Let v denote the set of ordered triples (x, y, z) and define addition in v as in
r3.for each of the following definitions of scalar multiplication, decide whether v is a vector space.
(a) a(x, y,z)=(ax, y,az)
(b) a(x, y,z)=(ax,0,az)
(c) a(x, y,z)=(0,0,0)
(d) a(x, y,z)=(2ax,2ay,2az)

a) No

b) No

c) No

d) No

Step-by-step explanation:

Remember, a set V wit the operations addition and scalar product is a vector space if the following conditions are valid for all u, v, w∈V and for all scalars c and d:

1. u+v∈V

2. u+v=v+u

3. (u+v)+w=u+(v+w).

4. Exist 0∈V such that u+0=u

5. For each u∈V exist −u∈V such that u+(−u)=0.

6. if c is an escalar and u∈V, then cu∈V

7. c(u+v)=cu+cv

8. (c+d)u=cu+du

9. c(du)=(cd)u

10. 1u=u

let's check each of the properties for the respective operations:

Let

Observe that  

1. u+v∈V

2. u+v=v+u, because the adittion of reals is conmutative

3. (u+v)+w=u+(v+w). because the adittion of reals is associative

4.

5.

then regardless of the escalar product, the first five properties are met for a), b), c) and d). Now let's verify that properties 6-10 are met.

a)

6.

7.

8.

Since 8 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product

b)  6.

7.

8.

9.

10

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product

c) Observe that

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product .

d)  Observe that

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product .

tube or goog would be a great resource to with that question

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)