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.
Answer:
0.447 sec
Explanation:
velocity = distance/ time
distance 'h' is a function of time 't' given by
⇒ differentiating equation
⇒ velocity = 
⇒ acceleration = 
= 24 
deceleration due to gravity = -9.8 m/
= - 9.8 x 3.281 = - 32.154 ft/
Net deceleration = 24 - 32.154 = - 8.15 ft/
at maximum height v = 0
= 
+ 2ah
⇒ 0 = 
+ 2 (-8.15)h
⇒ h = 118.72 ft⇒
118.72 = 12
+ 44t + 16
⇒ 
- 102.72 = 0
⇒ t = -44 ± 
/ (2 x 12)
t = 0.447 sec
Cultural environment, hope this helps ;)
B. Weighted GPA
You see the unweighted is considered the college GPA and removes all AP, IB, or Honors clasess and puts all college applicants on the same level, cumulative GPA is wrong and doesn't make any sense with relation to the question, and the semester GPA wouldn't technically include the extra 10 points that these classes give.
