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Vlad [161]
3 years ago
7

Hey help me its due today

Mathematics
1 answer:
Alona [7]3 years ago
6 0
Answer 105 meters . because each second is 21 meters so the you just do 21 times 5 then you get 105 meters
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Explain how to use the distributive property to find the product 3 x 4 1/5
NikAS [45]

Answer:

12.6

Step-by-step explanation:

The answer is 12.6 and here is the explanation of using the Distributive Property. Because we are multiplying a whole number by a mixed number, we have to multiply the whole number by the whole number in the mixed number and then multiply the whole number by the fraction. In this case, we have 3 x 4 1/5. We multiply 3x4=12 first, and then we multiply 3x1/5 to get 0.6. 12+0.6=12.6. Hope it helps!

5 0
3 years ago
Can someone helpp pleaseee, ill paypal you 15$
svetoff [14.1K]

Step-by-step explanation:

The angles on the same side of the parallel lines and the transversal are called corresponding angles.

Corresponding angles are congruent

So 11x+1 = 10x+10

x = 10-1 = 9

11*9+1 = 100

So both are 100 degrees

7 0
2 years ago
Please help me! Screenshot of question is attached. 15 Points and Brainliest to the correct answer.
ehidna [41]

Hello,

I would love to answer but there is no screenshot to see!


7 0
3 years ago
What is the perimeter of a parallelogram, if its area is 24 cm^2 and and the distances between the point of intersection of the
Stells [14]
Answer is 20 cm i guess

4 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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