The image of the linear function y = KX + B and the positive scale function y = kbx in the same coordinate system may be

The image of the linear function y = KX + B and the positive scale function y = kbx in the same coordinate system may be

The image of the linear function y = KX + B and the positive scale function y = kbx in the same coordinate system may be
Intersect or parallel, but not coincident
1 given the complete set u = a ∪ B = {x ∈ n ∣ 0 ≤ x ≤ 10}, a ∩ (complement b) = {1,3,5,7}, try to find the set B
I think this problem has bugs, if a = {1,7} then B = {0-10 arbitrary value}.
A ∩ (complement b) = {1,3,5,7}
Explain that complement B contains {1,3,5,7}
Then B does not contain {1,3,5,7}
So B can have a variety of situations
But the maximum set is b = {0,2,4,6,8,9,10}
I hope I can help you. I wish you progress in your study_ ∩)O
As shown in the figure, the image of the first-order function y = KX + B and the image of the inverse scale function y = MX intersect at points a (- 2, - 5), C (5, n), Y-axis intersects at point B, and X-axis intersects at point D. (1) find the expression of the inverse scale function y = MX and the first-order function y = KX + B; (2) connect OA, OC, and find the area of △ AOC
(1) ∵ the image of the inverse scale function y = MX passes through the points a (- 2, - 5), ∵ M = (- 2) × (- 5) = 10 ∵ the expression of the inverse scale function is y = 10x. ∵ point C (5, n) on the image of the inverse scale function, ∵ n = 105 = 2, ∵ the coordinates of C are (5, 2). ∵ the image of the first-order function passes through the points a, C
Given the complete set u = a ∪ B = {x ∈ n | 0 ≤ x ≤ 10}, a ∩ (∁ UB) = {1, 3, 5, 7}, find the set B
U = a ∪ B = {x ∈ n | 0 ≤ x ≤ 10} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, {1, 3, 5, 7} ⊆ a, but B does not contain {1, 3, 5, 7}, which is represented by Venn graph, as shown in the figure ∪ B = {0, 2, 4, 6, 8, 9, 10}
It is known that the image of inverse scale function K / X intersects with the image of linear function y = KX + m at points (2,3), and the relationship between the two functions is obtained
Because the image of inverse scale function K / X passes through point (2,3)
So k = 2 * 3 = 6
So the analytic expression of inverse proportion function is y = 6 / X
Because y = 6x + m passes through (2,3)
So 3 = 12 + M
m=-9
So the analytic expression of a function is y = 6x-9
Because the image of inverse scale function K / X passes through point (2,3)
So k = 2 * 3 = 6
So the analytic expression of inverse proportion function is y = 6 / X
Because y = 6x + m passes through (2,3)
So 3 = 12 + M
m=-9
So the analytic expression of a function is y = 6x-9
Set u = {x ≤ 10, X ∈ n} A is the proper subset of u, B is the proper subset of u, and a ∩ B = {4,5,6}, (complement b) ∩ a = {2,3}, (complement a) ∩ (complement b) = {7,8}, find sets a and B
Set a = {2,3,4,5,6} B = {1,4,5,6,9,10}
A=2.3.4.5.6；B=1.4.5.6.9.10
Given that the image of a linear function y = KX + B passes through (- 1,1) and (2,3), the relation of this linear function is
Substituting (- 1,1), (2,3) to y = KX + B
We get B-K = 1
2k+b=3
The solution is b = 5 / 3, k = 2 / 3
2/3X+5/3=y
1=-k+b
3=2k+b
k=2/3
b=5/3
y=2/3x+5/3
y=2/3x+5/3
Substituting 1 = - K + B
3=2k+b
Simultaneous evaluation
k=2/3
b=5/3
Given the complete set u = a ∪ B = {x ∈ n | 0 ≤ x ≤ 10}, a ∩ (∁ UB) = {1, 3, 5, 7}, find the set B
U = a ∪ B = {x ∈ n | 0 ≤ x ≤ 10} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, {1, 3, 5, 7} ⊆ a, but B does not contain {1, 3, 5, 7}, which is represented by Venn graph, as shown in the figure ∪ B = {0, 2, 4, 6, 8, 9, 10}
It is known that the image of a linear function y = KX + B (K ≠ 0) passes through the point (0,1), and Y increases with the increase of X. please write a function that meets the above conditions______ (the answer is not unique)
As long as K ＞ 0, B ＞ 0 and passing through point (0, 1), we can get the function formula that meets the above conditions, such as y = x + 1 (the answer is not unique)
If the complete set is a real number set R and the set a = {x | Log & # 189; (2x-1) > 0}, then what is the complement of a in R
If the complete set is a real number set R and the set a = {x | Log & # 189; (2x-1) > 0}, then what is the complement of a in R?
If 2x minus one is greater than one, X is greater than one. So the complement is less than or equal to one