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It is known that, as shown in the figure, points D and E are points on AB and AC on both sides of equilateral triangle ABC, and ad = CE proves CD = be

In triangle ADC and triangle CEB, AC = CB, ad = CE, ∠ DAC = ECB, so triangle ADC and triangle CEB are congruent, so CD = be

It is known that 1 + I is a complex root of the real coefficient equation x2 + ax + B = 0 of X. 1 is the value of a and B. 2 is the judgment
It is known that 1 + I is a complex root of the real coefficient equation x2 + ax + B = 0 about X. 1 finds the value of a and B, and 2 judges whether 1 + I is the root of the equation

Solution 1 by 1 + I is a complex root of the real coefficient equation x2 + ax + B = 0 with respect to X
Then (1 + I) ^ 2 + a (1 + I) + B = 0
That is, 2I + A + AI + B = 0
That is, a + B + (a + 2) I = 0
Solutions a + B = 0 and a + 2 = 0
The solution is a = - 2, B = 2
2 from (1), the equation is x ^ 2-2x + 2 = 0
Let the other of the equation follow t
From the relationship between root and coefficient
t（1+i）=-b/a=2
That is, t = 2 / (1 + I) = 1-i
That is, 1-I is the root of the equation

High school mathematics known function f (x) = LNX - (AX ^ 2) / 2 + (A-1) x, where the real number | a|

According to Lagrange mean value theorem [f (x1) - f (x2)] / (x1-x2) = f '(B1) (B1 ∈ (x1, x2)) x1, X2 ∈ [1,2] B1 ∈ (x1, x2) ∈ (1,2) if f' (x) in the range of (0,1) contains f '(x) in the range of (1,2), then if, 2) F '(B2) = f' (B1) f '(x) = 1 / x-ax + A-1 f' (0) = ∞ f '(1) = 0 f' (2) = - 1 / 2-A f '(x) = - 1 / x ^ 2-A such that f' (x) = 0 - 1 / x ^ 2-A = 0 x ^ 2 = - 1 / a a > 0 f '(x) = 0 has no solution, that is, f' (x) is a monotone function f '(1)

If the waist length of an isosceles triangle is 10 cm and the height of its bottom is 8 cm, what is the area of the triangle?

Bottom edge = 2 √ (10 * 10-8 * 8) = 2 * 6 = 12
S = 12 * 8 / 2 = 48 square centimeter

Excuse me: in RL Series AC circuit, r = 3 Ω. XL = 4 Ω, what is the circuit impedance at this time?

5 Euro

Known set a = {X_ 3 less than or equal to x less than or equal to three}, B = {X! 2m_ 1 is less than or equal to x, less than or equal to m + 2}, if a intersects B = B, find the value range of M

A ={x| -3 ≤ x ≤ 3}
B ={x|2m -1≤ x≤ m+2}
A ∩ B = B, there are two cases
Situation 1:
B is an empty set
2m -1 > m + 2
m > 3
Case 2:
-3 ≤ 2m-1 ≤ m+2 ≤ 3
The solution is: - 1 ≤ m ≤ 1
In conclusion: - 1 ≤ m ≤ 1 or M > 3

In △ ABC, ∠ ACB = 90 °, ad bisects ∠ BAC, de ⊥ AB in E. it is proved that the straight line ad is the vertical bisector of CE
The proof of the vertical bisector in the second grade of junior high school

Because, ∠ ACB = 90; de ⊥ ab
So △ ACD is equal to △ ade
So DC = De, ∠ CDA = ∠ EDA
So △ CDO is equal to △ deo (SAS)
So co = EO, ∠ cod = ∠ DOE = 90 degrees
Therefore, it has been proved

Let z = 3-ai, | Z be complex numbers|

z=3-ai,
|z|=√[3^2+(-a)^2]