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3 years ago

Find the sum of the geometric series.

Mathematics

1 answer:

sveticcg [70]3 years ago

7 0

Answer:

40/3 option B

Step-by-step explanation:

20-10+5-5/2+ ...

We need to find the sum of infinite geometric series

$S_n = \frac{a}{1-r}$

a is the first term (20)

r is the common ratio

To find common ratio we divide second term by first term

r= $\frac{-10}{20} = -\frac{1}{2}$

now we plug in the values

$S_n = \frac{20}{1-(-\frac{1}{2}}$

$S_n = \frac{20}{1+\frac{1}{2}}$

$S_n = \frac{20}{\frac{3}{2}}$

$S_n = 20 * \frac{2}{3}$

sum = 40/3

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The sum of three consecutive odd integers is 17 less than four times the smallest integer. Find the integers.

lisabon 2012 [21]

Answer:

23, 25, 27

Step-by-step explanation:

n + (n+2) + (n+4) = 4n - 17

3n + 6 = 4n - 17

3n - 4n = -17 - 6

-n = -23

n = 23

therefore,

23, 25, 27 are the 3 consecutive odd integers

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2 years ago

2*10^3 in scientific notation

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3 years ago

Find the degree of the monomial 3x4y3

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2 years ago

Please solve this<br> (Rational numbers 8th grade)

pochemuha

Question # 1

Answer:

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}$

Step-by-step explanation:

Given the expression

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}$

$=-\frac{2}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}$ ∵ $\frac{-2}{3}\times \frac{3}{5}=-\frac{2}{5}$

$=-\frac{2}{5}+\frac{1}{4}$ ∵ $\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=\frac{1}{4}$

$\mathrm{Least\:Common\:Multiplier\:of\:}5,\:4:\quad 20$

$=-\frac{8}{20}+\frac{5}{20}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-8+5}{20}$

$\mathrm{Add/Subtract\:the\:numbers:}\:-8+5=-3$

$=\frac{-3}{20}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{3}{20}$

Therefore,

$\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}\times \frac{3}{5}\times \frac{1}{6}=-\frac{3}{20}$

Question # 2

Answer:

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}$

Step-by-step explanation:

Given

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}$

$=-\frac{6}{35}-\frac{1}{6}\times \frac{3}{2}\times \frac{2}{5}\times \frac{1}{14}$ ∵ $\frac{2}{5}\times \frac{-3}{7}=-\frac{6}{35}$

$=-\frac{6}{35}-\frac{1}{140}$ ∵ $\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=\frac{1}{140}$

$\mathrm{Least\:Common\:Multiplier\:of\:}35,\:140:\quad 140$

$=-\frac{24}{140}-\frac{1}{140}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-24-1}{140}$

$\mathrm{Subtract\:the\:numbers:}\:-24-1=-25$

$=\frac{-25}{140}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{25}{140}$

$\mathrm{Cancel\:the\:common\:factor:}\:5$

$=-\frac{5}{28}$

Therefore,

$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}\times \frac{1}{14}\times \frac{2}{5}=-\frac{5}{28}$

4 0

3 years ago

The difference between 22 and x

lukranit [14]

Answer:

22-x

Step-by-step explanation:

you dont know what x is, so you would just do 22-x

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2 years ago