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Ann [662]
3 years ago
13

Determine which of the sets of vectors is linearly independent. A: The set where p 1(t) = 1, p 2(t) = t 2, p 3(t) = 1 + 5t B: Th

e set where p 1(t) = t, p 2(t) = t 2, p 3(t) = 2t + 5t 2 C: The set where p 1(t) = 1, p 2(t) = t 2, p 3(t) = 1 + 5t + t 2 A and C C only B only A only all of them
Mathematics
1 answer:
denis-greek [22]3 years ago
6 0

Answer:

(A)A and C

Step-by-step explanation:

In each case, represent each of the p_i as a column vector where  each row corresponds to the constant term, coefficient of t and t^2 respectively.

A= The set where p_1(t)=1 p_2(t)=t^2 p_3(t)=1+5t

A=\left[\begin{array}{ccc}1&0&1\\0&0&5\\0&1&0\end{array}\right]

|A|=\left|\begin{array}{ccc}1&0&1\\0&0&5\\0&1&0\end{array}\right|=-5

B: The set where p_1(t)=t, p_2(t)=t^2, p_3(t)=2t+5t^2

B=\left[\begin{array}{ccc}0&0&0\\1&0&2\\0&1&5\end{array}\right]\\|B|=0

C: The set where p_1(t)=1, p_2(t)=t^2, p_3(t)=1+5t+t^2

C=\left[\begin{array}{ccc}1&0&1\\0&0&5\\0&1&1\end{array}\right]\\|C|=-5

Since the determinants of A and C are not 0, the set of vectors in A and C are linearly independent.



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3 years ago
X^2+6x-6=0 rewritten by completing the square
Pie

Hello from MrBillDoesMath!

Answer:

(x+3)^2 -15 = 0



Discussion:

To complete the square add (and then subtract) the square of one-half the coefficient of x:


x^2 + 6x -6 = 0 =>

                 -- complete the square part --

x^2 + 6x +   (6/2)^2 - (6/2)^2                    - 6 = 0 =>

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Thank you,

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