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3 years ago

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

$a_1= 2$
$a_2=2(-4)$
$a_3=2(-4)(-4)=2(-4)^{2}$
$a_4=2(-4)(-4)(-4)=2(-4)^{3}$
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.
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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

Doss [256]

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

Total time taken in 2003- total time taken in 2005

$=2\frac{17}{24}-\frac{3}{8}$

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

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

$=\frac{65-9}{24}$

$=\frac{56}{24}$

$\frac{7}{3}$

$=2\frac{1}{3}$ minutes

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3 years ago

Read 2 more answers

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Hi,
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