel being tested should be stressed:
How to complete task No. 8
1. Eliminate alternating words from the list. They are not checked by stress, but are written according to the rule.
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Alternating letters A-O |
Alternating letters I-E |
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gar-gor |
ber-bir |
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clan-clone |
der-dir |
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creature |
mer-world |
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zar-zor |
per-feast |
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grow-grow-grow |
ter-tyr |
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log-log |
blesmt-blist |
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swim-swim |
stealth-steel |
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jump-skoch |
zheg-zhig |
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mak-mok |
cheat-cheat |
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equal-equal |
|
|
cas-cos |
A (i) - im, in (borrow - borrow) |
|
(understand – understand) |
2. Eliminate from the list words with an untestable vowel in the root. These words are easy to find - these are mainly words of foreign language origin:
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 |
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3. The remaining word will be the answer. Don't forget to check the accent on this word to be sure of the correct answer.
Train more, do tests and exercises. Task options No. 8 are given on our website.
GOOD LUCK!
Melnikova Vera Alexandrovna
Task No. 8 The Unified State Examination in the Russian language tests the skills of spelling roots. To complete this task, you need to analyze 5 words and determine which of them is missing the unstressed vowel of the root being tested.
Example task 8:
in..arthuos
ch..ruyuschiy
application
p..norama
dammit
Answer: _______________
The vowel at the root of a word can be verifiable, unverifiable, or alternating. As a rule, all three types are offered among the answer options.
Verified vowel in the root
To check the unstressed vowel in the root, you need to choose a word with the same root or change the form of the word so that the vowel is stressed.
1. Selection of a cognate word: om A fly - well A lky, din A micny - din A Mika
2. Changing the form of a word: in O yes - in O bottom, d O ska - d O ski
Important! You cannot check the vowel at the root with verbs with the suffixes -IVA- (-YVA-), since alternation O/A (op O give - n O ok, not ok A building yva t);
Words with the combinations -OLO-/-ORO- and -RA-/-LA- are checked only by words with the same combinations (to check the word CITY you cannot use the word GRAD...)
Unverifiable vowel at the root
There are a large number of words in which the vowel spelling is not checked by stress. Their spelling is checked by a dictionary, which is why they are called vocabulary.
For example: vinaigrette, conductor, avant-garde, dilemma, design, contrast, dependent, carnival, disaster, president, parody, comfort, blizzard, stepmother, summary, blasphemy, satire, self-interest, overpass...
Alternating vowels in the root
In the Russian language there are a number of roots in which the vowel can alternate depending on certain conditions (o/a, e/i):
Algorithm for completing task 8:
1) Highlight the roots in all words
2) Discard words with an alternating vowel in the root and vocabulary words that are familiar to you
3) For the remaining words, try to find test words in which the stress falls on the place of the gap
4) Write down the answer by inserting a vowel in the place of the blank
So, let's turn to the task:
Let's highlight the roots in all words: in?mertuoso, h?r cozy, with l?f tion, p?noram A, bl?st at;
thanks to this, we can identify two words with alternation in the root - application (lozh/lag) and shine (blist/blest), we exclude them.
We need to emphasize the vowels in the remaining words: ch?ruyuschiy - chary; words virtuoso And panorama it is impossible to check, they are dictionary.
ANSWER: charming.
2nd example:
8. Identify the word in which the unstressed vowel of the root being tested is missing. Write out this word by inserting the missing letter.
sk..chock
cr..branches
h..sink
updated
time to listen
1. Let’s highlight the roots in all words: sk..h OK, cr..branch And, h..kan it, about present day lassed, according to r..in yat; among these words we see 2 with an alternating vowel - jump (jump/skoch, we write A, because it is exclusive), level (even/equal, we write O, because the meaning is: even, straight);
2. Let's try to check the rest of the words with stress. The words mint and shrimp cannot be checked, they are dictionary words.
UPDATED - new, in a new way; write O
Answer: updated.
Unified State Exam. Russian language. How to easily complete task number 8?
Task No. 8 has 3 options.
