Stop only watching, and start writing dude. thanks

Logan saves the same amount of money each month for college. His current total savings is 300m2 + 120m + 180 dollars. Which fact

nlexa [21]

If we let x as the amount of money saved per month and m be the number of months:

xm = 300m2 + 120m 180

The quadractic equation would result to:
300m2 + (120-x)m + 180 = 0

Using the quadratic formula we have the following factorization:
(m - ( -120 + x + sqrt (x2 - 240x - 2145600) ) / 600) (m - ( -120 +x - sqrt (x2 - 240x - 2145600) ) / 600) = 0

3 0

4 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

Plz answer this question

Rina8888 [55]

Answer:

ANSWER IS 2 ND OPINION

Step-by-step explanation:

MARK ME BRAINLIEST ❤

8 0

3 years ago

Read 2 more answers

If anyone knows how to obtain the answer, please notify me.

Tems11 [23]

Answer:

Area =

b1 + b2

×h

2

=

2 + 6

× 5

2

=

20 centimeters2

Step-by-step explanation:

6 0

3 years ago

Represent the system of linear equations 3x+y-5=0 and 2x-y-5=0 graphically. From the graph write solution of the system and also

Wittaler [7]

Answer:

<h2>The solution is (2, -1)</h2><h2>The area of the triangle formed is 10 square units.</h2>

Step-by-step explanation:

The given system is

$3x+y-5=0\\2x-y-5=0$

First, you need to graph both lines. To do so, you just need to find the interceptions with both axis.

$3x+y-5=0$

For $x=0 \implies y=5$

For $y=0 \implies x=\frac{5}{3}$

Then, you draw both points to have the straight line.

Repeat the process for the second line. The image attached shows both lines.

Remember, the solution of a linear system of equation is the common point between lines. In this case, we can observe that the solution is (2, -1).

On the other hand, to find the area of the triangle formed, we need to use the length of its base and its height.

Its base is 10 units long.
Its height is 2 units long.

Now, we use the area formula for triangles

$A=\frac{bh}{2}=\frac{10(2)}{2}= 10 \ u^{2}$

Therefore, the area of the triangle formed is 10 square units.

8 0

3 years ago