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earnstyle [38]
3 years ago
14

How to find vertical asymptotes of cotangent function?

Mathematics
1 answer:
Wittaler [7]3 years ago
3 0

Vertical asypmtotes are line of the form x = k, where k is a point not in the domain of the function.

Since the definition of the cotangent function is

\cot(x) = \dfrac{\cos(x)}{\sin(x)}

The function is not defined where the denominator is zero, i.e.

\sin(x) = 0 \iff x = k\pi,\ k \in \mathbb{Z}

So, every line with equation x = k\pi is a vertical asymptote for the cotangent function. Some examples may be

x = 0,\ x = \pi,\ x = -3\pi,\ x = 152\pi,\ x = -234\pi,\ldots

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Which are the solutions of x2 = –5x + 8?
Nostrana [21]

Answer:

As exact answers:

x=\frac{-5-\sqrt{57} }{2}  and  x=\frac{-5+\sqrt{57} }{2}

As decimal answers:

x = -6.2749172 ≈-6.3

x = 1.27491722 ≈  1.3

Step-by-step explanation:

For quadratic equations, you can use the quadratic formula. Rearrange the equation to standard from, which is ax² + bx + c = 0.

x² = –5x + 8

x² + 5x - 8 = 0

State the values for "a", "b" and "c",

a = 1; b = 5; c = -8

x=\frac{-b±\sqrt{b^{2}-4ac} }{2a}            (Ignore the Â, it's a formatting error)

x=\frac{-5±\sqrt{5^{2}-4(1)(-8)} }{2(1)}           Simplify

x=\frac{-5±\sqrt{25-(-32)} }{2}               Two negatives make a positive

x=\frac{-5±\sqrt{57} }{2}  

Split the equation at the ± for plus and minus:

x=\frac{-5+\sqrt{57} }{2}  = 1.27491722 ≈ 1.3

x=\frac{-5-\sqrt{57} }{2}  = -6.2749172 ≈ -6.3

Therefore the solutions are 1.3 and -6.3.

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Answer:

k = 1.75

Step-by-step explanation:

5 - 4k = -7

k = 1.75, because if you subtract 4 from both sides, you are left with 1.75 = -7.

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1 year ago
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Answer:

are you sure this is high school level stuff? its 9% of 100

Step-by-step explanation:

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3 years ago
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