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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2 years ago

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1. A Babylonian tablet found in 1936 shows that the Babylonians sometimes used a more accurate value of π than 3. They determine

madreJ [45]

Answer:

Step-by-step explanation:

Area of a circle is given as πr² and its circumference is expressed as 2πr.

If the babylonians determined the area of a circle by taking it as equal to the square of the circle’s circumference then;

Area of circle = (circumference of a circle)²

πr² = (2πr)²

πr² = 4π²r²

Dividing both sides of the equation by πr² we have;

$\frac{\pi r^{2} }{\pi r^{2} } =\frac{4\pi^{2} r^{2} }{\pi r^{2} }\\1 = 4\pi \\\pi =\frac{1}{4}$

The value of π this yields is 1/4

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2 years ago