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3 years ago

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Two intersecting segments form the diagonals of a quadrilateral. What additional information do you need to

1 answer:

Aleksandr-060686 [28]3 years ago

7 0

Answer:

B. They must bisect each other

Step-by-step explanation:

Found the answer on quizziz

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Determine whether the series is convergent or divergent by expressing sn as a telescoping sum.

Goshia [24]

Answer:

The sum converges at: $\frac{10}{3}$

Step-by-step explanation:

Given

$\sum\limits^{\infty}_{n =2} \frac{8}{n^2 - 1}$

Express the denominator as difference of two squares

$\sum\limits^{\infty}_{n =2} \frac{8}{(n - 1)(n+1)}$

Express 8 as 4 * 2

$\sum\limits^{\infty}_{n =2} \frac{4 * 2}{(n - 1)(n+1)}$

Rewrite as:

$4 * \sum\limits^{\infty}_{n =2} \frac{2}{(n - 1)(n+1)}$

Express 2 as 1 + 1 + 0

$4 * \sum\limits^{\infty}_{n =2} \frac{1+1+0}{(n - 1)(n+1)}$

Express 0 as n - n

$4 * \sum\limits^{\infty}_{n =2} \frac{1+1+n - n}{(n - 1)(n+1)}$

Rewrite as:

$4 * \sum\limits^{\infty}_{n =2} \frac{(n + 1)-(n - 1)}{(n - 1)(n+1)}$

Split

$4 * \sum\limits^{\infty}_{n =2} \frac{(n + 1)}{(n - 1)(n+1)}-\frac{(n - 1)}{(n - 1)(n+1)}$

Cancel out like terms

$4 * \sum\limits^{\infty}_{n =2} \frac{1}{(n - 1)}-\frac{1}{(n+1)}$

In the above statement, we have:

$a_3 + a_5 = 4[(\frac{1}{2} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{6})]$

$a_3 + a_5 = 4[(\frac{1}{2} - \frac{1}{6})]$

Add $a_7$

$a_3 + a_5 + a_7= 4[(\frac{1}{2} - \frac{1}{6}) + (\frac{1}{7 - 1} - \frac{1}{7+1})]$

$a_3 + a_5 + a_7= 4[(\frac{1}{2} - \frac{1}{6}) + (\frac{1}{6} - \frac{1}{8})]$

$a_3 + a_5 + a_7= 4[(\frac{1}{2} - \frac{1}{8})]$

Notice that the pattern follows:

$a_3 + a_5 + a_7 + ...... + a_{k}= 4[(\frac{1}{2} - \frac{1}{k+1})]$

The above represent the odd sums (say S1)

For the even sums, we have:

$4 * \sum\limits^{\infty}_{n =2} \frac{1}{(n - 1)}-\frac{1}{(n+1)}$

In the above statement, we have:

$a_4 + a_6 = 4[(\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7})]$

$a_4 + a_6 = 4[(\frac{1}{3} - \frac{1}{7})]$

Add $a_8$ to both sides

$a_4 + a_6 +a_8 = 4[(\frac{1}{3} - \frac{1}{7}) + \frac{1}{7} - \frac{1}{9}]$

$a_4 + a_6 +a_8 = 4[\frac{1}{3} - \frac{1}{9}]$

Notice that the pattern follows:

$a_4 + a_6 + a_8 + ...... + a_{k}= 4[(\frac{1}{3} - \frac{1}{k+1})]$

The above represent the even sums (say S2)

The total sum (S) is:

$S = S_1 + S_2$

$S =4[(\frac{1}{2} - \frac{1}{k+1})] + 4[(\frac{1}{3} - \frac{1}{k+1})]$

Remove all k terms

$S =4[(\frac{1}{2}] + 4[(\frac{1}{3}]$

Open bracket

$S =\frac{4}{2} + \frac{4}{3}$

$S =\frac{12 + 8}{6}$

$S =\frac{20}{6}$

$S =\frac{10}{3}$

<em>The sum converges at: </em> $\frac{10}{3}$

7 0

3 years ago

What is the surface area of the following cylinder in square centimeters? 25 mm 1 cm 471 cm 2 2 31.4 cm 47.1 cm 2 54.95 cm 2

Fittoniya [83]

Answer:

54.95 cm 2.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Write Three equations that have the same solution as 25x + (-15) = -85<br> Pls answer

Bumek [7]

Answer:

275

Step-by-step explanation:

25x15=375-15=360-85=274

4 0

3 years ago

I cant figure out the answer for this lol

sveta [45]

Probably d? ........

8 0

3 years ago

PLS I NEED HELP THIS IS THE LAST QUESTION FOR MY GEOMETRY WORKSHEET ANYTHING IS APPERCIATED

mr_godi [17]

Answer:

4.64

Step-by-step explanation:

X^2= r/s

8 0

3 years ago