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igomit [66]
4 years ago
14

Explain how the Quotient of Powers was used to simplify this expression. 2 to the fifth power, over 8 = 22

Mathematics
1 answer:
vlabodo [156]4 years ago
3 0
\bf ~~~~~~~~~~~~\textit{negative exponents}\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad 
a^n\implies \cfrac{1}{a^{-n}}\\\\
-------------------------------\\\\
\cfrac{2^5}{8}\qquad \boxed{8=2^3}\qquad \cfrac{2^5}{2^3}\implies 2^5\cdot 2^{-3}\implies 2^{5-3}\implies 2^2
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In the typing world,80 words per minute is considered acceptable.how many words per 30 minutes is this
lukranit [14]
W = words per minute = 80
t = time in minutes = 30

The total number of words per 30 minutes is w*t = 80*30 = 2400 words
8 0
4 years ago
Error Analysis Your classmate tries to find an equation for a line parallel to y = 3x --
Arada [10]

Answer:

The classmate's only error is that the slope of the parallel line should be 3 aswell.

Step-by-step explanation:

Parallel lines always have the same slope to each other

7 0
3 years ago
Solve for x. Round your answer to the nearest tenth.
zlopas [31]

Answer:

By Pythagorean theorem:

c^2=a^2+b^2

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7 0
3 years ago
Jennifer bought a printer that cost $240. there was a 7% sales tax on the printer. how much sales tax did jennifer pay?
lilavasa [31]
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5 0
3 years ago
Read 2 more answers
Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
Aleks [24]

Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

5 0
4 years ago
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