If the image of inverse scale function y = 2 / X intersects with the image of primary function y = kx-4, the value range of K is obtained

k> - 2 and K is not equal to 0

If there is an intersection between the inverse scale function y = KX and the positive scale function y = 2x, then the value range of K is If the inverse scale function y = KX intersects with the image of the first-order function y = KX + 2, then the value range of K is ．

(1) (2) when k > 0, y = KX is in the first and third quadrants, y = KX + 2 is in the first, second and third quadrants

As shown in the figure, it is known that a & nbsp; (4, a), B & nbsp; (- 2, - 4) are the intersection of the image of the first-order function y = KX + B and the image of the inverse scale function y = MX. (1) find the analytic expressions of the inverse scale function and the first-order function; (2) find the area of △ a0b

(1) A & nbsp; (4, a), B & nbsp; Substituting (- 2, - 4) two-point coordinates into y = MX, 4A = (- 2) × (- 4) = m, a = 2, M = 8, substituting a (4, 2), B (- 2, - 4) into y = KX + B, 4K + B = 2 − 2K + B = − 4, k = 1b = − 2, the analytic expression of the inverse proportion function is y = 8x, and the solution of the first-order function is y = X-2; (2) let the line AB intersect the y-axis at C, and the analytic expression of the line AB is y = X-2, then C (0, - 2), s △ is obtained AOB=S△AOC+S△BOC=12×2×4+12×2×2=6．

It is known that the ordinate of an intersection of the inverse scale function y = K / X and the linear function y = 2x + k is - 4, and the value of K is obtained

Because if you substitute y = - 4 for y = K / x, you get
-4 = k/x
Substituting y = - 4 into y = 2x + K, we get
-4 =2x+k
xy=y-2x,y=-4
x=2
k=xy=-8

As shown in the figure, it is known that the two vertices of RT △ OAB are a [6,0], B [0,8], O is the origin, and △ OAB rotates 90 ° clockwise around point A. point O arrives at point o ', and point B arrives at point B'?

The coordinates of o are (6,6), and the coordinates of B are (14,6). You can draw the picture by yourself

As shown in the figure, it is a part of the image of an inverse scale function, and points a (1,10) and B (10,1) are its endpoints. (1) find the analytic expression of the function, and write out the value range of the independent variable x; (2) please give an example of life that can be described by the functional relationship of this problem

(1) Let y = KX, ∵ a (1,10) in the image, ∵ 10 = K1, that is, k = 1 × 10 = 10, ∵ y = 10x, where 1 ≤ x ≤ 10; (2) the answer is not unique. For example, if Xiaoming's home is 10km away from school and goes to school at the speed of vkm / h every day, then the time required for Xiaoming to go to school from home is t = 10V

The image of inverse scale function y = - 1 x is approximately ()
A. B. C. D.

In the inverse scale function y = - 1 x, k = - 1 < 0, the image is in the second and fourth quadrant, so select D

The tangent equation of the curve y ^ 2 = x at the point (1,1) is x-2y + 1 = 0 to find the area of the figure and the volume of a circle around the X axis
The tangent equation of curve y ^ 2 = x at point (1,1) is x-2y + 1 = 0. What is the area of the plane figure enclosed by the above curve, tangent and x-axis, and what is the volume of the above plane figure rotating around the x-axis,
Er, we need to use the method of definite integral

See figure
The intersection of tangent and x-axis is B (- 1, & nbsp; 0), and the intersection of & nbsp; and y-axis is C (0, & nbsp; 1 / 2)
The area bounded by tangent (Y & nbsp; = & nbsp; (x + 1) / 2) and parabola and coordinate axis is green and purple respectively
The cross-sectional area at x is:
f(x) = π[(x+1)/2]² = π[(x+1)²/4 　 (-1 ≤ x < 0)
= π[(x+1)/2]² - π(√x)² = π[(x+1)²/4 - πx = π[(x-1)²/4 (0 ≤ x ≤ 1)
V & nbsp; = & nbsp; ∫ & nbsp; π (x + 1) & # 178; DX / 4 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (from - 1 to 0)
&Nbsp; + & nbsp; ∫ & nbsp; π [(x-1) &# 178; DX / 4 & nbsp; & nbsp; & nbsp; & nbsp; (from 0 to 1)
=&Nbsp; (π / 4) ∫ & nbsp; (x + 1) &# 178; D (x + 1) & nbsp; (from - 1 to 0)
+&Nbsp; & nbsp; (π / 4) ∫ & nbsp; (x-1) & # 178; D (x-1) & nbsp; & nbsp; & nbsp; (from 0 to 1)
=&Nbsp; π (x + 1) & # / 12 & nbsp; & nbsp; & nbsp; (from - 1 to 0)
+&Nbsp; π (x-1) & # / 12 & nbsp; & nbsp; & nbsp; (from 0 to 1)
= π/12 + π/12
= π/6

Find the area of the figure surrounded by the curve y = 2x2, y = x2 and the straight line x = 1

Use definite integral to find their area respectively and then subtract them
The answer is one third

If the image of inverse scale function y = 2 / X intersects with the image of primary function y = kx-4, the value range of K is obtained