How i did? In livada sunt 12 mere am cules 6. Cate mera au ramas?

erica [24]

$\alpha ) \\ \frac{(8x10^9)}{(2x10^4)} =4x10^5 \\ \\ \alpha ) \\ \frac{(3x10^{12})}{(6x10^9)} = \frac{1}{2}x10^{3} \\ \\ \alpha ) \\ \frac{(7.2x10^9)}{(4.1x10^-4)} = \frac{14}{4.10^{13}} \\ \\ \alpha )) \\ (3.2x10^6)+(4.1x10^8)=>> \\ \\ 
 6.10^6+4.10^8=>> \\ \\ 6000000+100000000==>>10^0$

$\framebox[1.1\width]{Good Luck !} \par
$

8 0

3 years ago

What is the cube root of 3?

PtichkaEL [24]

The cube root of 3 is 1.44224957031

7 0

3 years ago

Read 2 more answers

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

A swimming pool had 30 gallons of water in it. Then water was added to the pool at a rate of 5 gallons per second. The function

alexdok [17]

................The answer is C

5 0

3 years ago

These box plots show daily low temperatures for a sample of days in two

Elan Coil [88]

Answer:

The answer is B.

Step-by-step explanation:

As you can see, box plot A is symmetrical so that applies to answers A and B. However, box B is very skewed to the right which makes it positively skewed so answer B is the only one that matches the correct description of the two box and whisker plots.

8 0

3 years ago

Read 2 more answers