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sergiy2304 [10]

2 years ago

8

In circle O, radius OQ measures 9 inches and arc PQ measures 6π inches. Circle O is shown. Line segments P O and Q O are radii w

ith length of 9 inches. Angle P O Q is theta. What is the measure, in radians, of central angle POQ? StartFraction 2 pi Over 3 EndFraction radians StartFraction 3 pi Over 4 EndFraction centimeters StartFraction 4 pi Over 3 EndFraction radians StartFraction 3 pi Over 2 EndFraction radians

2 answers:

kupik [55]2 years ago

5 0

The answer is A. 2pi/3 radians.

nikklg [1K]2 years ago

4 0

Answer:

First Option

Explanation:

A) 2pi/3 radians

Hope this helps :)

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marusya05 [52]

Answer:

See below

(B) and (C) are correct.

Step-by-step explanation:

We have the following limit

$\lim \limits_{n\rightarrow \infty} \left(\frac{n^n(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right)}{n!(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right)^{\dfrac{x}{n} }, \forall x>0$

I am not sure about methods concerning the quotient, but in this type of question I would try to convert this limit into integration.

Considering the numerator, we have

$(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right) = \prod_{k=1}^n \left(x+\dfrac{n}{k} \right)$

- I didn't forget about $n^n$

Considering the denominator, we have

$(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right) = \prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right)$

- I didn't forget about $n!$

Therefore,

$\left(\frac{n^n(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right)}{n!(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right)^{\dfrac{x}{n} } = \left(\dfrac{n^2 \prod_{k=1}^n \left(x+\dfrac{n}{k} \right)}{ n!\prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right)} \right)$

$= \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}$

Now we have

$\lim \limits_{n\rightarrow \infty} \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}, \forall x>0$

This is just the notation change so far.

What I want to do here is apply definite integrals using Riemann Integrals (We will write the limit as an definite integral). A nice way to do it is using logarithms. Therefore, we can apply the natural logarithm in both sides.

Now, recall two properties of logarithms:

$\boxed{\log_a mn = \log_a m + \log_a n}$

$\boxed{\log_a m^p = p\log_a m}$

$\boxed{\log_a \left(\dfrac{m}{n} \right) = \log_a m- \log_a n}$

Thus,

$\ln f(x) = \lim \limits_{n\rightarrow \infty} \ln \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n}\ln\left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[\ln \left(n^n \prod_{k=1}^n \left(x+\dfrac{n}{k} \right) \right)-\ln \left( n!\prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right) \right) \right]$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[\ln n^n + \prod_{k=1}^n \ln \left(x+\dfrac{n}{k} \right)-\ln n! -\prod_{k=1}^n \ln\left(x^2+\dfrac{n^2}{k^2} \right) \right]$

Considering

$\lim \limits_{n\rightarrow \infty} \frac{x}{n} (\ln n^n - \ln n! ) = \lim \limits_{n\rightarrow \infty} \frac{x}{n} (n\ln n - \ln n! )= \lim \limits_{n\rightarrow \infty} \frac{x \cdot\ln\frac{n^n}{n!} }{n}$

Using Stirling's formula

$\dfrac{n^n}{n!}\underset{\infty}{\sim} \dfrac{n^n}{\sqrt{2n \pi}\left(\dfrac{n}{e}\right)^n}=\dfrac{e^n}{\sqrt{2n \pi}}$

then

$\ln\left(\frac{n^n}{n!}\right)\underset{\infty}{=}n\ln\left(e\right)-\frac{1}{2}\ln\left(2n\pi\right)+o\left(1\right)$

$\implies \frac{\ln\left(\frac{n^n}{n!}\right)}{n}=1-\frac{\ln(2n\pi)}{2n}+o\left(1\right)$

This shows our limit equals 1 as $\frac{\log(2\pi n)}{2n} \rightarrow 0$ and $\ln(e)=1$

Employing a Riemann sum in the main limit, we have

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[ \sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right) - \sum_{k=1}^n\ln \left(x^2+\dfrac{n^2}{k^2} \right) \right]$

Now dividing the terms inside the parenthesis by $\dfrac{n}{k}$ in $\sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right)$

we have

$\sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right) = \sum_{k=1}^n \ln \left(\frac{kx}{n} +1\right)$

Now dividing the terms inside the parenthesis by $\dfrac{n^2}{k^2}$ in $\sum_{k=1}^n \ln \left(x^2+\dfrac{n^2}{k^2} \right)$

we have

$\sum_{k=1}^n \ln \left(x^2+\dfrac{n^2}{k^2} \right) = \sum_{k=1}^n \ln \left(\frac{(kx)^2}{n^2} +1\right)$

Therefore

$= \frac xn\sum_{k=1}^n\ln\dfrac{z+1}{z^2+1}$

for $\dfrac{kx}{n} = z$

Using Riemann Integral,

$\lim \limits_{n\rightarrow \infty} \int_0^1\ln\frac{z+1}{z^2+1}dz$

From

$\frac{f'(x)}{f(x)}=\ln\frac{z+1}{z^2+1}$

We can see that the function is increasing for , but because of the denominator, it is negative for .

Therefore,

(A) is false because $\dfrac{1}{2} < 1$

(B) is true because

(C) is true the slope is negative at that point

(D) is false, just consider $\ln\frac{z+1}{z^2+1}$ for $z=1$ and $z=2$

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Tcecarenko [31]

Answer:

326

Step-by-step explanation:

Add a nails: 65+40+45+176= The answer

6 0

2 years ago

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