The perimeter of a square is 2 mm, its side length is () DM, and its area is () DM square meter

The perimeter of a square is 2 mm, its side length is () DM, and its area is () DM square meter

0.5,0.25

How many tons is a kilogram?

One ton equals 1000 kg
One kilogram is equal to 0.001 ton

The function y = log a (X-2) - 1, a > 0, a is not equal to 1

loga1=0
So, x = 3, y = - 1
Namely (3, - 1)

For an isosceles triangle, the middle line of a waist divides the circumference of the triangle into two parts: 2:1, and the length of the bottom edge is 5?

Suppose the length of a waist is X
(X+x/2):(5+X/2)=2:1
X=20

Can the three factors KX, cos and Tan be interchanged?

The fake big empty plane, the TV set, the visibility of style, the dozing process

Given the set a = {x | 3 ≤ x ＜ 7}, B = {x | 2 ＜ x ＜ 10}, C = {x | 5-a ＜ x ＜ a}. (1) find a ∪ B, (∁ RA) ∩ B; (2) if C ⊆ (a ∪ B), find the value range of A

(1) From the graph, a ∪ B = {x | 2 ＜ x ＜ 10}, ∁ RA = {x | x ＜ 3 or X ≥ 7}, ∁ RA) ∩ B = {x | 2 ＜ x ＜ 3 or 7 ≤ x ＜ 10} (6 points) (2) from (1) know a ∪ B = {x | 2 ＜ x ＜ 10}, ① C ⊆ (a ∪ b), then 5-a ≥ a

Given that there are two points c and D on the line AB, and AC: CD: DB = 2:3:4, e and F are the midpoint of a, C and DB respectively, EF = 2.4cm, the length of line AB is calculated
Help, express in geometric language, thank you!

EF = EC + CD + DF
=0.5AC + CD + 0.5 DB
Because AC: CD: DB = 2:3:4
So let AC = 2x; CD = 3x; DB = 4x
Then 0.5ac = x,; 0.5dB = 2x
So 2.4cm = EF = x + 3x + 2x = 6x
So x = 0.4cm
AB=9x=3.6cm

If a and B are normal matrices and ab = Ba, how to prove that AB and Ba are normal matrices

A matrix A is normal if and only if there exists a matrix X such that x * ax is a diagonal matrix, where x * is the conjugate transpose of X
So there exists a matrix X, y such that x * AX = k, y * by = J, where K and j are diagonal matrices, and K = diag (K1, K2,..., KK) can be recorded, where the diagonal elements of KI and kJ are different from each other, Ki = AIE, e is unit matrix
K (x * yjy * x) = (x * yjy * x) K. by dividing x * yjy * x into blocks, we can see that x * yjy * x is a block diagonal matrix, and diagonal blocks can be diagonalized
So K (x * yjy * x) = (x * yjy * x) K can be diagonalized, that is, ab = x (KX * yjy * x) x * can be diagonalized and is a normal matrix. Similarly, it can be proved that Ba is a normal matrix

For any x ∈ R, the function f (x) represents the larger of − x + 3, 32x + 12, X2 − 4x + 3, then the minimum value of F (x) is ()
A. 2B. 3C. 8D. -1

Make the images of − x + 3, 32x + 12, X2 − 4x + 3 respectively, as shown in the figure: (the curve ABCDE corresponding to the shadow part), then from the image, we can see that the function f (x) obtains the minimum value at C, and from y = − x + 3Y = 32x + 12, we can get x = 1y = 2, that is, the minimum value of (x) is 2

The function (SiNx) / X has no meaning on 0, so how can we do definite integration from 0 to 1?
Does the definition of definite integral require the function to be defined in a closed interval?

It is easy to know that when x ---- > 0 ＋, (SiNx) / X. ---- > 1
X = 0 is a removable discontinuity
Definition: when x = 0, SiNx / x = 1
That's it