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2 years ago

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The Cute Kitten Shelter has $486$ cute kittens at the start of Monday. On each day, $\frac13$ of the kittens available at the st

art of the day are adopted, and no new kittens are added to the shelter. How many kittens remain at the Cute Kitten Shelter at the end of Tuesday ($2$ days later)?

2 answers:

nadya68 [22]2 years ago

6 0

Answer:

216

Step-by-step explanation:

We are told that 1/3 of the available kittens are adopted at the start of the day and no new kittens are added to the shelter. This means 2/3 of the available kittens will remain at the shelter at the end of every day.

Lat us find number of kittens left at the end of Monday.

$486*\frac{2}{3} =162*2=324$

Now we will find 2/3 of 324 to find kittens left at the shelter at the end of Tuesday.

$324*\frac{2}{3} =108*2=216$

Therefore, the number of kittens left at The Cute Kitten Shelter at the end of Tuesday will be 216.

Kryger [21]2 years ago

3 0

Answer:

216

Step-by-step explanation:

<em>486 * 2/3 = 162 * 2</em>

<em>324 * 2/3 = 108 * 2 = 216</em>

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Let

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Differentiating twice gives

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$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

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Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

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$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

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$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

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pantera1 [17]

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