If a = {x | x | is less than or equal to 1, X belongs to R}, B {y | y = the square of X, X belongs to R}, then a intersects B=

If a = {x | x | is less than or equal to 1, X belongs to R}, B {y | y = the square of X, X belongs to R}, then a intersects B=

|x|

Given that the point P (x, y) moves on the circle (X-2) ^ 2 + y ^ 2 = 1, then the maximum value of Y / (x + 1) is

Let z = Y / (x + 1), then y = Z (x + 1),
The line y = Z (x + 1) passes through the point (- 1,0) with a slope of Z
If the maximum value is required, it depends on the maximum slope of the intersection point between the line y = Z (x + 1) and the circle,
zmax=√2/4
The maximum value of Y / (x + 1) is √ 2 / 4

Find the volume of the body of revolution generated by a circle (X-5) ^ 2 + y ^ 2 = 16 rotating around the Y axis

A: x = 5 ± √ (16-y ^ 2) and symmetric about X axis, so v = 2 π ∫ 0 to 4 [(5 + √ (16-y ^ 2)) ^ 2 - (5 - √ (16-y ^ 2)) ^ 2] dy = 2 π ∫ 0 to 4 20 √ (16-y ^ 2) dy = 40 π ∫ 0 to 4 √ (16-y ^ 2) dy let y = 4sint, then t integral region is 0 to π / 2, then 40 π ∫ (16-y ^ 2) dy = 40 π * 16 ∫ (...)

Given that the line L: y = KX + 2 (k is a constant) passes through the upper vertex B and left focus F of ellipse x2a2 + y2b2 = 1 ((A & gt; B & gt; 0), the chord length of line L cut by circle x2 + y2 = 4 is D, (1) if d = 23, find the value of K, (2) if D ≥ 455, find the value range of eccentricity e of ellipse

(1) Let m be the middle point of the string and OM be connected. From the knowledge of plane geometry, OM = 1, OM = 2k2 + 1 = 1. The solution is K2 = 3, k = ± 3. ∵ if the line passes F, B, ∵ K & gt; 0, then k = 3. (2) let m be the middle point of the string and OM be connected, then om2 = 41 + K2, D2 = 4 (4-41 + K2) ≥ (455) 2, then K2 ≥ 14. E2 = c2a2 = (2k) 24 + (2k) 2 = 11 + K2 ≤ 45, ∵ 0 & lt; e ≤ 255

The circumference of a rectangle is 72 cm, and its length is twice its width. What's its area?

Length: 72 △ 2 * 2 / 3 = 24
Width: 24 △ 2 = 12
Area: 24 * 12 = 288 (square centimeter)

Simplification 3. Cos (27 + α) cos (33 - α) - sin (27 + α) sin (33 - α)
Simplification of COS (27 + α) cos (33 - α) - sin (27 + α) sin (33 - α)

cos(27°+α)cos(33°-α)-sin(27°+α)sin(33°-α)
=cos[(27°+α)+(33°-α)]
=cos60°
=1/2

If there is only one circle passing through points a (0,1) and B (4, m) and tangent to the X axis, find the value of the real number m and the equation of the circle
The simplest way to mark clear symbols

Let the center of the circle be (x0, Y0), because it is tangent to the x-axis, then the radius is Y0, (x-x0) 2 + (y-y0) 2 = Y02, which is simplified as (x-x0) 2 + y2-2y0 = 0 to bring in point a, and X02 + 1-2y0 = 0. Y0 = (X02 + 1) / 2 to bring in point B, (4-x0) 2 + (m-y0) 2 = 0. Take Y0 into the above formula and get the equation: 5 / 4 * X02 + (- 8 + (m-0.5) * 0.5 * 2) x0 + 16 + (m-0.5) 2 = 0 triangle = (m-8.5) 2-4 (5 / 4 * (15 + 2m)) = 0 get m = - 11.5. Get x0 = 20 / 5 * 2 = 8, Y0 = 9 / 2, so the equation is: (X-8) 2 + (y-4.5) 2 = 20.25

How to find the value range of independent variable x for quadratic function
I'll have it before tomorrow morning
Y = AX2 + C (a is not equal to 0, but B is equal to 0)
Y = - 3 / 4x2-3 / 4... Find the value range of independent variable x
Y is equal to minus 3 / 4 times the square of x minus 3 / 4

Quadratic function has no limit on the value of X
You can take any real number

Use the equation to solve: the sum of 8 times of a number and 1 / 4 of it is 66. What is the number?
It's got to be the equation

Let this number be X
8x+x1/4=66
x=8

Given the function f (x) = 2x-1, then f (x + 1) is equal to
Forget why to love (x + 1) as 2x-1

f(x)=2x-1
Then f (x + 1) is to replace X in F (x) with x + 1
So f (x + 1) = 2 (x + 1) - 1 = 2x + 1