549 vexhicals on 9 acres

Tara and Jodi's bedroom are shaped like rectangles Tara bedroom is 9 feet long and 8 feet wide Jody's bedroom is 7 ft long and 1

brilliants [131]

Tara has a bigger bedroom since 9x8 =72 and ten x seven =seventy

7 0

3 years ago

Read 2 more answers

More series issues!

choli [55]

It can't be that $p$ makes the series converge, because this would introduce a zero in the denominator when $n=1$ . For a similar reason, $p=0$ would involve an indeterminate term of $0^0$ .

That leaves checking what happens when $p>0$ . First, consider the function

$f(x)=\dfrac{(\ln x)^p}x$

and its derivative

$f'(x)=\dfrac{p(\ln x)^{p-1}-(\ln x)^p}{x^2}=\dfrac{(\ln x)^{p-1}}{x^2}(p-\ln x)$

$f(x)$ has critical points at $x=1$ and $x=e^p$ . (These never coincide because we're assuming $p>0$ , so it's always the case that $e^p>1$ .) Between these two points, say at $c=\dfrac{e^p}2$ , you have $f'(c)=\dfrac{4\ln2}{e^{2p}}(\ln2)^{p-1}$ , which is positive regardless of the value of $p$ . This means $f(x)$ is increasing on the interval $(1,e^p)$ .

Meanwhile, if $x>e^p$ - and let's take $c=2e^p$ as an example - we have $f'(c)=\dfrac{(\ln2+p)^{p-1}}{4e^{2p}}(-\ln2)^{p-1}$ , which is negative for all $p>0$ . This means $f(x)$ is decreasing for all $x>e^p$ , which shows that $\dfrac{(\ln n)^p}n$ is a decreasing sequence for all $n>N$ , where $N$ is any sufficiently large number that depends on $p$ .

Now, it's also the case that for $p>0$ (and in fact all $p\in\mathbb R$ ),

$\displaystyle\lim_{n\to\infty}\dfrac{(\ln n)^p}n=0$

So you have a series of a sequence that in absolute value is decreasing and converging to 0. The alternating series then says that the series must converge for all $p>0$ .

For the second question, recall that

$h_n=\displaystyle\sum_{k=1}^n\frac1k=1+\frac12+\cdots+\frac1{n-1}+\frac1n$
$s_n=\displaystyle\sum_{k=1}^n\frac{(-1)^{k-1}}k=1-\frac12+\cdots-\frac1{n-1}+\frac1n$

(note that the above is true for even $n$ only - it wouldn't be too difficult to change things around if $n$ is odd)

It follows that

$h_{2n}=\displaystyle\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k=1+\frac12+\cdots+\frac1{2n-1}+\frac1{2n}$
$s_{2n}=\displaystyle\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k=1-\frac12+\cdots+\frac1{2n-1}-\frac1{2n}$

Subtracting $h_{2n}$ from $s_{2n}$ , you have

$\displaystyle s_{2n}-h_{2n}=(1-1)+\left(-\frac12-\frac12\right)+\left(\frac13-\frac13\right)+\left(-\frac14-\frac14\right)+\cdots+\left(\frac1{2n-1}-\frac1{2n-1}\right)+\left(-\frac1{2n}-\frac1{2n}\right)$
$s_{2n}-h_{2n}=-1-\dfrac12-\cdots-\dfrac2{2n}$
$s_{2n}-h_{2n}=-\left(1+\dfrac12+\cdots+\dfrac1n\right)$
$s_{2n}-h_{2n}=-h_n$
$\implies s_{2n}=h_{2n}-h_n$

as required. Notice that assuming $n$ is odd doesn't change the result; the last term in $h_{2n}$ ends up canceling with the corresponding term in $s_{2n}$ regardless of the parity of $n$ .

5 0

4 years ago

A publisher shipped 15 boxes of books to a bookstore each box contained 32 books how many books did the publisher ship to the bo

viktelen [127]

480 books were shipped to the bookstore

$15 \times 32 = 480$

4 0

4 years ago

What on a number line are the fractions before 1

AVprozaik [17]

The most used fractions on a number line used before 1 are 1/4, 1/2, 1/3, 2/4, 2/3, and 3/4.

4 0

4 years ago

A local hamburger shop sold a combined total of 575 hamburgers and cheeseburgers on Sunday. There were 75 fewer cheeseburgers so

lara31 [8.8K]

Answer:

250 Hamburgers

Step-by-step explanation:

575 is total

Hamburger is x

Cheesburgers is (x-75) as it is 75 less than the number of hamburgers

x + (x-75) = 575

2x -75 = 575

2x = 500

x = 250

<em>Feel free to mark this as brainliest! :D</em>

5 0

3 years ago