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Vikki [24]
2 years ago
14

The typical pattern of costs for a monopoly can be analyzed by using: I) total cost II) fixed cost III) variable cost IV) margin

al cost V) average cost VI) average variable cost
Mathematics
1 answer:
avanturin [10]2 years ago
4 0

Answer:

ALL of the above

Step-by-step explanation:

The pattern of cost for a monopoly can be analyzed within the same framework as the cost of a perfectly competitive firm, by using total cost, fixed cost, variable cost, marginal cost, average cost, average variable cost

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Please help I’m stuck!!!
Anika [276]

Answer:

The value of angle B is 33.69°

Step-by-step explanation:

You can find out using Tangent Rule, tanθ = oppo./adj. where opposite and adjacent are the length of triangle :

oppo. = AC = 6 units

adj. = BC = 9 units

θ = ∠B

tan ∠B = AC/BC

tan ∠B = 6/9

∠B = tan^(-1) (6/9)

= 33.69° (near. hundredth)

3 0
3 years ago
Lin solved the equation
Alekssandra [29.7K]

Answer: x=1/2

she was wrong, it's supposed to be

Step-by-step explanation: 8(x−3)+7=2x(4−17)

Step 1: Simplify both sides of the equation.

8(x−3)+7=2x(4−17)

(8)(x)+(8)(−3)+7=2x(4−17)(Distribute)

8x+−24+7=−26x

(8x)+(−24+7)=−26x(Combine Like Terms)

8x+−17=−26x

8x−17=−26x

Step 2: Add 26x to both sides.

8x−17+26x=−26x+26x

34x−17=0

Step 3: Add 17 to both sides.

34x−17+17=0+17

34x=17

Step 4: Divide both sides by 34.

34x/34=17/34 x=1/2

3 0
2 years ago
3x^2+ 4x + 5 for x = -2
Alexus [3.1K]
The answer is 33
3(-2)^2+4(-2)+5
-6^2+ (-8) + 5
36 + (-8) + 5
28+5
33
7 0
3 years ago
Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{
SVETLANKA909090 [29]

y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

-----------------------

  • Note 1: A more general solution is y = Ce^x for some constant C.
  • Note 2: It might be tempting to say the general solution is y = e^x+C, but that is not the case because y = e^x+C \to y' = e^x+0 = e^x and we can see that y' = y would only be true for C = 0, so that is why y = e^x+C does not work.
6 0
3 years ago
What is a correct name for the angle shown?
Masteriza [31]

Answer:

the answer is option B. angle S.

when naming an angle we place the vertex of the angle in the middle. here the angle is RST. But that option is unavailable. very often when there are no other angles interfering with the parent angle, we represent it using one letter that is the mid letter, the vertex. here in this case it is S.

4 0
3 years ago
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