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Crazy boy [7]
3 years ago
14

true or false The typical measurement of hotel demand is the room night (RN). When we say that each room night is perishable , w

e mean that once each room goes unoccupied/unsold for any night, it cannot be "stored away" and then sold on another date.
Mathematics
1 answer:
weqwewe [10]3 years ago
4 0

Answer:

True

Step-by-step explanation:

Perishable is a term for a product that has limited time to sell. A product like meat, fruit, and most fresh food will have a shelf life and you have to sell the product before the shelf life expires. Some preservation techniques can help to extend food products like freezing or curing them.

Room night of a hotel or any rent product is also considered as perishable since you can't store the rent time and sell it another day. So, every unoccupied room is treated as expired food. Unlike food products, room nights for the hotel can't be preserved. There are other strategies to maximize the revenue of rent products such as overselling or selling promise/reservation.

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−3x + y = 4,<br><br> −9x + 5y = −1<br> find x and y
katrin [286]

Answer:

9x - 3y = -12

-9x +5y = -1

2y = -13

y = -13/2

-3x - 13/2 = 8/2

-3x = 21/2

-6x = 21

x = -21/6 = -7/2

(-7/2, -13/2)

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3 years ago
Partial fraction decomposition<br> 20x+9/25x^2+20x+4
ozzi

Step-by-step explanation:

We have

\frac{20x + 9}{25 {x}^{2} + 20x + 4 }

Let's factor the denomiator first,

the denomaitor is a perfect square so we get

\frac{20x + 9}{(5x + 2) {}^{2} }

Now, we must think of two fractions that

\frac{a}{(5x + 2) {}^{2} }  +  \frac{b}{5x + 2}

We use a perfect square term for one fraction, then a linear one for the next, because if we set both of the denomiator to the same factor, we would get a inconsistent system.

So right now, we have

\frac{a}{(5x + 2) { }^{2} }  +  \frac{b}{5x + 2}  =  \frac{20x +9 }{25 {x}^{2} + 20x + 4 }

a + (5x  + 2)b = 20x + 9

5b (x)= 20

a + 2b = 9

b = 4

so that means that a is

a + (2)(4) = 9

a + 8 = 9

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So our equation is

\frac{1}{(5x + 2) {}^{2} }  +  \frac{4}{5x + 2}

6 0
2 years ago
Use the method of reduction of order to find a second solution to t^2y' + 3ty' – 3y = 0, t&gt; 0 Given yı(t) = t y2(t) = Preview
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Let y_2(t)=tv(t). Then

{y_2}'=tv'+v

{y_2}''=tv''+2v'

and substituting these into the ODE gives

t^2(tv''+2v')+3t(tv'+v)-3tv=0

t^3v''+5t^2v'=0

tv''+5v'=0

Let u(t)=v'(t), so that u'(t)=v''(t). Then the ODE is linear in u, with

tu'+5u=0

Multiply both sides by t^4, so that the left side can be condensed as the derivative of a product:

t^5u'+5t^4u=(t^5u)'=0

Integrating both sides and solving for u(t) gives

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Integrate again to solve for v(t):

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tv=y_2=C_1t^{-5}+C_2t

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