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elena-s [515]
2 years ago
13

Simplify the expression by combining like terms: 3+ v + 4v2 + 2 - V2 + 3v ]

Mathematics
1 answer:
const2013 [10]2 years ago
7 0

Answer:

I believe the answer would be 3v2+4v+5

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Step-by-step explanation:

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Simplify the expression<br> (-3+4k) - (-2+ 7k)
DedPeter [7]

Answer: −1−3k

Step-by-step explanation:

Distribute:

(−3+4k)−1(−2+7k)

(-3+4k)+{2-7k}

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(−3+4k)+2−7k

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4 0
3 years ago
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:

Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{6} } =\frac{1}{\frac{5}{6} } =\frac{6}{5}

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
3 years ago
Read 2 more answers
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