(2009, Anhui) if the plane region represented by the system of inequalities x ≥ 0x + 3Y ≥ 43x + y ≤ 4 is divided into two equal parts by the line y = KX + 43, then the value of K is () A. 73B. 37C. 43D. 34

(2009, Anhui) if the plane region represented by the system of inequalities x ≥ 0x + 3Y ≥ 43x + y ≤ 4 is divided into two equal parts by the line y = KX + 43, then the value of K is () A. 73B. 37C. 43D. 34

Satisfy the constraint conditions: X ≥ 0 x + 3 y ≥ 43 x + y ≤ 4, the plane area is as shown in the figure: from the figure, the straight line y = KX + 43 passes through the point a (0, 43). When the straight line y = KX + 43 passes through the midpoint D (12, 52) of BC, the plane area is divided into two parts with equal area by the straight line y = KX + 43. When x = 12, y = 52, the equation of the straight line y = KX + 43 is substituted into k = 73, so a is selected
If you know the degree of one angle of a triangle, how can you find the degree of the other two angles?
For example, for a triangle, I know the degree of one angle. How can I find the degree of the other two angles
If the answer is clear, there is a clear answer,
The idea is good, but it is impossible. The sum of three angles ABC, you know, is 180, plus the degree of an angle, which is equivalent to two formulas solving three unknowns. The solution is infinite, and there is no unique solution
It's impossible to find a B C triangle B + C > = a b-c
If the plane region represented by the inequality system X ≥ 0 x + 3Y ≥ 4 3x + y ≤ 4 is divided into two equal parts by the straight line y = KX + 4 / 3, then K is ()
Why is it when the line y = KX + 4 / 3 passes through the midpoint of the line 3x + y = 4;
The plane is a triangle, and y = KX + 4 / 3 goes through a vertex of the triangle, so just go through the midpoint of the side opposite the vertex. The middle line divides the area of the triangle equally
If in △ ABC, 2 (∠ a + ∠ C) = 3 ∠ B, then what is the external angle degree of ∠ B ()
A. 36B. 72C. 108D. 144
∫ a + B + C = 180 °, ∫ 2 (∫ a + B + C) = 360 °, ∫ 2 (∫ a + C) = 3 ∫ B, ∫ B = 72 °, ∫ B's external angle degree is 180 ° - B = 108 °, so select C
Given the vector a = (2coswx, 1), B = (sinwx + coswx, - 1), w ∈ R, w > 0, Let f (x) = a * B (x ∈ R), if the minimum positive period of F (x) is π / 2
1. Find the value of W
2. Find the monotone interval of F (x)
It is known that: (1) f (x) = (2coswx, 1) (sinwx + coswx, - 1) = 2coswx (sinwx + coswx) - 1 = 2coswxcoswx-1 + sin2wx = cos2w + sin2wx = √ 2 / 2 * sin (2wx + π / 4), then 2 π / 2W = π / 2, so w = 2 (2) because w = 2, f (x) = √ 2 / 2 * sin (4x + π / 4) when 4x + π
It is known that, as shown in the figure △ ABC, ab = AC, ∠ a = 90 °, the bisector CD of ∠ ACB intersects AB at point E, ∠ BDC = 90 °, and the verification is CE = 2bd
It is proved that: as shown in the figure, extend the extension line of BD intersection CA to F, ∵ - BAC = 90 °
Given the vector M = (2coswx, - 1), n = (sinwx coswx, 2), where w > 0, the period of function E (x) = m times N + 3 is beat
Given the vector M = (2coswx, - 1), n = (sinwx coswx, 2), where w > 0, the period of the function E (x) = m times N + 3 is beat, the value of W is obtained
m*n + 3 = 2coswx(sinwx-coswx) + 3
= sin 2wx - 2cos^2 wx + 3
= sin2wx - 1 - cos2wx + 3
= sin2wx - cos2wx + 2
= sin(2wx - pi/4) + 2
w = 1
Adaptation of chromium milk powder
As shown in the figure, BD and CE are bisectors of ∠ ABC and ∠ ACB respectively, and ∠ DBC = ∠ ECB = 31 °. Find the degree of ∠ ABC and ∠ ACB, are they equal? (write a simple process)
If BD and CE are bisectors of ∠ ABC and ∠ ACB, we can get ∠ abd = ∠ DBC = 12 ∠ ABC, ∠ ace = ∠ ECB = 12 ∠ ACB, from ∠ DBC = ∠ ECB = 31 °, we can get ∠ ABC = ∠ ACB = 62 ° and ∠ ABC = ∠ ACB
Given the vector a = (sinwx, - cosw). B = (sinwx, - 3coswx), C = (- coswx, sinwx), (W > 0), Let f (x) = a · (B + C). X ∈ R
1) Finding the maximum value of function f (x)
f(x)=a(b+c)
=ab+ac
=sin²wx-3cos²wx-sinwxcoswx-sinwxcoswx
=(1-cos2wx)/2-3(1+cos2wx)/2-sin2wx
=-1-(sinwx+2coswx)
=-1-√5sin(wx+ψ)
The maximum value is - 1 + √ 5
Seeking the same answer
As shown in the figure, BD and CE are bisectors of ∠ ABC and ∠ ACB respectively, and ∠ DBC = ∠ ECB = 31 °. Find the degree of ∠ ABC and ∠ ACB, are they equal? (write a simple process)
If BD and CE are bisectors of ∠ ABC and ∠ ACB, we can get ∠ abd = ∠ DBC = 12 ∠ ABC, ∠ ace = ∠ ECB = 12 ∠ ACB, from ∠ DBC = ∠ ECB = 31 °, we can get ∠ ABC = ∠ ACB = 62 ° and ∠ ABC = ∠ ACB