The piston in the cylinder of an internal combustion engine is called a stroke from ---One stroke is called a work cycle, and the ratio of the heat released by the part of the fuel used for the work cycle is called heat engine efficiency

The piston in the cylinder of an internal combustion engine is called a stroke from ---One stroke is called a work cycle, and the ratio of the heat released by the part of the fuel used for the work cycle is called heat engine efficiency

The piston in an internal combustion engine cylinder moves from one end of the cylinder to the other end, which is called a stroke;
Each stroke is called a work cycle, and the ratio of the heat released from the complete combustion of the part of the fuel used to do useful work is called the heat engine efficiency

How to understand and use cross multiplication? What is the principle

1. The method of cross multiplication: multiplication on the left side of cross equals the coefficient of quadratic term, multiplication on the right side equals the constant term, cross multiplication and addition equals the coefficient of primary term
2. The use of cross phase multiplication: (1) to decompose the factor by cross phase multiplication. (2) to solve quadratic equation with one variable by cross phase multiplication
3. Advantages of cross phase multiplication: the speed of solving problems by cross phase multiplication is relatively fast, which can save time, and the amount of calculation is not large, so it is not easy to make mistakes
4. The disadvantages of cross multiplication: 1. Some problems are easy to solve by cross multiplication, but not every problem is easy to solve by cross multiplication. 2. Cross multiplication is only suitable for quadratic trinomial problems. 3. Cross multiplication is difficult to learn
5. Examples of solving cross phase multiplication problems:
1) Use cross multiplication to solve some simple and common problems
Example 1 decomposes M & sup2; + 4m-12 into factors
Analysis: in this problem, the constant term - 12 can be divided into - 1 × 12, - 2 × 6, - 3 × 4, - 4 × 3, - 6 × 2, - 12 × 1. When - 12 is divided into - 2 × 6, it is in line with this problem
Because 1-2
1 ╳ 6
So M & sup2; + 4m-12 = (m-2) (M + 6)
Example 2 decomposes 5x & sup2; + 6x-8 into factors
Analysis: in this problem, 5 can be divided into 1 × 5, - 8 can be divided into - 1 × 8, - 2 × 4, - 4 × 2, - 8 × 1
Because 1 2
5 ╳ -4
So 5x & sup2; + 6x-8 = (x + 2) (5x-4)
Example 3 solves the equation x & sup2; - 8x + 15 = 0
Analysis: if X & sup2; - 8x + 15 is regarded as a quadratic trinomial about X, then 15 can be divided into 1 × 15,3 × 5
Because 1-3
1 ╳ -5
So the original equation is deformable (x-3) (X-5) = 0
So X1 = 3, X2 = 5
Example 4. Solve the equation 6x & sup2; - 5x-25 = 0
Analysis: take 6x & sup2; - 5x-25 as a quadratic trinomial of X, then 6 can be divided into 1 × 6,2 × 3, - 25 can be divided into - 1 × 25, - 5 × 5, - 25 × 1
Because 2-5
3 ╳ 5
So the original equation can be changed to form (2x-5) (3x + 5) = 0
So X1 = 5 / 2, X2 = - 5 / 3
2) Use cross multiplication to solve some difficult problems
Example 5: Factoring 14x & sup2; - 67xy + 18y & sup2
Analysis: if 14x & sup2; - 67xy + 18y & sup2; is regarded as a quadratic trinomial about X, then 14 can be divided into 1 × 14,2 × 7,18y & sup2; and y.18y, 2y.9y, 3y.6y
Because 2 - 9y
7 ╳ -2y
So 14x & sup2; - 67xy + 18y & sup2; = (2x-9y) (7x-2y)
Example 6 decomposes 10x & sup2; - 27xy-28y & sup2; - x + 25y-3 into factors
Analysis: in this problem, we should arrange this polynomial into the form of quadratic trinomial
Solution 1: 10x & sup2; - 27xy-28y & sup2; - x + 25y-3
=10x²-（27y+1）x -（28y²-25y+3） 4y -3
7y ╳ -1
=10x²-（27y+1）x -（4y-3）（7y -1）
=[2x -（7y -1）][5x +（4y -3）] 2 -（7y – 1）
5 ╳ 4y - 3
=（2x -7y +1）（5x +4y -3）
Note: in this problem, 28y & sup2; - 25y + 3 is decomposed into (4y-3) (7y-1) by cross phase multiplication, and then 10x & sup2; - (27y + 1) x - (4y-3) (7y-1) is decomposed into [2x - (7y-1)] [5x + (4y-3)]
Solution 2: 10x & sup2; - 27xy-28y & sup2; - x + 25y-3
=（2x -7y）（5x +4y）-（x -25y）- 3 2 -7y
=[（2x -7y）+1] [（5x -4y）-3] 5 ╳ 4y
=（2x -7y+1）（5x -4y -3） 2 x -7y 1
5 x - 4y ╳ -3
Note: in this problem, first, 10x & sup2; - 27xy-28y & sup2; is decomposed into (2x - 7Y) (5x + 4Y) by cross phase multiplication, and then (2x - 7Y) (5x + 4Y) - (x - 25y) - 3 is decomposed into [(2x - 7Y) + 1] [5x - 4Y) - 3] by cross phase multiplication
Example 7: the solution of X equation: X & sup2; - 3ax + 2A & sup2; – ab - B & sup2; = 0
Analysis: 2A & sup2; – ab-b & sup2; can be factorized by cross phase multiplication
x²- 3ax + 2a²–ab -b²=0
x²- 3ax +（2a²–ab - b²）=0
x²- 3ax +（2a+b）（a-b）=0 1 -b
2 ╳ +b
[x-（2a+b）][ x-（a-b）]=0 1 -（2a+b）
1 ╳ -（a-b）
So X1 = 2A + B, X2 = a-b
The relationship of quadratic function between two related variables can be expressed in three different forms: general, vertex and intersection
Intersection type
By using the collocation method, the general expression of quadratic function is transformed into
Y=a[(x+b/2a)^2-(b^2-4ac)/4a^2]
The right end is factorized by the square difference formula, and the result is obtained
Y=a[x+b/2a+√b^2-4ac/2a][x+b/2a-√b^2-4ac/2a]
=a[x-(-b-√b^2-4ac)/2a][x-(-b+√b^2-4ac)/2a]
Because the two equations ax ^ 2 + BX + C = 0 are x1,2 = (- B ± √ B ^ 2-4ac) / 2A respectively
So the above formula can be written as y = a (x-x1) (x-x2), where X1 and X2 are the two roots of the equation AX ^ 2 + BX + C = 0
Because x 1 and x 2 are the abscissa of the intersection (x 1,0), (x 2,0) of the function image and X axis, we call the function y = a (x-x 1) (x-x 2) the intersection of functions
In solving the problems related to the image of quadratic function and the coordinate of x-axis intersection, it is more convenient to use the intersection formula
The intersection formula of quadratic function can also be obtained by the following deformation methods:
Let ax ^ 2 + BX + C = 0 be X1 and X2 respectively
According to the relationship between root and coefficient, X1 + x2 = - B / A, x1x2 = C / A,
There are B / a = - (x1 + x2), a / C = x1x2
∴y=ax^2+bx+c=a[x^2+b/a*x+c/a]
=a[x^2-(x1+x2)x+x1x2]=a(x-x1)(x-x2)

