Respuesta :
The ratio RS:ST is 2:3.
We will use the distance formula:
[tex]d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]
The distance from R to S is:
[tex]d=\sqrt{(-5--3)^2+(-6--2)^2} \\ \\=\sqrt{(-5+3)^2+(-6+2)^2} \\ \\=\sqrt{(-2)^2+(-4)^2} \\ \\=\sqrt{4+16}=\sqrt{20}=\sqrt{4*5}=2\sqrt{5}[/tex]
The distance from S to T is:
[tex]d=\sqrt{(-3-0)^2+(-2-4)^2} \\ \\=\sqrt{(-3)^2+(-6)^2}=\sqrt{9+36}=\sqrt{45} \\ \\=\sqrt{9*5}=3\sqrt{5}[/tex]
The ratio of RS to ST is then
[tex]\frac{2\sqrt{5}}{3\sqrt{5}}[/tex]
Since √5 cancels out on the top and bottom, we are left with
2/3 = 2:3
We will use the distance formula:
[tex]d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]
The distance from R to S is:
[tex]d=\sqrt{(-5--3)^2+(-6--2)^2} \\ \\=\sqrt{(-5+3)^2+(-6+2)^2} \\ \\=\sqrt{(-2)^2+(-4)^2} \\ \\=\sqrt{4+16}=\sqrt{20}=\sqrt{4*5}=2\sqrt{5}[/tex]
The distance from S to T is:
[tex]d=\sqrt{(-3-0)^2+(-2-4)^2} \\ \\=\sqrt{(-3)^2+(-6)^2}=\sqrt{9+36}=\sqrt{45} \\ \\=\sqrt{9*5}=3\sqrt{5}[/tex]
The ratio of RS to ST is then
[tex]\frac{2\sqrt{5}}{3\sqrt{5}}[/tex]
Since √5 cancels out on the top and bottom, we are left with
2/3 = 2:3
Answer:
Which of the following coordinates is the solution to this system? The coordinate grid below represents a system of linear equations.A.(0, 3)B.(0, -3)C.(3, -2)D.(-3, 2
Step-by-step explanation:
The answer is A Because the plane is 3/0 wide and the length is way longer then that so its 3,0
