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gladu [14]
3 years ago
7

Mark the statement either true (in all cases) or false (for at least one example). If false, construct a specific example to sho

w that the statement is not always true. Such an example is called a counterexample to the statement. If the statement is true, give a justification. If v1,…,v4 are in R4 and {v1,v2,v3} is linearly dependent then {v1,v2,v3,v4} is also linearly dependent.
Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

(A) True. Because v3 = 2v1 + v2, v4 must be the zero vector. Thus, the set of vectors is linearly dependent.
(B) True. The vector v3 is a linear combination of v1 and v2, so at least one of the vectors in the set is a linear combination of the others and set is linearly dependent.
(C) True. If c1 =2, c2 = 1, c3 = 1, and c4 = 0, then c1v1 + ........... + c4v4 =0. The set of vectors is linearly dependent.
(D) False. If v1 = __, v2 =___, v3 =___, and v4 = [1 2 1 2], then v3 = 2v1 + v2 and {v1, v2, v3, v4} is linearly independent.
Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
8 0

Answer with Step-by-step explanation:

We are given that v_1,v_2,..,v_4 are in R^4 and v_1,v_2,v_3 is linearly dependent then {v_1,v_2,v_3,v_4}[/tex] is also linearly dependent.

We have to find that given statement is true or false.

Dependent vectors:Dependent vectors are those vectors in which atleast one vector is  a linear combination of other given vectors.

Or If we have vectors x_1,x_2,....x_n

Then their linear combination

a_1x_1+a_2x_2+.....+a_nx_n=0

There exist at least one scalar which is not zero.

If v_1,v_2,v_3 are dependent vectors then

a_1v_1+a_2v_2+a_3v_3=0 for scalars a_1,a_2,a_3

Then , by definition of dependent  vectors

There exist a vector which is not equal to zero

If vector v_3 is a linear combination of v_1\;and \;v_2, So at least one of vectors in the set is a linear combination of others and the set is linearly dependent.

Hence, by definition of dependent vectors

{v_1,v_2,v_3,v_4} is linearly dependent.

Option B is true.

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pantera1 [17]

Answer:

M=a2-a1/b2-b1

Step-by-step explanation:

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7 0
2 years ago
Water is pumped out of a holding tank at a rate of 6-6e^-0.13t liters/minute, where t is in minutes since the pump is started. I
astraxan [27]
Procedure:

1) Integrate the function, from t =0 to t = 60 minutues to obtain the number of liters pumped out in the entire interval, and

2) Substract the result from the initial content of the tank (1000 liters).

Hands on:

Integral of (6 - 6e^-0.13t) dt  ]from t =0 to t = 60 min =

= 6t + 6 e^-0.13t / 0.13 = 6t + 46.1538 e^-0.13t ] from t =0 to t = 60 min =

6*60 + 46.1538 e^(-0.13*60) - 0 - 46.1538 = 360 + 0.01891 - 46.1538 = 313.865 liters

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Answer: 613.135 liters



 

3 0
3 years ago
I need to know the answer to the second one.
Debora [2.8K]

Travis got 43 dollars

Robin collected 35 percent more so we can model this with

Travis + (Travis x .35)

43 + 15.05 =58.05

Jessica colelcted 20 percent less so

Travis - (Travis x .20)

43 - 8.6 = 34.3

34.4 + 58.05 + 43 = 135.45

C is the closest

7 0
3 years ago
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Answer:

The mode is the number that is repeated most often, but all the numbers in this list appear only once, so there is no mode. The largest value in the list is 7, the smallest is 1, and their difference is 6, so the range is 6.

Step-by-step explanation:

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3 years ago
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Answer: Both numbers 6/5 and 3/2 do not have a common denominator. Just as how 2 cannot go into 5 evenly.

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