Answer:
56
Step-by-step explanation:
1 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
2 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
4 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
7 + 4 + 5 + 6 + 7 + 8 + 9 + 10
11 + 5 + 6 + 7 + 8 + 9 + 10
16 + 6 + 7 + 8 + 9 + 10
22 + 7 + 8 + 9 + 10
29 + 8 + 9 + 10
37 + 9 + 10
46 + 10
56
Hope this helps!
Also hi
Answer:
true and twelve
Step-by-step explanation:
i think
Answer:
a.35.25
b. 42.56
c.23.06
Step-by-step explantion
Just subsitute 30, 15, and 5 for x each time for each question
ex:
y=-0.0975(5)^(2)+3.9(5)+6
Answer: ok
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
