# How do you convert #3=(x+4y)^2+3x# into polar form?

Please see the explanation for a detailed description of how the conversion is done.

Expand the square, using the F.O.I.L. method:

This can be written as a quadratic in r:

Use the quadratic formula:

where:

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To convert the equation (3 = (x + 4y)^2 + 3x) into polar form, we use the conversion formulas (x = r \cos(\theta)) and (y = r \sin(\theta)).

Substitute these values into the equation and simplify:

(3 = ((r \cos(\theta)) + 4(r \sin(\theta)))^2 + 3(r \cos(\theta)))

(3 = (r \cos(\theta) + 4r \sin(\theta))^2 + 3r \cos(\theta))

Expand the squared term:

(3 = (r^2 \cos^2(\theta) + 8r^2 \cos(\theta) \sin(\theta) + 16r^2 \sin^2(\theta)) + 3r \cos(\theta))

Combine like terms:

(3 = r^2(\cos^2(\theta) + 8 \cos(\theta) \sin(\theta) + 16 \sin^2(\theta)) + 3r \cos(\theta))

Simplify the expression inside the parentheses using trigonometric identities:

(3 = r^2(\cos^2(\theta) + 2\cos(\theta) \sin(\theta) + \sin^2(\theta) + 7\sin^2(\theta) + 7\cos^2(\theta)) + 3r \cos(\theta))

(3 = r^2(1 + 7\sin^2(\theta) + 7\cos^2(\theta)) + 3r \cos(\theta))

Since (1 = \sin^2(\theta) + \cos^2(\theta)), we can simplify further:

(3 = r^2(1 + 7) + 3r \cos(\theta))

(3 = 8r^2 + 3r \cos(\theta))

So, the equation in polar form is (3 = 8r^2 + 3r \cos(\theta)).

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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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