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adelina 88 [10]
3 years ago
15

Anna plans a business model to compete with two video stores, where she hopes to draw in customers from one store but not lose m

oney on the deal.
Movie Mania charges a subscription fee of $30 and an additional $5 per movie, x. Movie Time charges a subscription fee of $25 and an additional $6 per movie, x.

Based on this information, which system of inequalities could be used to determine how many movies need to be rented for a customer on Anna’s plan, y, to pay her more than they would at Movie Time, but less than they would at Movie Mania?
Mathematics
2 answers:
babymother [125]3 years ago
4 0
Let x be the number of movies, m(x) be how much Movie Mania charges and t(x) be how much Movie Time charges and a(x) be Anna's plan.

<span>Movie Mania charges a subscription fee of $30 and an additional $5 per movie
</span>
m(x)=30+5x 

<span>Movie Time charges a subscription fee of $25 and an additional $6 per movie

</span>t(x)=25+6x 

<span>How many movies need to be rented for a customer on Anna’s plan, y, to pay her more than they would at Movie Time, but less than they would at Movie Mania?

That's 

</span>t(y) < a(y) < m(y) 

Expanding it out,

25+6y< a(y) < 30+5y 
  
Is there any room there?

25+6y  <  30+5y

y < 5

There's no possible plan that will do what Anna wants for 5 or more movies, because in that domain Movie Time costs more than Movie Mania.  She can squeeze in there between 1 and 4 movies.

I write the answer as:

25+6y< a(y) < 30+5y 

shepuryov [24]3 years ago
4 0

Answer:

The person below is right but they just explained it alittle oddly so I'll clarify for them because it is correct :)

y<5x+30/x

y>6x+25/x

Step-by-step explanation:


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Solve the system of equations:x + 3y - z = -4 2x - y + 2z = 13 3x - 2y - z = -9
tatiyna

Answer:

The solution to the system of equations is

\begin{gathered} x=\frac{179}{13} \\  \\ y=-\frac{279}{39} \\  \\ z=-\frac{48}{13} \end{gathered}

Explanation:

Giving the system of equations:

\begin{gathered} x+3y-z=-4\ldots\ldots\ldots\ldots\ldots\ldots..........\ldots\ldots\ldots\ldots.\ldots\text{.}\mathrm{}(1) \\ 2x-y+2z=13\ldots\ldots...\ldots\ldots\ldots\ldots..\ldots..\ldots\ldots\ldots\ldots\ldots.(2) \\ 3x-2y-z=-9\ldots\ldots\ldots.\ldots\ldots\ldots\ldots....\ldots\ldots.\ldots\ldots\ldots\text{.}\mathrm{}(3) \end{gathered}

To solve this, we need to first of all eliminate one variable from any two of the equations.

Subtracting (2) from twice of (1), we have:

5y-4z=-21\ldots\ldots\ldots\ldots\ldots.\ldots.\ldots..\ldots..\ldots\ldots.\ldots..\ldots\text{...}\mathrm{}(4)

Subtracting (3) from 3 times (1), we have

3y-5z=-3\ldots\ldots...\ldots\ldots..\ldots\ldots\ldots\ldots\ldots.\ldots\ldots\ldots\ldots\ldots..\ldots\ldots(5)

From (4) and (5), we can solve for y and z.

Subtract 5 times (5) from 3 times (4)

\begin{gathered} 13z=-48 \\  \\ z=-\frac{48}{13} \end{gathered}

Using the value of z obtained in (5), we have

\begin{gathered} 3y-5(-\frac{48}{13})=-3 \\  \\ 3y+\frac{240}{13}=-3 \\  \\ 3y=-3-\frac{240}{13} \\  \\ 3y=-\frac{279}{13} \\  \\ y=-\frac{279}{39} \end{gathered}

Using the values obtained for y and z in (1), we have

\begin{gathered} x+3(-\frac{279}{39})-(-\frac{48}{13})=-4 \\  \\ x-\frac{279}{13}+\frac{48}{13}=-4 \\  \\ x-\frac{231}{13}=-4 \\  \\ x=-4+\frac{231}{13} \\  \\ x=\frac{179}{13} \end{gathered}

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