Step-by-step explanation:
(f+g)(x) means f(x) + g(x).
(f−g)(x) means f(x) − g(x).
So all you have to do is add them and subtract them.
1. (f+g)(x) = f(x) + g(x)
(f+g)(x) = (3x − 7) + (2x − 4)
(f+g)(x) = 5x − 11
2. (f−g)(x) = f(x) − g(x)
(f−g)(x) = (3x − 7) − (2x − 4)
(f−g)(x) = 3x − 7 − 2x + 4
(f−g)(x) = x − 3
3. (f+g)(x) = f(x) + g(x)
(f+g)(x) = (2x + 3) + (x² + ½ x − 7)
(f+g)(x) = x² + 2½ x − 4
4. (f−g)(x) = f(x) − g(x)
(f−g)(x) = (2x + 3) − (x² + ½ x − 7)
(f−g)(x) = 2x + 3 − x² − ½ x + 7
(f−g)(x) = -x² + 1½ x + 10
Answer:
The given statement is false.
Step-by-step explanation:
Reason
let D be a directed graph with 'n' no of vertices and 'E' edges.
where 'n'=1. thus D =(n,E).
In degree: in directed graph the number of incoming edges on a vertex is known as indegree.
it is denoted as deg ⁺(n).
And now we know that in a directed graph
if deg ⁻(n)= deg ⁺(n) for each vertex n.
Equation
Because expressions don't contain equal signs
Answer:
Step-by-step explanation:
1/5 + 1/5 = 2/5
1/7 + 1/7 = 2/7
1/3 + 1/3 = 2/3
There are an infinite number of these fractions. They must be 1 and 1 in the numerator, and the denominator must be relatively prime to 2. The examples I have picked are prime in the denominator, but the rule is not without many exceptions. For example
1/9 + 1/9 = 2/9
I don't think you can pick an even denominator because it will reduce when put with two. Oh wait 2/18 + 2/18 = 4/18 = 2/9 But these could be reduced before adding. Still, it might count. It depends on who is marking the question.
What about an odd and even denominator?
1/9 + 1/18 = 3/18 = 1/6 There must be something that works, but I can't come up with an example.
