Answer: 8/9 is rational
.2593 irrational
square root of 7: irrational
square root of .25 is rational
square root of 14: irrational
0 is a: rational
square root of 280: irrational
square root of 35: is irrational
.2222: is rational
square root of 3r is a rational number......
Step-by-step explanation:
Answer:
18 cm
Step-by-step explanation:
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: Choice C. 
============================================
Explanation:
The domain is the set of all allowed x inputs.
Here we see that x = -3 is the smallest x value possible as it is from the left-most point. The graph goes on forever to the right, so there is no largest x value. Effectively infinity is the largest x value even though infinity is not a number.
We can say the domain is
which can be simplified to
or 
In short: x can be -3 or larger.