Han Jinjian s question 2, Han Jinjian s problem

Yes, you have to study physics in college, and you can apply for a theoretical physics major in the future.

This question is a categorical counting problem from the meaning of the question, first determine the number of 2 and 3 in the number, when there is 1 2 in the number, 3 3, there are a total of C4 1 = 4 kinds of results, when there are 2 2 in the number, 2 3, a total of C4 2 = 6 kinds of results, when there are 3 2 in the number, 1 3, a total of C4 1 = 4 kinds of results, according to the principle of classification and addition, a total of 4 + 6 + 4 = 14 kinds of results, so it is 14.

1) V1 + AT=V2, 60 + 2A = 48, acceleration a = -6 m seconds squared.

2) 60-6t=0, t=10s, the plane only taxied for 10s after landing, and then it was stationary, so the displacement s=60*12-1 2*6*12*12=288m

3)v=s/t=288/10=;

4) The whole process is regarded as a uniform linear motion of the aircraft starting from stationary with an acceleration of 6 meters per second square, and the displacement of taxiing within 4s is s=1 2*a*t*t=1 2*6*4*4=48m

Hello! y=kx-k+1

y=(x+2k)/k

Solve the system of equations in a row together.

x=-(k^2+1)/(k^2-1)

by 0x>0Determine the intersection in the first and fourth quadrants.

y=kx-k+1

kx>0

k+1>0

If there are other questions after this question, send and click on my avatar to ask me for help, it is not easy to answer the question, please understand, thank you.

Your adoption is the driving force for my service.

Good luck with your studies!

Simultaneous solution: kx-y=k-1

ky-x=2k②

k 2x-ky=k 2-k, add (k 2-1)x=k 2+k to get x=k (k-1) 0, k 2y-kx=2k 2, add (k 2-1)y=2k 2+k-1=(2k-1)(k+1), get y=(2k-1) (k-1) 0, the intersection is in the second quadrant, choose b,

The answer should be b, which can be obtained by two formulas: x=1+1 (k-1).

y=2+1/（k-1)

Since the value of k is known, it can be judged that x<0, y>0, and z are in the second quadrant.

Multiple choice questions can do this:

When k=0, solve the system of equations y=1,x=0;

So the cd is excluded.

k=1 2 x-2y=-1,y-2x=2 x=-1,y=0 excludes ad

So it should be b

First of all, we can think of k as known, find x,y by concatenating the two equations, and then bring in the range of k to see the symbols of x,y.

b The coordinates of the intersection are x=-k (1+k)<0

and y=2-1 (1+k)>1>0

Two quadrants.

c Why, adopt, let's talk.