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melomori [17]
3 years ago
7

7x-3x= what is x = x=

Mathematics
2 answers:
Zielflug [23.3K]3 years ago
8 0

Answer:

4x

Step-by-step explanation:

pochemuha3 years ago
7 0
7X = 3X

Solving
7X = 3X

Solving for variable 'X'.

Move all terms containing X to the left, all other terms to the right.

Add '-3X' to each side of the equation.
7X + -3X = 3X + -3X

Combine like terms: 7X + -3X = 4X
4X = 3X + -3X

Combine like terms: 3X + -3X = 0
4X = 0

Divide each side by '4'.
X = 0

Simplifying
X = 0
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4 years ago
What does 5/6 mean in math <br> p.s 5/6 in fraction
SOVA2 [1]

In math the fraction 5/6 means 5 and 6 tenths.

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3 years ago
The difference between two numbers is 9 and the product of the numbers is 162. Find the two numbers
fiasKO [112]
Let the no be X and Y

acc to ques....

x-y=9 .........1

xy=162 ..........2

substituting value from 1 in 2 we get;

x=9+y

[9+y][y] = 162

y^2+9y = 162

y^2 + 9y - 162=0

y^2 + 18y - 9y - 162=0

y[y+18] + 9[y+18]=0

[y+9][y+18}

y= -9.................................3


y= -18......................................4



case 1 :

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7 0
3 years ago
Lagrange multipliers have a definite meaning in load balancing for electric network problems. Consider the generators that can o
Ivahew [28]

Answer:

The load balance (x_1,x_2,x_3)=(545.5,272.7,181.8) Mw minimizes the total cost

Step-by-step explanation:

<u>Optimizing With Lagrange Multipliers</u>

When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.

Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.

The cost of each generator is given by the formula

\displaystyle C_i=3x_i+\frac{i}{40}x_i^2

It means the cost for each generator is expanded as

\displaystyle C_1=3x_1+\frac{1}{40}x_1^2

\displaystyle C_2=3x_2+\frac{2}{40}x_2^2

\displaystyle C_3=3x_3+\frac{3}{40}x_3^2

The total cost of production is

\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2

Simplifying and rearranging, we have the objective function to minimize:

\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)

The restriction can be modeled as a function g(x)=0:

g: x_1+x_2+x_3=1000

Or

g(x_1,x_2,x_3)= x_1+x_2+x_3-1000

We now construct the auxiliary function

f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)

\displaystyle f(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)-\lambda (x_1+x_2+x_3-1000)

We find all the partial derivatives of f and equate them to 0

\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0

\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0

\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0

f_\lambda=x_1+x_2+x_3-1000=0

Solving for \lambda in the three first equations, we have

\displaystyle \lambda=3+\frac{2}{40}x_1

\displaystyle \lambda=3+\frac{4}{40}x_2

\displaystyle \lambda=3+\frac{6}{40}x_3

Equating them, we find:

x_1=3x_3

\displaystyle x_2=\frac{3}{2}x_3

Replacing into the restriction (or the fourth derivative)

x_1+x_2+x_3-1000=0

\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0

\displaystyle \frac{11}{2}x_3=1000

x_3=181.8\ MW

And also

x_1=545.5\ MW

x_2=272.7\ MW

The load balance (x_1,x_2,x_3)=(545.5,272.7,181.8) Mw minimizes the total cost

5 0
4 years ago
Help with this please!
slega [8]

Hello,

No, this is NOT a proportional relationship.

Have a great day/night and stay safe! :)

8 0
3 years ago
Read 2 more answers
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