7、 When is 5 (x + 2) = 5-9 (x-3) 8, 1-2 (2x-5) = 3 (3-x) 9 and X, the value of formula 4 (x-1) + 2-x and 2 (4-x) - 6 are equal

7、 When is 5 (x + 2) = 5-9 (x-3) 8, 1-2 (2x-5) = 3 (3-x) 9 and X, the value of formula 4 (x-1) + 2-x and 2 (4-x) - 6 are equal

7、 5 (x + 2) = 5-9 (x-3)
5x+10=5-9x+27
5x+9x=5+27-10
14x=22
x=22÷14
X = 1 and 4 / 7
8、 1-2 (2x-5) = 3 (3-x)
1-4x+10=9-3x
4x-3x=1+10-9
x=2
9、 When is x, the value of formula 4 (x-1) + 2-x and 2 (4-x) - 6 are equal
4（x-1)+2-x=2（4-x)-6
4x-4+2-x=8-2x-6
4x-x+2x=8-6+4-2
5x=4
x=4÷5
X = 4 / 5

3 ^ x = 4 3 ^ y = 6 find the value of 9 ^ 2x + y + 27 ^ x + y

9^(2x+y)+27^(x+y)=(3^2)^(2x+y)+(3^3)^(x+y)=(3^x)^4×（3^y）^2+[3^(x+y)]^3=4^4×6^2+12^3=256×36+1728=10944

It is known that the vertex coordinates of the image of quadratic function are (1, - 4) and pass through points (2, - 2)
(1) Find the analytic expression of the quadratic function
(2) If the image of the quadratic function is shifted several units to the left, the translated image can pass through the origin coordinate? And the analytic expression of the quadratic function corresponding to the translated image can be obtained
(saying

1) According to the vertex, let y = a (x-1) ^ 2-4
Substituting point (2, - 2), we get - 2 = A-4
That is, a = 2
So y = 2 (x-1) ^ 2-4
2) Shift a unit to the left, then the function becomes y = 2 (x-1 + a) ^ 2-4
To make it cross the origin, substitute (0,0) in:
0=2(a-1)^2-4
The positive value of solution: a = √ 2 + 1
That is, shift to the left √ 2 + 1 unit

If the square of Y minus 8y plus m minus 1 is a complete square, then M equals 1

y²-8y+m=y²-2•y•4+m
m=4×4=16

Make a square on each side of the right triangle whose right sides are 3 and 4, and connect the three squares with the hexagon ABCDEF as shown in the figure, then the area of the hexagon is______ ．

Let F 1 and F 3 be the two focal points of hyperbolic square / a - y square / b square = 1. If F 1. F 2. P (0,2b) is the three vertices of an equilateral triangle, then the two focal points are
Let F 1 and F 2 be the two focuses of hyperbola (x squared divided by a squared) - (y squared divided by B squared) (a > 0, b > 0). If F 1. F 2. P (0, 2b) are the three vertices of an equilateral triangle, then the eccentricity of hyperbola is?
Because: F1, F2 are hyperbolic respectively
Two focuses of x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1
Then: F1 (- C, 0) F2 (C, 0)
Then: F1F2 = 2C
Also: f1.f2.p (0,2b) is the three vertices of an equilateral triangle
And O is the midpoint of F1F2
Po is perpendicular to F1F2
Then: Po = (√ 3 / 2) F1F2
Namely: 2B = (√ 3 / 2) (2C)
Then there is: 4B ^ 2 = 3C ^ 2
4(c^2-a^2)=3c^2
c^2=4a^2
c^2/a^2=4
Then: e = C / a = 2
How did (√ 3 / 2) come from

∵△ f1f2p is an equilateral triangle with a side length of 2C
∴sin∠PF1F2=sin60°=PO/F1F2
∴PO=sin60°*F1F2=(√3/2)F1F2
In other words (√ 3 / 2) is the value of sin60

Economic significance of second order mixed partial derivatives
If GDP is a function of P and K, what is the economic significance of the mixed second partial derivative of GDP to P and K;

If GDP is a function of P and K,
Let f (P, K) be the marginal function of GDP with respect to P,
The marginal function of F (P, K) with respect to K is h (P, K)
Then the mixed second partial derivative of GDP to P and K = H (P, K)
Therefore, the economic significance of the mixed second-order partial derivative of GDP to P and K is obvious
The marginal function of GDP with respect to P and the marginal function with respect to K

Write the words as they are
Example: standing on top of the sky
Example: seven channels and eight beams
There are many calamities

Example: standing on the top of the sky
Example: seven channels and eight beams (seven ups and eight Downs) (seven ups and eight Downs) (seven mouths and eight tongues) (seven emotions and six desires) (seven hands and eight feet)
Example: all kinds of calamities (all kinds of holes) (all kinds of beauties) (all kinds of twists and turns)

The image of function y = KX, y = K / X in the same coordinate system

y=kx,y=k/x
(1) If k = 0, then the same line y = 0 (x axis);
(2) When k is not equal to 0, the two images intersect,
Let the coordinates of the intersection be (x, y),
Then KX = K / x,
The solution is x = 1, y = K,
Or x = - 1, y = - K,
That is to say, the images all pass through points (1, K) and (- 1, - K),
And the quadrant is the same (but y = KX goes through the origin, and y = K / X is not the origin)

Judge the parity of function f (x) = ax + 1 / AX-1 (a > 0, a ≠ 1). Note: ax is the x power of A

Odd function
Calculate f (- x) = a (- x) + 1 / a (- x) - 1
Because a (- x) = 1 / ax, the reduced f (- x) = ax + 1 / 1-ax
F (x) = - f (- x), so it is an odd function