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marusya05 [52]
3 years ago
9

A small publishing company is planning to publish a new book. The production costs will include one-time fixed costs (such as ed

iting) and variable costs (such as printing). There are two production methods it could use. With one method, the one-time fixed costs will total $22,427, and the variable costs will be $19.25 per book. With the other method, the one-time fixed costs will total $53,962 , and the variable costs will be $10.50 per book. For how many books produced will the costs from the two methods be the same?
NEED HELP QUICK
Mathematics
1 answer:
monitta3 years ago
7 0

Answer:

3,604 books

Step-by-step explanation:

We have 2 situations:  situation A and situation B.  What we are looking for is the number of books that has situation A equal to situation B.  So the game plan is to write the equations for A and B, set them equal to each other, and then solve for the unknown number of books, x.

Situation A:  If each book costs 19.25 to produce and the number of books is x, we express that as 19.25x.  The fixed cost to use that company, regardless of the number of books it produces for you, is 22,427.  Which means it is going to charge you 22,427 whether you produce 1000000 books or no books at all.  The equation for A is:

C(A) = 19.25x + 22,427

Situation B uses the exact same reasoning, with the cost of each book being 10.50x and the flat rate cost of 53,962.  Therefore, the equation for B:

C(B) = 10.50x + 53,962

We need the number of books where A = B, so we set the equations equal to each other and solve for x:

19.25x + 22,427 = 10.50x + 53,962 so

8.75x = 31,535 and

x = 3604

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Line segment AB has endpoints A (-4, 4) and B (2, 4). What is the length of line segment AB?
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If we digitized music in stereo using 24 bits per sample and sampling the analog signal 96,000 times per second, this would gene
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6 0
2 years ago
What is the solution?
Ilia_Sergeevich [38]

Answer:

A. {3+7i,3-7i}

Step-by-step explanation:

The given equation  is  x^2-6x+58=0

Use the quadratic formula with a=1,b=-6 and c=58

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x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}

We substitute the given values to get;

x=\frac{--6\pm \sqrt{(-6)^2-4(1)(58)} }{2(1)}

x=\frac{6\pm \sqrt{36-232} }{2}

x=\frac{6\pm \sqrt{-196} }{2}

Recall that;

\sqrt{-1}=i

x=\frac{6\pm 14i}{2}

x=3\pm 7i

x=3+7i or x=3-7i

8 0
3 years ago
Read 2 more answers
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