Let's assume that x is just a real number. They write it like this: x∈ℝ - read x belongs to the set of real numbers. To the whole set, element 0 is also included there. And there is no trick here with the elements of x: infinity is not part of the set to which x belongs, and x is not an antiderivative.
I will offer another solution, not yet mentioned here:
First, let's introduce some definition:
A group is a set of elements of arbitrary nature with a (single!) operation introduced between them (denoted in this case by +), which has the following properties:
(Let's denote our group by the letter G)
1) Closedness: ∀x,y ∈ G ⇒ x+y ∈ G. It reads like this: for any two elements x and y from the group G it follows that their sum is also an element of the group G
2) Associativity: ∀x,y,z ∈ G ⇒ (x+y)+z = x+(y+z). It reads like this: for any three elements x,y,z belonging group G it follows that you can first apply the group operation to the elements x and y, and as a result obtain some element (x+y) ∈ G, and then apply the group operation to the elements (x+y) and z. The resulting element must be equal to the element that was obtained by applying the operation first to y and z, and then to x and (y+z). That is, to put it simply, rearranging the brackets does not change the result: (x+y)+z = x+(y+z)
3) ∀x ∈ G ⇒ ∃e ∈ G: x + e = e+ x = x. It reads like this: The group must have an element e (called the group unit), such that if we apply the group operation e + x, and then x + e - the same element x should be obtained. That is, the group unit, when added to the left and right, does not “shift” the group element.
4) ∀x ∈ G ⇒ ∃x⁻¹: x + x⁻¹ = x⁻¹ + x = e. It reads like this: Any element x in the group G has an inverse such that the result of the operation between x and x⁻¹ is on the left and right equal to one groups.
You need to understand that the + operation in a group can be absolutely anything. The + symbol is just a designation for this operation. It is most correct to say that x+y = f(x,y)
where f is some function that returns an element of the group.
Examples of groups and non-groups. (You can skip this paragraph):
For example, the set ℤ (integers) is a group if we introduce into it the operation of the usual addition we are all familiar with. It is closed, for any x ∈ ℤ the inverse is the element -x, since x + (-x) = 0. The unit of the group is 0. And associativity, of course, holds.
However, if we consider the set ℤ with the operation of standard multiplication introduced on it, then such a structure will no longer be a group - despite the fact that there is a unit: x*1 = x and it is associative: (x*y)*z = x*(y *z), in the set of integers there are no inverse elements for any elements except one. Really. For example, the inverse element for multiplication for the number 4 is 1/4, because 4 * (1/4) = 1. But 1/4 is not included in the set of integers. 1/4 is a rational number.
But if we remove the element 0 from the set ℚ (rational numbers), then if we introduce the operation of standard multiplication on ℚ, then ℚ will be a group, because there are inverses and unity, and it is associative and closed.
So let's try to see. what should be the operation on the set ℝ (of real numbers) so that a solution to the equation x⊕1=x exists in it. Where ⊕ is the group operation designation.
Let us introduce the operation x⊕y = x+y-1
Then the unit of our group will be element 1, because x⊕1 = 1⊕x = x + 1 - 1 = x.
That is, x⊕1 = x.
The inverse element will be: (2-x), because x⊕(2-x) = (2-x)⊕x = x + (2-x) - 1 = 1 (group unit)
Associativity, obviously. completed:
(x⊕y)⊕z = (x + y - 1)⊕z = x + y - 1 + z - 1 = x + y + z - 2
x⊕(y⊕z) = x⊕(y+z-1) = x + y + z - 1 - 1 = x + y + z - 2 = (x⊕y)⊕z
In addition, it is easy to see that our group is closed with respect to the introduced operation.
So, we have checked that the set with the operation we introduced is a group, let’s see how our equation fares there if x belongs to the group we constructed.
x⊕1 = x. But when we checked whether the structure we constructed was a group, we already found out that 1 in our group is the unit of the group and the property x⊕1=x in it is obviously satisfied for any elements of the group we constructed.
It’s interesting that in the group we constructed 0⊕0 = -1 :)
The author obviously hinted that the + operation is addition. But from the point of view of group theory, the set ℝ with the operation of ordinary addition introduced in it is no different from the set ℝ/(0) (the set of real numbers, but one element was removed from it - 0) with the operation of usual multiplication introduced in it. And in algebra + usually means that the set ℝ is used (without throwing out zero). The solution takes this into account - at the very beginning I mentioned that 0∈ℝ.
If not for this condition, one could simply assume that x⊕y = x*y.
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)
What's happened "quadratic inequality"? No question!) If you take any quadratic equation and replace the sign in it "=" (equal) to any inequality sign ( > ≥ < ≤ ≠ ), we get a quadratic inequality. For example:
1. x 2 -8x+12 ≥ 0
2. -x 2 +3x > 0
3. x 2 ≤ 4
Well, you understand...)
It’s not for nothing that I linked equations and inequalities here. The point is that the first step in solving any quadratic inequality - solve the equation from which this inequality is made. For this reason, the inability to solve quadratic equations automatically leads to complete failure in inequalities. Is the hint clear?) If anything, look at how to solve any quadratic equations. Everything is described there in detail. And in this lesson we will deal with inequalities.
The inequality ready for solution has the form: left - quadratic trinomial ax 2 +bx+c, on the right - zero. The inequality sign can be absolutely anything. The first two examples are here are already ready to make a decision. The third example still needs to be prepared.
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
Things to consider for those looking for a reliable bookmaker:
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For example, total number search queries in Yandex for October 2018 for the words “1xBet”, “1khbet”, “1xstavka”, “1xstavka” amounted to 3,395,544, which is much more than the number of search queries for other bookmakers. (Similar dynamics characterize the Google search engine.)
