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MaRussiya [10]
1 year ago
14

Which of the following statements are true?

Mathematics
1 answer:
Varvara68 [4.7K]1 year ago
4 0

I and IV only are statements that are true out of all the statements.

<h3>How to solve an equation?</h3>

To solve linear equations, utilize the steps that are provided below.

  1. Remove parentheses from each side of the equation and combine similar terms to make it simpler.
  2. To separate the variable term on one side of the equation, use addition or subtraction.
  3. To find the variable, use division or multiplication.
  4. The common denominator can be multiplied by each side of the equation to eliminate fractions.

Lets take an example of solve z for 7z – (3z – 4) = 12

Here is no multiplication or division, so this will be easy to solve

⇒ 7z – (3z – 4) = 12

⇒ 7z – 3z + 4 = 12

⇒ 4z = 12 – 4

⇒ 4z = 8

⇒ z = 8/4

⇒ z = 2

See, that was so simple, using the mentioned methods, every linear equation can be solved easily.

<h3 />

I. -(-6)=6 and -(-4)>-4

⇒ 6 = 6 and 4 > -4

IV. 17>2 or 6<9

Learn more about  linear equations

brainly.com/question/14323743

#SPJ4

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Answer:

A: -5, -1, 5, 6, 7

B: -12.4, -12.1, -2, 1/2, 12, 12 1/4, 12 1/2

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F(x) = 8cos(2x) on (0,2π)
Leviafan [203]

Answer:

Step-by-step explanation:

First put the lower limit, i.e., x=0,

F(x)=8cos[2(0)]=8cos(0)=8(1)=8

;cos(0)=1

Now,put the upper limit of given interval, i.e., x = π,

F(x)=8cos[2(π)]=8cos(2π)=8(1)=8

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7 0
4 years ago
What is the product ?
rjkz [21]

Answer:

Step-by-step explanation:

[4   2] *  -2   5               = [ 4*-2 + 2*7       4*5+2*-1]

             7     -1

         

        = [-8+14        20-2]

        = [6    18]

             

6 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.

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