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
Answer:
Step-by-step explanation:
Given:
ΔCAD and ΔCBD
∠A ≅∠B (Angle)
AD ≅BD (Side)
From the graph we see that
CD≅CD (Side)
because of reflexive propriety ( a line segment is congruent with itself)
If you put in order those congruencies we have SSA witch does NOT prove congruence.
we not use SAS because the angle between the sides is not congruent
Given : f(x)= 3|x-2| -5
f(x) is translated 3 units down and 4 units to the left
If any function is translated down then we subtract the units at the end
If any function is translated left then we add the units with x inside the absolute sign
f(x)= 3|x-2| -5
f(x) is translated 3 units down
subtract 3 at the end, so f(x) becomes
f(x)= 3|x-2| -5 -3
f(x) is translated 4 units to the left
Add 4 with x inside the absolute sign, f(x) becomes
f(x)= 3|x-2 + 4| -5 -3
We simplify it and replace f(x) by g(x)
g(x) = 3|x + 2| - 8
a= 3, h = -2 , k = -8
