If the image of the linear function y = (m-1) x + m-5 does not pass through the second quadrant, then the value range of M is

If the image of the linear function y = (m-1) x + m-5 does not pass through the second quadrant, then the value range of M is

Because it's a linear function, M is not equal to 1
It is known that:
M-1 > 0 and m-5
Because it does not pass through the second quadrant, k > 0, B < 0
That is, M-1 > 0, m-5 < 0
The solution is 1 < m < 5
First of all, we find the crossing point (- 1, - 4), and then we find that the critical state is just a straight line crossing the origin, so m is less than or equal to 5
m-1＞0 m-5＜0
m＞1 m＜5
The value range of M is 1 < m < 5
M is greater than or equal to 1 and less than 5
The function can pass through the origin or on the negative half axis of the y-axis
They are all increasing functions
So M-1 > 0, m-5 is less than or equal to 0
So m is greater than 1 and less than or equal to 5
Let a and B be two nonempty sets, and define the difference set of a and B as A-B = {x | x ∈ a, and X ∉ B}, then a - (a-b) is equal to ()
A. AB. BC. A∩BD. A∪B
∩ A and B are two nonempty sets, A-B = {x | x ∈ a, and X ∉ B}, ∩ A-B represents the part of a that removes a ∩ B, ∩ a - (a-b) = a ∩ B
If the image of the first-order function y = (M + 2) x + (m-1) does not pass through the second quadrant, then the value range of M is larger__________
Without going through the second quadrant
So m + 2 > 0 and M-1 ≤ 0
So - 2
M-1 < 0 and M + 2 > 0
So - 2 < m < 1
M is greater than minus 2 and less than 1
Let a and B be two nonempty sets, and define the difference set of a and B as A-B = {x | x ∈ a, and X ∉ B}, then a - (a-b) is equal to ()
A. AB. BC. A∩BD. A∪B
∩ A and B are two nonempty sets, A-B = {x | x ∈ a, and X ∉ B}, ∩ A-B represents the part of a that removes a ∩ B, ∩ a - (a-b) = a ∩ B
If the image of the linear function y = 3x + m + 3 does not pass through the second quadrant, what is the value range of M?
There should be a process
= =b
Firstly, the original function is an increasing function, so when the constant term (M + 3) ≤ 0, the image does not pass through the second quadrant,
The solution m + 3 ≤ 0 is m ≤ - 3
When x = 0, y is less than or equal to 0, so m ≤ - 3
Because the image doesn't go through the second quadrant
So the image must intersect the Y-axis and the negative half axis
That is m + 3
Let m and p be two nonempty sets, define the difference set M-P = {XLX ∈ m, and X ￠ P} of M and P, then M - (M-P) is equal to
Let m and p be two nonempty sets, define the difference set M-P = {XLX ∈ m, and X does not belong to p}, then M - (M-P) is equal to
1.P
2.M
3. M and u
4. M to p
For any set a and B, B 'is the complement of B, then according to the definition
A-B = A∩B'
so
M-(M-P) = M∩(M-P)' = M∩(M∩P')' = M∩(M'∪P) = (M∩M')∪(M∩P) = M∩P
If the image of the linear function y = 3x + M-1 does not pass through the second quadrant, then the value range of M is______ .
∵ if the image of the first-order function y = MX + 2m-1 does not pass through the second quadrant, ∵ M-1 ≤ 0, the solution is m ≤ 1
Let a and B be two nonempty sets, and define the difference set of a and B as A-B = {x | x ∈ a, and X ∉ B}, then a - (a-b) is equal to ()
A. AB. BC. A∩BD. A∪B
∩ A and B are two nonempty sets, A-B = {x | x ∈ a, and X ∉ B}, ∩ A-B represents the part of a that removes a ∩ B, ∩ a - (a-b) = a ∩ B
If point a (- 3,4) is on the image of the linear function y = - 3x-5d, the intersection of the image and the Y axis is B (0, - 5), then the area of △ AOB is?
7.5. Make the normal of y-axis from point a, and the intersection point (0,4) is point D. use △ ADB - △ ADO to be the area of △ AOB. Both △ ADB and △ ADO are right triangles, then we can get the answer
Let a and B be two nonempty sets, and define the difference set of a and B as A-B = {x | x ∈ a, and X ∉ B}, then a - (a-b) is equal to ()
A. AB. BC. A∩BD. A∪B
∩ A and B are two nonempty sets, A-B = {x | x ∈ a, and X ∉ B}, ∩ A-B represents the part of a that removes a ∩ B, ∩ a - (a-b) = a ∩ B