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lisabon 2012 [21]
3 years ago
12

Your firm has a price of ​$5​, an average total cost of ​$7​, and an average variable cost of ​$4. In the short​ run, you should

________(operate/shut down) because ________exceeds ________ average variable cost price . In the long​ run, you should _________(stay in/exit) the market because ________ average total cost price exceeds__________average variable cost price average total cost .
Mathematics
1 answer:
djverab [1.8K]3 years ago
5 0

Answer:

Step-by-step explanation:

Given that a firm has  a price of ​$5​, an average total cost of ​$7​, and an average variable cost of ​$4

Price = 5

Var cost = 4

Contribution = 1 dollar per unit

Since contribution is positive, there is scope for getting profit by increasing production.

In the short​ run, you should __operate______(operate/shut down) because __Price______exceeds ________ average variable cost price . In the long​ run, you should __exit______(stay in/exit) the market because ________ average total cost price exceeds____price.______average variable cost price average total cost

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146 feet below the surface

Step-by-step explanation:

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1 year ago
Solve the equatuon<br><br> x/2-(2/(x+1))=1<br><br> x=?
Snezhnost [94]

\frac{x}{2}-\frac{2}{x+1}=1

We have 2 denominators that we need to get rid of. Whenever there are the denominators, all we have to do is multiply all whole equation with the denominators.

Our denominators are both 2 and x+1. Therefore, we multiply the whole equation by 2(x+1)

\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]

Then shorten the fractions.

\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]\\x(x+1)-2(2)=1(2x+2)

Distribute in all.

x^2+x-4=2x+2

We should get like this. Because the polynomial is 2-degree, I'd suggest you to move all terms to one place. Therefore, moving 2x+2 to another side and subtract.

x^2+x-4-2x-2=0\\x^2-x-6=0\\

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(x-3)(x+2)=0\\x=3,-2

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7 0
2 years ago
Sally has a discount card that reduces the price of her grocery bill in a certain grocery store
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0.95c

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3 years ago
100 POINTS TO RIGHT AWNSER!!!!
makvit [3.9K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3000586

——————————

The answer is option D)  r < 5  or  r > – 1.

I'm going to graph each inequality below on a number line.


A)  r > 5  or  r > – 1.

\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}

The result is found just by joining those two intervals together. Actually that compound inequality only implies

r > – 1

which does not represent all real numbers.

—————

B)  r > 5  or  r < – 1.

\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r5~~or~~r

Numbers between – 1  and  5 (including them) are not included in the union, so you don't have all real numbers represented there either.

—————

C)  r < 5  or  r < – 1.

\large\begin{array}{cl} \mathsf{r

Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.

—————

D)  r < 5  or  r > – 1.

\large\begin{array}{cl} \mathsf{r-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r-1}&\qquad\mathsf{\overset{~~~**************************~~~}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}

Now all real numbers are included in the union. So this is the right choice.


Answer:  option D)  r < 5  or  r > – 1.


I hope this helps. =)

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How do you times fraction by whole numbers
ehidna [41]
To multiply whole numbers and fractions, multiply the numerator by the whole number. Example: 1/3×4= 4/3=1 1/3
5 0
2 years ago
Read 2 more answers
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