Hey there, Dudedan!
In a#b = a+2b, we have the first term, a, plus 2 times the second term, b.
Let's try that with 3#4.
3#4 = 3+2(4) =
3 + 8 = 11.
Hope this helps!
Answer:
![Union = 15](https://tex.z-dn.net/?f=Union%20%3D%2015)
Step-by-step explanation:
Given
![n(A) = 5; n(B) = 10; n(C) = 15](https://tex.z-dn.net/?f=n%28A%29%20%3D%205%3B%20n%28B%29%20%3D%2010%3B%20n%28C%29%20%3D%2015)
Solving (a): Possible union elements
This is the represented by element with the largest number of the sets
i.e.
![Union = 15](https://tex.z-dn.net/?f=Union%20%3D%2015)
Solving (b): Possible intersection elements
This is the represented by element with the least number of the sets
i.e.
![Intersection = 5](https://tex.z-dn.net/?f=Intersection%20%3D%205)
Answer:
whats yo paypal?
Step-by-step explanation:
Answer:
Step-by-step explanation:
The answer is 33.4
Answer: 14p +70
Step-by-step explanation: