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sveticcg [70]
3 years ago
12

(1-сos mx)

Mathematics
1 answer:
Maru [420]3 years ago
7 0
22xsgjkbcgllllllllll
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Can the following side measures make a triangle: 11in., 12in. and 13in?
Aloiza [94]

Answer:

Yes

Step-by-step explanation:

Hello There!

Remember the sum of the two shorter sides has to be greater than the largest side in order for it to be a triangle

11+12 must be greater than 13

11+12=23

23>13 therefore the side lengths can form a triangle

4 0
3 years ago
948/10 to the third power
Fantom [35]
The answer is 0.948  
Enjoy your day/night!
8 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
A 20° sector in a circle has an area of 684 yd2.
nasty-shy [4]
Set up a proportion:

We know that a circle equals 360 degrees, 20 degrees is equal to 684 yards, so we want to know the area of 360 degrees:

20/684 = 360/x

Cross multiply:

20x = 246240

Divide 20 to both sides:

x = 12312
3 0
4 years ago
ABCD is an isosceles trapezoid with legs AB and CD, and base BC. If the length of AB = 6y +5, the length of BC= 4y - 6, and the
sertanlavr [38]
<span>ABCD is an isosceles trapezoid with legs AB and CD
∴ AB = CD
</span>
<span>AB = 6y +5
</span>
<span>CD= 2y +1

∴ 6y + 5 = 2y + 1
   6y - 2y = 5 - 1
   4y = 4

 ∴  y = 4/4 = 1
</span>
8 0
4 years ago
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