# What is the distance between the following polar coordinates?: # (5,(3pi)/4), (3,(11pi)/8) #

The distance is approximately

Here we have:

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To find the distance between two polar coordinates ( (r_1, \theta_1) ) and ( (r_2, \theta_2) ), we use the formula:

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

Given the polar coordinates ( (5, \frac{3\pi}{4}) ) and ( (3, \frac{11\pi}{8}) ), we substitute the values into the formula:

[ d = \sqrt{5^2 + 3^2 - 2(5)(3)\cos\left(\frac{11\pi}{8} - \frac{3\pi}{4}\right)} ]

Solving inside the cosine function:

[ \frac{11\pi}{8} - \frac{3\pi}{4} = \frac{22\pi - 12\pi}{16} = \frac{10\pi}{16} = \frac{5\pi}{8} ]

Substituting back into the formula:

[ d = \sqrt{25 + 9 - 30\cos\left(\frac{5\pi}{8}\right)} ]

Now, calculate ( \cos\left(\frac{5\pi}{8}\right) ) using a calculator, then substitute it into the formula:

[ d = \sqrt{25 + 9 - 30 \cdot \cos\left(\frac{5\pi}{8}\right)} ]

[ d = \sqrt{25 + 9 - 30 \cdot \cos\left(\frac{5\pi}{8}\right)} ]

[ d = \sqrt{25 + 9 - 30 \cdot \left(-\frac{\sqrt{2+\sqrt{2}}}{2}\right)} ]

[ d = \sqrt{34 + 15\sqrt{2+\sqrt{2}}} ]

This is the distance between the given polar coordinates.

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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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