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Olegator [25]
3 years ago
14

Brainliest answer

Mathematics
1 answer:
Flura [38]3 years ago
5 0

A geometric sequence is a sequence of numbers that follows a pattern where the next term is found by multiplying by a constant called the common ratio, q:

a_n=a_{n-1}\cdot q=(a_{n-2}\cdot q)\cdot q=a_{n-2}\cdot q^2=...,\\ a_n=a_1\cdot q^{n-1}.

Given the functions f(n)=11 and g(n)=\left(\dfrac{3}{4}\right)^{n-1}, you can consider a_n=11\cdot \left(\dfrac{3}{4}\right)^{n-1} as n-th term of the geometric sequence with first term a_1=11 and common ratio q=\dfrac{3}{4}.

For n=9,

a_9=11\cdot \left(\dfrac{3}{4}\right)^{9-1} =11\cdot \left(\dfrac{3}{4}\right)^8=1.101.

Answer: the correct choice is B.

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3 years ago
The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.
Taya2010 [7]
Given a solution y_1(x)=\ln x, we can attempt to find a solution of the form y_2(x)=v(x)y_1(x). We have derivatives

y_2=v\ln x
{y_2}'=v'\ln x+\dfrac vx
{y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}

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w'x\ln x+(2+\ln x)w=0

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w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0

and noting that

\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x

we can write the ODE as

\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0

Integrating both sides with respect to x, we get

wx(\ln x)^2=C_1
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Now solve for v:

v'=\dfrac{C_1}{x(\ln x)^2}
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So you have

y_2=v\ln x=-C_1+C_2\ln x

and given that y_1=\ln x, the second term in y_2 is already taken into account in the solution set, which means that y_2=1, i.e. any constant solution is in the solution set.
4 0
3 years ago
An edging was placed around a circular garden. How long was the edging if the radius of the garden was 3m. Use TT = 3.14
Tom [10]
Edging is 18.84m. to find the edging you  must do the radis times 2, then that answer by pi (3.14). So you do 3x2=6, then 6x3.14=18.84m
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