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Katen [24]
3 years ago
12

How do you write expanded from for 0.68 with decimals

Mathematics
1 answer:
BigorU [14]3 years ago
3 0

Answer:

0.6 + 0.08

Step-by-step explanation:

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Solve the Quetion fine with the variable equals to make it true<br> (x+3)(x-5)=0 show your work
xenn [34]

Answer:

x=-3

Step-by-step explanation:

(-3+5) (-3-5)=0

(0) (-8)=0

0=0

7 0
3 years ago
4.05 LC)
Goshia [24]

Answer:

r

Step-by-step explanation:

we know this because it is the only part of the graph that is moving downwards, meaning its decreasing. P is an example of increasing, Q is 'neutral' and s is also increasing.

5 0
3 years ago
A)<br> 2x + y - 12 = 0<br> x + y + 3 = 0
AVprozaik [17]

Answer:

x=15   and   y= -18

Step-by-step explanation:

2x + y - 12 = 0            Equation 1

x + y + 3 = 0                Equation 2

From two given equations, we can find the value of both x and y by finding the value of one variable from any one of the two equations in terms of other variable and putting it to other equation.

Solving equation 1

Finding 'y' in terms of 'x'           2x+y=12\\y=12-2x                 Equation 3

Putting above equation in equation 3 in 2

                                       x + y + 3 = 0

                                      x+12-2x+3=0

                                         x=15

Putting value of 'x' in equation 3 to find the value of 'y'

                                      y=12-2(15)\\y=12-30\\y= -18

6 0
4 years ago
How do I solve 7x+3=59​
lesya692 [45]

Answer:7x+3=59

59-3

7x=56

Then you divide them both by 7

7x/7=56/7

Then you get your answer

X=?

I'll let you do the last part tho

Step-by-step explanation:

7 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
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