Answer:
1) False
2) False
3) True
4) False
Step-by-step explanation:
1) Flase, {v1,v2,v3, ..., vp} is a base for H when they span H and also they are linearly independent.
2) False. A single nonzero vector is linearly independent , not dependent. There is not null linear combination that gives 0 as a result involving that vector.
3) True, if the columns werent linearly independent, we could triangulate the matrix and obtain 0, so the matrix wouldnt be invertible. This means that the columns should be linearly independent for the matrix to be invertible and as a consecuence, they will spam a subspace of R^n of dimension n, which means that they will spam all R^n and therefore, they form a basis of R^n.
4) False. A basis is a spanning set that is as small as possible. Larger spanning sets will have extra elements apart from those who can form a base toguether. Those elements will make the set linearly dependent.
As a formula, you're saying 4*(n-7)=32
This solves to n-7 = 8, so n=15
the first one ((x+8, y+2), r x-axis)
Answer:
6.67408 × 10^-11 m^3 kg^-1 s^-2
Step-by-step explanation:
F = G M 1 M 2 d 2 , where F is the gravitational force between two point masses, M1 and M2; d is the distance between M1 and M2; G is the universal gravitational constant, usually taken as 6.670 × 1011 m3/(kg)(s2) or 6.670 × 10−8 in centimeter–gram–second units. a = F M 1 = G M 2 r 2 .
