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3 years ago

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Suppose that the current equilibrium price of silver is $34 per ounce. If silver is produced under conditions of perfect competi

tion and the industry is in long-run equilibrium, the average total cost of producing silver: a. is less than $34 per ounce. b. is $34 per ounce. c. is indeterminate. d. exceeds $34 per ounce.

1 answer:

Bond [772]3 years ago

6 0

Answer:

b. is $34 per ounce

Step-by-step explanation:

If the production cost were less, a competitor would drive the price down. If the production cost were more, the supplier would go out of business.

Since we're at equilibrium, the production cost must be equal to $34 per ounce.

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System of Equations <br> Need help<br> Trig

Yanka [14]

Looks like the system is

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We can eliminate $y$ by taking

$(3x+2y+5z)-2(-2x+y+6z)=16-2(1)$

$\implies3x+2y+5z+4x-2y-12z=16-2$

$\implies7x-7z=14$

$\implies x-z=2$

so that $z=x-2$ , and

$(7x+3y-4z)-3(-2x+y+6z)=11-3(1)$

$\implies7x+3y-4z+6x-3y-18z=11-3$

$\implies13x-22z=8$

Substitute $z=x-2$ into this last equation and solve for $x$ :

$13x-22(x-2)=8$

$\implies13x-22x+44=8$

$\implies-9x=-36$

$\implies x=4$

Then

$z=x-2$

$\implies z=4-2$

$\implies z=2$

Plug these values into any one of the original equation to solve for $y$ :

$-2x+y+6z=1$

$\implies-2(4)+y+6(2)=1$

$\implies-8+y+12=1$

$\implies y=-3$

Hence the solution is x = 4, y = -3, and z = 2.

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4 years ago

How many U.S. quarts are contained in 12 L? A. 13.638 qt B. 11.352 qt C. 10.56 qt D. 12.684 qt

adoni [48]

(12)1.057=12.684

The answer is D. 12.684qt

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3 years ago

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Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

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3 years ago

Powers of 10 with positive exponents are:<br> less than 1<br> greater than 1

nata0808 [166]

Answer:

greater than 1

Step-by-step explanation:

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3 years ago

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Nuts Berry Farm sells nuts for $60 for 4 pounds. Today, and just for today only,

tino4ka555 [31]

Answer:

$10.50

Step-by-step explanation:

First, find the cost of a pound of nuts.

Divide the price by the number of pounds:

60/4

= 15

Calculate the price with the 30% discount by multiplying this by 0.7:

15(0.7)

= 10.5

So, after the discount, a pound of nuts will cost $10.50

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3 years ago