I. To divide a number by a decimal fraction, you need to move the commas in the dividend and divisor as many digits to the right as there are after the decimal point in the divisor, and then divide by the natural number.
Primary.
Perform division: 1) 16,38: 0,7; 2) 15,6: 0,15; 3) 3,114: 4,5; 4) 53,84: 0,1.
Solution.
Example 1) 16,38: 0,7.
In the divider 0,7 there is one digit after the decimal point, so let’s move the commas in the dividend and divisor one digit to the right.
Then we will need to divide 163,8 on 7 .
We divide as natural numbers are divided. How to remove the number 8 - the first digit after the decimal point (i.e. the digit in the tenths place), so immediately put a comma in the quotient and continue dividing.
Answer: 23.4.
Example 2) 15,6: 0,15.
We move commas in the dividend ( 15,6 ) and divisor ( 0,15 ) two digits to the right, since in the divisor 0,15 there are two digits after the decimal point.
We remember that you can add as many zeros as you like to the decimal fraction on the right, and this will not change the decimal fraction.
15,6:0,15=1560:15.
Perform division natural numbers.
Answer: 104.
Example 3) 3,114: 4,5.
Move the commas in the dividend and divisor one digit to the right and divide 31,14 on 45 By
3,114:4,5=31,14:45.
In the quotient we put a comma as soon as we remove the number 1 in the tenth place. Then we continue dividing.
To complete the division we had to assign zero to the number 9 - differences between numbers 414 And 405 . (we know that zeros can be added to the right side of a decimal fraction)
Answer: 0.692.
Example 4) 53,84: 0,1.
Move the commas in the dividend and divisor to 1 number to the right.
We get: 538,4:1=538,4.
Let's analyze the equality: 53,84:0,1=538,4. Pay attention to the comma in the dividend in in this example and a comma in the resulting quotient. We notice that the comma in the dividend has been moved to 1 number to the right, as if we were multiplying 53,84 on 10. (Watch the video “Multiplying a decimal by 10, 100, 1000, etc..") Hence the rule for dividing a decimal fraction by 0,1; 0,01; 0,001 etc.
II. To divide a decimal by 0.1; 0.01; 0.001, etc., you need to move the decimal point to the right by 1, 2, 3, etc. digits. (Dividing a decimal by 0.1, 0.01, 0.001, etc. is the same as multiplying that decimal by 10, 100, 1000, etc.)
Examples.
Perform division: 1) 617,35: 0,1; 2) 0,235: 0,01; 3) 2,7845: 0,001; 4) 26,397: 0,0001.
Solution.
Example 1) 617,35: 0,1.
According to the rule IIdivision by 0,1 is equivalent to multiplying by 10 , and move the comma in the dividend 1 digit to the right:
1) 617,35:0,1=6173,5.
Example 2) 0,235: 0,01.
Division by 0,01 is equivalent to multiplying by 100 , which means we move the comma in the dividend on 2 digits to the right:
2) 0,235:0,01=23,5.
Example 3) 2,7845: 0,001.
Because division by 0,001 is equivalent to multiplying by 1000 , then move the comma 3 digits to the right:
3) 2,7845:0,001=2784,5.
Example 4) 26,397: 0,0001.
Divide a decimal by 0,0001 - it's the same as multiplying it by 10000 (move the comma by 4 digits right). We get:
II. To divide a decimal fraction by 10, 100, 1000, etc., you need to move the decimal point to the left by 1, 2, 3, etc. digits.
Examples.
Perform division: 1) 41,56: 10; 2) 123,45: 100; 3) 0,47: 100; 4) 8,5: 1000; 5) 631,2: 10000.
Solution.
Moving the decimal point to the left depends on how many zeros after the one are in the divisor. So, when dividing a decimal fraction by 10
we will carry over in the dividend comma to the left one digit; when divided by 100
- move the comma left two digits; when divided by 1000
convert to this decimal fraction comma three digits to the left.
In examples 3) and 4) we had to add zeros before the decimal fraction to make it easier to move the comma. However, you can assign zeros mentally, and you will do this when you learn to apply the rule well II to divide a decimal fraction by 10, 100, 1000, etc.
