# What is the Cartesian form of #(-64,(5pi)/12)#?

(If you had three coordinates, you would have to specify spherical or cylindrical coordinates.)

In polar coordinates, recall that

Now we can use the additive angle formulas

to get

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To convert the polar coordinates ((-64, \frac{5\pi}{12})) to Cartesian coordinates ((x, y)), you can use the following formulas:

[ x = r \cos(\theta) ] [ y = r \sin(\theta) ]

Where ( r ) is the radius (magnitude) and ( \theta ) is the angle in radians.

Given: [ r = -64 ] [ \theta = \frac{5\pi}{12} ]

Using the formulas:

[ x = (-64) \cos\left(\frac{5\pi}{12}\right) ] [ y = (-64) \sin\left(\frac{5\pi}{12}\right) ]

Calculating the values:

[ x = (-64) \cos\left(\frac{5\pi}{12}\right) ] [ x = (-64) \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) ] [ x = -16\sqrt{6} - 16\sqrt{2} ]

[ y = (-64) \sin\left(\frac{5\pi}{12}\right) ] [ y = (-64) \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) ] [ y = -16\sqrt{6} + 16\sqrt{2} ]

So, the Cartesian form of ((-64, \frac{5\pi}{12})) is ((-16\sqrt{6} - 16\sqrt{2}, -16\sqrt{6} + 16\sqrt{2})).

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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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