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3 years ago

6

Please help! Will mark brainliest! Simplify.

tle="(3a^{2}*-3a)" alt="(3a^{2}*-3a)" align="absmiddle" class="latex-formula">

1 answer:

PilotLPTM [1.2K]3 years ago

8 0

<em>S</em><em>t</em><em>e</em><em>p</em><em>-</em><em>b</em><em>y</em><em>-</em><em>s</em><em>t</em><em>e</em><em>p</em><em> </em><em>e</em><em>xplanation</em><em>:</em>

<em>Since</em><em> </em><em>-3a</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>said</em><em> </em><em>to</em><em> </em><em>have</em><em> </em><em>a</em><em> </em><em>power</em><em> </em><em>of</em><em> </em><em>1</em><em>,</em><em> </em><em>we</em><em> </em><em>add</em><em> </em><em>the</em><em> </em><em>powers</em><em> </em><em>together</em><em> </em><em>and</em><em> </em><em>multiply</em><em> </em><em>the</em><em> </em><em>coefficients</em>

-9a³

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Answer:

The 8th term of the sequence is 896/2187.

Step-by-step explanation:

We want to find the 8th term of a geometric sequence whose common ratio is 2/3 and whose first term is 7.

We can write a direct formula. Recall that the direct formula of a geometric sequence is given by:

$\displaystyle x_{n} = a\left(r\right)^{n-1}$

Where <em>a</em> is the initial term and <em>r</em> is the common ratio.

Substitute:

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To find the 8th term, let <em>n</em> = 8. Substitute and evaluate:

$\displaystyle \begin{aligned} x_{8} &= 7\left(\frac{2}{3}\right)^{(8) - 1} \\ \\ &= 7\left(\frac{2}{3}\right)^{7} \\ \\ &= 7\left(\frac{128}{2187}\right) \\ \\ &= \frac{896}{2187} = 0.4096...\end{aligned}$

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