The requirement is that every element in the domain must be connected to one - and one only - element in the codomain.
A classic visualization consists of two sets, filled with dots. Each dot in the domain must be the start of an arrow, pointing to a dot in the codomain.
So, the two things can't can't happen is that you don't have any arrow starting from a point in the domain, i.e. the function is not defined for that element, or that multiple arrows start from the same points.
But as long as an arrow start from each element in the domain, you have a function. It may happen that two different arrow point to the same element in the codomain - that's ok, the relation is still a function, but it's not injective; or it can happen that some points in the codomain aren't pointed by any arrow - you still have a function, except it's not surjective.
Answer:
8 units
Step-by-step explanation:
did the assignment
Hello!
You have to find what a and b are
3a + 4a = 14
b + 6b = 21
Lets do a first
3a + 4a = 14
Combine like terms
7a = 14
Divide both sides by 7
a = 2
Now lets do b
b + 6b = 21
Combine like terms
7b = 21
Divide both sides by 7
b = 3
Put a and b into the third equation
7( 2 + 3) =
Do what is in parenthesis first
7(5) =
Multiply
35
The answer is 35
Hope this helps!
Answer: 2/3
Step-by-step explanation:
As the base of the two sides are equal, we can say that
3x = 2 which gives
x = 2/3
