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3 years ago

5

PLEASE HELP I WILL MARK YOU BRAINLIEST. Ms. Simmons is using sticky notes to write reminders to herself. Each sticky note is 5/6

inches in length and 2/3 inches in width. If Ms. Simmons places 9 sticky notes on her board, what is the total area, in square inches, of Ms. Simmons’s sticky notes?

1 answer:

Masja [62]3 years ago

7 0

Answer:

5 square inches

Step-by-step explanation:

First, we need to find the area of one sticky note:

The formula is Lenth*Width

Area of one sticky note:

5/6*2/3=10/18

=5/9 square inches

Now, we find the area of 9 sticky notes:

5/9*9=5

5 square inches

Hope this helps! Have an amazing day!

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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:

$g: x_1+x_2+x_3=1000$

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

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kobusy [5.1K]

Answer:

The answer is below

Step-by-step explanation:

An equation shows the relationship between two or more variables. An equation is a statement that shows the equality between expressions. An equation with infinitely many solution is when all numbers are solutions, that is there is no one solution. Example is: x + 3 = x + 3.

When each side of an equation has been simplified, equations that have the same coefficients and the same constants on each side have infinitely many solutions

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Evgesh-ka [11]

Answer:

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Step one: Add 4a to -3a

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Step two: Add 7 to 1

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Answer:

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Answer:

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Step-by-step explanation:

x/-9=3

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