Step-by-step explanation:
(a)
Using the definition given from the problem
![f(A) = \{x^2 \, : \, x \in [0,2]\} = [0,4]\\f(B) = \{x^2 \, : \, x \in [1,4]\} = [1,16]\\f(A) \cap f(B) = [1,4] = f(A \cap B)\\](https://tex.z-dn.net/?f=f%28A%29%20%3D%20%5C%7Bx%5E2%20%20%5C%2C%20%3A%20%5C%2C%20x%20%5Cin%20%5B0%2C2%5D%5C%7D%20%3D%20%5B0%2C4%5D%5C%5Cf%28B%29%20%3D%20%5C%7Bx%5E2%20%20%5C%2C%20%3A%20%5C%2C%20x%20%5Cin%20%5B1%2C4%5D%5C%7D%20%3D%20%5B1%2C16%5D%5C%5Cf%28A%29%20%5Ccap%20f%28B%29%20%3D%20%5B1%2C4%5D%20%20%3D%20f%28A%20%5Ccap%20B%29%5C%5C)
Therefore it is true for intersection. Now for union, we have that
![A \cup B = [0,4]\\f(A\cup B ) = [0,16]\\f(A) = [0,4]\\f(B)= [1,16]\\f(A) \cup f(B) = [0,16]](https://tex.z-dn.net/?f=A%20%5Ccup%20B%20%3D%20%5B0%2C4%5D%5C%5Cf%28A%5Ccup%20B%20%29%20%3D%20%5B0%2C16%5D%5C%5Cf%28A%29%20%3D%20%5B0%2C4%5D%5C%5Cf%28B%29%3D%20%5B1%2C16%5D%5C%5Cf%28A%29%20%5Ccup%20f%28B%29%20%3D%20%5B0%2C16%5D)
Therefore, for this case, it would be true that 
.
(b)
1 is not a set.
(c)
To begin with
Therefore
Now, given an element of 
it will belong to both sets, therefore it also belongs to 
, and you would have that
, therefore 
.
(d)
To begin with 
, therefore
35000, rounding down from five.
If two tangent segments to a circle share a
common endpoint outside a circle, then the two segments are congruent. This
is according to the intersection of two tangent theorem. The theorem states
that given a circle, if X is any point
within outside the circle and if Y and Z are points such that XY and XZ are
tangents to the circle, then XY is equal to XZ.
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In the right triangle ABC wherein AB is the hypotenuse, BC is the opposite and CA is the adjacent tanA=0.45. The approximate length of AB which is the hypotenuse is 22, Opposite (BC) is 9 and Adjacent (CA) is 20. You need to use pythagorean formula in getting the length of AB.
YOUR ANSWER IS (22).
