It is known that in ∠ ABC, D is a point on the bisector of ∠ ABC, e and F are on the sides AB and AC respectively, and de = DF. (points D, e and F do not coincide with point B) judge the relationship between ∠ bed and ∠ BFD, and explain the reason

It is known that in ∠ ABC, D is a point on the bisector of ∠ ABC, e and F are on the sides AB and AC respectively, and de = DF. (points D, e and F do not coincide with point B) judge the relationship between ∠ bed and ∠ BFD, and explain the reason

The reason is as follows: as shown in the figure, DM ⊥ AB is in M, DN ⊥ BC is in N, and ⊥ DEM and △ DFN are right triangles. ∵ BD is the bisector of ∵ ABC, DM = DN. In RT △ DEM and RT △ DFN, de = DFDM = DN, ≌ RT △ DEM ≌ RT △ DFN (HL), ≌ RB ≌ DT ≌ DFN (HL) and ≌ RB ≌ DT ≌ DFN are right triangles

The 0 th power of 2 + the 1 th power of 2 + the square of 2 + ·· + the 1000 th power of 2

Sn=a1(1-q^n)/(1-q)
=1(1-2^1001)/(1-2)
=2^1001 -1
If you don't understand this question, you can ask,

Is not log (1 / 2) ^ (3x-2) ≥ 0, 3x-2 greater than or equal to 1, x greater than or equal to 1?

You are right, but you are wrong
Logarithmically significant, 3x-2 > 0
Log (1 / 2) (3x-2) ≥ 0
0

The circumference of hula hoop and table tennis increased by 10 cm, the radius increased more
process
thank you

The formula of perimeter is 2 Π R
Because the perimeter is the same, the radius is the same

In RL Series sinusoidal circuit, if the resistance voltage is 30 V and the inductance voltage is 40 V, then the total voltage is 30 v

Because there is a 90 ° phase angle between the voltage of the resistor and the voltage of the inductor, the resultant voltage is 50V

Let the character set D = {a, B, C, D, e}, and the frequency of each character w = {10,2,5,6,4}. Draw the corresponding Huffman tree when Huffman encoding characters, and give the encoding of each character. Is there only one possibility

The frequency is w = {10,2,5,6,4}. You can calculate the usage probability of each symbol according to this. The basic idea of Huffman coding is: for symbols with higher usage frequency, use shorter codewords to encode, and for symbols with lower usage frequency, use longer codewords to encode, which makes the coding efficiency very high

As shown in the figure, C, D and E are the three points on the line AB, and AC = half of CD, e is the midpoint of BD, de = one fifth of 2, so the length of CE can be obtained

De = 1 / 5ab = 2cm, e is the midpoint of BD
BD=2DE=2×2=4cmAB=10cm
Ad = ab-de = 6cm ∵ AC = 1 / 2CD, that is CD: AC = 2:1
CD=6÷(2+1)×2=4cm∴CE=CD+DE=4+2=6(cm)

Do positive definite matrices have to be symmetric? Please explain why and where?

The definition of positive definite matrix in linear algebra comes from positive definite quadratic form
Tongji fourth edition linear algebra p.136 definition 10
Let f (x) = x ^ tax, if f (x) > 0 for any x ≠ 0, then f is said to be positive definite quadratic form, and symmetric matrix A is said to be positive definite
So the positive definite matrix of linear algebra is symmetric

If the tangent of circle (x-1) 2 + (y + 3) 2 = 1 is made through point P (2,4), then the tangent equation is______ ．

When the tangent slope does not exist, the tangent equation is x = 2. When the tangent slope exists, let the tangent equation be y-4 = K (X-2), that is, kx-y + 4-2k = 0, and then according to the distance from the center of the circle (1, - 3) to the tangent equal to the radius, we can get | K + 3 + 4 − 2K | K2 + 1 = 1 In conclusion, the tangent equation of a circle is 24x-7y-20 = 0, or x = 2, so the answer is: 24x-7y-20 = 0, or x = 2

1-cosx/xsinx
LIM (n → 0) (1-cosx) / xsinx solution: LIM (n → 0) (1-cosx) * SiNx / x = LIM (n → 0) SiNx / x-lim (n → 0) cosx * SiNx / x = 1-lim (n → 0) 2 * cosx * SiNx / 2x = 1-lim (n → 0) sin2x / 2x = 1-1 = 0 why not? That step is not right?
The answer to this limit is 1 / 2. I'm wrong. I want to know what's wrong?

Change (1-cosx) into 2 [sin (x / 2)] ^ 2