Determine if the described set is a subspace. The subset of Rn (n even) consisting of vectors of the form v = v1 vn , such that v1 − v2 + v3 − v4 + v5 − − vn = 0. The set is a subspace. The set is not a subspace.

The set is a subspace

Step-by-step explanation:

We need to check 3 things: whether the 0 vector is in the set, whether the sum of 2 elements of the set is an element of the set and whether the product of an element of the set for a real scalar is an element of the set.

Yes: the 0 vector (0, 0, ..., 0) satysfies the set property: 0-0+0-0-0 = 0.

Yes: Note that (v1+w1)-(v2+w2)+(v3+w3)- - (vn+wn) = v1-v1+v3 - ... - vn + w1 - w2 + w3 - ... - wn = 0+0 = 0.

Yes: By taking r as common factor, we have rv1 - rv2 + rv3 - ... - rvn = r * (v1-v2+v3 - ... - vn) = r*0 = 0.

Thus, the described set is effectively a subspace.

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.