Answer:
Closer to 0 than 0.5, or under zero
Step-by-step explanation:
On a vertical number line, anything above zero is positive, and anything under zero is negative. If 0.5 is the greater number, then it has to be closer to zero, or under it.
Answer:

Step-by-step explanation:
Given


Required
Find x such that: 
This gives:

Collect like terms


Expand

Factorize

Factor out x - 1

Solve:

Answer: The vectors are linearly dependent.
The solution is in the attachment