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2 years ago

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Is it possible to arrange 152 books on a shelf so that the first shelf has 7 fewer books than the second shelf and 11 more than

the third shelf?

1 answer:

Zepler [3.9K]2 years ago

7 0

Using a system of equations, it is found that <u>since the solution of the system does not involve decimal numbers</u>, it is possible to arrange the books in this way.

For the system, we consider that:

x is the number of books on the first shelf.

y is the number of books on the second shelf.

z is the number of books on the third shelf.

In total, there are 152 books, hence:

$x + y + z = 152$

The first shelf has 7 fewer books than the second shelf, hence:

$x = y - 7$

Also, it has 11 more books than the third shelf, hence:

$x = z + 11$

Then:

$y - 7 = z + 11$

$z = y - 18$

Then, replacing in the first equation:

$x + y + z = 152$

$y - 7 + y + y - 18 = 152$

$3y = 177$

$y = \frac{177}{3}$

$y = 59$

The number of books in each shelf has to be countable, that is, cannot be decimal.

Then, <u>since the solution of the system does not involve decimal numbers</u>, it is possible to arrange the books in this way.

To learn more about system of equations, you can take a look at brainly.com/question/14183076

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The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

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So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

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Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

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By plugging the values of $y_1$ and

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$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

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