Respuesta :
Answer:
[tex]\displaystyle d = 5\sqrt{2}[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
Algebra I
- Coordinates (x, y)
Algebra II
- Distance Formula: [tex]\displaystyle d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Step-by-step explanation:
Step 1: Define
Identify
Point (-2, 1)
Point (5, 0)
Step 2: Find distance d
Simply plug in the 2 coordinates into the distance formula to find distance d
- Substitute in points [Distance Formula]: [tex]\displaystyle d = \sqrt{(5--2)^2+(0-1)^2}[/tex]
- [√Radical] (Parenthesis) Subtract: [tex]\displaystyle d = \sqrt{(7)^2+(-1)^2}[/tex]
- [√Radical] Evaluate exponents: [tex]\displaystyle d = \sqrt{49+1}[/tex]
- [√Radical] Add: [tex]\displaystyle d = \sqrt{50}[/tex]
- [√Radical] Simplify: [tex]\displaystyle d = 5\sqrt{2}[/tex]
Answer:
Solution given:
(x1,y1)=(-2,1)
(x2,y2)=(5,0)
distance=?
we have
distance=[tex] \sqrt{(x2-x1) ²+(y2-y1) ²}[/tex]
d=[tex] \sqrt{(5+2)²+(0-1) ²}[/tex]
d=[tex] \sqrt{50}[/tex]
d=[tex] 5\sqrt{2}[/tex]units
[tex] 5\sqrt{2}[/tex]units I travelled.