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worty [1.4K]
2 years ago
8

How many DVDs. A small independent movie company determines the profit P for producing n DVD copies of a recent release is P = −

0.02n2 + 3.40n − 16. P is the profit in thousands of dollars and n is in thousands of units.
a. How many DVDs should the company produce to maximize the profit?

should the company produce to maximize the profit?
Mathematics
1 answer:
MrRissso [65]2 years ago
3 0

Answer:

128.5 thousand dollars

Step-by-step explanation:

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Percy solved the equation x2 7x 12 = 12. his work is shown below. is percy correct? explain. 1. (x 3)(x 4) = 12 2. x 3 = 12 or x
stepladder [879]

Percy solved the equation by expanding the expression on the left-hand side, and this is incorrect

<h3>How to determine the true solution?</h3>

The equation is given as:

x^2 + 7x + 12 = 12

The first step in solving the equation is to equate the equation to 0.

i.e.

x^2 + 7x + 12 - 12 = 0

This given

x^2 + 7x = 0

From the question, Percy solved the equation by expanding the expression on the left-hand side

Hence, Percy's solution is incorrect

Read more about equations at:

brainly.com/question/2972832

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2 years ago
Question below here you can rear if not aks me please and thank you ​
Vanyuwa [196]
180-48.2-75=56.8 thank you
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3 years ago
What is the decimal number of 2 quaters
Sergio039 [100]
I think 0.50 Hope that helped! Have a good day!
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3 years ago
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A tank contains 180 gallons of water and 15 oz of salt. water containing a salt concentration of 17(1+15sint) oz/gal flows into
Stels [109]

Let A(t) denote the amount of salt (in ounces, oz) in the tank at time t (in minutes, min).

Salt flows in at a rate of

\dfrac{dA}{dt}_{\rm in} = \left(17 (1 + 15 \sin(t)) \dfrac{\rm oz}{\rm gal}\right) \left(8\dfrac{\rm gal}{\rm min}\right) = 136 (1 + 15 \sin(t)) \dfrac{\rm oz}{\min}

and flows out at a rate of

\dfrac{dA}{dt}_{\rm out} = \left(\dfrac{A(t) \, \mathrm{oz}}{180 \,\mathrm{gal} + \left(8\frac{\rm gal}{\rm min} - 8\frac{\rm gal}{\rm min}\right) (t \, \mathrm{min})}\right) \left(8 \dfrac{\rm gal}{\rm min}\right) = \dfrac{A(t)}{180} \dfrac{\rm oz}{\rm min}

so that the net rate of change in the amount of salt in the tank is given by the linear differential equation

\dfrac{dA}{dt} = \dfrac{dA}{dt}_{\rm in} - \dfrac{dA}{dt}_{\rm out} \iff \dfrac{dA}{dt} + \dfrac{A(t)}{180} = 136 (1 + 15 \sin(t))

Multiply both sides by the integrating factor, e^{t/180}, and rewrite the left side as the derivative of a product.

e^{t/180} \dfrac{dA}{dt} + e^{t/180} \dfrac{A(t)}{180} = 136 e^{t/180} (1 + 15 \sin(t))

\dfrac d{dt}\left[e^{t/180} A(t)\right] = 136 e^{t/180} (1 + 15 \sin(t))

Integrate both sides with respect to t (integrate the right side by parts):

\displaystyle \int \frac d{dt}\left[e^{t/180} A(t)\right] \, dt = 136 \int e^{t/180} (1 + 15 \sin(t)) \, dt

\displaystyle e^{t/180} A(t) = \left(24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t)\right) e^{t/180} + C

Solve for A(t) :

\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) + C e^{-t/180}

The tank starts with A(0) = 15 oz of salt; use this to solve for the constant C.

\displaystyle 15 = 24,480 - \frac{66,096,000}{32,401} + C \implies C = -\dfrac{726,594,465}{32,401}

So,

\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) - \frac{726,594,465}{32,401} e^{-t/180}

Recall the angle-sum identity for cosine:

R \cos(x-\theta) = R \cos(\theta) \cos(x) + R \sin(\theta) \sin(x)

so that we can condense the trigonometric terms in A(t). Solve for R and θ :

R \cos(\theta) = -\dfrac{66,096,000}{32,401}

R \sin(\theta) = \dfrac{367,200}{32,401}

Recall the Pythagorean identity and definition of tangent,

\cos^2(x) + \sin^2(x) = 1

\tan(x) = \dfrac{\sin(x)}{\cos(x)}

Then

R^2 \cos^2(\theta) + R^2 \sin^2(\theta) = R^2 = \dfrac{134,835,840,000}{32,401} \implies R = \dfrac{367,200}{\sqrt{32,401}}

and

\dfrac{R \sin(\theta)}{R \cos(\theta)} = \tan(\theta) = -\dfrac{367,200}{66,096,000} = -\dfrac1{180} \\\\ \implies \theta = -\tan^{-1}\left(\dfrac1{180}\right) = -\cot^{-1}(180)

so we can rewrite A(t) as

\displaystyle A(t) = 24,480 + \frac{367,200}{\sqrt{32,401}} \cos\left(t + \cot^{-1}(180)\right) - \frac{726,594,465}{32,401} e^{-t/180}

As t goes to infinity, the exponential term will converge to zero. Meanwhile the cosine term will oscillate between -1 and 1, so that A(t) will oscillate about the constant level of 24,480 oz between the extreme values of

24,480 - \dfrac{267,200}{\sqrt{32,401}} \approx 22,995.6 \,\mathrm{oz}

and

24,480 + \dfrac{267,200}{\sqrt{32,401}} \approx 25,964.4 \,\mathrm{oz}

which is to say, with amplitude

2 \times \dfrac{267,200}{\sqrt{32,401}} \approx \mathbf{2,968.84 \,oz}

6 0
2 years ago
Find all the roots of the polynomial equation<br><br> x^3-2x^2+5x-10=0
saul85 [17]

Step-by-step explanation:

x^3+x^2+4x+4=0

x³+x²+4x=-4

x(x²+x+4)=-4

either

x=-4

or

x²+x+4=-4

x²+x+4+4=0

x²+x+8=0

comparing with ax² +bx+c

a=1

b=1

c=8

x={-1±√{1²-4×1×8}}/2×1

x={-1±√{-31}}/2

7 0
3 years ago
Read 2 more answers
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