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3 years ago

I don’t quite get the concept yet and i really don’t wanna bother the teacher again :/ please help

Mathematics

1 answer:

IrinaVladis [17]3 years ago

7 0

Answer:

they are asking which claims are true vs false

Step-by-step explanation:

I totally get not wanting to bother our teachers again,... (that is why we're here)

m (AB = 20

m<ABD = 140

m (AD = 60

I think they want to know which ones (based on the "graph") are true,...

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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:

$g: x_1+x_2+x_3=1000$

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

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

3 years ago

Solve (x – 3)2 = 49. Select the values of x. –46 -4 10 52

sergiy2304 [10]

After using the algebraic equation, the required value of x = 55/2.

WHAT IS ALGEBRIC EQUATION?
A mathematical statement wherein two expressions have been set equal to one another is known as an algebraic equation. A variable, coefficients, and constants make up an algebraic equation in most cases. Equations, or the equal sign, simply indicate equality. Equating each quantity with another is what equations are all about. Equations act as a scale of balance. If you've ever seen a balance scale, users know that for the scale to be deemed "balanced," an equal amount of weight must be applied to each side. The scale will tip to one side if we only add weight to one side, and the two sides will no longer be equally weighted.

(x-3)2 = 49

= x-3 = 49/2

= x = (49/2) + 3

= x = 55/2

So, the required value of x = 55/2.

To know more about algebraic equation click on the below given link

brainly.com/question/24875240

#SPJ1

4 0

1 year ago

The difference of twice a number and 6 is less than - 18.

oee [108]

Answer:

2x-6<-18

Step-by-step explanation:

4 0

3 years ago

How to solve X^4-19x^2+48=0

ZanzabumX [31]

Two numbers that add up to -19 and multiply to 48 are -16 and -3:

$(x^4-19x^2+48)=0\\(x^2-16)(x^2-3)=0\\(x+4)(x-4)(x^2-3)=0$

So, the solutions come from each parentheses: x+4=0, x-4=0, and x^2-3=0.

x+4=0
x = -4

x-4=0
x = 4

x^2-3=0
x^2 = 3
x = +/- √3

So, the solutions are -4, -√3, √3, and 4.

4 0

3 years ago

$70 with a 25% discount

stealth61 [152]

70(.25) = 17.25
70 - 17.25 = 52.50

3 0

2 years ago