Ask question

Ask question

All categories

3 years ago

10

I need to answer three questions about y = x^(-2) on the interval [-4, -1/2]. I have attached my questions and my answers as scr

eenshots:
Part 1 - ANSWER C
Part 2 - ANSWER A
Part 3 - ANSWER B

I am referring to each part in the order that I uploaded the screenshots. You will see that my answers are highlighted in blue. I like second opinions!

1 answer:

77julia77 [94]3 years ago

5 0

I wasn't sure how to order things, so I decided to write directly on top of your screenshots. Have a look at the three attachments. Two of which are correct. There is one incorrect answer though you're fairly close.

You might be interested in

Clarissa sells handmade bracelets for each. The graph below shows her profits after expenses.

ser-zykov [4K]

Answer:

It represents how many bracelets she sold and the profit she makes from it.

3 0

3 years ago

‎‎‎‎‎‎‎‎‎‎ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Rasek [7]

Answer:

ok

Step-by-step explanation:

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%7B%5Chuge%7B%5Cunderline%7B%5Csmall%7B%5Cmathbb%7B%5Cpink%7BREFER%20%5C%20TO%20%5C%20THE%20%5

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 $z=2$

7 0

2 years ago

If A and B are independent events with P(A)=0.1 and P(B)=0.8, find P(A AND B).Give your answer as a decimal rounded to two decim

astra-53 [7]

Answer:

P(A and B)= P(B) P(A*B). = 0.3 * 0.4. = 0.12 g. If A and B are independent. (i hope that this helps :p)

Step-by-step explanation:

7 0

3 years ago

Twice a number increased by 12 is equal to 31 less than three times the number. find the number

dmitriy555 [2]

Step-by-step explanation:

12x31

do that and you get the answer

7 0

3 years ago