Let v1 = -4 -1
-2

v2 = -3
1
-2

v3= 1
-5
2

and H = Span{v1,v2,v3} . Note that v3 = 2v1 - 3v2.

Which of the following sets form a basis for the subspace H, i. e., which sets form an efficient spanning set containing no unnecessary vectors?

a. {V1, V2, V3}
b. {V1, V2}
c. {V1,V3}
d. {V2,V3}

We're told (and we can confirm) that , so is a linear combination of the other two vectors.

This means H is sufficiently spanned by ; no need for the third vector.

But this also means we can write either as a linear combination of , and as a lin. com. of . So any set of these three vectors taken two at a time will span the subspace H. Hence all of b, c, and d are acceptable.

step-by-step explanation:

to find if any function is an inverse of the other, put the first equation in as x in the next one. then simplify and solve. if at the end you get the statement y = x, then it is an inverse. if you get anything else, it is not.

let y equal to

the total fare and x is the mile of taxi ride. so the equation is

y = 0.55x + 1.75

since susie has

$10 to spend for a taxi cab, so he can have

10 = 0.55x +

1.75

x = 15 miles of

taxi ride

so the system of

inequality is

10 < 0.55x +

1.75

x > 2