Answer:
Step-by-step explanation:
The formula for the length of a vector/line in your case.
![L = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{[4 - (-1)]^2 + [2 -(-3)]^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}](https://tex.z-dn.net/?f=L%20%3D%20%5Csqrt%7B%28x_2-x_1%29%5E2%20%2B%20%28y_2-y_1%29%5E2%7D%20%3D%20%5Csqrt%7B%5B4%20-%20%28-1%29%5D%5E2%20%2B%20%5B2%20-%28-3%29%5D%5E2%7D%20%3D%20%5Csqrt%7B5%5E2%20%2B%205%5E2%7D%20%3D%20%5Csqrt%7B50%7D%20%3D%205%5Csqrt%7B2%7D)
Answer:
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Step-by-step explanation:
Answer:
4/3
Step-by-step explanation:
The tangent of any angle (θ) in standard position that has point (x, y) on its terminal ray is ...
tan(θ) = y/x
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For the given point on the terminal side, the tangent is ...
tan(θ) = (-4)/(-3) = 4/3
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<em>Additional comment</em>
There are several ways this can be explained. One of them makes use of the relation between rectangular and polar coordinates:
(x, y) = (r·cos(θ), r·sin(θ))
Then the ratio y/x is ...
y/x = (r·sin(θ))/(r·cos(θ)) = sin(θ)/cos(θ) = tan(θ)
Answer:
About what?
Step-by-step explanation:
