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Doss [256]
2 years ago
15

I need help if anyone could help me solve these problems that would be very helpful!

Mathematics
1 answer:
nikklg [1K]2 years ago
4 0

Answer:Q9 is 12.2. Q10 is C. Q11 is C. Q12 Is 2.8. The level 3-4 thing the hight is 24.

Step-by-step explanation: a^2+b^2=c^2 pythagoras theorem or whatever. Hope this helped.

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Use the distributive property to find the equivalent expression.<br><br> 6(2 + 5t) = ?+?t
IgorLugansk [536]

6(2 + 5t) = 6*2 + 6*5t = 12 + 30t

Answer is 12 +30t.

5 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
WHO KNOWS THIS PLEASE HELP
Scorpion4ik [409]
The answer for this is the rules is x n y
4 0
3 years ago
Please help with this question
yaroslaw [1]

Answer:

25%

Step-by-step explanation:

find how many possible combinations (20) find how many combinations with tomato sauce(5) divide 5/20 which is .25 and to find this in a percent multiply by 100 to find 25%

3 0
3 years ago
Using the segment addition postulate, which is true?
gulaghasi [49]
The answer is BC + CD = BD
7 0
3 years ago
Read 2 more answers
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