Question

On your answer sheet, answer the following questions:
1. What type of song is "Leron, Leron Sinta" and "Pamulinawen"?
2. What type of song is "Brother John"
3. What is the texture of partner and round songs?

pasagot po ang pamilian po ay monophonic,polyphonic,and homophonic

I be brainlist ko po ang masagot plss ​

Answer 1

$\large \pmb{\bf{\underline{\gray{Solution :-}}}}$

$\sf {Assume\:\:\displaystyle \lim_{x\to\infty}\left ( \dfrac{x!}{x} \right )^{\dfrac{1}{x}} = L}$

Take log both sides, we get

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( \dfrac{x!}{x} \right )}$

$\pmb{\sf{\gray{ Put\ value\ of\ x! }}}$

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( \dfrac{x(x-1)!}{x} \right )}$

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( (x-1)! \right )}$

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} \dfrac{ln \left ( (x-1)! \right )}{x}}$

$\pmb{\tt{Multiplying \ with\ ( x - 1 )\ in\ numerator\ and\ denominator\, we\ get\ }}$

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} (x-1)\dfrac{ln \left ( (x-1)! \right )}{x(x-1)}}$

$\pmb{\sf{We\ know\ that}}$

$\sf {\displaystyle \lim_{x\to\infty} \dfrac{ln \left ( x! \right )}{x}=\infty}$

$\sf {ln(L)=(\infty) \displaystyle \lim_{x\to\infty} \dfrac{x-1}{x}}$

$\sf {ln(L)=(\infty) \displaystyle \lim_{x\to\infty} \left(1-\dfrac{1}{x}\right)}$

Put value of limits,

$\sf {ln(L)=(\infty) \left(1-\dfrac{1}{\infty}\right)}$

$\sf {ln(L)=(\infty) \left(1-0\right)}$

$\sf {ln(L)=\infty}$

$\sf{L=e^{\infty}}$

$\sf{L=\infty}$

Answer 2

Answer:

Assume

x→∞

lim

(

x

x!

)

x

1

=L

Take log both sides, we get

\sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( \dfrac{x!}{x} \right )}ln(L)=

x→∞

lim

(

x

1

)ln(

x

x!

)

\pmb{\sf{\gray{ Put\ value\ of\ x! }}}

Put value of x!

Put value of x!

\sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( \dfrac{x(x-1)!}{x} \right )}ln(L)=

x→∞

lim

(

x

1

)ln(

x

x(x−1)!

)

\sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( (x-1)! \right )}ln(L)=

x→∞

lim

(

x

1

)ln((x−1)!)

\sf {ln(L)=\displaystyle \lim_{x\to\infty} \dfrac{ln \left ( (x-1)! \right )}{x}}ln(L)=

x→∞

lim

x

ln((x−1)!)

\pmb{\tt{Multiplying \ with\ ( x - 1 )\ in\ numerator\ and\ denominator\, we\ get\ }}

Multiplying with (x−1) in numerator and denominatorwe get

Multiplying with (x−1) in numerator and denominatorwe get

\sf {ln(L)=\displaystyle \lim_{x\to\infty} (x-1)\dfrac{ln \left ( (x-1)! \right )}{x(x-1)}}ln(L)=

x→∞

lim

(x−1)

x(x−1)

ln((x−1)!)

\pmb{\sf{We\ know\ that}}

We know that

We know that

\sf {\displaystyle \lim_{x\to\infty} \dfrac{ln \left ( x! \right )}{x}=\infty}

x→∞

lim

x

ln(x!)

=∞

\sf {ln(L)=(\infty) \displaystyle \lim_{x\to\infty} \dfrac{x-1}{x}}ln(L)=(∞)

x→∞

lim

x

x−1

\sf {ln(L)=(\infty) \displaystyle \lim_{x\to\infty} \left(1-\dfrac{1}{x}\right)}ln(L)=(∞)

x→∞

lim

(1−

x

1

)

Put value of limits,

\sf {ln(L)=(\infty) \left(1-\dfrac{1}{\infty}\right)}ln(L)=(∞)(1−

∞

1

)

\sf {ln(L)=(\infty) \left(1-0\right)}ln(L)=(∞)(1−0)

\sf {ln(L)=\infty}ln(L)=∞

\sf{L=e^{\infty}}L=e

∞

\sf{L=\infty}L=∞.

Explanation:

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