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The expression “most likely” causes many difficulties with punctuation, since it may or may not require commas depending on its role in the sentence (context). However, learning to determine whether separation is necessary in a given situation is not a difficult matter.
Introductory construction
To correctly place punctuation marks, you need to determine whether the expression “most likely” is an introductory phrase.
What does it mean?
Introductory word (or stable combination words) is a construction that is not a member of a sentence and is not syntactically related to any of its members. It is impossible to ask a question to her either from the subject, or from the predicate, or from minor members, it is also impossible for her to ask questions to other members.
Introductory words can, for example, convey the emotional coloring of a sentence (“fortunately,” “unfortunately”), express confidence (“of course,” “of course”) or uncertainty (“probably,” “maybe”) of the author, or indicate a reference to someone's opinion (“in my opinion”, “they say”).
“Most likely” is highlighted with commas if this is an introductory phrase with the meaning of uncertainty, since an introductory word or expression always requires isolation.
How to determine this?
- Introductory turnover can be rearranged into any part of the sentence without losing meaning. If “most likely” is at the beginning of a sentence, then it can be used at the end or middle, and the essence of the sentence will remain unchanged.
- The introductory phrase can be replaced by any other synonymous introductory construction. You should try to replace the introductory expression “most likely” introductory words"probably" or "maybe" construction. If “most likely” is an introductory word, then the degree of confidence will change, but the meaning of the statement will not disappear.
- The introductory turnover can be excluded. The sentence must remain grammatically correct.
If the conditions are met, “most likely” is separated by commas.
A phrase consisting of an adjective and a pronoun
The word "more likely" can be an adjective in comparative degree and be part of the predicate. Then "total" is dependent word also as part of the predicate, is attributive pronoun.
How to determine this?
It is enough to check the same three conditions.
If the conditions are not met, that is, when discarded, moved to another part of the sentence or replaced with introductory constructions “maybe”, “probably” the sentence loses its meaning or becomes grammatically incorrect, “most likely” is not separated by commas.
Examples
Consider two similar proposals:
This behavior was most likely predicted in advance.
This behavior was most likely.
In the first case, to understand whether commas are needed, we move them to the beginning of the sentence “most likely”:
Most likely, this behavior was predicted in advance.
Replace the phrase with “probably”:
This behavior was probably predicted in advance.
Now let's try to discard the phrase in question:
This behavior was predicted in advance.
In all three cases, the sentence retained its meaning and remained grammatically correct. It can be concluded that in this proposal"Most likely" - introductory construction. Separate with commas on both sides. Of course, except at the very beginning or end of a sentence, when a comma on one side is enough.
Let's move on to the second sentence.
Let's move "most likely" to the beginning of the sentence.
Most likely this was the behavior.
As you can see, the result is a phrase that is extremely inconvenient to understand. But to be sure, let's check the other two signs.
Let's replace it with "probably":
This kind of behavior probably happened.
The meaning is completely lost.
If we discard “most likely”, we are left with:
There was such behavior.
In this case, too, the meaning is completely lost.
Conclusion: in the considered sentence, “most likely” is not an introductory word. This means we don’t separate “most likely” with commas.
To fold decimals, need to: 1) equalize the number of decimal places in these fractions; 2) write them one below the other so that the comma is written under the comma; 3) perform the addition without paying attention to the comma, and put a comma in the sum under the commas in the added fractions.
Examples. Add decimals.
1) 0,07+13,23.
Solution. Let's apply the commutative law of addition: 0.07 + 13.23 = 13.23 + 0.07 and write the fractions under each other so that the comma is under the comma. Add it together, ignoring the comma. In the resulting amount, put a comma under the commas in the terms. The zero at the end of the resulting result 13.30 can be discarded.
13,23+0,07=13,3.
2) 11,21+9,3.
Solution. We write these fractions one below the other so that the comma is under the comma. We equalize the number of decimal places in the terms. To do this, we add a zero to the right of the fraction 9.3. We add, not paying attention to the commas, and put a comma under the commas in the terms in the total.
11,23+9,3=20,51.
3) Calculate in a rational way. 1,245+(0,755+3,02).
Solution. We use commutative and associative laws addition.
1,245+(0,755+3,02)=(1,245+0,755)+3,02=2+3,02=5,02.
