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Dimas [21]
3 years ago
7

In each of Problems 7 through 11, determine whether the members of the given set of vectors are linearly independent. If they ar

e linearly dependent, find a linear relation among then. The vectors are written as row vectors to save space but may be considered as column vectors that is, the transposes of the given vectors may be used instead of the vectors themselves. x^(1) = (1, 1, 0), x^(2) = (0, 1, 1), x^(3) = (1, 0, 1) x^(1) = (2, 1, 0) x^(2) = (0, 1, 0), x^(3) = (-1, 2, 0) x^(1) = (1, 2, 2, 3), x^(2) = (-1, 0, 3, 1), x^(3) = (-1, 2, 0) x^(4) = (-3, t-13) x^(1) = (1, 2, -1, 0), x^(2) = (2, 3, 1, -1), x^(3) = (-1, 0, 2, 2), x^(4) = (3, -1, 1, 3) Suppose that each of the vectors x^(1), ..., x^(m) has n components, where n < m. x^(1), ...x^(m) are linearly dependent. In each of Problems 13 and 14, determine whether the member of the given of vectors linearly independent for -infinity < t < infinity. If they are linearly dependent, find the linear among them. As in Problems 7 through 11, the vectors are written as row vectors to save space x^(1) (t) = (e^-t, 2e^-t), x^(2) (t) = (e6-t, e^(-t), x^(3) (t) = (3e^-t, 0) x^(1) (t) = (2 sin t, sin t), x^(2) (t) = (sin t, 2 sin t)

Mathematics
1 answer:
Ad libitum [116K]3 years ago
5 0

Answer:

Step-by-step explanation:

Check attachment

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4 years ago
There are some goats, cows and, sheep. 2/5 of the animals were goats. There were three times as many sheep than cows. If there w
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Answer:

81 sheep were there on the farm

Step-by-step explanation:

Consider the provided information.

Let x represents the number of goats.

y represents the number of cows.

z represents the number of sheep.

Therefore, the total number of animals are x+y+z

2/5 of the animals were goats. This can be written as:

\frac{2}{5}(x+y+z)=x

There were three times as many sheep than cows.

z=3y

There were 45 more goats than cows,

x=45+y

Substitute the value of x and z in equation 1.

\frac{2}{5}(45+y+y+3y)=45+y

\frac{2}{5}(45+5y)=45+y

2(9+y)=45+y

18+2y=45+y

y=27

Hence, there are 27 cows.

Substitute the value of y in x=45+y

x=45+27

x=72

Therefore, there are 72 goats.

Substitute the value of y in z=3y

z=3(27)

z=81

Thus, there are 81 sheep.

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