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andriy [413]
3 years ago
13

I NEED HELP PLS I WILL MARK BRAINLIEST

Mathematics
1 answer:
ANEK [815]3 years ago
8 0

Answer:

\approx 11.2\:\mathrm{in}

Step-by-step explanation:

The swing of the tip of pendulum is creating an arc. The question is actually asking for the length of this arc. The length of an arc is given by 2r\pi\cdot \frac{\theta}{360}, where r is the radius of the circle and \theta is the angle of the arc.

In this problem, we're given:

\theta = 40^{\circ},\\r=16

Substituting given values, we get:

2\cdot 16\cdot \pi \cdot \frac{40}{360}=32\pi\cdot \frac{1}{9}\approx \boxed{11.2\:\mathrm{in}}

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Sent a picture of the solution to the problem (s). You can always break a number down to something you understand.

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What is the best estimate of the lateral area of a cube with edges that are 2.1 inches long?​
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The best estimate of the lateral area of the cube with edges that are 2.1 inches long is 4.2
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Help me I don’t understand this
Georgia [21]
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4 0
3 years ago
Read 2 more answers
If the length of a rectangle is three times it’s width and it’s parameter is 24cm what is the area
katrin [286]

Answer:

Area = 27 cm²

Step-by-step explanation:

let L be the length of the rectangle

    w the width

    P the perimeter

    A the area

<u>Formulas</u> :

P = 2×(L + w)

A = L × w

<u>We are Given</u> :

L = 3w

P = 2×(L + w)

⇔ 24 = 2×(3w + w)

⇔ 24 = 2×4w

⇔ 24 = 8w

⇔ w = 24/8 = 3

Then

L = 3w = 3×3 = 9

We obtain L = 9 and w = 3.

Then

A = L × w

  = 9 × 3

  = 27

6 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:

<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
3 years ago
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