After braking, the moving car moves in a straight line with uniform deceleration and stops in 3.5 seconds. The ratio of the displacement that it passes in 1 second, 2 seconds and 3 seconds after braking is () A. 1:3:5B. 3:5:7C. 1:2:3D. 3:5:6

After braking, the moving car moves in a straight line with uniform deceleration and stops in 3.5 seconds. The ratio of the displacement that it passes in 1 second, 2 seconds and 3 seconds after braking is () A. 1:3:5B. 3:5:7C. 1:2:3D. 3:5:6


The schematic diagram is as shown in the figure. The car's uniformly decelerated linear motion with the final speed of 0 from a to e is reversed and converted into a uniformly accelerated linear motion with the initial speed of 0 from e to a for equivalent treatment. Because the acceleration is the same before and after the reversal, the motion displacement and speed time before and after the reversal are symmetrical. Therefore, it is known that the displacement ratio of the car in the same time is 1:3 :5:… So xde: XCD: XBC: XAB = 1:8:16:24, so XAB: XAC: XAD = 3:5:6. So option D is correct



After braking, the moving car moves in a straight line with uniform deceleration and stops in 3.5 seconds. The ratio of the displacement that it passes in 1 second, 2 seconds and 3 seconds after braking is ()
A. 1:3:5B. 3:5:7C. 1:2:3D. 3:5:6


The schematic diagram is as shown in the figure. The car's uniform deceleration linear motion with the final speed of 0 from a to e is reversed and converted into uniform acceleration linear motion with the initial speed of 0 from e to a for equivalent treatment. Because the acceleration is the same before and after reversing, the motion displacement and speed time before and after reversing are symmetrical



The master and the apprentice worked together to make a batch of parts. The apprentice made a total of 27 parts, 21 less than the master. How many parts are there in this batch?


A: there are 49 parts in this batch



Is the phase difference of sinusoidal signal of the same frequency fixed
If the frequency changes, will the phase difference change again? Why is the phase difference unchanged when I do the experiment


As for the frequency, phase and phase difference, it is easy to understand by vector rotation. Frequency is the speed of vector rotation, and the number of rotations per second is Hz. Phase is the angle between the vector and the horizontal axis, and phase difference is the angle between two vectors. Two signals with the same frequency have the same rotation speed, If there is a frequency difference (even if it is very small), the phase difference will become very large with the accumulation of time
In addition, the amplitude of sinusoidal signal is the projection of the vector on the vertical axis



There are 385 teachers and students who rent cars for tourism. There are 42 seats and 60 seats in the taxi company. The rent of 42 seats is 320 yuan and that of 60 seats is 460 yuan
If you want to rent 8 buses of two kinds at the same time (you can't sit enough), and it's cheaper than using one vehicle in a single group, please choose the most economical scheme


The problem is two equations
Let the number of 42 seat cars be x and the number of 60 seat cars be y
x+y=8
42x+60y>=385
The solution is: x = 5, y = 3
Therefore, the most economical car rental scheme is 5 cars with 42 seats and 3 cars with 60 seats, with a total cost of 320 * 5 + 460 * 3 = 2980 yuan



Let the square matrix a satisfy the square - A - E = 0 of a, prove that a is invertible, and find the negative power of A


A ^ 2-a-e = 0, then a (A-E) - E = 0, a (A-E) = e
So a is reversible, a ^ (- 1) = a-e



If (M + mi) ^ 6 = - 64i, then M is equal to?


(m+mi)^6=-64i
m^6*(1+i)^6=-64i
(1+i)^6=-8i
m^6=8
m=±√2



When a class of students go out for a tour, some students take a group photo. The cost of one color negative is 0.6 yuan, and the cost of one color photo is 0.4 yuan
How many students will take a group photo if one color photo is in advance and the cost per person is not more than 0.5 yuan?


Let's have at least x people
1+0.6(X-1)=0.8X
1+0.6X-0.6=0.8X
0.4=0.2X
X=2
A: at least two



Find the limit LIM (1-2xy) / (x ^ 2 + y ^ 2) x tends to 0, y tends to 1


Substituting x = 0, y = 1,
Molecules 1-2xy tend to 1,
The denominator x ^ 2 + y ^ 2 also tends to 1,
that
The limit is one



The definition field of function y = (X-2) ^ (3 / 2) + (3x-7) ^ 0 is ()
The answer is [2,7 / 3] ∪ (7 / 3, + ∞)


Make the function y = (X-2) ^ (3 / 2) + (3x-7) ^ 0 meaningful
It must be X-2 ≥ 0, that is, X ≥ 2 and 3x-7 is not equal to zero
That is, X is not equal to 7 / 3
So the definition field of function y = (X-2) ^ (3 / 2) + (3x-7) ^ 0 is
[2,7/3)∪(7/3,+∞)