Given the set a = {(x, y) | x + y ≥ 1}, B = {(x, y) | X & # 178; + Y & # 178; ≤ 1}, then the area of the graph represented by a ∩ B is equal to

Given the set a = {(x, y) | x + y ≥ 1}, B = {(x, y) | X & # 178; + Y & # 178; ≤ 1}, then the area of the graph represented by a ∩ B is equal to

0.25π-0.5

It is known that the length of the longest side of the obtuse triangle ABC is 2, and the length of the other two sides is a, B. then the area of the plane figure represented by the set P = {(x, y) | x = a, y = B} is ()
A. 2B. 4C. π-2D. 4π-2

From the length of the longest side of the obtuse triangle ABC is 2, and the other two sides are a and B. from the cosine theorem, we can get A2 + B2 < 4, and then from the sum of the two sides is greater than the third side, we can get a + b > 2, and a > 0, b > 0. The range of point P is shown in the figure below. From the graph, we can get the area of feasible region is π - 2

It is known that the longest side of the obtuse angle △ ABC is 2, and the lengths of the other two sides are a and B, then the area of the plane figure represented by the set P = {(x, y) | x = a, y = B is equal to

From the meaning of the title:
a+b>2,
cosC=(a^2+b^2-4)/(2ab)0,b>0
Namely:
x+y>2,
x^2+y^20,y>0
Draw a flat area: bow
S=1/4*π*4-1/2*2*2=π-2.

Given a = {the square of Y / y = x-4x + 3, X ∈ r}, B = {the square of Y / y = x-2x + 2, X ∈ r}, then a intersection B is equal to ()

The square of y = x-4x + 3 = the square of x-4x + 4-1 = (X-2) & # - 1
So assemble
A is y ≥ - 1
The square of y = x - 2x + 2 = the square of X - 2x + 1 + 1 = (x-1) & # 178; + 1
So set B is y ≥ 1
So a cross B = y ≥ 1

Let a = {X / | X-2 | ≤ 2, X ∈ r}. B = {Y / y = the square of negative x, - 1 ≤ x ≤ 2}, then Cr (a ∩ b) is equal to
The answer is {X / X ∈ R, X ≠ 0} why is x not equal to 0?

Because a ∩ B = 0
A={0≤x≤4}
B={-4≤x≤0}

Let a = {X / / X-2 / ≤ 2, X ∈ r}, B = {Y / y = - x ^ 2, - 1

Hello, of course, a and B can take the intersection. X in a is different from X in B. If you solve | X-2 | ≤ 2, you will know the element of A. then, if you solve y = - x ^ 2, you will get y, and you will get the element of B
The specific methods are as follows:
A={x|x≤4,x∈R}=（-∞,4]
B={y|y=-x^2,-1≤x≤2}=[-4,0]
A∩B=（-∞,4]∩[-4,0]=[-4,0]
Cr (a, b) = (- ∞, - 4) ∪ (0, + ∞)
Although X and y are not the same, the most important part of a set is to solve the elements of the surface, so it doesn't matter
I hope I can help you

Let a = {x | X-2 | ≤ 2, X ∈ r}, B = {y | y = - x2}, then what is Cr (a ∩ b) equal to
Ask someone who can do it to write down the process

A=〔0,4〕 B=〔-8,0〕
A ∩ B = empty set
CR（A∩B）=R

Let a = {x | x + 1 | ≤ 2, X ∈ r}, B = {y | = X2, - 1 ≤ x ≤ 2}, then Cr (a ∩ b) is equal to?

If a = {x | - 3 ≤ x ≤ 1}, B = {y | y = x & # 178;, - 1 ≤ x ≤ 2} = {y | 0 ≤ y ≤ 4}, then:
CR(A∩B)={y|x1}

If a = {x | X-2 | 1, X belongs to R}, B = {y | y = - x ^ 2, - 1 | x | 1}, then Cr (A and b) is equal to

In this paper, we want to be the 124\\\\\\\\\\\\\\124\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\} Cr (a ∪ b) = {x|

The known set a = {1, 2, 3, K}, B = {4, 7, A4, A2 + 3A}, and a ∈ n *, X ∈ a, y ∈ B, so that the element Y = 3x + 1 in B corresponds to the element X in a, then a=______ ，k=______ ．

If x ∈ a, y ∈ B, so that the element Y = 3x + 1 in B corresponds to the element X in a, then when x = 1, y = 4; when x = 2, y = 7; when x = 3, y = 10; when x = k, y = 3K + 1; and from a ∈ n *, ｜ A4 ≠ 10, then A2 + 3A = 10, A4 = 3K + 1, a = 2, k = 5, so the answer is: 2, 5