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Ksenya-84 [330]
3 years ago
15

Find the series shown.

Mathematics
1 answer:
solmaris [256]3 years ago
8 0

Answer:

C 11+13+15+17+...

Step-by-step explanation:

Consider the series

\sum\limits_{n=3}^{\infty}(2n+5)

The nth term of series is a_n=2n+5

The bottom index tells you that n starts changing from 3, so

a_3=2\cdor 3+5=11\\ \\a_4=2\cdot 4+5=13\\ \\a_5=2\cdot 5+5=15\\ \\a_6=2\cdot 6+5=17\\ \\...

Thus, the sum of all terms is

11+13+15+17+...

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Given: KLMN is a trapezoid m∠K = 90°, m∠N = 45° LK = LM = 10 Find: KN, Area of KLMN
Ilya [14]
Let KLMN be a trapezoid (see added picture). From the point M put down the trapezoid height MP, then quadrilateral KLMP is square and KP=MP=10.
A triangle MPN is right and <span>isosceles, because
</span>m∠N=45^{0}, m∠P=90^{0}, so m∠M=180^{0}-45^{0}-90^{0}=45^{0}.Then PN=MP=10.
The ttapezoid side KN consists of two parts KP and PN, each of them is equal to 10, then KN=20 units.
Area of KLMN is egual to A= \frac{LM+KN}{2} *MP= \frac{10+20}{2} *10=150 sq. units.


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