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Shkiper50 [21]
3 years ago
11

Please help me- ill be your friend if you do- hurry please-

Mathematics
1 answer:
ololo11 [35]3 years ago
3 0

Answer: B) 400/3 Pi ft^3

Step-by-step explanation:

V=πr2h

3=π·52·16

3≈418.87902

and

400 /3

(3.141593)

=418.87902

so its B

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5 divided 630 in partial qoutients
KIM [24]
I think it is 0.007936
5 0
4 years ago
Let production be given by P = bLαK1−α where b and α are positive and α < 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
What is the slope of the line on the graph <br> ОА. 5/9<br> OB. 5/7<br> OC. 7/5<br> OD. 9/7
Luda [366]

Answer: The answer is B. Hope this helps :)

Step-by-step explanation:

3 0
4 years ago
Read 2 more answers
The quantity of milk consumed in five households in a week is 10L.12.13 L. 11 L and
lorasvet [3.4K]

Answer:

12

Step-by-step explanation:

Add 10l to 12l to13l to11l to 14l=60l the divide 60l by the number of houses which will be 12 and there is your correct answer

4 0
3 years ago
What is the completely factored form of x^4+8x^2-9
denis-greek [22]

Remark

You'll see it a whole lot easier if you make a substitution so that it looks like something you have already seen

Solution

let y = x^2

x^4 = x^2 * x^2

x^4 = y * y

x^4 = y^2

Now the expression becomes

y^2 + 8 y - 9  =  z

(y + 9)(y - 1) = z

Now put the x^2 back in.

(x^2 + 9) ( x^2 - 1) = z

x^2 - 1 becomes x + 1 and x - 1. At this level x^2 + 9 can't be factored.

Answer

(x^2 + 9) (x + 1)(x - 1)


3 0
4 years ago
Read 2 more answers
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