If we know that the image of a linear function y = 2x + A, y = - x + B passes through a (- 2,0) and intersects with y axis at two points B and C respectively, then the area of △ ABC is () A. 4B. 5C. 6D. 7

If we know that the image of a linear function y = 2x + A, y = - x + B passes through a (- 2,0) and intersects with y axis at two points B and C respectively, then the area of △ ABC is () A. 4B. 5C. 6D. 7

Substituting the coordinates of a into the linear function y = 2x + A, y = - x + B, we can get a = 4, B = - 2, then the coordinates of B and C are: B (0,4), C (0, - 2), so the area of △ ABC is: BC × OA △ 2 = 6 × 2 △ 2 = 6
Let a = {x | y = log2 (x-1)}, B = {y | y = - x2 + 2x-2, X ∈ r} (1) find the set a, B; (2) if C = {x | 2x + a ＜ 0}, and B ∪ C = C, find the value range of real number a
(1) A = {x | y = log2 (x-1)} = {x | y (x-1) > 0} = (1, + ∞), B = {y | y = - x2 + 2x-2, X ∈ r} = {y | y = - (x-1) 2-1, X ∈ r} = (- ∞, - 1). (2) set C = {x | 2x + a ＜ 0} = {x | x ＜ - A2}, ∪ C = C, ⊆ B ⊆ C, ＞ − A2} a ＜ 2, ∪ the value range of real number a}
If we know that the image of a linear function y = 2x + A, y = - x + B passes through a (- 2,0) and intersects with y axis at two points B and C respectively, then the area of △ ABC is ()
A. 4B. 5C. 6D. 7
Substituting the coordinates of a into the linear function y = 2x + A, y = - x + B, we can get a = 4, B = - 2, then the coordinates of B and C are: B (0,4), C (0, - 2), so the area of △ ABC is: BC × OA △ 2 = 6 × 2 △ 2 = 6
Set a = {X / y = log2 (x-1)} set B = {X / y = - x ^ 2 + 2x-2} set C = {X / x ^ 2 - (m-1) x + 2m = 0}
1. Find the set a, B
2. If a intersection C is not an empty set and B intersection C is not an empty set, then the range of M is?
3. Whether there is a real number m such that (AUB) intersection C is an empty set. If there is, find the range of M
Y = log2 (x-1)} the difference between base number and true number is not clear
It is known that the image of the first-order function is parallel to the positive proportion y = - 2 / 3x, and the expression of the first-order function can be obtained through M (0,4) (1). (2) if the point (- 8, m) is a positive proportion, the expression of the first-order function can be obtained
It is known that the image of the first-order function is parallel to the positive proportion y = - 2 / 3x, and it passes through M (0,4)
(1) Try to find the expression of a function?
(2) If the points (- 8, m) and (n, 5) are on the graph of first-order function, try to find the value of M, n
Because it is parallel to - 2 / 3, so k = - 2 / 3, when x = 0, y = 4 is substituted, so B = 4, so the analytic formula is y = - 2 / 3x + 4
When x = - 8, y = 28 / 3,
When y = 5, x = - 3 / 2
(1)y=-2/3X+4
(2)m=28/3;n=-3/2
The expression is y = - 2 / 3x + 4 M = 28 / 3 N = - 3 / 2
(1) It is known that the image of a function is parallel to the positive proportion y = - 2 / 3x, and the expression of this function is y = - 2 / 3x + 4 by M (0,4). This is the oblique section of a straight line, I don't know how to check it online.
(2) Just go in and solve it.
Let I = R, a = {x │ x ≥ - 2}, B = {x │ x}
C
First, because the complete set I = R
And set a = {x | x ≥ - 2}
So the complement of a (CIA) is {x | x ＜ - 2}
So the complement union of a and B is B ∪ (CIA) = {x | x ＜ 3}
Choose C
But all of these options seem to be wrong except a
CIA={x│x
The image of a given function is intersected with the line y = 3x-4 by (- 2,5) and the Y axis,
The intersection of the line y = 3x-4 and the Y axis is (0, - 4), and the analytic formula of the first-order function is y = - 4.5x-4
y=-4.5x-4
Let the primary function be y = KX + B
Let x = 0, then y = - 4
So there is a function that goes through two points (- 2,5), (0, - 4)
-2k+b=5
b=-4
Solution
k=-9/2
The linear function is y = - 9 / 2X-4
y=-4.5x-4
If I = {x | x ∈ r}, a = {x | x ≤ 1 or X ≥ 3}, B = {x | K ＜ x ＜ K + 1, K ∈ r}, and (CIA) NB = empty set, then
Given the complete set I = {x | x ∈ r}, set a = {x | x ≤ 1 or X ≥ 3}, set B = {x | K < x | K + 1, K ∈ r}, and (CIA) NB = empty set, then the value range of real number K is
∵ a = {x | x ≤ 1 or X ≥ 3}
∴CIA=｛x|1＜x＜3｝
And ∵ (CIA) NB = empty set
∴①k≥3②k+1≤1→k≤0
The value range of real number k is {K | K ≥ 3 or K ≤ 0}
The sequence of n elements into the stack is 1,2,3., N, and its output sequence is P1, P2, P3. PN. If P1 = 3, then the value of P2 is?
A may be 2 B must be 2 C may be 1 D must be 1
1 into the stack, 2 into the stack, 3 into the stack, out of the stack, and then naturally 2 out of the stack (may also be 4 into the stack out of the stack), can not choose B, can only choose a
If we know the complete set I, set a and B, we can find CIA ∩ B
(CIA is the complement of set a in complete set I)
(1) I = {pair of real numbers (x, y)}, a = {(x, y) | (y-4) / (X-2) = 3}, B = {(x, y) | y = 3x-2}
(2) I = R, a = {a | quadratic equation AX ^ 2-x + 1 = 0 has real roots}, B = {a | quadratic equation x ^ 2-ax + 1 = 0 has real roots}
(1) Y = 3x-2 and (y does not = 3x-2 or x = 0) CIA ∩ B = {(0, - 2)} (2) CIA ∩ B = {a | quadratic equation AX ^ 2-x + 1 = 0 has no real roots and quadratic equation x ^ 2-ax + 1 = 0 has real roots} this is really simple
(1) A∩B={(x,y)|y=3x-2, x≠2}
(2) A={a|a=4}
A∩B={a|a