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Lorico [155]
3 years ago
13

Tell whether the sequence is arithmetic. If it is, what is the common difference?

Mathematics
1 answer:
Ilya [14]3 years ago
7 0
The answer is B. Yes; 4

Arithmetic sequence is the type of sequence of numbers which difference between each number will always remain constant
in the sequence above we could see the difference between one number with another is exactly 4:

15 + 4 = 19

19 + 4 = 23

23 + 4 = 27


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What is BC?<br><br> BC= units
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ΔXYZ was reflected to form ΔLMN.
goldfiish [28.3K]

ΔXYZ was reflected to form ΔLMN, hence ΔXYZ ≅ ΔLMN, ∠X ≅ ∠L and XZ ≅ LN

<h3>What is transformation?</h3>

Transformation is the movement of a point from its initial location to a new location. Types of transformation are <em>rotation, translation, reflection and dilation.</em>

Reflection is a rigid transformation, hence it produces congruent figures with congruent angles.

∠Y + ∠X + ∠Z = 180 (angle in a triangle)

86 + 38 + ∠Z = 180

∠Z = 56°

Also:

∠N + ∠M + ∠L = 180 (angle in a triangle)

86 + 56 + ∠L = 180

∠L = 38°

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Find out more on transformation at: brainly.com/question/4289712

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7 0
2 years ago
Explain why it does not matter what letter or symbol is used to find an unknown number
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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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1 year ago
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