1 version of task No. 8
Answer algorithm:
Eliminate all words with alternating vowel in the root. How to define them, read below (in this example these are words for MEP on CLONE happen)
Eliminate all words with unverifiable vowel in the root. How to find them, read below. (In this example, this is the word F E DERAL)
Don't confuse a letter missing from the root with a letter missing in the suffix(in this example it is WRONG AND FLAX)
Be sure to check the vowel in the root that you think has unstressed verifiable(approx. And ryat -m AND R).
Difficult. Don't confuse the alternating vowel with the one being checked. The main difference between an alternating root is that such a root always has pair with another letter, and these are words with the same root, so the meaning of the root is approximately the same.
Compare:
at WORLD at-at MEP net is a pair of words with different letters in the root, the meaning of the root is approximately the same.
BUT!!! etc WORLD five friends - this root meaning “peace” will never be written with a vowel E.
Difficult. Find the root correctly. For example, in the word NA SLAD The root is not crap(you could then easily write FALSE), and SLAD, so there is no alternating root here. This is the vowel being tested at the root of words - SWEET yy.
ANSWER in this example : reconcile enemies.
Write down the answer only when you were able to find a word in which the given letter is stressed. A correctly found word, but misspelled, is incorrect unswer!
Option 2 of task No. 8
Answer algorithm:
Eliminate all words with alternating vowel in the root (voz GROWTH,race STEL it)
Eliminate words with verifiable vowel in the root (dr. ABOUT beat - dr ABOUT b. development E waving flag E et)
Remember that most often words with unverifiable vowel - these are foreign words, that is, their meaning needs to be explained, it may be incomprehensible (to ABOUT institutional).However, there are words of non-foreign origin (def. E divide)
Answer : constitutional
Option 3 of task No. 8
Answer algorithm:
Eliminate words with verifiable vowel in the root (embedded) E tea-built E cha)
Eliminate all words with unverifiable vowel in the root (apl ABOUT dirate, art AND Lleriya, d AND rector)
Alternating roots memorize visually, just learn them and remember that they always have pair, root value they contain approximately same(with BIR at-so BER yeah)
In the task it is necessary not only to find the given word, but also to write it CORRECTLY. So teach rules, remember exceptions.
Answer: gather
Remember the roots with alternation:
|
Without accent - O gar-gor clan-clone creature Without accent - A zar-zor swim-swim Depends on the next letter in the root grow-grow-grow log-log jump-skoch Depends on the suffix A after the root cas+A - braid Depends on value equal-equal mak-mok |
Depends on the suffix A. If there is a suffix after the root A, then the root is written I ber-bir der-dir mer-world per- at ter-tyr best-blist stealth-steel zheg-zhig cheat-cheat Letters in the root them (borrow - borrow) in (start - start) |
Remember exceptions and write them correctly
Let's look at examples
Example 1
Sample reasoning
I find verifiable roots: op AND sanie-op AND shet, entertaining E chenie- rzvl E whose, vych AND slyat-h AND sla
I find alternating roots: you tvOR yat
Answer: l E pour
Example 2
Sample reasoning
I find verifiable roots(in this version they are not)
I find unverifiable roots (ek ABOUT lgia, g AND mnazist,s AND sympathy, this AND ketka)
I find alternating roots (beginning AND nashy)
Example 3
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Identify the word in which the unstressed word is missing tested vowel root Write out this word by inserting the missing letter. manufacturing ph..logy spread out offer E-mail |
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.):
| 1 2 3 4 5 | var k, s: integer ; begin s: = 512 ; k: = 0 ; while s |
var k,s:integer; begin s:=512; k:=0; while s
✍ Solution:
- 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:
| 1 2 3 4 5 6 7 8 9 10 11 | 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 . |
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.
✍ Solution:
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 ?
| 1 2 3 4 5 6 7 8 9 10 11 12 | 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. |
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.
✍ Solution:
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:
| 1 2 3 4 5 6 7 8 9 10 | var k, s: integer ; begin s: = 3 ; k: = 1 ; while k ‹ 25 do begin s: = s+ k; k: = k+ 2 ; end ; write(s); end. |
var k, s: integer; begin s:=3; k:=1; while k< 25 do begin s:=s+k; k:=k+2; end; write(s); end.
✍ Solution:
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)
| 1 2 3 4 5 6 7 8 9 10 | 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 . |
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.
