Answer:
Dimensions of printed area
w = 8.95 cm
h = 13.44 cm
A(max) = 120.28 cm²
Step-by-step explanation:
Lets call " x " and "y" dimensions of the poster area ( wide and height respectively) . Then
A(t) = 180 cm² = x*y y = 180/ x
And the dimensions of printed area is
A(p) = ( x - 2 ) * ( y - 3 ) then as y = 180/x we make A function of x only so
A(x) = ( x - 2 ) * ( 180/x - 3 ) ⇒ A(x) = 180 - 3x - 360/x +6
A(x) = - 3x - 360 /x + 186
Taking derivatives on both sides of the equation we get:
A´(x) = -3 + 360/ x²
A´(x) = 0 -3 + 360/ x² = 0 -3x² + 360 = 0
x² = 120 ⇒ x = √120 x = 10.95 cm
And y = 180 / 10.95 ⇒ y = 16.44 cm
Then x and y are the dimensions of the poster then according to problem statement
w of printed area is x - 2 = 10.95 - 2 = 8.95 cm
and h of printed area is y - 3 = 16.44 - 3 = 13.44 cm
And the largest printed area is w * h = ( 8.95)*(13.44)
A(max) = 120.28 cm²
Answer:
first 1 second one x
Step-by-step explanation:
And you are nor R-tarted
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, ...}
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