Let z = KX + y, where the real numbers x and y satisfy x + Y-2 ≥ 0x-2y + 4 ≥ 02x-y-4 ≤ 0. If the maximum value of Z is 12, then the real number k = 0___ .

Let z = KX + y, where the real numbers x and y satisfy x + Y-2 ≥ 0x-2y + 4 ≥ 02x-y-4 ≤ 0. If the maximum value of Z is 12, then the real number k = 0___ .

The feasible region is as follows: a (4,4) is obtained from x-2y + 4 = 02x-y-4 = 0. Similarly, B (0,2), z = KX + y, that is, y = - KX + Z, divided into two cases: k > 0 and K < 0. When k > 0, the objective function z = KX + y takes the maximum value at point a, that is, the intercept Z of the line z = KX + y on the Y axis is the largest, that is, 12 = 4K + 4, and K = 2
Given 2sinb = sin (2a + b), find the value of Tan (a + b): tanb
Don't forget Baidu's original process, thank you
Well, the title is really correct. This is on page p55 of one lesson one practice. I don't know if it's right
2 sin (a + B-A) = sin (a + B-A) = > sin (a + b) cosa = 3cos (a + b) Sina = > Tan (a + b) = 3tanatanb = Tan (a + B-A) = (Tan (a + b) - Tana) / (1 + Tana · Tan (a + b)) = (3-1) Tana / (1 + 3 (Tana) ^ 2) so tan (a + b) / tanb = 3 /
It seems to me that you have typed the question wrong, otherwise you can't work it out.
It should be the value of Tan (a + b): Tana!
From the title:
2sin｛（a + b ）- a｝ = sin｛（a + b）+ a｝
That is: 2Sin (a + b) cosa - 2cos (a + b) Sina = sin (a + b) cosa + cos (a + b) Sina
Ψ sin (a + b) cosa = 3cos (... Expansion)
It seems to me that you have typed the question wrong, otherwise you can't work it out.
It should be the value of Tan (a + b): Tana!
From the title:
2sin｛（a + b ）- a｝ = sin｛（a + b）+ a｝
That is: 2Sin (a + b) cosa - 2cos (a + b) Sina = sin (a + b) cosa + cos (a + b) Sina
∴sin（a + b）cosa = 3cos(a + b)sina
Divide the two sides of the equation by cos (a + b) cosa
The results show that Tan (a + b) = 3tana
х: Tan (a + b): Tana = 3 х
Three
If the lines X + 2y-1 = 0 and y = KX are parallel to each other, then the value of the real number k is______ .
∵ the lines X + 2y-1 = 0 and y = KX are parallel to each other ∵ the slopes of the two lines are equal, k = - 12, so the answer is: - 12
Given Tan (π / 4 + a) = 1 / 2, find the value of sin (2a + 2 π) - Sin & # 178; (π / 2-A) / 1-cos (π - 2A)?
[sin（2a+2π）-sin²（π/2-a)]/[1-cos(π-2a）]=[sin2a-cos^2a]/[1+cos2a]=cosa[2sina-cosa]/2cos^2a=tana-1/2=-1/3-1/2=-5/6tan（π/4+a）=(1+tana)/(1-tana) tana=-1/3
If the images of positive scale functions y = KX and y = 2x are symmetric about X axis, then the value of K is equal to
k=-2
(1,2) is a point on the function y = 2x,
In fact, you can take any point on it. You can take (2,4) or (3,6)
It is proved that: (1-cos4 α) / SiN4 α * Cos2 α / (1 + Cos2 α) = Tan α
(1-cos4α)/sin4α*cos2α/(1+cos2α)
=2 (sin2 α) ^ 2 / (2sin2 α Cos2 α) * Cos2 α / (1 + Cos2 α) [(sin2 α) ^ 2 represents the square of sin2 α]
=sin2α/(1+cos2α)
=2sinαcosα/[2(cosα)^2]
=tanα
For any real number k, the positional relationship between the line (3K + 2) x-ky-2 = 0 and the circle x2 + y2-2x-2y-2 = 0 is______
The equation of the circle is changed into the standard form: (x-1) 2 + (Y-1) 2 = 22. It can be seen that the radius of the circle is equal to 2, and the distance between the center of the circle and the straight line d = | 2K | (3K + 2) 2 + K2 ≤| 2K | K2 = 2, so the straight line is tangent or intersect with the circle
If α and β are acute angles and cos α / sin β + cos β / sin α = 2, it is proved that α + β = π / 2
The formula is sin α (COS α - sin β) = sin (sin α - cos β), α and β are acute angles, so if α! = β, the symbols on both sides are different, so only α = β = 45 degrees, so it is true
It is known that the family of lines L: kx-y-4k + 3 = 0, and the fixed circle C: x ^ 2 + y ^ 2-6x-8y + 21 = 0. Try to determine the position relationship between the moving line L and the fixed circle C and prove it
First, the equation of circle is transformed into standard equation, and (x-3) ^ 2 + (y-4) ^ 2 = 4
So the coordinate of the center of a circle is (3,4), find the distance from the center of a circle to the straight line | - 3k-4-4k + 3 | / sqrt (k ^ 2 + 1 ^ 2) = D (| - is the absolute value sign, sqrt means the open root sign, / is the division sign, D is the distance from the center of a circle to the straight line, for example: the center of a circle is (x1, Y1), the linear equation is ax + by + C = 0, then the distance from the point to the straight line is d = | - ax1 + by1 + C | / sqrt (a ^ 2 + B ^ 2))
The solution here is | - k-1 | / sqrt (k ^ 2 + 1) = D, then the two sides of the formula are squared, and the value range of D is obtained. Compared with the radius, if it is greater than the radius, then it is separated, equal to: tangent, less than: intersection (secant)
The standard equation of a circle is (x-3) ^ 2 + (y-4) ^ 2 = 4, and the coordinates of its center are (3,4)
The equation of the family of lines l can be reduced to Y-3 = K (x-4), which always passes through the point (3,4) (i.e. the center of the circle)
So the moving line l always bisects the circle
Evaluation: 2cos π / 2-tan π / 4 + 3 / 4tan & sup2; π / 6-sin π / 6 + cos & sup2; π / 6 + sin 3 π / 2 - 1 / 4cos π
2cosπ/2-tanπ/4+3/4tan²π/6-sinπ/6+cos²π/6+sin3π/2 -1/4cosπ
=2×0-1+3/4×(√3/3)²-1/2+(√3/2)²+(-1)-1/4×(-1)
=0-1+1/4-1/2+3/4-1+1/4
=-5/4