Explanation: the terms 1.245 and 0.755 have the same number of decimal places (three digits each), therefore, it is convenient to add them verbally, like adding whole numbers, and then separate three digits to the right with a comma, as was the case in the terms. It turned out to be 2,000. We discard three zeros after the decimal point, we get the number 2. We added 3.02 and got 5.02.
1,245+(0,755+3,02)=5,02.
- One hundredth part is called a percentage.
- To express percentages as a fraction or a natural number, you need to divide the percentage by 100%. (4%=0.04; 32%=0.32).
- To express a number as a percentage, you need to multiply it by 100%. (0.65=0.65·100%=65%; 1.5=1.5·100%=150%).
- To find the percentage of a number, you need to express the percentage as a common or decimal fraction and multiply the resulting fraction by the given number.
- To find a number by its percentage, you need to express the percentage as an ordinary or decimal fraction and divide the given number by this fraction.
- To find what percentage the first number is from the second, you need to divide the first number by the second and multiply the result by 100%.
Example 1. Express percentages as a fraction or natural number: 130%, 65%, 4%, 200%.
- 130% =130%:100%=130:100=1,3 ;
- 65% =65%:100%=65:100=0,65 ;
- 4% =4%:100%=4:100=0,04 ;
- 200% =200%:100%=200:100=2 .
Example 2. Write the following numbers as a percentage: 1; 1.5; 0.4; 0.03.
- 1 =1·100%= 100% ;
- 1,5 =1.5·100%= 150% ;
- 0,4 =0.4·100%= 40% ;
- 0,03 =0.03·100%= 3% .
Example 3. Find 15% of the number 400.
1) 15%=15%:100%=15:100=0,15;
2) 0.15·400=60.
Example 4. Find a number if 18% of it is 900.
1) 18%=18%:100%=18:100=0,18;
2) 900:0,18=90000:18=5000.
Answer: 5000.
Example 5. Determine what percentage the number 320 is from the number 1600.
(320:1600)·100%=0.2·100%=20%.
Answer: 20%.
- The method is plotting each equation included in this system, one coordinate plane and finding the intersection points of these graphs V. Coordinates of this point (x; y) and will appear decision of this system of equations.
- If straight intersect, then the system of equations has the only thing solution.
- If straight, which are graphs of the system equations, parallel, then the system of equations has no solutions.
- If straight, which are graphs of the system equations, match, then the system of equations has infinite many solutions.
Examples. Decide graphically system of equations.
The graph of each equation is a straight line, to construct which it is enough to know the coordinates two points. We have compiled tables of values X And at for each of the system equations.
The straight line y=2x-3 was drawn through the points (0; -3) and (2; 1).
The straight line y=x+1 was drawn through the points (0; 1) and (2; 3).
Graphs of these system equations 1) intersect at point A(4; 5). That's what it is only decision of this system.
Answer: (4; 5).
We express at through X from each equation of the system 2)
, and then create a table of variable values X And at for each of the resulting equations.
We draw the straight line y=2x+9 through the points (0; 9) and (-3; 3). We draw the straight line y=-1.5x+2 through the points (0; 2) and (2; -1).
Our lines intersected at point B(-2; 5).
Answer: (-2; 5).
1) Square of the sum of two expressions equal to square the first expression plus twice the product of the first expression and the second plus the square of the second expression.
(a+b) 2 = a 2 +2ab+b 2
A) (x + 2y ) 2 = x 2 + 2 x 2y + (2y ) 2 = x 2 + 4xy + 4y 2
b) (2k + 3n) 2 = (2k) 2 + 2 2k 3n + (3n) 2 = 4k 2 + 12kn + 9n 2
2) Square of the difference of two expressions is equal to the square of the first expression minus twice the product of the first expression and the second plus the square of the second expression.
