How to solve the equation (1-5 / 9) x = 36?

How to solve the equation (1-5 / 9) x = 36?

(1-5 / 9) x = 36
4/9X=36
X=36*9/4
X=81

How to solve the equation of 1 / 8 x 5x = 120

1 / 8 x 5x = 120
5x = 120
5x=120（x-8）
5x=120X-960
115X=960
X = 8.3478

Given the real number T, if there exists t ∈ [1 / 2,3] such that the inequality | T-1 | - | 2t-5 | ≥ | X-1 | + | X-2 |, the value range of real number x is obtained

The left side becomes | T-1 | - 2 | T-5 / 2 |, that is, t ∈ [1 / 2,3], the distance from t to 1 plus 2 times the distance from t to 5 / 2. When t = 5 / 2, the minimum value is | T-1 | - | 2t-5 | ≥ | X-1 | + | X-2 | if it is true, the minimum value on the left side is greater than or equal to the maximum value on the right side, that is, 3 / 2 | X-1 | + | X-2 |, that is the distance from X to 1 plus

For the formula - (- 8), the following understanding: (1) it can express the opposite number of - 8; (2) it can express the product of - 1 and - 8; (3) it can express the absolute value of - 8; (4) the result of operation is equal to 8. The number of misinterpretations is ()
A. 0B. 1C. 2D. 3

Only two numbers with different symbols are opposite to each other, so (1) is correct; the product of - 1 and - 8 can also be expressed as - (- 8), so (2) is correct; the absolute value of - 8 can be expressed as | - 8 |. According to the fact that the absolute value of negative number equals its opposite number, we can know that (3) is correct; - (- 8) = 8, so (4) is correct

Given the function f (x) = KX + 1, where the real number k is randomly selected from the interval [- 2,1]. For ∀ x ∈ [0,1], the probability of F (x) ≥ 0 is ()
A. 13B. 12C. 23D. 34

From the meaning of the problem, we know that the problem is a geometric probability, the ratio of the corresponding length of the value of probability, ∵ - 2 ≤ K ≤ 1, its interval length is 3 and ∵ for ∀ x ∈ [0, 1], f (x) ≥ 0, and f (x) is a function of linear type about X, on [0, 1], it is monotone ∀ f (0) ≥ 0, f (1) ≥ 0 − 2 ≤ K ≤ 1 ≤ K ≤ 1, its interval length is 2 ∀ P = 23, so we choose C

If the absolute value of (k-1) x k + 21 = 0 is a linear equation with one variable, then k = what

If the absolute value of (k-1) x k + 21 = 0 is a linear equation with one variable, then k-1 ≠ 0, absolute value k = 1, then K ≠ 1, k = ± 1, so k = - 1

What is the inverse function of y = 4x-1 / 2?

y=(4x-1)/2
2y=4x-1
4x=2y+1
x=(2y+1)/4
So the inverse function is y = (2x + 1) / 4

It is known that for the X function f (x) = (1-A) x2 + (a + 2) x-4, a is a real number. Find: (1) the function f (x) has only one zero point in [- 2,1], find the value range of a; (2) all zeros of function f (x) are greater than 0, find the value range of A

(1) ∵ function has only one zero point in [- 2,1], ∵ f (- 2) × f (1) ≤ 0 (2 points) (when the equal sign is different), that is: - 1 (- 6a-4) ≤ 0, the solution is: a ≤ − 23, that is, the value range of a is (- ∞, - 23]. & nbsp (5 points) (2) when 1-A = 0, that is, a = 1, then 3x-4 = 0

Let a ∈ R, the quadratic function f (x) = ax2-2x-2a. If the solution set of F (x) > 0 is a, B = {x | 1 ＜ x ＜ 3}, a ∩ B ≠ 0, find the value range of real number a

The second function a ≠ 0, f (x) = 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, f (x) 0, and f (x) 0, two of its two solutions are X1 = 1 = 1A = 1a {1a {1A {1a ∩ B ∩ B ϕ the necessary and sufficient condition of a ϕ necessary and sufficient condition is x2 < 3, that is 1A + 2 + 2 + 2 + 2 + 2 + 2 + 1A + 2 + 2 + 2 + 2 + 1A + 2 + 2 + 2 + 2 + 2 + 1A + 2 + 2 + 2 + 3, the solution of a solution of a isx2 > 1, that is 1A + 2 + 1A2 > 1, the solution is a < - 2 The value range of a is (− ∞, − 2) ∪ (67, + ∞)

Linear Algebra: square matrix A of order n is an orthogonal matrix, and it is proved that a * is an orthogonal matrix

Because the square matrix A of order n is an orthogonal matrix,
So a'a = e, a ^ - 1 = a'reversible! And ia'ai = ia'iiai = IAI ^ 2 = IEI = 1
A^-1=A*/IAI
A*=IAIA^-1=IAIA'
So (a *)'a * = (IAIA ')'iaia'
=IAIA IAIA'
=IAI^2 AA'
=IAI^2 E
=1*E
=E
So a * is an orthogonal matrix