✍ Solution:
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 ?
| 1 2 3 4 5 6 7 8 9 10 11 | 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 . |
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.
✍ Solution:
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.
| 1 2 3 4 5 6 7 8 9 10 11 | var s, n: integer ; begin s := 260 ; n:=0; while s › 0 do begin s : = s - 15 ; n : = n + 2 end ; writeln (n) end . |
var s, n: integer; begin s:= 260; n:= 0; while s > 0 do begin s:= s - 15; n:= n + 2 end; writeln(n)end.
✍ Solution:
- 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:
| 1 2 3 4 5 6 7 8 9 10 11 | var s, i: integer ; begin i := 1 ; s := 105 ; while s › 5 do begin s : = s - 2 ; i : = i + 1 end ; writeln (i) end . |
var s, i: integer; begin i:= 1; s:= 105; while s > 5 do begin s:= s - 2; i:= i + 1 end; writeln(i)end.
✍ Solution:
- 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.
| 1 2 3 4 5 6 7 | a: =- 5; c: = 1024 ; while a‹ › 0 do begin c: = c div 2 ; a: = a+ 1 end ; |
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.
✍ Solution:
- The loop condition depends on the variable n, which changes in a loop according to obtaining powers of two:
Let's consider the algorithm:
Write down the number that will be printed as a result of executing the following program:
Pascal:
| 1 2 3 4 5 6 7 8 9 10 11 | var s, n: integer ; begin s := 522 ; n:=400; while s - n > 0 do begin s : = s - 20 ; n := n - 15 end ; write (s) end . |
var s, n: integer; begin s:= 522; n:= 400; while s - n > 0 do begin s:= s - 20; n:= n - 15 end; write(s)end.
✍ Solution:
- The algorithm contains a loop. To understand the algorithm, let’s trace the initial iterations of the loop:
- We see that in the condition the difference between the values is 5 :
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:
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.
| A | avant-garde, adventure, lawyer, almanac, abstract, anomaly, antagonism, apartments, applause, appeal |
| B | luggage, boycott |
| IN | vacancy, magnificent, veterinarian, vinaigrette |
| G | dimensions, garrison, horizon |
| D | deserter, declaration, deficit, amateur, directive, thoroughly |
| AND | ignore, dependent, intelligent, inquisition |
| TO | quotes, pun, calendar, closet, carnival, disaster, insidious, conjure, overalls, competent, compose, compromise, constitutional, burner, luminary, cosmetology, criterion |
| L | cherish |
| M | meridian, philanthropist, motivation |
| N | obsession, nostalgia |
| ABOUT | original |
| P | front garden, panorama, paradox, pessimist, foam rubber, preliminary, fastidious, privilege, primitive, priority, pedestal |
| R | rehabilitation, regulations, residence, rehearsal, restore |
| WITH | seminar, certificate, lilac, scholarship, strive, sovereignty |
| U | compact |
| F | faculty, philharmonic, festival |
| Sh | chocolate, chauvinism, highway, parade |
| E | crew, exhibit, experiment, excavator, element, operation, extreme, expedition, erudition |
Algorithm for completing the task
- 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).
- We insert the missing vowels into each word given in the task, determine the rule on which the spelling of each word is based.
- 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
- m..tsenat
- look...see
- mountainous (terrain)
- grow up
- comp..nent
Execution algorithm:
- 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
- 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.
- exp...dication
- hydrogen...sli
- k...lendary
- k...shaky
- post...pour
Execution algorithm:
- An unstressed checkable vowel is a vowel that can be checked by changing the word and placing it under stress: MOUNTAIN – MOUNTAINS.
- 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.
- 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.
- v..rsy
- adventure
- g..roar
- accept..mother
- adv..cat
Execution algorithm:
- Roots with alternating vowels: ber - bir, kas - kos, lag - lodge, etc.
- 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).
- 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.
- op..sanie
- l...lay
- dry out
- entertainment
- calculate
Execution algorithm:
- An untestable vowel at the root of a word is a vowel whose spelling needs to be remembered (for example: the vinaigrette).
- 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.
- Cherish- a word with an unverifiable vowel at the root; its spelling must be remembered.