Set a = {x | x ≥ 2}, B = {y | y = 2 x power + 1}, then a ∪ B= The answer is (negative infinity, - 2] ∪ (1, positive infinity). Isn't the x value of 2 in B arbitrary

Set a = {x | x ≥ 2}, B = {y | y = 2 x power + 1}, then a ∪ B= The answer is (negative infinity, - 2] ∪ (1, positive infinity). Isn't the x value of 2 in B arbitrary

The description object of set B is y, which refers to the range of Y, not the range of X
2 ^ x > 0, then: Y > 1
∴B=(1,+oo)

Mathematical problems sin ^ (α + π) cos (- α + π) / Tan (α + π) Tan (α + 2 π) cos ^ (- α - π)
Simplification: 1. Sin ^ (α + π) cos (- α + π) / Tan (α + π) Tan (α + 2 π) cos ^ (- α - π)
Calculation: 1.sin35cos25 + sin55cos65
2.cos28cos73+cos62cos17

1.sin35cos25+sin55cos65
=sin35cos25+cos35sin25
=sin(35+25)
=Sin60 = (radical 3) / 2
2.cos28cos73+cos62cos17
=cos28sin17+sin28cos17
=sin(28+17)
=sin45
=(radical 2) / 2

A mathematical problem about sin θ, cos θ, Tan θ
Tan 1 ° Tan 2 ° Tan 3 °... Tan 87 ° Tan 88 ° Tan 89 ° / cos & # 178; 1 ° + cos & # 178; 2 ° + cos & # 178; 3 ° +... + cos & # 178; 87 degrees + cos & # 178; 88 ° + cos & # 178; 89 °
It can be calculated by the following formula:
tanθ=sinθ / cosθ
sin²θ+cos²θ=1
sin(90°-θ)=cosθ
cos(90°-θ)=sinθ
tan(90°-θ）=1 / tanθ
Solution! ~ &;

(Tan 1 ° Tan 2 ° Tan 3 °... Tan 87 ° Tan 88 ° Tan 89 °) / (COS & # 178; 1 ° + cos & # 178; 2 ° + cos & # 178; 3 ° +... + cos & # 178; 87 ° + cos & # 178; 88 ° + cos & # 178; 89 °)
=[TAN1 ° tan2 ° tan3 °. Tan45 * (1 / tan44) * (1 / tan43). * (1 / TAN1)] / [cos & # 178; 1 ° + cos & # 178; 2 ° + cos & # 178; 3 ° +... + cos & # 178; 45 ° + Sin & # 178; 44 ° + Sin & # 178; 43 °... + Sin & # 178; 3 ° + Sin & # 178; 2 ° + Sin & # 178; 1 °]
=1/(44+1/2)
=1/(89/2)
=2/89
Molecular application: Tan (90 ° - θ) = 1 / Tan θ
The use of denominator: Sin & # 178; θ + cos & # 178; θ = 1
sin(90°-θ)=cosθ

Through the intersection of line L1: 3x + 4y-5 = 0 and line L2: 3x-4y-13 = 0, the linear equation with slope 2 is

Y=2x-5

How to see the slope of the linear equation 3x-4y + 5 = 0

Turn the equation into a standard oblique section
y=ax+b
Where a is the slope and B is the intercept on the y-axis
3x-4y+5=0
y=(3/4)x+5/4
Slope k = 3 / 4

Let a (- 2,3), the right focus of ellipse 3x2 + 4y2 = 48 is f, and point P moves on the ellipse. When | AP | + 2 | PF | is the minimum, the coordinates of point P are ()
A. （0，23）B. （0，-23）C. （23，3）D. （-23，3）

According to the second definition of ellipse, if a vertical line passing through a as the right guide line intersects with point B, then the minimum value of | AP | + 2 | PF | = | AP | + 1E | PF |, the ordinate of | AP | + 2 | PF |, the ordinate of | AP | + 2 | PF |, ∵ a (- 2,3), P is 3, and the abscissa of P is 23, | P (23,3)

Let a (- 2, √ 3), f be the right focus of the ellipse 3x ^ 2 + 4Y ^ = 48, and point P move on the ellipse. When | AP | + 2 | PF | takes the minimum value, the coordinates of point P will change

It is transformed into standard form X ^ 2 / 16 + y ^ 2 / 12 = 1; a = 4; C = 2
It is easy to know that a (- 2, √ 3), in the ellipse, P leads a vertical line to its right guide line, and the vertical foot is h
A draw the vertical line to the right, and the vertical foot is K
So PF / pH = e = 1 / 2, so 2pF = pH
It is equivalent to the minimum value of AP + pH!
AP + pH ≥ ah ≥ AK, when p is the intersection of line segment AK and ellipse, the minimum value is taken
At this time, the ordinate of P √ 3 shows that P (2 √ 3, √ 3)

Focus coordinates of ellipse 3x * x + 4Y * y = 1

3x*x+4y*y=1
x*x / (√3/3*√3/3) + y*y /(1/2*1/2) = 1
a = √3/3 ,b = 1/2
The major axis is on the x-axis
c = √(a^2-b^2) = √(1/3-1/4) = 1/√12 = √3/6
Focus coordinates (- √ 3 / 6,0), (√ 3 / 6,0)

Given that the line L: 3x + 4Y + C = 0, the circle C: X & # 178; + Y & # 178; - 2x + 4Y + 1 = 0, find the equation of the line tangent to the circle C and perpendicular to the line L

Circle C: (x-1) &# 178; + (y + 2) &# 178; = 4, center C (1, - 2), radius r = 2
Let the linear equation be: 4x-3y + M = 0
Then the distance from the center of the circle to the straight line is
d=|4+6+m|/√(4²+3²)=r=2
Then M = 0 or - 20
Ψ 4x-3y = 0 or 4x-3y-20 = 0

If the line y = KX + 2 and the ellipse 2x & # 178; + 3Y & # 178; = 1 have a common point, then the value of K is

Substituting: 2x ^ 2 + 3 (KX + 2) ^ 2 = 12x ^ 2 + 3K ^ 2x ^ 2 + 12kx + 12-1 = 0 (2 + 3K ^ 2) x ^ 2 + 12kx + 11 = 0, straight line y = KX + 2 and ellipse 2x & # 178; + 3Y & # 178; = 1 have common points, then the equation has real roots, B ^ 2-4ac > > 0 (12K) ^ 2-4 * (2 + 3K ^ 2) * 11 > > 0144k ^ 2-88-132k ^ 2 > > 012k ^ 2 > > 88k ^ 2 > > 66 / 9K > (...)