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:
24
Step-by-step explanation:
6*6=36
6+10=16
2*10=20
Flower=10-6+20=24
There’s no attachment, i can try to help tho what does the problem look like
Answer:
B
Step-by-step explanation:
Let us set up some variabe:
Use the known information:
Now lets find the area
Area = (1/2) * b *h = (1/2) * h * (2h + 8)
Hope that helps!
