Determine whether the set W is a subspace of R3 with the standard operations. If not, state why. W = {(0, x2, x3): x2 and x3 are real numbers}
W is a subspace of R3.
W is not a subspace of R3 because it is not closed under addition.
W is not a subspace of R3 because it is not closed under scalar multiplication.

Step-by-step explanation:

From the given information.

R3 is a vector space over the field R, where R is the set of real numbers.

Where;

The set W = {(0, x2, x3): x2 and x3 are real numbers} is the subspace of ³

To proof:

(0,0,0) ∈ W ⇒ W ≠ ∅

Suppose u and v is an element of W;

i.e.

u,v ∈ W (which implies that) ⇒ u (0,x2,x3) and v = (0,y2,y3) are  real numbers.

Then

u+v = (0,x2,x3) +(0, y2, y3)

u+v = (0,x2+y2, x2+y3) ∈ W

⇒ u+v ∈ W  ----- (1)

Now, if we take any integer to be an element  of the real number

i.e

∝ ∈

∝* = ∝(0,x2,x3)

∝* = (0,x2,x3) ∈ W

⇒ ∝*u ∈ W ------(2)

Thus from (1) and (2), W is a subspace of ³

wholesale price is the initial price before markup. retail price comes after the markup.

a price of 8 bucks that's marked up by 100%   is

formula: initial price + (percent markup in fraction/decimal)(initial price)

8+(1)(8)

8+8

16

retail price is $16.