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AURORKA [14]
2 years ago
12

X

Mathematics
2 answers:
xz_007 [3.2K]2 years ago
8 0

Answer:

9 miles

Step-by-step explanation:

Given scale:  2 in = 3 miles

⇒ 1 in = 1.5 miles

If the width of the plot of land is 6 inches, then to calculate the actual width multiply 6 by 1.5:

actual width = 6 × 1.5 = 9 miles

arlik [135]2 years ago
7 0

Answer:

9

Step-by-step explanation:

6 divided by 2 equals 3 then you multiply 3 by 3 which equals 9.

Hope this helps!!!!!!!!!!

Be safe!!!!!!!

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Marco can finish a jigsaw puzzle in 3 hours. Working together with Cliff, it takes them 1 hour. How
vovikov84 [41]

Macro can finish the puzzle in 3 hours. We can say that her speed for working is 1/3 or 1/3 puzzle per hour.

Now if they work together, this turns into 1 puzzle per hour.

That must mean that Cliff would be doing the other 2/3 of the puzzle while Macro does his 1/3 in that hour. If Cliff has a speed of doing them 2/3 per hour. Then it would take 1 and a half hour to finish 1 whole puzzle.

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[2(3+5)-2(4+1)]5 what is the value?
Jet001 [13]
<span>[2(3+5)-2(4+1)]5
=</span><span>[2(8)-2(5)]5
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3 years ago
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Please Help!!!!!! Using the rest of my points for this!
navik [9.2K]

It’s the first one on the top left corner

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3 years ago
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A 180 pound weight is supported by two ropes each at 30 degrees, as shown in the figure. Find the tension in each rope.
Anit [1.1K]

Answer:

103.9 lb.

Step-by-step explanation:

T cos 30°+T cos 30°=180

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2 years ago
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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

5 0
3 years ago
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