It is known that the image with positive scale function y = KX passes through point a (x, 4), and O is the origin of the coordinate (1) If OA = 5, then X= (2) If the angle between OA and X axis is 60 degrees, then x= (1) The answer to this question is ± 3; the answer to this question is 4 / 3 × root 3

It is known that the image with positive scale function y = KX passes through point a (x, 4), and O is the origin of the coordinate (1) If OA = 5, then X= (2) If the angle between OA and X axis is 60 degrees, then x= (1) The answer to this question is ± 3; the answer to this question is 4 / 3 × root 3

The point a, O and the point projected by a on the x-axis form a triangle. Then the abscissa and ordinate of a are the lengths of the two right angle sides of the triangle, and OA is the length of its oblique side
X ^ 2 + 4 ^ 2 = 5 ^ 2 = x = ± 3
The angle between OA and X axis is 60 degrees, tan60 = root 3, tan60 = ordinate divided by abscissa, so the abscissa of a is x = 4 divided by root 3, which is 4 / 3 × root 3

There is no intersection between the image of a certain function and the line y = 3x-5, and the abscissa of the intersection with the line 2x-y = 1 is 2. Find the analytic expression of this function

If there is no intersection between the image of a function and the line y = 3x-5, then the slope of the function K = 3
If the abscissa of the intersection point with the line 2x-y = 1 is 2, then the coordinates of the intersection point are (2, 3)
Let the analytic expression of the function be y = 3x + B. after substituting the coordinates of the intersection (2, 3): 3 = 3 * 2 + B
The solution is b = - 3
So the analytic expression of the function is y = 3x-3

It is known that the abscissa of the intersection of the image of the first-order function y = KX + B and the straight line y = 2x + 1 is 2, and there is no intersection with the straight line y = 3x + 2

When x = 2, y = 2x + 1 = 5
The intersection of the image of the linear function y = KX + B and the line y = 2x + 1 is (2,5)
∵ the line y = KX + B has no intersection with the line y = 3x + 2 ∵ k = 3
Substituting (2,5) into y = 3x + B, B = - 1
The analytic expression of the first-order function is y = 3x-1

Given the linear function y = 3x + m of 2 and y = LX + n of negative 2, the image is all right
After passing through point a (- 2,0), and intersecting with Y-axis at point B and point C respectively, calculate the coordinates of B and C bright spot

Substituting point a, calculating m and N, and then substituting x equal to 0 to calculate y is required. I can't understand what you write, so I can't calculate it