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3 years ago

Determine the 10th term of the sequence 3, 6, 12,

Mathematics

1 answer:

Charra [1.4K]3 years ago

4 0

Answer:

$1536$

Step-by-step explanation:

In order to find the answer to this question you need to find the rule then apply that rule in till you find the 10th term of the sequence.

Rule = ×2

$3\times2=6\times2=12\times2=24\times2=48\times2=96\times2=192\times2=384\times2=768\times2=1536$

$=1536$

Hope this helps

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$\bf \textit{quad-angle identities}\\\\\\

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\\\\\\
sin(4\theta)=
\begin{cases}
8sin(\theta)cos^3(\theta)-4sin(\theta)cos(\theta)\\\\
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\end{cases}\\\\
-----------------------------\\\\
$

$\bf sin(5\theta)\iff sin(4\theta + \theta)
\\\\\\
sin(4\theta)cos(\theta)+cos(4\theta)sin(\theta)
\\\\\\\
[4sin(\theta)cos(\theta)-8sin^3(\theta)cos(\theta)]cos(\theta)\\\\ + [8cos^4(\theta)-8cos^2(\theta)+1]sin(\theta)$

$\bf -----------------------------\\\\
cos(5\theta)\iff cos(4\theta + \theta)
\\\\\\
cos(4\theta)cos(\theta)-sin(4\theta)sin(\theta)\\\\\\\
[8cos^4(\theta)-8cos^2(\theta)+1]cos(\theta)\\\\
-
[4sin(\theta)cos(\theta)-8sin^3(\theta)cos(\theta)]sin(\theta)$

distribute... and simplify :)

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