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vlabodo [156]
3 years ago
5

Determine which relation is a function. A. Number graph ranging from negative four to four on the x and y axes. Points are drawn

on the graph at (negative four, three), (negative two, one), (negative one, three), (negative one, negative one), (one, one) and (two, negative one). B. Number graph ranging from negative four to four on the x and y axes. Points are drawn on the graph at (negative two, one), (negative two, negative two), (negative one, three), (zero, one), (two, zero) and (two, negative two). C. Number graph ranging from negative four to four on the x and y axes. Points are drawn on the graph at (negative two, one), (negative one, three), (zero, two), (zero, negative one), (one, three) and (two, one). D. Number graph ranging from negative four to four on the x and y axes. Points are drawn on the graph at (negative four, three), (negative three, negative one), (negative two, negative two), (negative one, zero), (two, zero) and (three, two).
Mathematics
1 answer:
Soloha48 [4]3 years ago
8 0
A) is not a function because you have two X values of -1

B) is not a function because you have two X values of -2

C) is not a function because you have two X values of 0

D) This is a function because all X values are different.
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Find the radius of convergence, then determine the interval of convergence
galben [10]

The radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series. This can be obtained by using ratio test.  

<h3>Find the radius of convergence R and the interval of convergence:</h3>

Ratio test is the test that is used to find the convergence of the given power series.  

First aₙ is noted and then aₙ₊₁ is noted.

For  ∑ aₙ,  aₙ and aₙ₊₁ is noted.

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = β

  • If β < 1, then the series converges
  • If β > 1, then the series diverges
  • If β = 1, then the series inconclusive

Here a_{k} = \frac{(x+2)^{k}}{\sqrt{k} }  and  a_{k+1} = \frac{(x+2)^{k+1}}{\sqrt{k+1} }

   

Now limit is taken,

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }/\frac{(x+2)^{k} }{\sqrt{k} }|

= \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }\frac{\sqrt{k} }{(x+2)^{k}}|

= \lim_{n \to \infty} |{(x+2) } }{\sqrt{\frac{k}{k+1} } }}|

= |{x+2 }|\lim_{n \to \infty}}{\sqrt{\frac{k}{k+1} } }}

= |{x+2 }| < 1

- 1 < {x+2 } < 1

- 1 - 2 < x < 1 - 2

- 3 < x < - 1

 

We get that,

interval of convergence = (-3, -1)

radius of convergence R = 1

Hence the radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series.

Learn more about radius of convergence here:

brainly.com/question/14394994

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