Answer:
The flying distance between the greenhouse and the stadium = 5 units
Explanations:
The coordinates of the greenhouse: (-6, 0)
The coordinates of the stadium: (-2, 3)
The distance between two points of coordinates (x₁, y₁) and (x₂, y₂) is given as:
[tex]D\text{ = }\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
For the flying distance between the greenhouse and the stadium:
x₁ = -6, y₁ = 0, x₂ = -2, y₂ = 3
Substitute these values into the distance equation given above:
[tex]\begin{gathered} D\text{ = }\sqrt[]{(-2-(-6))^2+(3-0)^2} \\ D\text{ = }\sqrt[]{(-2+6)^2+3^2} \\ D\text{ = }\sqrt[]{4^2+3^2} \\ D\text{ = }\sqrt[]{16+9} \\ D\text{ = }\sqrt[]{25} \\ D\text{ = 5} \end{gathered}[/tex]
The flying distance between the greenhouse and the stadium = 5 units