It should be taken into account that there is practically no difference between the sites of bookmakers 1XSTAVKA and 1XBET (in terms of interface, functionality, events offered for betting, odds, etc.). You can find out more about the differences between them
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What is total in sports betting?
Total (from the English “total” = “sum, total”) in sports betting is the number of goals scored in a sporting event (in football and hockey), points scored (in basketball and volleyball) or games played (in tennis). In addition to those listed, in bookmakers you can bet on the total on a large number of other various events, for example, on the number of corners in football, misses of one or another participant in biathlon, the number of penalty minutes in hockey, the number of points in a game in tennis, the number of games played (in volleyball), etc. But, once again, it should be noted that in betting on football in the main line, the total is always the number of goals scored during the main time of the football match. And in the future in this post, betting on totals in football will mean betting on the number of goals.
Total bets are divided into two types:
1. Bets on total “over” (or TB) – i.e. bets placed on the fact that more goals will be scored in a match than the total indicated on the line.
2. Bets on total “under” (or TM) – i.e. bets placed on the fact that fewer goals will be scored in a match than the total indicated on the line.
Total under 1 and total over 1 in football
Total less than 1 (or Under 1)
A bet on TM 1 wins if no goals are scored in the match (a 0:0 draw is recorded) and loses if more than one goal is scored in the match. And if one goal is scored in the match, then the bet amount is returned (i.e. in this case, the bet on TM 1 did not lose or win).
Thus, if the match ends with a score of 0:0, then the TM 1 bet will win.
If two or more goals are scored in the match (for example, 2:0, 1:2, 3:1, 2:2, etc.), then the bet on TM 1 will lose.
Moreover, in the case when only one goal is scored in the match (with one of the following possible outcomes: 1:0 or 0:1), such a bet is calculated with a coefficient equal to 1 (K=1). In other words, then the player who bet on TM 1 gets the bet amount returned.
Total over 1 (or TB 1)
A bet on Over 1 wins if more than one goal is scored in the match and loses if no goals are scored in the match (if it ends with a score of 0:0). And if one goal is scored in the match, then the bet amount is returned (i.e. in this case there is no loss or win).
Accordingly, if the match is goalless and ends with the score 0:0, then the TB 1 bet will lose.
If two or more goals were scored in the match (for example, with the score 1:4, 2:0, 3:1, 0:3, etc.), then the bet on Over 1 will win.
Moreover, in the case when one goal is scored in the match (with one of the following possible outcomes: 1:0 or 0:1), such a bet is calculated with a coefficient equal to 1 (K=1) and the player who bet on TB 1, the bet amount is returned.
Accordingly, with probable large quantities goals, a bet is placed on a higher total, and if there is a higher probability that there will be few goals scored, then a bet is placed on a lower total.
As a rule, bookmakers accept TM 1 or TB 1 bets on the result of one period of a football match or as in a match of one of the rival teams.
Below are the odds for the “Individual Total Sat” bet. Turkey: TB 1 (total over 1)" for the qualifying match of the national teams for the 2018 FIFA World Cup Ukraine - Turkey. (Date and time of the match: 09/02/2017 at 21:45 (Moscow time). Odds values are presented as of 19:00 (Moscow time) 09/02/2017)
|
Place |
Bookmaker name and corresponding |
Coefficient value Ind.TB 1 (Türkiye) |
|
2.26 |
||
|
2.22 |
||
|
2.16 |
||
|
bid is not accepted |
An example of calculating a bet on total over 1 (TB 1):
Consider the above match between the Ukrainian national team and the Turkish national team. Let's say a bet of 1000 rubles is made. for Turkey's individual total is more than 1 (Individual total of the second team in a pair of Over 1) with a coefficient of 2.26.
Thus, if the Turkish team does not score a single goal, then this bet will lose; if Türkiye scores two or more goals, this bet will win; and if Türkiye scores one goal, the bet amount will be refunded.
At the same time, the number of goals scored by the Ukrainian national team in the example under consideration does not matter, because The bet is placed on the individual total of the Turkish national team.
Let us consider the solution of trigonometric inequalities of the form tgx>a and tgx
To solve, we need a drawing of a unit circle and. The radius of the unit circle is equal to 1, therefore, plotting segments on the line of tangents whose length is equal to the radius, we obtain, respectively, points at which the tangent is equal to 1, 2, 3, etc., and downwards - -1, -2, -3 and etc. On the tangent line, tangent values greater than a correspond to the part located above point a. Shade the corresponding ray. Now we draw a straight line through point O - the origin - and point a on the tangent line. It intersects the circle at point arctan a. Accordingly, on the circle the solution to the inequality tgx>a corresponds to an arc from the point arctg a to n/2. To take into account all solutions (and there are an infinite number of them, taking into account the periodicity of the tangent), we add nn to each end of the interval, where n is an integer (n belongs to Z). To solve the inequality tgx>a, a semicircle from -n/2 to n/2 is quite sufficient. But if you need to find, for example, a solution to a system of inequalities with tangent and sine, then you need the entire circle. If the inequality is not strict, we include the point with arctan a in the answer (we shade it in the figure and write it in the answer with a square bracket). The point n/2 is never included in the answer, since it is not included in the area of definition of the tangent (the point is punctured, the parenthesis is round). To solve the inequality tgx>-a, we reason in the same way as for the inequality tgx>a. Since arctg (-a)=-arctg a, this is the only difference in the answer.
+ 
+