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Alik [6]
3 years ago
8

Help on horizontal asymptote!

Mathematics
2 answers:
user100 [1]3 years ago
7 0

Answer:

A. None

Step-by-step explanation:

They are no horizontal asymptotes. I get helped from another website when comes to graphs and looking for asymptotes. Good luck on your exam!

Westkost [7]3 years ago
5 0

Answer:  A. none

<u>Step-by-step explanation:</u>

There are three rules when finding the horizontal asymptote (H.A.)

Consider "m" as the degree of the numerator and "n" as the degree of the denominator.

Rules:

  1. m > n: no H. A (use long division to find the slant asymptote)
  2. m = n: H.A. is y = leading coefficient of m ÷ leading coefficient of n
  3. m < n: H.A. is y = 0

In the equation given, the numerator is x⁵ + ... and the denominator is x⁴ + ...

m = 5, n = 4

m > n so there is no H.A

<em>but you can use long division to find the Slant Asymptote.</em>

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Step-by-step explanation:

\large\underline{\sf{Solution-}}

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\rm :\longmapsto\:\dfrac{sin\theta }{1 + cos\theta }

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\rm \:  =  \: \dfrac{sin\theta }{1 + cos\theta }  \times \dfrac{1 - cos\theta }{1 - cos\theta }

We know,

\purple{\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) =  {x}^{2} -  {y}^{2} \: }}}

So, using this, we get

\rm \:  =  \: \dfrac{sin\theta (1  -  cos\theta )}{1 -  {cos}^{2}\theta  }

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\purple{\rm :\longmapsto\:\boxed{\tt{  {sin}^{2}x +  {cos}^{2}x = 1}}}

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\rm \:  =  \: \dfrac{sin\theta (1 -  cos\theta )}{{sin}^{2}\theta  }

\rm \:  =  \: \dfrac{1 - cos\theta }{sin\theta }

<u>Hence, </u>

\\ \red{\rm\implies \:\boxed{\tt{ \rm \:\dfrac{sin\theta }{1 + cos\theta }   =  \: \dfrac{1 - cos\theta }{sin\theta } }}} \\

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