Use the modifier to correct the following sentence After writing the composition, read it again and correct the wrong words 2. The morning glow reddened half of the sky, and there was no cloud in the sky 3. Everyone is in a neat line, flocking in one by one

Use the modifier to correct the following sentence After writing the composition, read it again and correct the wrong words 2. The morning glow reddened half of the sky, and there was no cloud in the sky 3. Everyone is in a neat line, flocking in one by one

After writing the composition, read it again and correct the wrong words
After writing the composition, read it again and correct the incorrect words
2. The morning glow reddened half of the sky, and there was no cloud in the sky
The morning glow reddened half of the sky, and the sky was covered with clouds
3. Everyone is in a neat line, flocking in one by one
Everyone in a neat line, one after another orderly into

The rescue work went on (in good order) without any dispute

Well organized,

The physical meaning of Hilbert transform

By observing the definition of Hilbert transform, we can find that the meaning of the result of Hilbert transform is that the input is the output of a linear non time varying system of S (T), and the impulse response of this system is 1 / (π T). Hilbert is actually an all pass phase-shifting network which makes the phase lag π / 2
Through Hilbert transform, it is possible for us to define and calculate the instantaneous parameters of short signal and complex signal, and realize the extraction of real instantaneous signal. Therefore, Hilbert transform plays a very important role in signal processing
It also has the following problems:
(1) Hilbert transform can only be applied to narrow-band signal approximately, that is, only applicable signal, where B (B is the signal bandwidth). But in practical application, there are many non narrow-band signals, and Hilbert transform can't do anything for these signals. Even for narrow-band signals, if the Hilbert transform condition can't be fully satisfied, the result will be wrong, It will make many signals which satisfy the condition of Hilbert transform can not satisfy completely;
(2) For any given time t, the result of Hilbert transform can only have one frequency value, that is, it can only process the signal with single frequency at any time;
(3) For a nonstationary data sequence, the result of Hilbert transform loses its original physical meaning

The graph of a function y = 2x + 3 and a quadratic function y = AX2 + BX + C intersects at two points a (m, 5) and B (3, n). When x = 3, the maximum value of the parabola is 9?
The graph of a function y = 2x + 3 and a quadratic function y = AX2 + BX + C intersects at two points a (m, 5) and B (3, n). When x = 3, the maximum value of parabola is 9
(1) Find the expression of quadratic function;
(2) When x is the value, the values of the first and second functions increase with the increase of X
(3) When x =?, the value of the first function is greater than that of the second function

(1) By substituting a (m, 5) and B (3, n) into y = 2x + 3, we get the following results:
2m + 3 = 5, so m = 1
6 + 3 = n, so n = 9 is a (1,5) B (3,9)
Substituting two points a and B into y = AX2 + BX + C, there is a + B + C = 5
9A + 3B + C = 9, the solution is 4A + B = 2
Because B = 3 of (- 2A) gives a = - 1, B = 6, C = 0, so y = - x2 + BX
(2) Because (3,9) is the vertex of quadratic function,
So, X3

If n is a positive integer, the value of the cubic power of n minus 6 must be divisible by 6

1 ^ 3-6 = - 5, divisible by 6?
The third power of odd number must be odd, minus 6 or odd, 6 is even, can you divide?
Is the title copied wrong?

Let a be a real number, f (x) = the cubic + ax power of X + the derivative function f '(x) of (A-3) x, and f' (x) be an even function, then the tangent equation at y = f (x) = (2, f (2)) is

f(x)=x^3+ax^2+(a-3)x
f'(x)=3x^2+2ax+(a-3)
And f '(x) is even function
∴f'(-x)=f(x)
3x^2+2ax+(a-3)=3x^2-2ax+(a-3)
2ax=-2ax
4ax=0
a=0
f(x)=x^3-3x
f(2)=2^3-3*2=2
f'(x)=3x^2-3
f'(2)=3*2^2-3=9
therefore
The tangent equation at y = f (x) = (2, f (2)) is y = 9x-16

1.a²+b²+___ =（a+b)²
2.(a+b)²+(a-b)²=
3.(a+b)²-(a-b)²=
4.x＋＿＿＝1＝（x+1）²
5. If the polynomial 9x & sup2; + 1 plus a monomial is the square of an integral, then the added monomial can be (fill in one)
6. Given (x + y) & sup2; = 7, (X-Y) & sup2; = 9, find X & sup2; + Y & sup2; and XY
7. Given x + 1 / x = 2, find X & sup2; + 1 / X & sup2;, X ＾ 4 + 1 / X ＾ 4
8. How to calculate easily: (3x + a) & sup2; (3x-a)
9. Calculate with two different methods
（1）（2a－3b）²－（2a＋3b）²
（2）（a+b+c）（a-b-c）
10. Given that real numbers a and B satisfy (a = b) & sup2; = 1, (a-b) & sup2; = 25, find the value of a & sup2; + B & sup2; + ab