(a-b) 2 = a 2 -2ab+b 2
a) (2a – c) 2 = (2a) 2 -2 2a c + c 2 = 4a 2 – 4ac + c 2
b) (3a – 5b) 2 = (3a) 2 -2 3a 5b + (5b) 2 = 9a 2 – 30ab + 25b 2
3) Difference of squares of two expressions is equal to the product of the difference between the expressions themselves and their sum.
a 2 –b 2 = (a–b)(a+b)
a) 9x 2 – 16y 2 = (3x ) 2 – (4y ) 2 = (3x – 4y )(3x + 4y )
b) (6k – 5n)(6k + 5n) = (6k) 2 – (5n) 2 = 36k 2 – 25n 2
4) Cube of the sum of two expressions equal to cube the first expression plus triple the product of the square of the first expression and the second plus triple the product of the first expression and the square of the second plus the cube of the second expression.
(a+b) 3 = a 3 +3a 2 b+3ab 2 +b 3
a) (m + 2n) 3 = m 3 + 3 m 2 2n + 3 m (2n ) 2 + (2n ) 3 = m 3 + 6m 2 n + 12mn 2 + 8n 3
b) (3x + 2y) 3 = (3x) 3 + 3 (3x) 2 2y + 3 3x (2y) 2 + (2y ) 3 = 27x 3 + 54x 2 y + 36xy 2 + 8y 3
5) Cube of difference of two expressions is equal to the cube of the first expression minus three times the product of the square of the first expression and the second plus three times the product of the first expression and the square of the second minus the cube of the second expression.
(a-b) 3 = a 3 -3a 2 b+3ab 2 -b 3
a) (2x – y ) 3 = (2x ) 3 -3 (2x ) 2 y + 3 2x y 2 – y 3 = 8x 3 – 12x 2 y + 6xy 2 – y 3
b) (x – 3n) 3 = x 3 -3 x 2 3n + 3 x (3n) 2 – (3n) 3 = x 3 – 9x 2 n + 27xn 2 – 27n 3
6) Sum of cubes of two expressions is equal to the product of the sum of the expressions themselves and the incomplete square of their difference.
a 3 +b 3 = (a+b)(a 2 –ab+b 2)
a) 125 + 8x 3 = 5 3 + (2x ) 3 = (5 + 2x )(5 2 – 5 2x + (2x ) 2) = (5 + 2x)(25 – 10x + 4x 2)
b) (1 + 3m)(1 – 3m + 9m 2 ) = 1 3 + (3m) 3 = 1 + 27m 3
7) Difference of cubes of two expressions is equal to the product of the difference between the expressions themselves and the partial square of their sum.
a 3 -b 3 = (a-b)(a 2 +ab+b 2)
a) 64s 3 – 8 = (4s) 3 – 2 3 = (4s – 2)((4s) 2 + 4s 2 + 2 2) = (4s – 2)(16s 2 + 8s + 4)
b) (3a – 5b)(9a 2 + 15ab + 25b 2) = (3a) 3 – (5b) 3 = 27a 3 – 125b 3
Dear friends! will help you choose the right topic.
There are systems oral counting allowing you to count orally quickly and rationally. We will look at some of the most commonly used techniques.
1) Multiplying a two-digit number by 11.
When multiplying a two-digit number by 11, the digits of this number are moved apart and the sum of these digits is placed in the middle.
Examples.
a) 23 11=253, because 2+3=5;
b) 45 11=495, because 4+5=9;
c) 57 11=627, because 5+7=12, the two was placed in the middle, and the one was added to the hundreds place;
d) 78 11=858, since 7+8=15, then the number of tens will be equal to 5, and the number of hundreds will increase by one and will be equal to 8.
And if we multiply decimal fractions, then we multiply without paying attention to the comma, and then in the resulting result we separate as many digits on the right with a comma as there were after the commas in both factors together.
a) 3, 8 0.11=0.418, because 38 11=418 and separate the 3 digits on the right with a comma (1+2);
b) - 0.32 1.1 = - 0.352. Product of numbers with different signs there is a negative number. 32 11 = 352 and separated the 3 digits on the right with a comma.
c) 0.062 1100 = 68.2. We multiplied 62 by 11, got 682, added 2 zeros, got 68200 and separated 3 digits on the right with a comma. It turned out 68.200=68.2;
d) - 730 (-0.011) = 8.03. Product of two negative numbers is a positive number. We multiply 73 by 11, it becomes 803, add zero to the right and separate 3 digits on the right with a comma.
