In the plane rectangular coordinate system, an isosceles right triangle ABC is first placed in the second quadrant, leaning on the two coordinate axes, and the point a (0,2), a, Point C (- 1,0), parabola y = ax & # 178; + AX-2 passes through point B. if the points on the parabola (except point B) make △ ACP an isosceles right triangle with AC as the right side, then the coordinate of point P is——

In the plane rectangular coordinate system, an isosceles right triangle ABC is first placed in the second quadrant, leaning on the two coordinate axes, and the point a (0,2), a, Point C (- 1,0), parabola y = ax & # 178; + AX-2 passes through point B. if the points on the parabola (except point B) make △ ACP an isosceles right triangle with AC as the right side, then the coordinate of point P is——

① If ∠ ACB = 90 ° and BD ⊥ CO is made through B, the perpendicular foot is D, and D is on the negative half axis of X axis. From ⊿ AOC ≌ ⊿ CDB, BD = OC = 1, CD = OA = 2, so the coordinates of point B are (- 3,1). The coordinates of another vertex of isosceles right triangle with AC as right side can be obtained by the same method as (2,1) and (- 2,3), (1, - 1)

1 minute = how many milliseconds?

One millisecond is equal to one thousandth of a second (10-3 seconds). 0.000 000 001 millisecond = 1 picosecond. 0.000 001 millisecond = 1 nanosecond. 0.001 millisecond = 1 microsecond. 1000 millisecond = 1 second

For any real number k, if the line y = K (x-1) + 1 and the square of ellipse X / M + the square of Y / 3 = 1 have a common point, then the range of real number m?
Rotate the square of the parabola y = 4x 90 degrees anticlockwise around the focus to obtain the Quasilinear equation of the parabola?

A straight line passes through a fixed point (1,1), and for any real number k, it always has a common point. Obviously (1,1) is in the interior of the ellipse, and in the interior, we get a relationship, where the square of X / M + the square of Y / 3 = 1, and (1,1) is substituted into the formula, which is less than 1, so as to get the range of M. of course, there are other solutions to this problem. You can combine the ellipse and the straight line, and get

Let x ∈ (0, π / 2), then the minimum value of function y = (2Sin ^ x + 1) / (sin2x) is

y=(2sin²x+1)/(sin2x)
=（2sin²x+sin²x+cos²x)/(2sinxcosx)
=3sinx/(2cosx)+cosx/(2sinx)
=3/2*tanx+1/2(tanx)
∵x∈（0,π/2）
∴tanx>0,
According to the mean value theorem
3/2*tanx+1/2(tanx)≥√3
[if and only if 3 / 2tanx = 1 / (2tanx)
That is, if TaNx = √ 3 / 3, take the equal sign]
That is, the minimum value is √ 3

Given that y = 3x & # 178; - 2 (3K + 1) x + 3K & # 178; - 1 and X-axis are on the left and right sides of the origin respectively, the range of K is calculated

First, △ 0
4(3k+1)^2-4*3*(3k^2-1)>0
4(9k^2+6k+1)-36k^2+12>0
36k^2+24k+4-36k^2+12>0
24k>-16
k>-2/3
Because the intersection is on the left side of the origin and the parabola opening is upward
f(0)>0
0+0+3k^2-1>0
k> Root 3 / 3 or K

The first question: y = x under the root sign & # 178; - 2x + 3. Find the function range. The second question: 2F (x) - f (- x) = 3x + 2, find the expression of F (x)
First question: y = x & # 178; - 2x + 3 under the root sign
Second question: 2F (x) - f (- x) = 3x + 2, find the expression of F (x)
Please answer in detail. If good, add!

Let t = x & # 178; - 2x + 3. Y = √ t  t = x & # 178; - 2x + 3. Y = {t  t = x & # 178; - 2x + 3 = (x-1) &# 178; + 2 ≥ 2  y = {t ᙧ 2} function range be [√ 2, + ∞) question 2: 2F (x) - f (- x) = 3x + 2, find the expression of F (x). 2F (x) - f (- x) = 3x + 2

A piece of aluminum can make a cuboid frame of 8cm in length, 5cm in width and 5cm in height. If this piece of aluminum is made into a cube frame, (see the supplement for specific problems)
A piece of aluminum can make a cuboid frame of 8cm in length, 5cm in width and 5cm in height. If this piece of aluminum is made into a cuboid frame, how long is the edge of the cuboid? Are they equal in volume?

Aluminum length 8 × 4 + 5 × 8 = 72
Square edge length 72 △ 12 = 6
Volume v = 6 × 6 × 6 = 216

Let f (x) = ax & sup2; + BX (a, B are constants, a ≠ 0) and f (2) = 0, the equation f (x) - x = 0 has two real roots
Let f (x) = ax & sup2; + BX (a, B are constants, a ≠ 0) and f (2) = 0. The equation f (x) - x = 0 has two real roots
(1) Finding the value of a and B
(2) If f (x) is defined in [M, M + 1], the maximum value of F (x) is g (m), find the analytic expression of G (m)
Correction: two unequal real roots

Insufficient conditions

If M and N are opposite to each other, then | M-6 + n|=_____ The absolute value of a and B is 9 / 4, a + B =?

6
9 / 2 or 0 or - 9 / 2

Given the function f (x) = x ^ 2 + AX-1 / 2a, try to find the value of constant a so that f'27 (x) = 0 and f (x) = 0
F '(x) denotes derivation
The number of F27 is wrong, which is f '(x)

f'(x)=0,
2X + a = 0. So x = - A / 2
Then (- A / 2) ^ 2-A * A / 2-1 / 2A = 0
Then - A ^ 2 / 4-1 / 2A = 0
a=0,-2