Relations are subsets of products <span><span>A×B</span><span>A×B</span></span> where <span>AA</span> is the domain and <span>BB</span> the codomain of the relation.
A function <span>ff</span> is a relation with a special property: for each <span><span>a∈A</span><span>a∈A</span></span> there is a unique <span><span>b∈B</span><span>b∈B</span></span> s.t. <span><span>⟨a,b⟩∈f</span><span>⟨a,b⟩∈f</span></span>.
This unique <span>bb</span> is denoted as <span><span>f(a)</span><span>f(a)</span></span> and the 'range' of function <span>ff</span> is the set <span><span>{f(a)∣a∈A}⊆B</span><span>{f(a)∣a∈A}⊆B</span></span>.
You could also use the notation <span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈f</span>]</span>}</span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈f</span>]</span>}</span></span>
Applying that on a relation <span>RR</span> it becomes <span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈R</span>]</span>}</span><span>{b∈B∣∃a∈A<span>[<span>⟨a,b⟩∈R</span>]</span>}</span></span>
That set can be labeled as the range of relation <span>RR</span>.
330-> -100 = 230, 230-> -100= 130
(I literally just watched a video how to do this, I never had to do it this way when I was in 2nd grade)
[130 final answer]
20 plus 24
Divided by 15
Which gets you to
44 over 15 or 2 wholes and 14 over 15
Answer:
The answers are C), D), and E) because these three answers satisfy the conditions above.
Step-by-step explanation:
Joseph could use C), D), and E) to make kebabs without any leftovers:
C) 30 pieces of meat and 40 pieces of vegetables
D) 45 pieces of meat and 60 pieces of vegetables
E) 72 pieces of meat and 96 pieces of vegetables