2) Work double digit numbers, which same number tens, and the sum of units is 10, i.e. 23 27; 34 36; 52 58 etc.
Rule: the tens digit is multiplied by the next digit in the natural series, the result is written down and the product of units is added to it.
a) 23 27=621. How did you get 621? We multiply the number 2 by 3 (the “two” is followed by “three”), it becomes 6 and next to it we add the product of ones: 3 7 = 21, it turns out 621.
b) 34 36 = 1224, since 3 4 = 12, we assign 24 to the number 12, this is the product of the units of these numbers: 4 6.
c) 52 58 = 3016, because we multiply the tens digit 5 by 6, it will be 30, we assign the product of 2 and 8, i.e. 16.
d) 61 69=4209. It is clear that 6 was multiplied by 7 and we got 42. Where does zero come from? The units were multiplied and we got: 1 9 = 9, but the result must be two-digit, so we take 09.
Just as in the previous examples, the multipliers can be decimal fractions, for example, 0.34 (-3.6) = - 1.224. (see example 2b))
3) Division of three-digit numbers consisting of identical numbers, to the number 37. Result equal to the sum these identical numbers three-digit number(or a number equal to three times the digit of a three-digit number).
a) 222:37=6. This is the sum 2+2+2=6.
b) 333:37=9, because 3+3+3=9.
c) 777:37=21, i.e. 7+7+7=21.
d) 888:37=24, because 8+8+8=24.
We also take into account that 888:24=37.
If we again take decimal fractions as factors, then the number of such examples becomes enormous! We also remember the rule for dividing a number by a decimal fraction: to divide a number by a decimal fraction, you need to move the decimal point in the dividend and divisor to the right by as many digits as there are after the decimal point in the divisor, and then divide by the natural number.
a) 77.7:0.37=7770:37=210;
b) - 0.444:3.7= - 4.44:37= - 0.12;
c) 9.99: (- 0.27) = - 999:27 = - 37;
d) - 5.55: (- 0.037) = 5550:37 = 150.
If you now come up with your own examples for each of the three rules above, you will learn these simple techniques better and will surprise your classmates and teachers by producing quite complex calculations without using a calculator! Good luck!
But as? Medicines for this disease are necessary knowledge! What knowledge? There are not so many of them:
1) Addition table within one ten (two tens).
Mentally imagine: from the sum of which two natural numbers can the number 10 be made.
1+9, 2+8, 3+7, 4+6, 5+5. Do we remember that rearranging the terms does not change the sum? Fine.
How to get 20?
1+19, 2+18, 3+17, 4+16, 5+15, 6+14, 7+13, 8+12, 9+11, 10+10. Wonderful.
2) Add the numbers bit by bit: units with ones, hundreds with hundreds, thousands with thousands, etc.
3) Multiplication table. Let’s not be ashamed to take a thin squared notebook with a multiplication table on the cover and repeat: twice two is four, etc.
4) Table of squares of two-digit numbers from 11 to 30.
11 2 =121, 12 2 =144, 13 2 =169, 14 2 =196, 15 2 =225, 16 2 =256,…,30 2 =900. If you compile this table yourself, remember it better.
5) Some powers of numbers 2, 3, 5, 7.
2 2 =4, 2 3 =8, 2 4 =16, 2 5 =32, 2 6 =64, 2 7 =128, 2 8 =256,2 9 =512, 2 10 =1024.
3 2 =9, 3 3 =27, 3 4 =81, 3 5 =243, 3 6 =729.
5 2 =25, 5 3 =125, 5 4 =625
7 2 =49, 7 3 =343.
6) Signs of divisibility of numbers.
If a number ends in an even digit (0, 2, 4, 6, 8), then the number is divisible by 2 without a remainder.
If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 3. For example, we find out whether the number 126795 is divisible by 3. We add the digits of the number: 1+2+6+7+9+5=30. The number 30 is divisible by 3, which means the number 126795 itself is divisible by 3.
If the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9.
If a number ends in “0” or “5,” then the number itself is divisible by 5 without a remainder. For example, the number 126795 is divisible by 5.
If a number ends in “0”, then the number is divisible by 10 without a remainder.
