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Anit [1.1K]
4 years ago
12

Domain: Ranges: Zeros: Discontinuities: Asymptotes:

Mathematics
1 answer:
trapecia [35]4 years ago
4 0

Domain: The set of possible values of x

Range: The set of possible values of y

Zeros: The x-intercepts (what value of x is on the x-axis)

Discontinuities: Where the graph stops (and restarts somewhere else)

Asymptotes: The graph cannot cross this line (vertical, horizontal, oblique)

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Please solve the problem with steps
Debora [2.8K]

Answer:

Infinite series equals 4/5

Step-by-step explanation:

Notice that the series can be written as a combination of two geometric series, that can be found independently:

\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}

The first one: (\frac{1}{2})^{n-1} is a geometric sequence of first term (a_1) "1" and common ratio (r) " \frac{1}{2} ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2

The second one: \frac{1}{6^{n-1}} is a geometric sequence of first term "1", and common ratio (r) " \frac{1}{6} ". Again, since the common ratio is smaller than one, we can find its infinite sum:

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now we simply combine the results making sure we do the indicated difference: Infinite total sum= 2-\frac{6}{5} =\frac{10-6}{5} =\frac{4}{5}

8 0
4 years ago
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Step-by-step explanation:

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4 years ago
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