The point representing the number a on the number axis must be to the right of the origin Judge right or wrong

The point representing the number a on the number axis must be to the right of the origin Judge right or wrong

If a > 0, that's right
Otherwise it's wrong

On the number axis, it is ()
A. A point B representing 0. A point C at the beginning. A point D in the middle of the axis. It is an endpoint on the axis

On the number axis, we define the origin as the point representing 0, so we choose a

When the values of K and M satisfy the condition (), the equations y = KX + m y = (2k-1) x + 4 & gt; have at least one set of solutions

y=kx+m=(2k-1)x+4
x=(m-4)/(k-1)
So K and M need to satisfy that K is not equal to 1 and M is not equal to 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

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

For which values of K, m, the equations y = KX + m, y = (2k-1) x + 4 have at least one set of solutions
Decline Ctrl + C Ctrl + V

To make the system of equations have at least one set of solutions,
That is KX + M = (2k-1) x + 4
That is: k = 2k-1; m = 4
K = 1; m = 4

Let f (x) = 2x square divide X-2 + ln (4-x) (1) find the domain of definition (2) find f (3)

X-2 ≠ 0. X ≠ 2.4-x ＞ 0. X ＜ 4. So the domain of definition is (negative infinity, 2) ∪ (2,4)
f（3）=18+ln1=18+0=18

If f (2x + 1) = the square of x-2x, then f (3)=

3=2X+1...X=1
Substituting, f (3) = f (2x + 1) x takes 1
=x^2—2x=1-2=-1

If f (2x + 1) = x-2x, then f (x)=

Let y = 2x + 1, then x = (Y-1) / 2
f(y)=(y-1)^2/2^2-(y-1)=1/4 y^2 - 3/2 y + 5/4

Given the function f (x) = (x square + 2x + a) / x, X ∈ [1, positive infinity),
(1) When a = 1 / 2, find the minimum value of function f (x)
(2) If f (x) > 0 holds for any x ∈ [1, positive infinity), try to find the value range of real number a

The original equation can be reduced to f (x) = ((x + 1) square) + A-1
So the axis of symmetry of the equation is x = - 1, that is, f (x) is the minimum when x = - 1
X = - 1 increases to both sides
(1) Because x ∈ [1, positive infinity), when x = 1, f (x) is the minimum of 3.5
(2) Because x ∈ [1, positive infinity), when x = 1, f (x) is the minimum
From F (x) = ((x + 1) square) + A-1
So let f (1) > 0
Then a ＞ - 3
So when a ＞ - 3, f (x) ＞ 0 holds