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bezimeni [28]
3 years ago
12

Order the following numbers from least to greatest. Put the lowest number on the left.

Mathematics
2 answers:
Viktor [21]3 years ago
8 0

Answer:

-4,0,1,3

Step-by-step explanation

Ummm, This lesson is in 6th grade and ur in college,LOL. Yea the answer is -4,0,1,3 because the negative numbers is below the numbers and zero is always above and beyond the point COLLEGE GRADER.

creativ13 [48]3 years ago
4 0

Answer: -4,0,1,3

Step-by-step explanation:

Negatives will always be less than zero

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y=-4x+2

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The width of a rectangle is 22 inches and the perimeter is at least 165 inches. what inequality could be used to find the minimu
Lera25 [3.4K]
The answer is D.

We know that a rectangle has two widths that are equal and two lengths that are equal. One width is 22, so the other one is also 22.

If you wanted to find the lengths, you would add both widths together (same as multiplying a width by two) and add that to the two lengths equaled to the perimeter.

So, 22 * 2 + 2x = perimeter of rectangle. We added all four sides together.

We know that the perimeter is at least 165, so 22 * 2 + 2x = 165. Here's the twist. They want the most minimum possible length. So, what answer choice gives you 165 or less for the most minimum or smallest length while still getting to 165?
That is D.
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Hope this helped!
6 0
3 years ago
Read 2 more answers
Write a doubles fact to find the sum 3+2
valentina_108 [34]

Answer:

5 or -5

Step-by-step explanation:

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3 0
3 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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