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stiv31 [10]
3 years ago
5

Suppose a firm in a competitive market earned $1,000 in total revenue and had a marginal revenue of $10 for the last unit produc

ed and sold. What is the average revenue per unit, and how many units were sold?
Mathematics
1 answer:
aleksklad [387]3 years ago
5 0

Answer:

      $5 and 50 units

Step-by-step explanation:

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T>12x. the total will be more than x multiplied by 12.
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What is the value of (10 x 0.5) squared + 4 cubed?
kotegsom [21]
First set up the equation:
(10*0.5)^{2}+4^{3}

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5^{2}+4^{3}

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25+64

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4 0
3 years ago
Read 2 more answers
Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4
anyanavicka [17]

Answer:

\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n

Step-by-step explanation:

Given

f(x) = \sin x\\

c = \frac{3\pi}{4}

Required

Find the Taylor series

The Taylor series of a function is defines as:

f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n

We have:

c = \frac{3\pi}{4}

f(x) = \sin x\\

f(c) = \sin(c)

f(c) = \sin(\frac{3\pi}{4})

This gives:

f(c) = \frac{1}{\sqrt 2}

We have:

f(c) = \sin(\frac{3\pi}{4})

Differentiate

f'(c) = \cos(\frac{3\pi}{4})

This gives:

f'(c) = -\frac{1}{\sqrt 2}

We have:

f'(c) = \cos(\frac{3\pi}{4})

Differentiate

f"(c) = -\sin(\frac{3\pi}{4})

This gives:

f"(c) = -\frac{1}{\sqrt 2}

We have:

f"(c) = -\sin(\frac{3\pi}{4})

Differentiate

f"'(c) = -\cos(\frac{3\pi}{4})

This gives:

f"'(c) = - * -\frac{1}{\sqrt 2}

f"'(c) = \frac{1}{\sqrt 2}

So, we have:

f(c) = \frac{1}{\sqrt 2}

f'(c) = -\frac{1}{\sqrt 2}

f"(c) = -\frac{1}{\sqrt 2}

f"'(c) = \frac{1}{\sqrt 2}

f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n

becomes

f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n

Rewrite as:

f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n

Generally, the expression becomes

f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n

Hence:

\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n

3 0
2 years ago
CHECK ANSWERS -
Tju [1.3M]
Both are correct.

Good job!
6 0
3 years ago
Figs come in 2.4 lb bags. Apples come in 7.2 lb bags. How many times as many pounds are the apple bags as the fig bags?
Helga [31]

Answer:

3

Step-by-step explanation:

7.2 ÷ 2.4 = 3

8 0
2 years ago
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