If a number made up of the last two digits given number, is divisible by 4, then the number itself is divisible by 4. For example, 2012 is divisible by 4, since 12 is divisible by 4. The number 345284 is divisible by 4, since 84 is divisible by 4.
These signs of division are enough to reduce fractions, for example.
And if a number is divisible by 3 and 5, then it is divisible by 15. Example: the number 126795 is divisible by 15.
Try to forget your calculator, at least for a while! Good luck!
Where do gaps in student knowledge come from?
Due to missing classes - you answer! And you will be right only 20%. If only it were that simple! If you think about this problem, you can remember cases when a student who missed a new topic, but mastered it at home on his own or with his parents, tutor, or others, knows it better than those who WAS at school and PRESENTED at the lesson. How did this happen? Let's try to figure it out.
The teacher explains a new topic. As a rule, students listen carefully. After one explanation from the teacher, few understand the topic (meaning the key topic of the program). An experienced teacher explains the topic again, using synonymous words. Several more students are added to the first to understand the new topic, but, unfortunately, not the whole class. Those who understand the topic (I remind you: there are few of them yet, but they are leaders) urge the teacher: “Let’s solve examples (problems)!” What does the teacher do? That's right - “surrender”. After all, the lesson is not “rubber”, and you need to reinforce the topic with examples. We started to decide. In the process of applying new theoretical knowledge in practice, several more students “got it” new topic, but most likely the knowledge acquired last group students will be formal: they will be able to solve only similar examples, i.e. This knowledge may already remain formal and will disappear immediately after completing the topic. But there were still those students who did not understand the topic either immediately or with subsequent examples. If they do not receive help at home, then there is a gap in their knowledge. But what about those “prosperous” children who understood everything in class? Are they immune to knowledge gaps on the topic? No, they will be in the “risk zone” until they INDEPENDENTLY fulfill the written homework and they won’t memorize the formulas (rules). If on this topic If at least three lessons are allocated, then an experienced teacher is able to organize work in the lessons so that not a single child remains in the “risk zone”. Then everything is fine? Yes, but only for a while. It’s not for nothing that they say: repetition is the mother of learning. And teachers are ready to repeat old material and explain new material, and then consolidate it and repeat everything again in order to eliminate gaps in the students’ knowledge, but we must remember that all our efforts will be justified only if the students themselves want to learn. That's why, Dear Guys, do not hesitate to ask questions to the teacher in class, ask for repeated explanations until you understand the essence of the topic. Be sure to learn all the new formulas, because after each lesson there are not many of them! Don’t accumulate problems, solve them as they arise. Don't neglect your homework: the teacher knows what and how much to assign so that you get solid knowledge. LEARN TO LEARN!
"Floating point" and "floating point"
Since in some, predominantly English-speaking and anglicized countries (see detailed list Decimal separator (English)) when writing numbers whole part separated from a fractional point, then in the terminology of these countries the name “floating point” appears. Since in Russia the integer part of a number is traditionally separated from the fractional part by a comma, the term “floating point” is used to denote the same concept.
origin of name
The name "floating point" comes from the fact that the comma in the positional representation of a number (decimal point, or, for computers, binary point - hereafter simply comma) can be placed anywhere relative to the digits in the string. This comma position is specified separately in the internal representation. Thus, representing a number in floating point form can be considered as a computer implementation of exponential notation for numbers.
The advantage of using a floating-point representation of numbers over a fixed-point (and integer) representation is that you can use a substantially larger range of values while maintaining the same relative precision. For example, in fixed-point form, a number occupying 8 integer places and 2 decimal places could be represented as 123456.78; 8765.43; 123.00 and so on. In turn, in floating point format (in the same 8 bits) you can write the numbers 1.2345678; 1234567.8; 0.000012345678; 12345678000000000 and so on.
The speed at which a computer performs operations with numbers represented in floating point form is measured in English units. FLOPS - number of floating point operations per second ),
Number structure
A floating point number consists of:
- Mantissa (expressing the value of a number without regard to order)
- Mantissa sign (indicating whether a number is negative or positive)
- Order (expressing the power of the base of the number by which the mantissa is multiplied)
- Sign of order
Normal form
Normal form a floating point number is a form in which the mantissa (without taking into account the sign) is located on the half-interval)