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erma4kov [3.2K]
3 years ago
8

Perform the indicated matrix row operation and write the new matrix

Mathematics
1 answer:
Ne4ueva [31]3 years ago
5 0
The given operation involves just swapping the two rows, so carrying out R_1\leftrightarrow R_2 on the matrix gives

\left[\begin{array}{cc|c}5&5&5\\1&-\dfrac23&5\end{array}\right]\to\left[\begin{array}{cc|c}1&-\dfrac23&5\\5&5&5\end{array}\right]
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What are the explicit equation and domain for a geometric sequence with a first term of 2 and a second term of −8?
Alexxx [7]
The common ratio in a geometric sequence is the ratio between 2 consecutive terms:

-8/2=-4, 

then the sequence is  2, -8, 32, -128, -512, 2048, ...

let a_n be the nth term of the sequence, then 

a_1= 2
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so clearly a_n=2(-4)^{n-1}

and, clearly n are integers >0, since we have a 1st term, a second term and so on... of a sequence (we do not have a "zero'th term"!

Answer:

<span>C. an=2(-4)^n-1; all integers where n>0</span>
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3 years ago
a total eclipse in 2003 lasted 2 17/24,minutes in 2005 it lasted 3/8 of a minute how much longer did it last in 2003
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Answer:

Total eclipse was 2\frac{1}{3} minutes longer than in 2005.

Step-by-step explanation:

It is given in the question that total eclipse in 2003 lasted 2\frac{17}{24} minutes and in 2005 it lasted 3/8 of a minute.

We have to calculate how much longer did it last in 2003.

So we have to subtract the duration in 2003 by duration in 2005

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\frac{48+17}{24}-\frac{3}{8}

=\frac{65}{24}-\frac{3}{8}

=\frac{65-9}{24}

=\frac{56}{24}

\frac{7}{3}

=2\frac{1}{3} minutes


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