Let's find the polar coordinates of 4 times of the polar circle?

Let's find the polar coordinates of 4 times of the polar circle?

From the shift of 3sina-4cosa = 4
3sinA=4+4cosA
Obviously, we want to make 3sina

If the polar coordinate equation of circle C is p = 2 (radical 2) Sina, then the equation of circle C is?

Because the polar equation of circle C is ρ = 2 √ 2sina
So ρ ^ 2 = 2 √ 2 * ρ Sina
So x ^ 2 + y ^ 2 = 2 √ 2Y
So x ^ 2 + (Y - √ 2) ^ 2 = 2

Given that the polar coordinate equation of a circle is p = 5 (radical 3) cosa-5sina, the polar coordinates and radius of the circle center are obtained

According to the relationship between polar coordinates and rectangular coordinates, x = pcosa, y = psina, as follows: x ^ 2 + y ^ 2 = 5 √ 3x-5yx ^ 2 + y ^ 2-5 √ 3x + 5Y = 0; therefore, the coordinates of the center of the circle are: (5 √ 3 / 2, - 5 / 2)

The distance between the centers of the two circles whose polar equations are p = cosa and P = sina is

P = cosa, P ^ 2 = pcosa, x ^ 2 + y ^ 2 = x, (x-1 / 2) ^ 2 + y ^ 2 = 1 / 4, so the center of the circle is (1 / 2,0)
Similarly, P = sina is transformed into x ^ 2 + (Y-1 / 2) ^ 2 = 1 / 4, and the center of circle is (0,1 / 2)
Therefore, the center distance is √ 2 / 2

Function f (x) = radical (- 2x2 + 3x-1) - 3 (1) find the definition domain of function f (x) (2) find the value range of F (x)

Nonnegative numbers under the arithmetic square root
f(x)=√(-2x²+3x-1)
(1)
∵-2x²+3x-1≥0 => 1/2≤x≤1
The definition domain of F (x) is x ∈ [1 / 2,1]
(2)
Obviously, f (x) ≥ 0
∵ - B / 2A = 3 / 4 = > the maximum value is f (3 / 4) = √ 2 / 4
The value range of F (x) is [0, √ 2 / 4]

Find the range of the following functions: (1) y = x ^ 2-3x + 1, X ∈ [0,3] (2) y = x + radical (1-2x) (3) y = (3x + 2) / (x + 2)

【-0.25 1】
【0.5 1 】
Negative infinity, positive infinity

Find the value range of the following function (1) y = 2x + 3 / x-3 (2) y = x + 2x + 1 (3) y = 1 / 2x ^ 2-3x + 1

1) Y = 2x + 3 / x-3 = (2x-6 + 9) / x-3 = 2 + (9) / x-3 is the inverse proportional function type. The value range is from negative infinity to 2 and from 2 to positive infinity. 2Y = x + squar (2x + 1) = ((squar (2x + 1) + 1) ^ 2) / 2-1, with squr ＂ 0, ＋ infinity), then the range is 〔 - 1 / 2, + infinity), and squar refers to the root sign 3) y = 1 / 2x ^ 2-3x + 1 = 1 / 2 (

The root field of (2x) - under (2x) - sign

-2X + 3 decline
So √ (- 2x + 3) decreases
And - √ (3x-4) decreased
So it's a minus function
-2x+3>=0,3x-4>=0
So 4 / 3

if x−1− 1 − x = (x + y) 2, then the value of X-Y is______ .

According to the meaning of the title,
x−1≥0
1 − x ≥ 0, x = 1;
Put x = 1 in
x−1−
1−x=（x+y）2，
The solution is y = - 1,
So, X-Y = 2

The value range of the square of (1-x) under the root sign of the function y = x + is as follows:

When x > 1, the original function can be changed into y = 2x-1 (Y > 1)
When x = 1, the original function can be replaced by y = x (y = 1)
When x < 1, the original function can be replaced by y = 1 (y = 1)
To sum up, the value range of Y is y greater than or equal to 1