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Charra [1.4K]
3 years ago
6

8x +2 factored using the GCF

Mathematics
1 answer:
wolverine [178]3 years ago
7 0

The answer is 2 because I solved it and checked it.


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What is the length of MN.
Natali5045456 [20]

Because the 2 bottom inside angles are identical ( they are both 50 degrees) this means that both sides LM and MN are the same length.


To find the length of MN, set both sides to equal, solve for x, then replace x in the equation.


4x = x+3

Subtract 1x from each side:

3x = 3

Divide both sides by 3:x = 3/3 = 1


Now we know x is equal to 1, we can solve for MN:

MN = x+3 = 1+3 = 4


The answer is D. 4

8 0
3 years ago
Read 2 more answers
Manny and Emma are both competing in a long jump competition at their school.Manny can jump three fifths of the distance that Em
Bingel [31]
The answer is D. say you have 21 split into 5 parts then you will have 5 4.2's. then take three of them and you get 12.6.
6 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
Help plsssssssssssssssssssssss
olga nikolaevna [1]

Explanation:

5.

The initial cost is $65.42

25% of 65.42 is 16.355

Add 16.355 + 65.42 and the total cost is: $81.78

6.

The initial cost is $12.50

3.2% of 12.50 is 0.4

Add 0.4 + 12.50 and the total cost is: $12.90

5 0
3 years ago
Is 1 (b+6) equivalent to (b+6)
astra-53 [7]

Answer:

Yes.

Step-by-step explanation:

Step 1: Write out expressions

1(b + 6) = b + 6

Step 2: Distribute 1

b + 6 = b + 6

6 0
3 years ago
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