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melomori [17]
3 years ago
8

Help any one please?

Mathematics
1 answer:
AfilCa [17]3 years ago
5 0
A = (1/2)(8yd)(15yd) = 60 yd²
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Could a set of three vectors in ℝ^4 span all ℝ^4? Explain. What about n vectors in ℝ^m when n is less than m?
Elena-2011 [213]

Answer:

Check the ecplanation

Step-by-step explanation:

A set of three vectors in R^{4} represents a matrix of 3 column vectors, and each vector containing 4 entries (that is, a matrix of 4 rows, and 3 columns).

Let A be that 4x 3 matrix. The columns of A span R^{m}. if and only if A has a pivot position in each row. So, there are at most 3 pivot positions in the matrix A, but the number of rows is 4, therefore, there exist at least one row not having a pivot position. If A does not have a pivot position in at least one row, then the columns of A do not span R^{4}. It implies that the set of 3 vectors of A does not span all of R^{4}.

In general, the set of n vectors in R^{m} represents a matrix of in rows, and n columns (an in x matrix). So, there are at most n pivot positions in the matrix A, but n is less than the number of rows. In therefore, there exist at least one row that does not contain a pivot position.

And, hence the set of n vectors of A does not span all of R^{m}. for n < m

8 0
3 years ago
Please help me how to solve steps by steps and answer.
ivolga24 [154]

84,000/15 = 5600

2 x 5600 = 11,200

The first person will get $11,200

4 0
2 years ago
Complete the following statement.
Viktor [21]

Answer: dividend

Step-by-step explanation:……..

6 0
2 years ago
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A culture started with 5000 bacteria. after 7 hours, it grew to 6500 bacteria. predict how many bacteria will be present after 1
spin [16.1K]

6500=7x+5000

6500-5000=7x+5000-5000

1500=7x

1500/7≈214

x≈214

y=19·214+5000

y=9066

answer=9066 bacteria

4 0
3 years ago
A tennis tournament has 2n contestants. We want to pair them up for the first round of singles matches. Show that the number of
ddd [48]

There are

\dbinom{2n}2 = \dfrac{(2n)!}{2! (2n-2)!}

ways of pairing up any 2 members from the pool of 2n contestants. Note that

(2n)! = 1\times2\times3\times4\times\cdots\times(2n-2)\times(2n-1)\times(2n) = (2n-2)! \times(2n-1) \times(2n)

so that

\dbinom{2n}2 = \dfrac{(2n)\times(2n-1)\times(2n-2)!}{2! (2n-2)!} = \boxed{n(2n-1)}

4 0
2 years ago
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