It is known that the image of a function y = KX + B and another function y = 3x + 2 intersect at point a on the y-axis, and point B (3, - 4) is obtained on the graph of a function y = KX + B

It is known that the image of a function y = KX + B and another function y = 3x + 2 intersect at point a on the y-axis, and point B (3, - 4) is obtained on the graph of a function y = KX + B

A (0,2), B (3, - 4) on the function y = KX + B. substituting y = - 2x + 2
∵ the function y = KX + B intersects the image of another linear function y = 3x + 2 on the Y axis
When x = 0, y = 2
The intersection of y = KX + B on the y-axis a (0,2)
Take a (0,2) B (3, - 4) into y = KX + B
SO 2 = B
There are - 4 = 3K + 2 K = - 2
So the linear function y = KX + B is y = - 2x + 2
The image of y = 3x + 2 intersects Y axis at point a
Ψ point a (0,2)
Substituting x = 0, y = 2, x = 3, y = - 4 into y = KX + B
We get 2 = B
-4=3k+b
The solution is k = - 2, B = 2
∴y=-2x+2
For set M, N, define M-N = {x | x belongs to m and X does not belong to n}, define m * n = (m-n) ∪ (n-m), let m = {y | y = x ^ 2, X belongs to R},
N = [- 3,3], then m * n =?
M = {y | y = x ^ 2, X belongs to R} = [0, + ∞),
M-N＝(3,+∞),N-M＝[-3,0),M*N=(3,+∞)∪[-3,0)
(3, positive infinity) ∪ [- 3, 0)
The intersection point of the image of the first-order function y = KX + 3 and y = 3x + 6 is found to be K on the x-axis
Do X in these two relations represent a value? Let's first set y = 3x + 6, y = 0, then find x = - 2, and then substitute y = KX + 3, k = 1.5. Is that right? If not, explain why x is a value here
That's what you did. You did it right
That's right
The line y = 3x + 6 has and only has one intersection with the X axis. In this case, the point must satisfy y = 0, then x can be calculated as - 2
The intersection point of the image of the first-order function y = KX + 3 and y = 3x + 6 is on the x-axis, which indicates that y = 0 in the intersection coordinate, so you first set y = 3x + 6, y = 0, and then find x = - 2, which is right. In this way, the intersection (- 2, 0) is obtained. Because it's the intersection, of course, it's on the first line. Substituting it into the first linear equation y = KX + 3, we get k = 1.5
So your whole process is absolutely right. ... unfold
The intersection point of the image of the first-order function y = KX + 3 and y = 3x + 6 is on the x-axis, which indicates that y = 0 in the intersection coordinate, so you first set y = 3x + 6, y = 0, and then find x = - 2, which is right. In this way, the intersection (- 2, 0) is obtained. Because it's the intersection, of course, it's on the first line. Substituting it into the first linear equation y = KX + 3, we get k = 1.5
So your whole process is absolutely right. Put it away
For nonempty sets a and B, define the operation: a ⊕ B = {x | x ∈ a ∪ B, and X ∉ a ∩ B}, M = {x | a ＜ x ＜ B}, n = {x | C ＜ x ＜ D}, where a, B, C and D satisfy a + B = C + D, ab ＜ CD ＜ 0, then m ⊕ n = ()
A. （a，d）∪（b，c）B. （c，a]∪[b，d）C. （c，a）∪（d，b）D. （a，c]∪[d，b）
We know that M = {x \\\124;a \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\∪ [D] ∪ [D] ∪ [D] ∪ [D] ∪ [D] So D
C
The graph of the first order function y = - 3x + 2 does not pass through the second order___ Quadrant
Because the analytic formula y = - 3x + 2, - 3 < 0, 2 > 0, the image passes through one, two, four quadrants, so the image does not pass through the third quadrant
Define the set operation a * b = {x | x ∈ a, and X &; B}, if a = {1,2,3,4}, B = {3.4.5.6}, then the sum of elements of a * B and B * a is————
A*B={1,2}
B*A={5.6}
The sum of them is {1,2,5.6}
The image of linear function y = - 3x passes through the fourth quadrant, and Y decreases with the increase of X
Through the second and fourth quadrants, y decreases with the increase of X
Dear landlord
The image of a function y = - 3x passes through the first quadrant, and Y decreases with the increase of X. I hope you will rise step by step. Thank you for your adoption
After the 24th quadrant, reduce the small
Define the operation m * n of set M and N = {x | x ∈ m or X ∈ n, and X ∉ m ∩ n}, then (m * n) * m = ()
A. M∩NB. M∪NC. MD. N
As shown in the figure, it can be seen from the definition that m * n is the shadow area in the figure, ■ (m * n) * m is the shadow area II and blank area in the figure, ■ (m * n) * m = n. so the answer is d
The graph of the first order function y = - 3x-2 does not pass through the second order______ Quadrant
For the first-order function y = - 3x-2, ∵ k = - 3 ＜ 0, ∵ the image passes through the second and fourth quadrants; and ∵ B = - 2 ＜ 0, ∵ the intersection of the image of the first-order function and the Y axis is below the X axis, that is, the function image also passes through the third quadrant, ∵ the image of the first-order function y = - 3x-2 does not pass through the first quadrant
Define a set operation a ⊗ B = {x | x ∈ a ∪ B, and X ∉ a ∩ B}, let m = {x | x | 2}, n = {x | x2-4x + 3 ＜ 0}, then the set represented by M ⊗ n is ()
A. （-∞，-2]∪[1，2）∪（3，+∞）B. （-2，1]∪[2，3）C. （-2，1）∪（2，3）D. （-∞，-2]∪（3，+∞）
And N = {x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\sob