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The equations 3y-2x = 48,3y + 2x = 90 are solved by addition and subtraction elimination method, and then the equations 3S + 2T = 4,2s-3t = 7 are solved by addition and subtraction elimination method
3y-2x=48（1）
3y+2x=90（2）
(2) - (1) get
4x=42
The solution is x = 21 / 2
Y = 23
3s+2t=4（3）
2s-3t=7（4）
(3) * 1.5
4.5s+3t=6（5）
(5) + (4) get
6.5s=13
The solution is s = 2
Substituting into the solution, t = - 1
The quadratic function y = x & # 178; - ax, when x < 1, y decreases with the increase of X, then the value range of a is______ ．
If the opening of Y is upward, then on the left side of the symmetry axis X = A / 2, y decreases with the increase of X
So x
Let the vertex of quadratic function y = (LGA -- 1) x ^ 2 -- 10x + C on the line x = 5 (1) find the value of a; (2) if it is always greater than 0, find the value range of C (please write the procedure)
Because the vertex is on x = 5
So the axis of symmetry is x = 5
So 5 / (lga-1) = 5
So a = 100
Here y = x ^ 2-10x + C is a parabola with the opening upward
If Y > 0 is constant
That is to make the parabola image always above the x-axis
So there must be a discriminant less than 0
That is 100-4c25
It is known that the quadratic function F X AX2 + BX (a) is not equal to 0 and satisfies 1
F (x) = ax ^ 2 + BX1 ≤ f (- 1) ≤ 2, i.e. 1 ≤ A-B ≤ 2, ① 3 ≤ f (1) ≤ 4, i.e. 3 ≤ a + B ≤ 4, ② ① * 10 / 3 + ② * 2 / 3, we get: 10 / 3 + 3 * 2 / 3 ≤ (10 / 3 + 2 / 3) a + (- 10 / 3 + 2 / 3) B ≤ 2 * 10 / 3 + 4 * 2 / 3, i.e. 16 / 3 ≤ 4a-2b ≤ 28 / 3, i.e. 16 / 3 ≤ f (- 2) ≤ 28 / 3
In △ ABC, the edge A.B.C opposite to the inner angle A.B.C satisfies 2B & # 178; = 3aC, and the angle B is equal to 60 degrees
Using cosine theorem, you wait, I'll take photos
In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, if cos (2B + C) + 2sinb
From this inequality, only one inequality can be obtained, because cos (2B + C) + 2sinb (sum difference product formula) = cos2bcosc-sin2bsinc + 2sinb (angle doubling formula) = (1-2 (SINB) ^ 2) cosc-2sinbcosbsincc + 2sinb = cosc-2sinb (sinbcosc + cosbsincc) + 2sinb = cosc-2sinbsi
In the triangle ABC, let the opposite sides of angle a, angle B and angle c be a, B and C respectively, and a plus B equals 2B, and C minus a equals 1 / 2B. Then what triangle is it
a+b=2b
A = B isosceles triangle
c-a=1/2b
c=a+1/2b=3/2b
cos C=(a^2+b^2-c^2)/2ab=(a^2+a^2-9/4a^2)/2a^2=-1/8
C is an obtuse angle
So ABC is an isosceles triangle with obtuse vertex angle
In the triangle ABC, the opposite sides of angle a, angle B and angle c are ABC respectively, and C plus a equals 2B, C minus a equals half B, then what is the shape of the triangle ABC?
C + a = 2B, C-A = 1 / 2B, so C = 5 / 4B. A = 3 / 4b, so a: B: C = 3:4:5. According to Pythagorean theorem, triangle ABC is a right triangle with angle c as right angle
The solution of 3 / m-2 / N = 25m-2n = 4 system
3/m-2/n=2 2m-3n=12 ①
5m-2n=4 ②
① 2 - 2 × 3
4m-15m=24-12
-11m=12
∴m=--12/11
n=--52/11
Solve the equations x-2y = 0,2x 3Y = 14 5m + 4N = 6,3m-2n = 8 3 (Y-2) = x + 1,
To solve the equations x-2y = 0,2x 3Y = 14
5m+4n=6,3m-2n=8
3（y-2）=x+1,2（x+1）=3（y+2）
1. X = 2Y, Dai Ren 2x + 3Y = 14: 4Y + 3Y = 1, y = 1 / 7, x = 2Y = 2 / 7. Multiply x = 2 / 7 {y = 1 / 72.3m-2n = 8 by 2 to get 6m-4n = 16, and the equation 5m + 4N = 6, 11m = 22, M = 2
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