Rotate a semicircle around a line through the endpoints of the semicircle.

Text? I go to much going on inside of my head..

IrinaK [193]

Answer:

If you need anything just let me know.

Step-by-step explanation:

Hope you have a good day.

6 0

3 years ago

X^2+3x+2 x^2+x-6 x^2-x-2

Varvara68 [4.7K]

Answer:

1. (+1)(+2)

2. (−2)(+3)

3. (−2)(+1)

Step-by-step explanation:

3 0

3 years ago

A length is measured at 21 cm correct to two significant figures, what is the lower bound of the length and upper bound?

Andreyy89

Answer:

20.5 and 21.5

Step-by-step explanation:

It says 2 significant figures so the least it can possibly be is 0.5 less and the most it can be is 0.5 more as if it was, lets say 20.4 it would round down to 20 not 21.

7 0

2 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:

Optimizing With Lagrange Multipliers

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

3 years ago

Find the difference and the next three terms -10 -20 -30

Dafna1 [17]

Answer:

- 40, - 50, - 60

Step-by-step explanation:

Note the common difference d between consecutive terms of the sequence.

- 20 - (- 10) = - 20 + 10 = - 10

- 30 - (- 20) = - 30 + 20 = - 10

Thus the difference d = - 10

To obtain a term in the sequence subtract 10 from the previous term

- 30 - 10 = - 40

- 40 - 10 = - 50

- 50 - 10 = - 60

The next 3 terms are - 40, - 50, - 60

6 0

3 years ago

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