1.a²+b²+2ab=（a+b)²
2.(a+b)²+(a-b)²=a²+2ab＋b²＋a²-2ab＋b²=2a²＋2b²
3.(a+b)²-(a-b)²=[（a+b）＋（a-b）][（a+b）－（a-b）]=2a·2b=4ab
4.x＋1＝1＝（x+1）²
5. The added monomial can be - 6x
6. Given (x + y) & sup2; = 7, (X-Y) & sup2; = 9, find X & sup2; + Y & sup2; and XY
∵（x＋y）²＝7,（x-y)²＝9
∴ x²＋y²＝7-2xy,x²＋y²＝9＋2xy
∴7-2xy＝9＋2xy
∴xy＝-1／2
7. Given x + 1 / x = 2, find X & sup2; + 1 / X & sup2;, X ＾ 4 + 1 / X ＾ 4
∵x+1/x＝2
∴（x+1/x）²＝2²
∴x²+1/x²＝2
And ∵ X & sup2; + 1 / X & sup2; = 2
∴（x²+1/x²）²＝2²
∴ x＾4＋1/x＾4＝2
8. How to calculate easily: (3x + a) & sup2; (3x-a)
The original formula = [(3x + a) (3x-a)] (3x + a)
＝[(3x)²-a²]（3x－a）
＝27x³-9ax²-3a²x+a³
9. Calculate with two different methods
Method 1, (1) (2a-3b) & sup2; - (2a + 3b) & sup2;
The original formula = [(2a-3b) + (2a + 3b)] [(2a-3b) - (2a + 3b)]
=4a·(-6b)
=-24ab
Method 2: the original formula is 4A & sup2; - 12ab + 9b & sup2; - 4A & sup2; - 12ab-9b & sup2;
=-24ab
（2）（a+b+c）（a-b-c）
Method 1. The original formula = [a + (B + C)] [a - (B + C)]
=a²-(b+c)²
=a²-b²-2bc-c²
Method 2: the original formula is a & sup2; - AB AC + ab-b & sup2; - BC + ac-bc-c & sup2;
=a²-b²-2bc-c²
10. Given that real numbers a and B satisfy (a = b) & sup2; = 1, (a-b) & sup2; = 25, find the value of a & sup2; + B & sup2; + ab
∵ real numbers a and B satisfy (AB) & sup2; = 1, (a-b) & sup2; = 25
∴ ab=1,ab=-1,a²-2ab＋b²=25
∴ a²＋b²=25＋2ab
∴ a²＋abb²=25＋ab
When AB = 1, a & sup2; + ABB & sup2; = 25 + AB = 26,
When AB = 1, a & sup2; + AB & sup2; = 25 + AB = 24

Let the quadratic function f (x) = ax (square) + BX + C satisfy f (x + 1) - f (x) = 2x, and find the value range of x such that f (x) > C + 6

f(x+1)-f(x)=2x
a(x+1)^2+b(x+1)+c -[ax^2+bx+c]
=ax^2+2ax+a+bx+b+c-ax^2-bx-c
=2ax+a+b=2x
2a=1 a+b=0
a=1/2 b=-1/2
f(x)=1/2x^2-1/2 x+c
f(x)>c+6
1/2x^2-1/2 x+c>c+6
1/2x^2-1/2 x -6>0
x^2-x-12>0
(x-4)(x+3)>0
x> 4 or X

The intersection coordinates of the line y = 2x + 2 and the parabola y = x2 + 3x are______ ．

∵ from the meaning of the problem, we get y = 2x + 2Y = x2 + 3x, and the solution is x = − 2Y = − 2 or x = 1y = 4. The intersection coordinates of the straight line y = 2x + 2 and the parabola y = x2 + 3x are (- 2, - 2), (1,4). So the answer is: (- 2, - 2), (1,4)

The vertex coordinates of the parabola whose axis of symmetry is parallel to the X axis are (2, - 9) and the coordinates of the parabola passing through the point (- 1,0) are obtained
The vertex coordinates of the parabola whose axis of symmetry is parallel to the X axis are (2, - 9) and the distance between the parabola and the two intersections of the X axis can be obtained by crossing the point (- 1,0)

Since the axis of symmetry is parallel to the X axis, and the fixed-point coordinates are (2, - 9), we can set the parabolic equation as y = a (X-2) ^ 2-9, which is brought into the equation by the parabolic over-current (- 1,0), and get 0 = a (- 1-2) ^ 2-9, and get a = 1, so the equation is y = (X-2) ^ 2-9, let y = 0, and get X1 = 5, X2 = - 1, so the distance between two points is / x1-x2 / = 6