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slavikrds [6]
3 years ago
9

Given: circle k(O), m∠A=(x+45)°, m KE =(x+20)°, m EW =(3x)° Find: m KW . HELP ASAP WILL GIVE BRAINLIEST!

Mathematics
1 answer:
egoroff_w [7]3 years ago
6 0

Answer:

<em>160°</em>

<em />

Step-by-step explanation:

We can use the central angle theorem for this. It says that an angle inscribed in a circle (on the circumference) measured half of the arc it intercepts (by 2 points on the circumference of the circle).

<em>that would mean that Angle KAW is </em><em>half </em><em>of measure of Arc KW.</em> Thus we can write:

m∠A = 0.5 (arc KE + arc EW)

x+45 =0.5 (x+20+3x)

x+45 = 0.5(4x+20)

x+45 = 2x + 10

45 - 10 = 2x - x

Thus, x = 35

<em>Since Arc KW = Arc KE + Arc EW and x = 35, we can say:</em>

<em>Arc KW = x + 20 + 3x = 35 + 20 + 3(35) = 160°</em>

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<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.

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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

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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)

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

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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

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

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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

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