Solution. To check whether the vectors are linearly independent, we must answer the following question: if a linear combination of the vectors is the zero vector, is it necessarily true that all the coefficients are zeros?
Suppose that
x 1 ⃗v 1 + x 2 ⃗v 2 + x 3 ( ⃗v 1 + ⃗v 2 + ⃗v 3 ) = ⃗0
(a linear combination of the vectors is the zero vector). Is it necessarily true that x1 =x2 =x3 =0?
We have
x1⃗v1 + x2⃗v2 + x3(⃗v1 + ⃗v2 + ⃗v3) = x1⃗v1 + x2⃗v2 + x3⃗v1 + x3⃗v2 + x3⃗v3
=(x1 + x3)⃗v1 + (x2 + x3)⃗v2 + x3⃗v3 = ⃗0.
Since ⃗v1, ⃗v2, and ⃗v3 are linearly independent, we must have the coeffi-
cients of the linear combination equal to 0, that is, we must have
x1 + x3 = 0 x2 + x3 = 0 ,
x3 = 0
from which it follows that we must have x1 = x2 = x3 = 0. Hence the
vectors ⃗v1, ⃗v2, and ⃗v1 + ⃗v2 + ⃗v3 are linearly independent.
Answer. The vectors ⃗v1, ⃗v2, and ⃗v1 + ⃗v2 + ⃗v3 are linearly independent.
A. the existence of other galaxies
1. .The longer the story is on the news, the more people thought it was important
<span>2. People's opinion shifted with tone used
</span><span>Effectively communicating a message is critical to political success. the key is gaining control over the political agenda, which involves getting one's priorities presented at the top of the daily news.
I think the bolded part is a better answer....
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Answer:b(x < 45; 100, 0.5) = 0.184
Explanation: b(x < 45; 100, 0.5) = b(x = 0; 100, 0.5) + b(x = 1; 100, 0.5) + + b(x = 45; 100, 0.5)
