Find the slope of the line represented by the table of values
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:
The product of 12 and k is 84
According to the given problem, the equation is
- <u>k = 7</u> is the right answer.
Problem 61
The nth triangular number is
T(n) = n(n+1)/2
I'll rewrite this into
T(n) = 0.5n(n+1)
The triangular number right after this is
T(n+1) = 0.5(n+1)(n+2)
I replaced every n with n+1 and simplified
Let's see what we get when we add up the two expressions
T(n) + T(n+1)
0.5n(n+1) + 0.5(n+1)(n+2)
0.5n^2+0.5n + 0.5(n^2+3n+2)
0.5n^2 + 0.5n + 0.5n^2 + 1.5n + 1
n^2+2n+1
(n+1)^2
This shows that the sum of any two consecutive triangular numbers results in a square number
Here's a few examples
- 0+1 = 1
- 1+3 = 4
- 3+6 = 9
- 6+10 = 16
- 10+15 = 25
Note each sum is a perfect square, which visually would plot out a square figure.
For quick reference, the set of the first few triangular numbers is {0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55,...}
<h3>Answer: Square number</h3>==========================================================
Problem 64
Let's say we go with n = 5.
This means,
T(n) = 0.5n(n+1)
T(n-1) = 0.5(n-1)(n-1+1)
T(n-1) = 0.5n(n-1)
T(5-1) = 0.5*5(5-1)
T(4) = 10
This says that when n = 5, the 4th triangular number is 10
Triple that result and add on n = 5
3*T(4) + n = 3*10+5 = 35
This result is beyond obvious which category of figurate number it belongs to. It's not a triangular number since it's not in the form n(n+1)/2. It's not a square number either.
Through a bit of trial and error, you should find it's a pentagonal number
Pentagonal numbers are of the form n(3n-1)/2
If you plugged n = 5 into that, it leads to 35
n(3n-1)/2 = 5*(3*5-1)/2 = 5*14/2 = 70/2 = 35
The diagram shown below represents the first few pentagonal numbers. The number of blue dots corresponds to the pentagonal number itself. Note the equal spacing when dealing with dots on each segment (eg: some interior blue dots are midpoints, others are quarter points, etc.)
<h3>Answer: Pentagonal number</h3>