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Montano1993 [528]
3 years ago
8

The region bounded by the given curve is rotated about the specified axis. find the volume v of the resulting solid by any metho

d.
x2 + (y - 22 = 4; about the y-axis
Mathematics
1 answer:
Blababa [14]3 years ago
4 0
x^2+(y-2)^2=4\implies y=2\pm\sqrt{4-x^2}

describes a circle of radius 2 centered at (0,2). A revolution of this region about the y-axis would give you a sphere of radius 2, so the volume would be \dfrac43\pi\times2^3=\dfrac{32}3\pi.
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Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4
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Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4}

if \lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b then y = b is horizontal asymptote

Then:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

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Answer:

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In order to find this, start with the base form of point-slope form.

y - y1 = m(x - x1)

Now input the slope for m and the point for (x1, y1)

y - -3 = 1/2(x - 6)

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