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2 years ago

Can somebody please help me? I’m stumped on this math problem 1x+9=11x-3

Mathematics

1 answer:

alexdok [17]2 years ago

6 0

Answer:

x = 6/5

Step-by-step explanation:

x = 6/5

Sorry I am not the best at explaining for math but I know it is the right answer.

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Determine determine whether the following geometric series converges or diverges. if the series converges find its sum.

Lilit [14]

For starters,

$\dfrac{3^k}{4^{k+2}}=\dfrac{3^k}{4^24^k}=\dfrac1{16}\left(\dfrac34\right)^k$

Consider the $n$ th partial sum, denoted by $S_n$ :

$S_n=\dfrac1{16}\left(\dfrac34\right)+\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\cdots+\dfrac1{16}\left(\dfrac34\right)^n$

Multiply both sides by $\frac34$ :

$\dfrac34S_n=\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\dfrac1{16}\left(\dfrac34\right)^4+\cdots+\dfrac1{16}\left(\dfrac34\right)^{n+1}$

Subtract $S_n$ from this:

$\dfrac34S_n-S_n=\dfrac1{16}\left(\dfrac34\right)^{n+1}-\dfrac1{16}\left(\dfrac34\right)$

Solve for $S_n$ :

$-\dfrac14S_n=\dfrac3{64}\left(\left(\dfrac34\right)^n-1\right)$

$S_n=\dfrac3{16}\left(1-\left(\dfrac34\right)^n\right)$

Now as $n\to\infty$ , the exponential term will converge to 0, since $r^n\to0$ if $0$ . This leaves us with

$\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n\frac{3^k}{4^{k+2}}=\sum_{k=1}^\infty\frac{3^k}{4^{k+2}}=\frac3{16}$

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Step-by-step explanation:

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