This question is incomplete, the complete question is;
Determine if the described set is a subspace. Assume a, b, and c are real numbers.
The subset of R³ consisting of vectors of the form
, where abc = 0
- The set is a subspace
- The set is not a subspace
Answer:
Therefore; The set is not a subspace
Step-by-step explanation:
Given the data the question;
the subset R³;
S = {
, where abc = 0 }
we know that a subset of R³ is a subspace if it stratifies the following properties;
- it contains additive identity
- it is closed under addition
- it is closed under scales multiplication
Looking at the properties, we can say that it is not a subspace
As;
u =
∈ S and v =
∈ S
As 1×1×0=0 0×1×1=0
But u+v =
∉ S as 1×2×1 ≠ 0
Hence, it is not closed under addition.
Therefore; The set is not a subspace
To work out how much something was before the discount, you have to divide the price (which is £90) by 75 (as this is the percentage that you paid). It's gives you £1.20. Therefore, 1% is £1.20. To find the original cost, which is 100%, you have to multiply it by 100. Multiplying £1.20 by 100 gives you £120.
Therefore, the cost of the TV before the 25% reeducation was £120.
Hope this helps :)
Volume of sphere =4/3*pi*r^3.
r^3=27
volume =113.04 =113.0 to the nearest tenth