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The common ratio of the geometric sequence is -1/2
<h3>How to determine the common ratio?</h3>
The geometric sequence is given as: -96, 48, -24, 12, -6, ...
The common ratio is calculated using:
r = T2/T1
From the sequence, we have:
T2 = 48
T1 = -96
This means that
r = 48/-96
Evaluate the quotient
r = -1/2
Hence, the common ratio of the geometric sequence is -1/2
Read more about common ratio at:
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Answer:
Step-by-step explanation:
bearing is the angle which a line makes with the north.
Not necessarily.
![\mathbf u](https://tex.z-dn.net/?f=%5Cmathbf%20u)
and
![\mathbf v](https://tex.z-dn.net/?f=%5Cmathbf%20v)
may be linearly dependent, so that their span forms a subspace of
![\mathbb R^2](https://tex.z-dn.net/?f=%5Cmathbb%20R%5E2)
that does not contain every vector in
![\mathbb R^2](https://tex.z-dn.net/?f=%5Cmathbb%20R%5E2)
.
For example, we could have
![\mathbf u=(0,1)](https://tex.z-dn.net/?f=%5Cmathbf%20u%3D%280%2C1%29)
and
![\mathbf v=(0,-1)](https://tex.z-dn.net/?f=%5Cmathbf%20v%3D%280%2C-1%29)
. Any vector
![\mathbf w](https://tex.z-dn.net/?f=%5Cmathbf%20w)
of the form
![(r,0)](https://tex.z-dn.net/?f=%28r%2C0%29)
, where
![r\neq0](https://tex.z-dn.net/?f=r%5Cneq0)
, is impossible to obtain as a linear combination of these
![\mathbf u](https://tex.z-dn.net/?f=%5Cmathbf%20u)
and
![\mathbf v](https://tex.z-dn.net/?f=%5Cmathbf%20v)
, since
![c_1\mathbf u+c_2\mathbf v=(0,c_1)+(0,-c_2)=(0,c_1-c_2)\neq(r,0)](https://tex.z-dn.net/?f=c_1%5Cmathbf%20u%2Bc_2%5Cmathbf%20v%3D%280%2Cc_1%29%2B%280%2C-c_2%29%3D%280%2Cc_1-c_2%29%5Cneq%28r%2C0%29)
unless
![r=0](https://tex.z-dn.net/?f=r%3D0)
and
![c_1=c_2](https://tex.z-dn.net/?f=c_1%3Dc_2)
.