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daser333 [38]
2 years ago
11

Find the 9th term of the geometric sequence whose common ratio is 1/3 and whose first term is 2.

Mathematics
1 answer:
raketka [301]2 years ago
7 0

Answer:

2/6561

Step-by-step explanation:

Geometric sequence formula : a_n=a_1(r)^n^-^1

where an = nth term, a1 = first term , r = common ratio and n = term position

given ratio : 1/3 , first term : 2 , given this we want to find the 9th term

to do so we simply plug in what we are given into the formula

recall formula : a_n=a_1(r)^n^-^1

define variables : a1 = 2 , r = 1/3 , n = 9

plug in values

a9 = 2(1/3)^(9-1)

subtract exponents

a9 = 2(1/3)^8

evaluate exponent

a9 = 2 (1/6561)

multiply 2 and 1/6561

a9 = 2/6561

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Step-by-step explanation:

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2 years ago
A play was attended by 984 persons. 75% of them were adults how many adults attend
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3 years ago
Suppose that an object moves along the y-axis so that its location is y=x2+3x at time x. (Here y is in meters and x is in second
vfiekz [6]

Answer:

a) 13 m/s

b) (15  + h) m/s

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Step-by-step explanation:

if the location is

y=x²+3*x

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Δy/Δx=[y(7)-y(3)]/(7-3)=[7²+3*7- (3²+3*3)]/4= 13 m/s

then the average velocity from x=6 to to x=6+h

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the instantaneous velocity can be found taking the limit of Δy/Δx when h→0. Then

when h→0 , limit Δy/Δx= (15 + h) m/s = 15 m/s

then v= 15 m/s

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7 0
3 years ago
What does the A equal to in a + (-7) = 3
Korolek [52]
A = 10

Step by step

3+7=10
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7 0
2 years ago
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