There is a moving point a on the number axis. It starts from the origin and moves along the number axis. It is only allowed to move one unit at a time. After 10 times of movement, point a moves to 6 points away from the origin Unit, how many ways does point a move

There is a moving point a on the number axis. It starts from the origin and moves along the number axis. It is only allowed to move one unit at a time. After 10 times of movement, point a moves to 6 points away from the origin Unit, how many ways does point a move


Move to 6 (on the far right), a total of 10 times, so right 8 times, left 2 times. Permutation C (10,2) = 45



On the number axis, the origin and the point to the right of the origin represent the number______ .


Because the number axis to the right is positive, the number represented by the origin and the point to the right of the origin is non negative



The number of common tangents between circle C1: x2 + y2-6x + 6y-48 = 0 and circle C2: x2 + Y2 + 4x − 8y − 44 = 0 is ()
A. 0 B. 1 C. 2 d. 3


∵ circle C1: x2 + y2-6x + 6y-48 = 0 is converted into the standard equation, and the center coordinates of circle C1 are (3, - 3), radius R1 = 8. Similarly, the center coordinates of circle C2 are (- 2,4), radius R2 = 8. Therefore, the center distance of two circles | C1C2 | = (− 2 − 3) 2 + (4 + 3) 2 = 74 ∵ R1-R2 | C1



The positional relationship between two circles C1: x ^ 2 + y ^ 2 = 1 and C2: x ^ 2 + y ^ 2-4x = 0


There are four kinds of positional relations between two circles: inclusion, tangency, intersection and separation
Let the radius of two circles be r and R respectively, and the distance between two circles be d
1. When R + R < D, the two circles are separated
2. When R + r = D, it is circumscribed
3. When R-R = D, it is inscribed
4. When R-R > D, it contains
5. When R-R ︱ D ︱ R + R, it intersects
From the equation of circle C1: x ^ 2 + y ^ 2 = 1, we can see that the center coordinate of circle is (0,0), and the radius is 1
The circle C2 is rewritten as: (X-2) ^ 2 + y ^ 2 = 4. The center coordinate of the circle is (2,0) and the radius is 2
The distance between the centers of two circles is 2-0 = 2, obviously 2-1 < 2 < 1 + 2, that is to say, the relationship between two circles conforms to the fifth case, and the two circles intersect



Point a is on the circle C1: x2 + Y2 + 2x + 8y-8 = 0, and point B is on the circle C2: x2 + y2-4x-4y-2 = 0. Find the maximum value of | ab |


Circle C1: x2 + Y2 + 2x + 8y-8 = 0, (x + 1) &# - 178; + (y + 4) &# - 178; = 25, Center (- 1, - 4) circle C2: x2 + y2-4x-4y-2 = 0, (X-2) &# - 178; + (Y-2) &# - 178; = 10, center distance of circle (2,2): √ 3 & # - 178; + 6 & # - 178; = 3 √ 5 < 5 + √ 10, i.e., the maximum value of AB at the intersection of two circles is: 5 + √ 10 + 3 √ 5



In the system of equations kx-y-4 = 0, ① 4x & # 178; + 9y & # 178; + 18y-18 = 0, ②, why is there only one real solution when k is the value?


y=kx-4
Substituting
(4+9k²)x²-48kx+234=0
So delta = 0
2304k²-3744-8424k²=0
Then K & # 178;



If the system of equations x ^ 2 + y ^ 2 = 16, X-Y = k with X, y as unknowns has a real number solution


I've written your formula here. It's for laziness!
It's actually two!
2+2=4
4+2=6
It's just to save trouble! After 2 + 2 = 4, add 2 directly!
It's usually on paper!
In fact, it is neither right nor wrong!
Sometimes I've done it!
For a ready-made number, don't bother to write it again!



When k is a real number, the system of equations, x square + y square = 16, X-Y = k, has a real solution


X ^ 2 + y ^ 2 = 16 X-Y = k substitute y = x-k into x ^ 2 + y ^ 2 = 16 to get 2x ^ 2-2kx + K ^ 2 = 16 has real number solution, then △≥ 0 △ = (- 2K) ^ 2-4 * 2K ^ 2 = 4K ^ 2-8k ^ 2 = - 4K ^ 2 ≥ 0, so k = 0



When k, m take______ Then the equations y = KX + my = (2k − 1) x + 4 have at least one solution


When k = 2k-1, M = 4, the line y = KX + m coincides with the line y = (2k-1) x + 4, that is, the system of equations has innumerable solutions, so k = 1, M = 4; when k ≠ 2k-1, M = 4, the line y = KX + m has an intersection with the line y = (2k-1) x + 4, that is, the system of equations has one solution, so K ≠ 1, M = 4



When k, m take______ Then the equations y = KX + my = (2k − 1) x + 4 have at least one solution


When k = 2k-1, M = 4, the line y = KX + m coincides with the line y = (2k-1) x + 4, that is, the system of equations has innumerable solutions, so k = 1, M = 4; when k ≠ 2k-1, M = 4, the line y = KX + m has an intersection with the line y = (2k-1) x + 4, that is, the system of equations has one solution, so K ≠ 1, M = 4. So when m = 4, the system of equations y = KX + my = (2k − 1) x + 4 has at least one solution