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Elodia [21]
2 years ago
12

El triángulo ABC es equilátero y L, M y N son los puntos medios de BC, AB y CA respectivamente. Si MN = 3, ¿cuál es el valor de

ML?

Mathematics
1 answer:
Flauer [41]2 years ago
4 0

The value of <u>ML = 3</u>, using the mid-point theorem of triangles.

According to the midpoint theorem, "the line segment of a triangle crossing the midpoints of two sides of the triangle is said to be parallel to its third side and also half the length of the third side."

In the question, we are given that triangle ABC is an equilateral triangle, and L, M, and N are the midpoints of BC, AB, and CA respectively.

Thus, by the midpoint theorem, we can say that:

  • MN || BC, and MN = (1/2)BC,
  • ML || AC, and ML = (1/2)AC, and
  • NL || AB, and NL = (1/2)AB.

Assuming AB = BC = AC = x units, we get:

  • MN = (1/2)BC = x/2,
  • ML = (1/2)AC = x/2, and
  • NL = (1/2)AB = x/2.


Thus, the triangle LMN is an equilateral triangle.

Thus, MN = ML = NL.

Given MN = 3, we can write the value of ML = 3.

Thus, the value of <u>ML = 3</u>, using the mid-point theorem of triangles.

Learn more about the mid-point theorem of triangle at

brainly.com/question/9635025

#SPJ1

The given question is in Spanish. The question in English is:

"Triangle ABC is equilateral and L, M, and N are the midpoints of BC, AB, and CA respectively. If MN = 3, what is the value of ML?"

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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

#SPJ1

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