# What is the distance the polar coordinates #(-2 ,( -3 )/8 )# and #(6 ,(-7 pi )/4 )#?

P(-2, -3/8pi) or, with positive r, P(2. -3/8pi-pi) =#

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To find the distance between two points given in polar coordinates, you can use the formula:

[ d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_2 - \theta_1)} ]

Where:

- ( r_1 ) and ( r_2 ) are the magnitudes of the polar coordinates
- ( \theta_1 ) and ( \theta_2 ) are the angles of the polar coordinates

Given the polar coordinates (-2, (-3π)/8) and (6, (-7π)/4):

- ( r_1 = -2 ) and ( \theta_1 = \frac{-3\pi}{8} )
- ( r_2 = 6 ) and ( \theta_2 = \frac{-7\pi}{4} )

Plugging these values into the formula:

[ d = \sqrt{(-2)^2 + 6^2 - 2(-2)(6)\cos\left(\frac{-7\pi}{4} - \frac{-3\pi}{8}\right)} ]

[ d = \sqrt{4 + 36 - 2(-12)\cos\left(\frac{-7\pi}{4} + \frac{3\pi}{8}\right)} ]

[ d = \sqrt{40 - 24\cos\left(\frac{-7\pi}{4} + \frac{3\pi}{8}\right)} ]

[ d = \sqrt{40 - 24\cos\left(\frac{-14\pi + 3\pi}{8}\right)} ]

[ d = \sqrt{40 - 24\cos\left(\frac{-11\pi}{8}\right)} ]

[ d = \sqrt{40 - 24\left(\frac{\sqrt{2}}{2}\right)} ]

[ d = \sqrt{40 - 12\sqrt{2}} ]

[ d \approx \sqrt{40} - \sqrt{12}\sqrt{2} ]

[ d \approx 6.3246 - 3.4641 ]

[ d \approx 2.8605 ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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