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attashe74 [19]
3 years ago
13

AHH AHH AHH AHH AHH AHH AHH AHH AHH

Mathematics
1 answer:
Alex3 years ago
3 0

ah... ahhh... ahhhhh... AHHHHHHHHHHH

- vine

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I have to explain the meaning of zero?<br> Please help, ill give brainly!
Klio2033 [76]

Answer:

0 ( zero) is a number, and the numerical digit used to represent that number in numerals. It fulfills a central role in mathematics as the additive identity of the integers, real numbers, and many other algebraic structures. As a digit, 0 is used as a placeholder in place value systems.

8 0
3 years ago
Let production be given by P = bLαK1−α where b and α are positive and α &lt; 1. If the cost of a unit of labor is m and the cost
Nana76 [90]

Answer:

The proof is completed below

Step-by-step explanation:

1) Definition of info given

We have the function that we want to maximize given by (1)

P(L,K)=bL^{\alpha}K^{1-\alpha}   (1)

And the constraint is given by mL+nK=p

2) Methodology to solve the problem

On this case in order to maximize the function on equation (1) we need to calculate the partial derivates respect to L and K, since we have two variables.

Then we can use the method of Lagrange multipliers and solve a system of equations. Since that is the appropiate method when we want to maximize a function with more than 1 variable.

The final step will be obtain the values K and L that maximizes the function

3) Calculate the partial derivates

Computing the derivates respect to L and K produce this:

\frac{dP}{dL}=b\alphaL^{\alpha-1}K^{1-\alpha}

\frac{dP}{dK}=b(1-\alpha)L^{\alpha}K^{-\alpha}

4) Apply the method of lagrange multipliers

Using this method we have this system of equations:

\frac{dP}{dL}=\lambda m

\frac{dP}{dK}=\lambda n

mL+nK=p

And replacing what we got for the partial derivates we got:

b\alphaL^{\alpha-1}K^{1-\alpha}=\lambda m   (2)

b(1-\alpha)L^{\alpha}K^{-\alpha}=\lambda n   (3)

mL+nK=p   (4)

Now we can cancel the Lagrange multiplier \lambda with equations (2) and (3), dividing these equations:

\frac{\lambda m}{\lambda n}=\frac{b\alphaL^{\alpha-1}K^{1-\alpha}}{b(1-\alpha)L^{\alpha}K^{-\alpha}}   (4)

And simplyfing equation (4) we got:

\frac{m}{n}=\frac{\alpha K}{(1-\alpha)L}   (5)

4) Solve for L and K

We can cross multiply equation (5) and we got

\alpha Kn=m(1-\alpha)L

And we can set up this last equation equal to 0

m(1-\alpha)L-\alpha Kn=0   (6)

Now we can set up the following system of equations:

mL+nK=p   (a)

m(1-\alpha)L-\alpha Kn=0   (b)

We can mutltiply the equation (a) by \alpha on both sides and add the result to equation (b) and we got:

Lm=\alpha p

And we can solve for L on this case:

L=\frac{\alpha p}{m}

And now in order to obtain K we can replace the result obtained for L into equations (a) or (b), replacing into equation (a)

m(\frac{\alpha P}{m})+nK=p

\alpha P +nK=P

nK=P(1-\alpha)

K=\frac{P(1-\alpha)}{n}

With this we have completed the proof.

5 0
3 years ago
The angle of elevation of a flagpole from the foot of a building is 65⁰, what is the angle of depression of the building from th
Vadim26 [7]

Answer:

115°

Step-by-step explanation:

The angles of elevation and depression are supplementary, thus

angle of depression = 180° - 65° = 115°

6 0
3 years ago
A car show has 22 cars competing for "Best in Show." Prizes are awarded for first, second and third place. In how many ways can
VikaD [51]
The answer is c. 66 if you multiply 22 by 3 you get 66

8 0
3 years ago
What is the value of the expression, written in standard form?
Ilia_Sergeevich [38]

Answer:

200

Step-by-step explanation:

= (6.6/3.3) × 10^(-2-(-4))

= 2 × 10^2 = 200

4 0
4 years ago
Read 2 more answers
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