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Anonymous users2024-02-05b=x-y then a+b= ?What about A-B?
Solution: a+b=2x; a-b=2y；
2After learning the addition and subtraction of integers, the teacher assigned the following problem: find the value of (5x 2+4x-3)-(x 2-3x-1)+(4-7x-6x 2), where x=2011, Xiao Ming carelessly regarded x=2011 as x=2012, but he calculated that the final result was the same as the correct answer, can you explain the reason?
Solution: (5x +4x-3)-(x -3x-1)+(4-7x-6x)=5x +x -6x +4x+3x-7x-3+1+4=2, its value is always equal to 2, and it has nothing to do with x, so it doesn't matter if x=2011 is regarded as x=2012, it does not affect the result.
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Anonymous users2024-02-04Is the title wrong, and finally (should be (4-7x-6x 2).
In that case, the answer is as follows:
5x +4x-3)-(x -3x-1)+(4-7x-6x)5x +4x-3 x 3x 1+4-7x-6x (5x x -6x)+4x 3x-7x)+(3 1+4) So the answer has nothing to do with x, so Xiao Ming's calculation is the same as the correct answer.
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Anonymous users2024-02-03Solution: 12 2 = 6 cm, 6 2 = 3 cm, 6 6 square centimeters, annular area = square centimeter.
cm, 12
centimeters, the length of the entire roll of paper is centimeters.
Please refer to it. Equations containing unknown quantities are equations, mathematics was first developed in counting, and about numbers and unknowns are combined by addition, subtraction, multiplication, division, and idempotency to form algebraic equations: unary equations.
Unary quadratic equations, binary quadratic square hidden rules.
Wait a minute. However, with the emergence of the concept of functions and the introduction of function-based differentiation and integration operations, the scope of equations has become more extensive, and unknown quantities can be mathematical objects such as functions and vectors, and operations are no longer limited to addition, subtraction, multiplication and division.
Equations occupy an important place in mathematics and seem to be an eternal topic in mathematics. The emergence of equations not only greatly expanded the scope of mathematical applications, enabling many problems that could not be solved by arithmetic problem solving, but also had a huge impact on the progress of mathematics in the future. In particular, many major discoveries in mathematics are closely related to it.
The formula for finding the root of a quadratic equation.
In middle school, the equations that I came into contact with are basically in this category, and the unknowns in the equation can appear in the fractions and integers in the equation.
Radicals and trigonometric functions.
in the independent variables of elementary functions such as exponential functions.
In secondary school, when you encounter problems with solving equations, in general, you can convert the equation into an integral equation; Generally, it is converted into a one-dimensional quadratic equation, or a system of multivariate one-dimensional equations.
Since mathematics from constants.
Mathematics is transformed into variable mathematics, and the content of equations is also enriched, because mathematics introduces more concepts, more operations, and thus more equations. The development of other natural sciences, especially the physics of the stove shed, has also directly put forward the need for equation solving and provided a large number of research topics.
Ordinary differential equation.
A differential equation is an equation that contains an unknown function and its derivative. The unknown quantity of this type of equation is a function, and unlike the function equation, there is a derivative operation for the unknown function, which can be a higher-order derivative.
However, if the unknown function in an equation contains only one independent variable, then a differential equation is an ordinary differential equation.
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Anonymous users2024-02-02The weight of the wine bought by one customer is exactly twice that of the other customer If the weight of the wine of the other customer = x kg, then the weight of the wine of the first customer = 2x kg The weight of the wine of two customers = 3x kg is a multiple of 3 The total weight of 6 barrels of wine is 18 + 15 + 16 + 19 + 20 + 31 = 119 kg is a multiple of 3 and a surplus of 2kg Of the 6 weights, only 20 kg is a multiple of 3 and 2 kg remains, and the rest add up to multiples of 3. The beer that the merchant left for himself was 20 kg
The weight of the wine bought by one customer is exactly twice that of the other customer If the weight of the wine of the other customer = x kg, then the weight of the first customer's wine = 2x kg The weight of the wine of two customers = 3x kg is a multiple of 3 The total weight of 6 barrels of wine is 18 + 15 + 16 + 19 + 20 + 31 = 119 kg is a multiple of 3, and the first 2 kg is a multiple of 3 Only 20 kg of the 6 weights are multiples of 3, and the rest add up to multiples of 3. The beer that the merchant left for himself was 20 kg
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Anonymous users2024-02-01Citrus is 3 more than honey and filial piety peaches, and 10-4 15=1 30
The liquid collapsed to 6 5 1 30 = 1 25 hectares.
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Anonymous users2024-01-311/15 of the two.
3/10 - 4/15 = 1/30 of a macro.
6/5 x 1/30 = 20/5 (ha) of the lease
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Anonymous users2024-01-302s = s car + s sound = v car t + v sound t = 10m s 5s + 340m s 5s
s=875m
Draw a picture and you'll understand.
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