Solve for the variable. 3.7x=4.44

1.2

2.12

1.19

2.1

1.2

2.12

1.19

2.1

Ramsey
2020-12-03
Answered

Solve for the variable. 3.7x=4.44

1.2

2.12

1.19

2.1

1.2

2.12

1.19

2.1

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broliY

Answered 2020-12-04
Author has **97** answers

3.7x=4.44

$\frac{3.7x}{3.7}=\frac{4.44}{3.7}$

x=1.2

$\frac{4.44}{3.7}=\frac{44.4}{37}\text{}so\text{}\frac{1.2}{44.4}$ long division

x=1.2

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I would like to find the closed form of the sequence given by

${a}_{n+2}=2{a}_{n+1}-{a}_{n}+{2}^{n}+2,\text{}\text{}\text{}\text{}n0\text{}\text{}\text{}\text{}and\text{}\text{}\text{}{a}_{1}=1,\text{}\text{}\text{}\text{}{a}_{2}=4$

This task is in the topic of differential and difference equation. I don't know how to start solving this problem and what are we looking for? (${a}_{n},{a}_{n+2}$)

I do know how to solve the following form

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using linear algebra as well. The actual problem I encountered the obstructionist term ${{2}^{n}+2}$

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${a}_{n+2}=2{a}_{n+1}-{a}_{n}+{2}^{n}+2,\text{}\text{}\text{}\text{}n0\text{}\text{}\text{}\text{}and\text{}\text{}\text{}{a}_{1}=1,\text{}\text{}\text{}\text{}{a}_{2}=4$

This task is in the topic of differential and difference equation. I don't know how to start solving this problem and what are we looking for? (${a}_{n},{a}_{n+2}$)

I do know how to solve the following form

${a}_{n+2}=2{a}_{n+1}-{a}_{n}$

using linear algebra as well. The actual problem I encountered the obstructionist term ${{2}^{n}+2}$

Are there some kind of variational constant method for recursive linear sequences,?

I only now this method for linear ODE with constant coefficient.

But I believe that such method could be doable here as well. Can any one provide me with a helpful hint or answer?.

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$\begin{array}{rl}{x}_{1}-{x}_{2}& \le 4\\ {x}_{1}-{x}_{5}& \le 2\\ {x}_{2}-{x}_{4}& \le -6\\ {x}_{3}-{x}_{2}& \le 1\\ {x}_{4}-{x}_{1}& \le 3\\ {x}_{4}-{x}_{3}& \le 5\\ {x}_{4}-{x}_{5}& \le 10\\ {x}_{4}-{x}_{3}& \le -4\\ {x}_{5}-{x}_{4}& \le -8\end{array}$

$\begin{array}{rl}{x}_{1}-{x}_{2}& \le 4\\ {x}_{1}-{x}_{5}& \le 2\\ {x}_{2}-{x}_{4}& \le -6\\ {x}_{3}-{x}_{2}& \le 1\\ {x}_{4}-{x}_{1}& \le 3\\ {x}_{4}-{x}_{3}& \le 5\\ {x}_{4}-{x}_{5}& \le 10\\ {x}_{4}-{x}_{3}& \le -4\\ {x}_{5}-{x}_{4}& \le -8\end{array}$

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4)$\frac{{\mathrm{log}}_{3}(6-x)}{{\mathrm{log}}_{3}\left(x\right)}=2$

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