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3 years ago

19 13 15 6 17 21 16 18 11 10 11 16 20 6 18 12 6 20 13 13 7 20 14 21 11 3 9 18

Mathematics

2 answers:

hoa [83]3 years ago

8 0

Answer:

13.08

Step-by-step explanation:

19 +13+ 15+ 6+ 17+ 21+ 16+ 18+ 11+ 10+ 11+ 16+ 20+ 6 +18 +12 +6 +20+ 13+ 13+ 7+ 20+ 14+ 21+ 11+ 3 +9 +18+12 +14+ 6 +11 +5 +18+ 8

485/35

13.08

Assoli18 [71]3 years ago

5 0

$19+13 +15 +6 +17 +21 +16 +18 +11+10+ 11 +16 +20+ 6 +18 +12 +6+ 20+ 13 +13+ 7+ 20+ 14 +21 11+3+9 +18+12+ 14+ 6 +11 +5 +18 +8=458\\458/35=13.08571429\\$

rounded to the nearest hundredth, it is 13.09

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Answer:

No solution

Step-by-step explanation:

24a-22= -4(1-6a)

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0-22= -4

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2 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

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Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

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$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

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