Prove: no matter what the value of K is, the straight line L: kx-y-4k + 3 = 0 and the curve C: x ^ 2 + y ^ 2-6x-8y + 21 = 0 always have several intersections 1. To make the line intersect the circle, the distance from the center of the circle to the line is | 3k-4-4k + 3 | / (radical K ^ 2 + 1) Radical (1 + 2K / (1 + K ^ 2))

Prove: no matter what the value of K is, the straight line L: kx-y-4k + 3 = 0 and the curve C: x ^ 2 + y ^ 2-6x-8y + 21 = 0 always have several intersections 1. To make the line intersect the circle, the distance from the center of the circle to the line is | 3k-4-4k + 3 | / (radical K ^ 2 + 1) Radical (1 + 2K / (1 + K ^ 2))

You're right. Of course, 2 is less than 2

A particle starts to move on the x-axis from the origin, with initial velocity V > 0 and acceleration a > 0. When the acceleration a decreases to 0, the ()
A. The velocity decreases and the displacement decreases
B. The velocity decreases and the displacement increases
C. When a = 0, the velocity reaches the maximum and the displacement increases
D. With the increase of velocity, the displacement reaches the maximum when a = 0
The speed is increasing and the displacement is increasing. I understand, but why does the speed reach the maximum when a = 0? Isn't the speed reach the maximum when a is the maximum?

No
When the acceleration is maximum, the force is unbalanced, the resultant force is maximum, and the velocity will increase (!)
When the acceleration a = 0, the force balance speed will not increase, and the uniform linear motion will be maintained

The circle x2 + y2-6x-8y + 21 = 0 and the line kx-y-4k + 3 = 0 are known

(1) It is proved that the equation of a circle is (x-3) 2 + (y-4) 2 = 4, and the distance from the center of the circle (3,4) to the straight line kx-y-4k + 3 = 0 is | =. To prove that there are always two different common points between a straight line and a circle, we only need to prove ＜ 2, that is, (K + 1) 2 ＜ 4 (1 + K2), that is, 3k2-2k + 3 > 0, and 3k2-2k + 3 = 3 (k -) 2 + > 0

A particle moves along a parabola 2Y = x ^ 2 from the origin. Its partial velocity on the x-axis is a constant 4.0m/s. Find the velocity and acceleration of the particle at x = 2m
Such as the title

A particle moves along a parabola 2Y = x & # 178 from the origin. Its partial velocity on the x-axis is a constant 4.0m/s. The velocity and acceleration of the particle at x = 2m can be calculated
The trajectory of the particle is a parabola y = (1 / 2) x & # 178;, the horizontal partial velocity VX = DX / dt = 4 is known, so x = ∫ 4dt = 4T + C, x = 0 when t = 0,
So C = 0, then x = 4T
Substituting (1) into parabolic equation, y = (1 / 2) (4T) &# 178; = 8t & # 178; (2)
The vertical partial velocity of the particle vy = dy / dt = 16t; so the velocity v = √ (V & # 178; X + V & # 178; y) = √ (16 + 256t & # 178;) = 4 √ (1 + 16t & # 178;)
Acceleration a = DV / dt = 4 [32t / 2 √ (1 + 16t & # 178;)] = 64t / √ (1 + 16t & # 178;)
When x = 2m, t = 2 / 4 = 0.5 s; the velocity V (0.5) = 4 √ (1 + 16 × 0.5 & # 178;) = 4 √ 5 (M / s),
Acceleration a (0.5) = (64 × 0.5) / √ (1 + 16 × (0.5) & # 178;) = 32 / √ 5 = 32 (√ 5) / 5 (M / S & # 178;)

When the line L: kx-y + 3 = 0 and the circle C (X-2) ^ 2 + y ^ 2 = 4, why is the line L tangent to the circle C?

The radius of the circle is r = 2 and the coordinate of the center of the circle is (2,0)
If the line L is tangent to the circle C, then the distance d from the center of the circle to the line L: kx-y + 3 = 0 is equal to the radius
That is: D = | 2K + 3 | / √ (K & # 178; + 1) = 2
|2k+3|=2√(k²+1)
The square of both sides is as follows:
4k²+12k+9=4k²+4
12k=-5
The solution is k = - 5 / 12
So when k is - 5 / 12, the line L is tangent to the circle C

A particle starts from the origin o of the x-axis and moves along the positive direction with the acceleration a
That's the question here. I'm the same as the landlord. The answers downstairs are all ambiguous
Wang ChuanChao's answer is quite reasonable. But I haven't learned infinite proportional sequence yet

d. It must be right. It's not difficult to understand. As long as you draw a sketch, after the particle first reaches the voto point, it can never go beyond this range, that is, when voto, the particle speed is 0, it also reaches the farthest place. The whole motion is asymmetric reciprocating motion. According to the limit theorem, it will stop at last

Given that the line L: kx-y-k + 4 = 0 is tangent to the circle C: (x-1) ^ 2 + y ^ 2 = 4, find the value of the real number K

The distance from the center of the circle to the tangent is equal to the radius
Center (1,0), radius = 2
So | k-0-k + 4 | / radical (k ^ 2 + 1) = 2
Root sign (k ^ 2 + 1) = 2
k^2+1=4
K = positive and negative root sign 3

The particle moves along the x-axis in a straight line. The relationship between the velocity and the coordinate is v = 1 + 2x. The initial time is at the origin of the coordinate. Find its position and the law of the velocity changing with time

V=dx/dt=1+2x
Separate variables, get
dx/(1+2x)=dt
Integral, get
1+2x=Ce^(2t)
from
When t = 0, x = 0
have to
C=1
so
x=(1/2){[e^(2t)] -1}
V=dx/dt=d(1/2){[e^(2t)] -1}/dt=e^(2t)
a=dV/dt=d[e^(2t)]/dt=2e^(2t)

The position relation between circle x Λ 2 + y Λ 2 = 4 and circle x Λ 2 + y Λ 2-6x + 8y-24 = 0 is

x∧2+y∧2=4
The center of the circle is (0,0) and the radius is 2
x∧2+y∧2-6x+8y-24=0
(x-3)^2+(y+4)^2=9+16+24=49
The center of the circle is (3, - 4) and the radius is 7
Center distance = 5
Radius difference = 7-2 = 5
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When an object moves in a straight line, the equation of motion is s = 3T ^ 2-5t, and the acceleration of the object at t = 2S is

The general form of S = - 5T + 3T ^ 2 formula is VT + 1 / 2at ^ 2