The definition of a circle uses the undefined term _______.
1 answer:
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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.
Answer:
Axis is a vertical line at x = 2
Vertex is (2, -1)
y-intercept is (0, 3)
Solutions are x = 1 and x = 3
Step-by-step explanation:
To draw the graph of the quadratic equation you must find at least 5 points lie on the graph by choose values of x and find their values of y
Let us do that
Use x = -1, 0, 1, 2, 3, 4, 5
∵ y = x² - 4x + 3
∵ x = -1
∴ y = (-1)² - 4(-1) + 3 = 1 + 4 + 3 = 8
→ Plot point (-1, 8)
∵ x = 0
∴ y = (0)² - 4(0) + 3 = 0 + 0 + 3 = 3
→ Plot point (0, 3)
∵ x = 1
∴ y = (1)² - 4(1) + 3 = 1 - 4 + 3 = 0
→ Plot point (1, 0)
∵ x = 2
∴ y = (2)² - 4(2) + 3 = 4 - 8 + 3 = -1
→ Plot point (2, -1)
∵ x = 3
∴ y = (3)² - 4(3) + 3 = 9 - 12 + 3 = 0
→ Plot point (3, 0)
∵ x = 4
∴ y = (4)² - 4(4) + 3 = 16 - 16 + 3 = 3
→ Plot point (4, 3)
∵ x = 5
∴ y = (5)² - 4(5) + 3 = 25 - 20 + 3 = 8
→ Plot point (5, 8)
→ Join all the points to form the parabola
From the graph
∵ The axis of symmetry is the vertical line passes through the vertex point
∵ x-coordinate of the vertex point is 2
∴ Axis is a vertical line at x = 2
∵ The coordinates of the vertex point of the parabola are (2, -1)
∴ Vertex is (2, -1)
∵ The parabola intersects the y-axis at point (0, 3)
∴ y-intercept is (0, 3)
∵ x² - 4x + 3 = 0
∵ The solutions of the equation are the values of x at y = 0
→ That means the intersection points of the parabola and the x-axis
∵ The parabola intersects the x-axis at points (1, 0) and (3, 0)
∴ Solutions are x = 1 and x = 3
