If the image of the function y = sin2x is shifted to the right by φ (φ > 0) units, and the obtained image is symmetric with respect to x = π 6, then the minimum value of φ is () A. 5 π 12b. 11 π 6C. 11 π 12D. None of the above is true

If the image of the function y = sin2x is shifted to the right by φ (φ > 0) units, and the obtained image is symmetric with respect to x = π 6, then the minimum value of φ is () A. 5 π 12b. 11 π 6C. 11 π 12D. None of the above is true

Let y = f (x) = sin2x, then f (x - φ) = sin2 (x - φ) = sin (2x-2 φ), and its image is exactly symmetric with respect to x = π 6, ∧ 2 × π 6-2 φ = 2K π + π 2 or 2 × π 6-2 φ = 2K π - π 2, K ∈ Z. ∧ φ = - K π - π 12 or φ = - K π + 5 π 12, K ∈ Z. moreover, the minimum value of ∧ φ is 5 π 12

Draw the graph of function y = 0.5x, and point out the independent variable and its value range

When x = 0, y = 0. When x = 2, y = 1. The line passes through points (0, 0), (2, 1), and its image is as shown in the figure. According to the figure, X takes any real number

1-2x & # 178; is it a linear equation with one variable
1-2x²
3x & # 178; - radical 2x = 7
x²+2x²=0
（3x-2）（x+6）=3x²-7

One, two, three, No
The fourth is
The first one is obviously quadratic, and there is no equal sign,
The second is the second time,
Although the third one can simply solve x = 0, its essence is a quadratic equation with one variable
Only the fourth is a linear equation of one variable

If the three eigenvalues of the third-order matrix A are - 2,1,3, then the value of the determinant | a ^ 2 + 2a-e | is equal to?

A is similar to a diagonal matrix! Then the sum above is similar to a diagonal matrix! Take the determinant on both sides to get it! Try it!

(M + 1) (m-1) (M & # 178; - 1) calculation

Original formula = (m ^ 2-1) (m ^ 2-1)
=m^4-2m^2+1

52-4x = 16 20x-0.5x-14.5 = 24.5

52－4x=16
4x=52-16
4x=36
x=36/4
x=9
20x－0.5x－14.5=24.5
20x-0.5x=14.5+24.5
19.5x=39
x=39/19.5
x=2

Known function f (x) = (4x ^ 2-7) / 2-x
Given the function f (x) = (4x ^ 2-7) / (2-x), X belongs to [0,1], (2) let a be 1, the function g (x) = x ^ 2-3a ^ 2x-2a, X belongs to [0,1]. If any x 1 belongs to [0,1], there is always x 0 belonging to [0,1], such that G (x 0) = f (x 1) holds, the value range of a is obtained
X0 belongs to [0,1], so that G (x0) = f (x1) holds, find the value range of A
It contains [- 4, - 3]. I think it contains [- 4, - 3], [1-2a-3a ^ 2]. Please help me

Please note:
It means that for f (x1), there is always g (x0) corresponding to it (x0, X1 belong to [0,1]), that is, the range of G contains the range of F!

In which quadrant is the image of inverse scale function y = (k2-k + 1) / x

One three quadrants
The proportional coefficient is greater than 0

1: If LAL = 8, LBL = 3 and a

b=3,a

0.2x + 3-0.4x-1 = 20

0.2 (0.1 x + 3) - 0.5 (0.4 x-1) = 20
5(0.1x+3)-2(0.4x-1)=20
0.5x+15-0.8x+2=20
-0.3x=3
x=-10