You know that the discrete metric only takes values of 1 and 0. Now suppose it comes from some norm ||.||. Then for any α in the underlying field of your vector space and x,y∈X, you must have that
∥α(x−y)∥=|α|∥x−y∥.
But now ||x−y|| is a fixed number and I can make α arbitrarily large and consequently the discrete metric does not come from any norm on X.
Step-by-step explanation:
hope this helps
Answer:
the area is 30 correct me if Im wrong
Step-by-step explanation:
1.5b=11-9
1.5b=2
b=1.3 repeating or 1 1/3
A shape with 4 equal sides.
T = 5, so after 5 years
p(t) = t^3 - 14t^2 + 20t + 120
Take derivative to find minimum:
p’(t) = 3t^2 - 28t + 10
Factor to solve for t:
p’(t) = (3t - 2)(t - 5)
0 = (3t - 2)(t - 5)
0 = 3t - 2
2 = 3t
2/3 = t
Plug 2/3 into original equation, this is a maximum. We want the minimum:
0 = t - 5
5 = t
Plug back into original:
5^3 - 14(5)^2 + 20(5) + 120
125 - 14(25) + 100 + 120
125 - 350 + 220
- 225 + 220
p(5) = -5