Let y = x √ x (x > 0). Find dy / DX

Y = x radical (x) (x > 0)
=x^(1+1/2)
=x^(3/2)
dy/dx=3/2*x^(3/2-1)
=3/2*x^(1/2)

How to convert cubic meter and liter

1 cubic meter = 1000 cubic decimeter = 1000 liter

Given that the real numbers a, B and C satisfy A-B + C = 7, AB + BC + B + C + 16 = 0, then what is the value of B of a
Given that a, B and C satisfy a + 2B = 7, B - 2C = - 1 and C - 6A = - 17, find the value of a, B and C

Let v be the region defined by the surface x + y + Z = 1, x = 0, y = 0, z = 0, then ∫ ∫ (V) 1 / (1 + X + y + Z) ^ 3dxdydz

How many kilometers per hour is 200 knots?
Such as the title

One knot is equal to 1 nautical mile / hour, i.e. 1.85 km / hour
200 knots is 370 km / h

Please write the appropriate company name in the brackets below 1 () - 1 () = 9 () 1 () - 1 () = 99 () 1 () - 1 () = 999
Please write the appropriate company name in the brackets below
1（ ）-1（ ）＝9（ ）
1（ ）-1（ ）＝99（ ）
1（ ）-1（ ）＝999（ ）

1m-1dm=9dm 1m-1cm=99cm 1m-1mm=999mm

We know that the radius of the bottom of the cylinder is 2 and the height is 3. If we use a plane to cut the cylinder, if we get an ellipse, then the eccentricity range of the ellipse
A【3/5,1） B（0,3/5】C【4/5,1） D（0,4/5】

This problem is not difficult. Obviously, C ^ 2 = a ^ 2-B ^ 2, dividing both ends by a ^ 2 at the same time, we know that the eccentricity only depends on B / A. then, B is the short axis, and a is the long axis. If we continue to cross the cylinder, we should pay attention not to be absolutely flat, because the eccentricity cannot be zero, then the section tends to be round, and B tends to be a, then the eccentricity tends to 0,
Consider that the maximum value should be the maximum difference between B and a, the radius of the cylinder is B, constant, then the maximum value of a appears in the oblique section of the cylinder, from top left to bottom right, according to the Pythagorean theorem, the oblique length is 5, and a = 5 / 2, so the eccentricity is 3 / 5
The answer is B

It is known that the area of a rectangle is 6m & # 178; + 60m-150, and the ratio of length to width is 3 ∶ 2

Suppose the length and width are 3a and 2A respectively, and the perimeter is 10a, then 3A * 2A = 6A ^ 2 = 6m & # 178; + 60m-150 a ^ 2 = m ^ 2 + 10m-25 a = √ (m ^ 2 + 10m-25), so the perimeter = 10 √ (m ^ 2 + 10m-25)