Let m.n.p be the points on the three sides of the triangle ABC, which make the vector BM = 1 / 3, the vector BC, the vector CN = 1 / 3, the vector Ca, the vector AP = 1 / 3, the vector AB = vector a, the vector AC = vector B, the vector a. the vector B represents the vector Mn. The vector NP. The vector PM

Let m.n.p be the points on the three sides of the triangle ABC, which make the vector BM = 1 / 3, the vector BC, the vector CN = 1 / 3, the vector Ca, the vector AP = 1 / 3, the vector AB = vector a, the vector AC = vector B, the vector a. the vector B represents the vector Mn. The vector NP. The vector PM

If ∠ ABC = 31 & ordm;, e is led out from the extension line of AC, and the bisector AE of ∠ BAC intersects the bisector ce of ∠ FCB at point E, ∠ AEC is ()

15.5°.
∠AEC=∠ECF-∠EAC=(1/2)∠BCF-(1/2)∠BAC=(1/2)(∠BCF-∠BAC)=(1/2)∠ABC=1/2·31°
=15.5°

1、 Given x ^ 2 + 3y-1 = 0, then the value of 2x ^ 2 + 6y-7 is
2、 It is known that a = 2A ^ 2 + 3ab-2a-1, B = - A ^ 2 + AB-1
Find 3A + 6B
If the value of 3A + 6B has nothing to do with the value of a, find the value of B

1、 The original formula = 2 (X & # 178; + 3Y) - 7 = 2 × 1-7 = - 5
2、 3A + 6B = 6A & # 178; + 9ab-6a-3-6a & # 178; + 6ab-6 = 15ab-6a-9
If 3A + 6B has nothing to do with a, then 15b-9 = 0, that is, B = 3 / 5

If the distance between a point P and the straight line x = - 25 / 4 on the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1 is 5 / 2, the distance between the point P and the right focus is 5 / 2

Easy to get: x = - 25 / 4 is the left standard line, eccentricity e = 4 / 5
Let the left focus be F1, the right focus be F2, and the distance from P to the right guide line be D1 = 5 / 2
According to the second definition of ellipse, Pf1 / D1 = e = 4 / 5
If D1 = 5 / 2 is substituted, Pf1 = 2 is obtained
According to the first definition of ellipse, Pf1 + PF2 = 2A = 10
If Pf1 = 2 is substituted, PF2 = 8 is obtained
So the distance from point P to the right focus is 8

In ABC, if AB = 4, AC = 8 and ad = 3, then the length of BC is

In ABC, if AB = 4, AC = 8 and ad = 3, then the length of BC is
AD^2+CD^2+AD^2+BD^2=AB^2+AC^2
BD = CD = radical 31
Then BC length = 2 times root sign 31

Given that x is a real number, y is a pure imaginary number, and satisfies (2x-1) + (3-y) I = Y-I, find x, y

Because y is a pure imaginary number, let y = Bi, (B ∈ R, and B ≠ 0). The original formula can be sorted as (2x-1) + (3-bi) I = bi-i, that is, (2x-1 + b) + 3I = (B-1) I. from the definition of complex number equality, we can get: 2x − 1 + B = 0b − 1 = 3, the solution is b = 4x = − 32, so x = − 32, y = 4I

The known function f (x) = 2Sin (π / 2 + x) sin (π / 3 + x), X ∈ R
Finding the minimum positive period of function f (x)

f(x)=2cosxsin(π/3+x)
=2(cosx)[√3/2cosx+1/2sinx]
=√3(cosx)^2+sinxcosx
=(√3/2)(1+cos2x)+(1/2)sin2x
=(1/2)sin2x+(√3/2)cos2x+(√3/2)
=sin(2x+π/3)+(√3/2)
T=2π/2=π

It is known that the hypotenuse ab of RT △ ABC is 13cm, and a right angle AC is 5cm. A geometry is obtained by rotating a circle around a straight line AB, then the surface area of the geometry is

AB = 13, AC = 5, so BC = 12, OC * AB = AC * BC, OC = 12 * 5 / 13 = 60 / 13,
The geometry consists of two cones, the upper and the lower. The formula of the side area of the cone is π Rl (R is the base radius, l is the generatrix)
Upper side area = π * OC * BC = π * 60 / 13 * 12 = 720 π / 13
The lower side area = π * OC * AC = π * 60 / 13 * 5 = 300 π / 13, so the surface area = 720 π / 13 + 300 π / 13 = 1020 π / 13

The original function of COS square x

Given the set a = {0.1.2.3} and the set B = {x | y = log3 (2-x)}, then a ∩ B =?

b={x|x