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bixtya [17]
2 years ago
11

Two pentagons are similar. the area of one of the Pentagon is 9 times that the other. determine the ratio of the length of the c

orresponding sides and the perimeter of the Pentagon
​
Mathematics
1 answer:
kolbaska11 [484]2 years ago
7 0
<h2><u><em>Answer:</em></u></h2><h2><u><em></em></u></h2><h2><u><em>The ratios of the lengths of the corresponding  sides of the two pentagons is 3:1</em></u></h2><h2><u><em></em></u></h2><h2><u><em>The ratios of the perimeters of the two pentagons is 3:1</em></u></h2><h2><u><em></em></u></h2><h2><u><em>Step-by-step explanation:</em></u></h2><h2><u><em></em></u></h2><h2><u><em>Let the first pentagon be X and the second pentagon be Y.</em></u></h2><h2><u><em></em></u></h2><h2><u><em>Ratio of area of the two pentagons is </em></u></h2><h2><u><em>9</em></u></h2><h2><u><em>Ratio of sides of the pentagon is equal to</em></u></h2><h2><u><em></em></u></h2><h2><u><em>3</em></u></h2><h2><u><em></em></u></h2><h2><u><em>Ratio of perimeters is equal to ration of sides</em></u></h2><h2><u><em></em></u></h2><h2><u><em>Hence,</em></u></h2><h2><u><em>The ratio of the perimeter is 3 as well</em></u></h2>

<u><em></em></u>

<em></em>

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Answer:

8

Step-by-step explanation:

Let's just call the number x for simplicity.

So, 7x is 8 less than x².

Putting this into an equation would look like this

x² - 8 = 7x

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x² - 7x -8 = 0

Now, we can proceed. To factor we first need to find the factors of -8.

The factors of -8 are

-2 ⋅ 4, -4 ⋅ 2, -1 ⋅ 8, 1 ⋅ -8.

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To find the solutions we will have to set them to 0 and solve each of these binomials individually.

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3 years ago
Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
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Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

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Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

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\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

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Vsevolod [243]
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