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marin [14]
3 years ago
13

Let f be defined by the function f(x) = 1/(x^2+9)

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
5 0

(a)

\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}

Substitute <em>x</em> = 3 tan(<em>t</em> ) and d<em>x</em> = 3 sec²(<em>t </em>) d<em>t</em> :

\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt

=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}

(b) The series

\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}

converges by comparison to the convergent <em>p</em>-series,

\displaystyle\sum_{n=3}^\infty\frac1{n^2}

(c) The series

\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}

converges absolutely, since

\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}

That is, ∑ (-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ converges absolutely because ∑ |(-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ| = ∑ (<em>n</em> ² + 9)/<em>e</em>ⁿ in turn converges by comparison to a geometric series.

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the mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.5 %. assume
zhuklara [117]
Given:
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Population standard deviation, σ  = 3.5% = 0.035
Sample size, n = 40
Confidence level = 95%

The standard error is
SE_{p} =  \sqrt{ \frac{p(1-p)}{n} } = \sqrt{ \frac{0.57*0.43}{40} } =0.0783

The confidence interval is 
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The sample proportion lies in the interval
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2 years ago
Two landscapers must mow a rectangular lawn that measures 100 feet by 200 feet. Each wants to mow no more than half of the lawn.
Citrus2011 [14]

The total area of the complete lawn is (100-ft x 200-ft) = 20,000 ft².
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When you think about it ... massage it and roll it around in your
mind's eye, and then soon give up and make yourself a sketch ...
you realize that if he starts along the length of the field, then with
a 2-ft cut, the lengths of the strips he cuts will line up like this:

First lap:
       (200 - 0) = 200
       (100 - 2) = 98
       (200 - 2) = 198
       (100 - 4) = 96    

Second lap:
       (200 - 4) = 196
       (100 - 6) = 94
       (200 - 6) = 194
       (100 - 8) = 92   

Third lap:
       (200 - 8) = 192
       (100 - 10) = 90
       (200 - 10) = 190
       (100 - 12) = 88 

These are the lengths of each strip.  They're 2-ft wide, so the area
of each one is (2 x the length). 

I expected to be able to see a pattern developing, but my brain cells
are too fatigued and I don't see it.  So I'll just keep going for another
lap, then add up all the areas and see how close he is:

Fourth lap:
       (200 - 12) = 188
       (100 - 14) = 86
       (200 - 14) = 186
       (100 - 16) = 84 

So far, after four laps around the yard, the 16 lengths add up to
2,272-ft, for a total area of 4,544-ft².  If I kept this up, I'd need to do
at least four more laps ... probably more, because they're getting smaller
all the time, so each lap contributes less area than the last one did.

Hey ! Maybe that's the key to the approximate pattern !

Each lap around the yard mows a 2-ft strip along the length ... twice ...
and a 2-ft strip along the width ... twice.  (Approximately.)  So the area
that gets mowed around each lap is (2-ft) x (the perimeter of the rectangle),
(approximately), and then the NEXT lap is a rectangle with 4-ft less length
and 4-ft less width.

So now we have rectangles measuring

         (200 x 100),  (196 x 96),  (192 x 92),  (188 x 88),  (184 x 84) ... etc.

and the areas of their rectangular strips are
           1200-ft², 1168-ft², 1136-ft², 1104-ft², 1072-ft² ... etc.

==> I see that the areas are decreasing by 32-ft² each lap.
       So the next few laps are 
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How much area do we have now:

             After 9 laps,    Area =   9,648-ft²
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And there you are ... Somewhere during the 10th lap, he'll need to
stop and call the company surveyor, to come out, measure up, walk
in front of the mower, and put down a yellow chalk-line exactly where
the total becomes 10,000-ft².   


There must still be an easier way to do it.  For now, however, I'll leave it
there, and go with my answer of:  During the 10th lap.

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