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3 years ago

Plz, Help I'll give brainlist to whoever answers all questions!

Mathematics

2 answers:

Tatiana [17]3 years ago

7 0

Answer:

Step-by-step explanation:

1. Area => 7.5 x 2 = 15 m² (Choice B)

2. Area => 5 x 11 = 55 in² (Choice B)

3. Area = (12 x 18) * 0.5 = 108 cm² (Choice A)

4. Triangular Pyramid (Choice D)

6. Area of Rectangle => 12 x 9 = 108

Area of Triangle => 7x6 = 42

Sum of Area => 150

7. Surface Area => (4x4)x6 = 96 in² (Choice D)

8. Volume = 3x15x14 = 630 in³ (Choice D)

GarryVolchara [31]3 years ago

5 0

1.B
2.B
3.A
4.D
5.150
6.D
7. D

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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:

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:

$g: x_1+x_2+x_3=1000$

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

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

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4 years ago

Help me with this math question please I'm giving away brainliests

anyanavicka [17]

The answer is A.All real numbers except -2.

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3 years ago

15 times a number increased

Minchanka [31]

Answer:

Step-by-step explanation:

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3 years ago

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Elodia [21]

Factors are: 0, 1, 2, 4, 8, 16

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4 years ago

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Cual es la. Longitud de un lápiz 2

asambeis [7]

Traduje su pregunta al Inglés y respondió a esto. La longitud de un lápiz estándar número 2 es de 7,5 pulgadas, y el lápiz tiene un diámetro de unos 7 milímetros. Espero que esto ayudó☻

I translated your question to english and answered this. The length of a standard number 2 pencil is 7.5 inches, and the pencil has a diameter of about 7 millimeters. Hope this helped ☻

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3 years ago