Determine which ordered pairs fall on the graph of f(x) = -2x-1 Show all of your work for each ordered pair. Ordered Pairs: (1,1
1 answer:
Answer:
(1, 1) is the only one on the graph
Step-by-step explanation:
To determine if an ordered pair lies on the graph, substitute the x-coordinate of the ordered pair into f(x) and evaluate. If the value equals the y-coordinate of the ordered pair then it lies on the graph
f(1) = (2 × 1) - 1 = 2 - 1 = 1 ⇒ (1, 1) lies on graph
f(- 1) = (2 × - 1) - 1 = - 2 - 1 = - 3 ≠ - 4 ⇔ (- 1, - 4 ) not on graph
f(- 2) = (2 × - 2) - 1 = - 4 - 1 = - 5 ≠ - ⇒ (- 2, -
) not on graph
f(0) = (2 × 0) - 1 = 0 - 1 = - 1 ≠ ⇒ not on graph
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Answer:
See the proof below.
Step-by-step explanation:
What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where . If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"
Proof
Since we have a if and only if w need to proof the statement on the two possible ways.
If X is linearly dependent, then a vector is a linear combination
We suppose the set is linearly dependent, so then by definition we have scalars
in C such that:
And not all the scalars are equal to 0.
Since at least one constant is non zero we can assume for example that , and we have this:
We can divide by c1 since we assume that and we have this:
And as we can see the vector can be written a a linear combination of the remaining vectors
. We select v1 but we can select any vector and we get the same result.
If a vector is a linear combination, then X is linearly dependent
We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select and we have this:
For scalars defined in C. So then we have this:
So then we can conclude that the set X is linearly dependent.
And that complet the proof for this case.