Answer:
Step-by-step explanation:
We have given:
-2x+y=4 ---------equation1
3x+4y=49 ---------equation 2
We will solve the 1st equation for y and substitute the value into the 2nd equation.
-2x+y=4 ---------equation1
Move the values to the R.H.S except y
y = 2x+4
Now substitute the value of y in 2nd equation:
3x+4y=49
3x+4(2x+4)=49
3x+8x+16=49
Combine the like terms:
3x+8x=49-16
11x=33
Now divide both the sides by 11
11x/11 = 33/11
x= 3
Now substitute the value of x in any of the above equations: We will substitute the value in equation 1:
-2x+y=4
-2(3)+y=4
-6+y=4
Combine the constants:
y=4+6
y = 10
Thus the solution set of (x,y) is {(3,10)}....
4x - 5
Or what kind of an answer are you looking for?
Only two real numbers satisfy x² = 23, so A is the set {-√23, √23}. B is the set of all non-negative real numbers. Then you can write the intersection in various ways, like
(i) A ∩ B = {√23} = {x ∈ R | x = √23} = {x ∈ R | x² = 23 and x > 0}
√23 is positive and so is already contained in B, so the union with A adds -√23 to the set B. Then
(ii) A U B = {-√23} U B = {x ∈ R | (x² = 23 and x < 0) or x ≥ 0}
A - B is the complement of B in A; that is, all elements of A not belonging to B. This means we remove √23 from A, so that
(iii) A - B = {-√23} = {x ∈ R | x² = 23 and x < 0}
I'm not entirely sure what you mean by "for µ = R" - possibly µ is used to mean "universal set"? If so, then
(iv.a) Aᶜ = {x ∈ R | x² ≠ 23} and Bᶜ = {x ∈ R | x < 0}.
N is a subset of B, so
(iv.b) N - B = N = {1, 2, 3, ...}
Answer:
91
Step-by-step explanation:
triangles always equals 180°
add 70 and 19
subtract that by 180
;)
