Y1 = 1 / 5x + 1, y2 = 2x + 1 / 4, when what is the value of X, Y1 and Y2 are opposite numbers?

Y1 = 1 / 5x + 1, y2 = 2x + 1 / 4, when what is the value of X, Y1 and Y2 are opposite numbers?

Y1 and Y2 are opposite numbers, indicating that Y1 + y2 = 0
1 / 5x + 1 + 2x + 1 / 4 = 0. By solving the equation about X, we can get x = - 25 / 44

Finding the minimum value of y = sin2x + 2 times root 2cos (π / 4 + x) + 3

y=sin2x+2√2cos(π/4+x）+3
=cos(2x-π/2)+2√2cos(π/4+x）+3
=1-2sin²(x-π/4)-2√2sin(x-π/4）+3
=4-2[sin²(x-π/4)+√2sin(x-π/4)]
Let sin (x - π / 4) = t, then - 1 ≤ t ≤ 1
y=4-2(t²+√2t)
y=4-2(t²+√2t+1/2)+2*1/2
=4-2(t-√2/2)²+1
=5-2(t-√2/2)² （-1≤t≤1）
When t = - 1, the function gets the minimum value
4-2[(-1)²+√2*(-1)]
=4-2(1-√2)
=2-2√2
So the minimum value is 2-2 √ 2

It is known that the solutions of the linear equation 2x-k = 5 and X-1 / 2 + 1 = x + 1 / 3 are the same?

First of all
(x-1) / 2 + 1 = (x + 1) / 3 multiply both sides of the equation by 6
3(x-1)+6=2(x+1)
3x-3+6=2x+2
x+3=2
x=-1
Substituting x = - 1 into the equation 2x-k = 5, we get
-2-k=5
k=-2-5=-7
So k = - 7

If the intersection coordinates of y = - x + A and y = x + B are (m, 8), then the value of a + B is ()
A. 32B. 24C. 16D. 8

∵ the intersection coordinates of the line y = - x + A and the line y = x + B are (m, 8), ∵ 8 = - M + A, 1, 8 = m + B, 2, 1 + 2, 16 = a + B, i.e. a + B = 16

It is known that the abscissa of the intersection of two straight lines Y & # 8321; = K & # 8321; X + B & # 8321;, Y & # 8322; = K & # 8322; X + B & # 8322; is X., and K & # 8321; > 0, K & # 8322; X
A.y₁=y₂
B.y₁>y₂
C.y₁

Choose B
Line y & #; = & nbsp; K &; & nbsp; X + & nbsp; B &; & nbsp;, Y &; = K &; & nbsp; X + B &; & nbsp; and K &; & gt; 0, K &; & lt; 0 & nbsp;
Y1 & nbsp; is an increasing function & nbsp; & nbsp; Y2 is a decreasing function & nbsp; & nbsp; after the intersection, Y1 goes up to the right & nbsp; and Y2 goes down to the right
So Yi & gt; Y2

It is known that P1 (x1, Y1) and P2 (X2, Y2) are two points on the same inverse scale function image. If x2 = X1 + 2 and 1Y2 = 1y1 + 12, then the expression of the inverse scale function is______ ．

Let y = KX, ∵ P1 (x1, Y1), P2 (X2, Y2) be two points on the same inverse scale function image, ∵ x1 · Y1 = x2 · y2 = k, ∵ 1y1 = X1K, 1Y2 = X2K, ∵ 1Y2 = 1y1 + 12, ∵ X2K = X1K + 12, ∵ 1K (x2-x1) = 12, ∵ x2 = X1 + 2, ∵ 1K × 2 = 12

It is known that the image of positive scale function y = k? X intersects with the image of linear function y = k? X + 2 at point P (1, - 3)
(1) Find out the values of K? And K?; (2) draw the two function images; (3) find out the area of the triangle formed by the two function images and the X axis

It is easy to get that K1 = - 3, K2 + 2 = - 3. Therefore, K1 = - 3, K2 = - 5 let the intersection of the first-order function and the x-axis be q (x, 0), and let the first-order function, - 5x + 2 = 0, x = 2 / 5. The bottom of OPQ is the absolute value of the abscissa of point Q, and the height is the absolute value of the ordinate of point P, so s = 2 / 5 × 3 △ 2 = 3 / 5

It is known that the image of inverse scale function y = m / X passes through point a (1, - 3), and the image of primary function y = KX + B passes through point a and point C (0, - 4), and the image of inverse scale function y = KX + B passes through point a and point C (0, - 4)
(1) Determine the expressions of these two functions
(2) Finding the coordinates of point B

Y = m / 3 through a
Then - 3 = m / 1
m=-3
y=kx+b
-3=k+b
-4=0+b
So k = 1, B = - 4
So y = - 3 / X and y = x-4
y=-3/x=x-4
x²-4x+3=0
(x-1)(x-3)=0
X = 1 is a
Then x = 3, y = - 3 / x = - 1
So B (3, - 1)

Simplify X & # 8321; &# 178; + X & # 8321; X & # 8322; + X & # 8322; &# 178; + 1
How to simplify it to (X & # 8321; + 1 / 2x & # 8322;) &# 178; + 3 / 4x & # 8322; &# 178; + 1

=x₁²+x₁x₂+1/4x₂²+3/4x₂²+1
=（x₁+1/2x₂）²+3/4x₂²+1
Take X & # 8322; &# 178; apart, and take X & # 8321; &# 178; + X & # 8321; X & # 8322; + 1 / 4x & # 8322; &# 178; square completely

The two focuses of hyperbola are F & # 8321; (0, √ 3), F & # 8322; (0, - √ 3), and one vertex is a (0,1). The standard equation of hyperbola is obtained

The focus is on the y-axis
c=√3
c²=3
A vertex on the y-axis is a (0,1)
∴a=1
a²=1
b²=c²-a²=2
The hyperbolic equation is
y²-x²/2=1