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Nezavi [6.7K]
2 years ago
5

2, en un curso de 37 estudiantes se proponen a elaborar un cohete con material reciclado, para poder hacer los cohetes, en el cu

rso se cuenta con 151 tubos de cartón que se recolectaron en una jornada de reciclable. A: si para hacer un cohete se necesitan 4 tubos de cartón, ¿Hay suficientes tubos para que cada estudiante elabore su cohete?
Mathematics
1 answer:
nlexa [21]2 years ago
7 0

Answer:

Yes, there enough tubes for each student to make their rocket.

Step-by-step explanation:

Given - In a course of 37 students they propose to make a rocket with recycled material, in order to make the rockets, in the course there are 151 cardboard tubes that were collected in a recyclable day.

To find - If 4 cardboard tubes are needed to make a rocket, are there enough tubes for each student to make their rocket ?

Proof -

Given that,

1 rocket is made with 4 cardboard tubes

Total cardboard tubes = 151

Now,

1 student will make 1 rocket

Total students = 37

⇒Total rockets made by the course = 37

Now,

1 rocket used 4 cardboard tubes

⇒37 rocket used 37×4 cardboard tubes

⇒37 rocket used 148 cardboard tubes.

As

148 < 151

⇒There are enough tubes for each student to make their rocket.

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<img src="https://tex.z-dn.net/?f=%20%20%5Crm%20%20%5Clim_%7Bk%20%5Cto%20%5Cinfty%20%7D%20%5Csqrt%5B%20%20k%5D%7B%20%5CGamma%20%
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We have

\sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)\right)}k\right) \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right)\right)+\ln\left( \Gamma\left(\dfrac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\dfrac kk\right)\right)}k\right)

and as k goes to ∞, the exponent converges to a definite integral. So the limit is

\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}

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