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sveticcg [70]
4 years ago
5

Suppose you write a book. The printer charges $4 per book to print it, and you spend $3500 on advertising. You sell the book for

$15 a copy. How many copies must you sell so that your income from sales is greater than your total cost?
Mathematics
1 answer:
Vesna [10]4 years ago
7 0

Answer:

I must sell more than 318 copies so my revenue is greater than my costs

Step-by-step explanation:

The cost function (C(x)) is divided in two: fixed costs and variable costs. In this case, we have both, the fixed costs are $3,500 that I will spent on advertising and the variable cost is the print cost that depends on the number of books i want to print. The cost function is:

C(x)= $3,500+$4x

The revenue function depends on the number of books I sell:

R(x)= $15x

If i want to know how many books I should sell to have a greater revenue than cost i must solve this inequality:

Revenue (R(x))>Cost (C(x))

$15x>$3,500+$4x

$15x-$4x>$3,500

$11x>$3,500

x>$3,500/$11

x> 318,18

I must sell more than 318 copies so my revenue is greater than my costs

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What are the roots of 3x2 + 10 = 4x?
sveticcg [70]

Answer:

Option C is correct.

roots of the given equation , x =\frac{2\pm i\sqrt{26}}{3}

Step-by-step explanation:

Given the equation: 3x^2+10 = 4x

We can write this equation as:

3x^2-4x + 10 =0

A quadratic equation is in the form of ax^2+bx+c =0      ......[1] where a,b ,c are the coefficient and x is the variable,

the solution of the equation is given by;

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On comparing given equation with equation [1] we get;

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So, the solution of the given equation is given by;

x = \frac{-(-4)\pm\sqrt{(-4)^2-4(3)(10)}}{2(3)}

or

x =  \frac{4\pm\sqrt{(16-120}}{6} = \frac{4\pm\sqrt{(-104)}}{6} = \frac{4\pm\sqrt{(-4 \times 26)}}{6}

or

x =\frac{4\pm2 \sqrt{(-26)}}{6} = \frac{4\pm2 i\sqrt{26}}{6}    [∴\sqrt{-1} = i

Simplify:

x =\frac{2\pm i\sqrt{26}}{3}

therefore, the roots of the given equation are; x =\frac{2\pm i\sqrt{26}}{3}

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Hope this helped :)

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