If the line y = KX (k > 0) and hyperbola y = 4 / X intersect at two points a (x1, Y1) and B (X2, Y2), then the value of 2x1y2-7x2y2

If the line y = KX (k > 0) and hyperbola y = 4 / X intersect at two points a (x1, Y1) and B (X2, Y2), then the value of 2x1y2-7x2y2

If this is a fill in question or multiple-choice question, you can think like this: y = KX and y = 4 / K are symmetrical figures relative to the origin (you can think about their sketches in the coordinate system in your mind), so as to judge that the intersection of the straight line and hyperbola is also symmetrical relative to the origin

When the speed of the car is 18m / s, the car makes a uniform deceleration linear motion after the emergency brake, and the acceleration is 6m / S2. The distance of the car passing through in 6S after the brake is calculated

It takes t = vt − v0a & nbsp; & nbsp; that is: T = 0 − 18m / s − 6m / S2 = 3S ＜ 6S. The car stops within 3S when it starts to move. Therefore, the distance passed within 6S is the distance passed within 3S. S = v0t + 12at2 & nbsp; that is: S = 18 × 3 + 12 × (− 6) × 32 = 27m. A: the distance passed within 6S after braking is 27m

It is known that real numbers x and y satisfy 2x + Y-2 > = 0, x-2y + k > = 0, X-1

Without drawing, we can get the result by simultaneous equations. The answer is 4
This is the experience of senior three

When a particle moves along the x-axis, its acceleration changes with time as a = 3 + 2T. If the velocity of the particle at the initial time is u = 5m / s, what is the velocity of the particle when t = 3

a=dv/dt=3+2t
Integral: v = T ^ 2 + 3T + C
When t = 0, v = 5m / S
The solution is C = 5m / s
Then: T = 3
v=23m/s

Let the real numbers x and y satisfy (x-1) ^ 2 + (y + 2) ^ 2 = 5, then the maximum value of x-2y

The real number x, y satisfies (x-1) ^ 2 + (y + 2) ^ 2 = 5. Let the parameter equation be x = 1 + √ 5cos θ y = - 2 + √ 5sin θ x-2y = 1 + √ 5cos θ - 2 (- 2 + √ 5sin θ) = 5 - (2 √ 5sin θ - √ 5cos θ) = 5 - √ (2 √ 5 ^ 2 + √ 5 ^ 2) = 5-5sin (θ + φ), then the minimum value of sin (θ + φ) is - 1 (x-2y) MA

When an object moves with a certain acceleration on a horizontal plane, the relation between its displacement and time is s = 24t-3t & sup2;, then what is the second when its velocity is zero

s = 24t - 3t²
v = s' = 24 - 2*3t = 24 -6t =0
t=4
Or according to s = v0t-1 / 2at ^ 2
v0=24,a=2*3=6
v=at
t=v/a=24/6=4s

If x = 5, y = 7 and kx-2y = 1, what is the value of K?
It is solved by a system of linear equations of two variables

Substituting the value of X, y directly into kx-2y = 1, we get k = 3

The motion law of the particle is y = the cube of root 3T + 2T, where y represents the displacement at time t, then when t = 2S, the acceleration of the particle is a? Required process

Take y twice as a derivative of T, and then take t = 2 to get the acceleration
The first derivative is the function of velocity with time. The second derivative is the function of acceleration with time

For different value ranges of real number k, the shape of curve represented by equation KX ^ 2 + y ^ 2-2x = 0 is discussed

When k = 0, it means parabola y ^ 2 = 2x
When k = 1, (x-1) ^ 2 + y ^ 2 = 1 denotes a circle
When 0

When a particle moves in a straight line along the x-axis, the relationship between the acceleration a and the position coordinate X is: a = 4 + 3x ^ 2 (SI). If the velocity of the particle at the origin is zero, try to find
Its velocity at any position
The answer says that according to the known conditions and the definition of acceleration, a = DX / dt = (DV / DX) * (DX / DT) = V * (DV / DX) = 4 + 3x ^ 2. I want to know why a = DX / dt = (DV / DX) * (DX / DT) = V * (DV / DX). The definition of acceleration is not only
A = DV / dt = (d ^ 2) R / dt ^ 2,

A = DX / dt = (DV / DX) * (DX / DT) this is obviously true, just divide DX and multiply DX
In a = DX / dt = (DV / DX) * (DX / DT), v = DX / DT, so) a = V * (DV / DX)