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
The domain is how far the graph stretches horizontally (on the x-axis).
Domain: ( -∞ , ∞ ) or -∞ < x < ∞
The range is how long the graph stretched vertically (on the y-axis).
Range: ( -∞ , 2 ] or -∞ < y ≤ 2
The zeros are where the graph intersects with the x-axis. In other words, the x-values when y=0.
Zeros: x = -1 and x = 3
Answer:
9
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Step-by-step explanation:
<u>Step 1: Define</u>
3(y + 2)² + 3x
x = 2
y = -1
<u>Step 2: Evaluate</u>
- Substitute: 3(-1 + 2)² + 3(2)
- Add: 3(1)² + 3(2)
- Exponents: 3(1) + 3(2)
- Multiply: 3 + 6
- Add: 9
2x+3y-10=0 (1)
+
4x-3y-2=0 (2)
____________
6x -12=0 (if you just want the resulting equation. It is 6x-12=0)
x=2
take x=2 and put it into equation (2)
4(2) -3y -2=0
-3y= 2-8
y= 2
(x=2,y=2)
Answer:
48
Step-by-step explanation:
6x4x2
