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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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Step-by-step explanation:

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Solve the formula for the variable h f=12gh
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You made a deal with Ms.Williams to read 100 pages every night. You know that it takes you 32 minutes to read 16 pages. How long
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Put the problem into a proportion to make it easier to solve.

32        x
      =
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Your equation should now look like this
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5 0
3 years ago
9/20
Stella [2.4K]

<u>Question:</u>

Find the number of real number solutions for the equation. x^2 + 5x + 7 = 0

<u>Answer:</u>

The number of real solutions for the equation x^{2}+5 x+7=0 is zero

<u>Solution:</u>

For a Quadratic Equation of form : a x^{2}+b x+c=0  ---- eqn 1

The solution is x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}  

Now , the given Quadratic Equation is x^{2}+5 x+7=0  ---- eqn 2

On comparing Equation (1) and Equation(2), we get

a = 1 , b = 5 and c = 7

In x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} , b^2 - 4ac is called the discriminant of the quadratic equation

Its value determines the nature of roots

Now, here are the rules with discriminants:

1) D > 0; there are 2 real solutions in the equation

2) D = 0; there is 1 real solution in the equation

3) D < 0; there are no real solutions in the equation

Now let solve for given equation

D= b^2 - 4ac\\\\D = 5^2 - 4(1)(7)\\\\D = 25 - 28 \\\\D = -3

Since -3 is less than 0, this means that there are 0 real solutions in this equation.

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Answer:

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b) written the same as a. except the c will replace the n and x will be (2y)^b.

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d) the answer would be (st)^v/u.

Hope this helps!

Brainliest?

3 0
3 years ago
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