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mina [271]
2 years ago
15

Given the triangle below, list its angles in order from smallest to largest.

Mathematics
2 answers:
Masja [62]2 years ago
5 0
J is the smallest, then K, then L
Rudiy272 years ago
3 0

Answer:

<J, <k, <L

Step-by-step explanation:

Automatically we know that L is the largest because it is a right triangle making it 90 degrees. Next we know J is the smallest because looking at the opposite side the 13 is the shortest side making that the smallest angle putting K in the middle.

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InWrite the equation of the line that passes through (7, - 4) and (- 1, 2) slope- intercept form.
blondinia [14]

Answer:

General Equation of a Line

General Equation \\(y - y_{1} )= m (x-x_{1}) \\Slope=m= \frac{x_{2}-x_{1} }{y_{2}-y_{1} } \\\\Given:\\p_{1}=(7,-4)\\p_{2}=(-1,2)\\Solution:\\m=(\frac{-1-7}{2-(-4)}) =\frac{-8}{6}\\y-(-4) = (\frac{-8}{6} )(x-7)\\y+4=\frac{-8x}{6} + 56\\y=\frac{-8x}{6} + 52 \\\\The\\  equation \\ is \\ y=\frac{-8x}{6} + 52

Step-by-step explanation:

Find the slope first and then place the values on the general formula. The final step is to solve for y

5 0
3 years ago
Solve x – 5 = 16 for x.
insens350 [35]
Get x all alone on its own side:
x=16+5
x=21
5 0
2 years ago
Read 2 more answers
Help plsssss!!!!!!!!!!!!!
lapo4ka [179]

Answer:

Percent, Part, whole, I am not sure the rest hope this helped though

Step-by-step explanation:

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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
Which is the best measure of central tendency for the type of data below–the mean, the median, or the mode? explain?
nirvana33 [79]
What type of data? Usually, mean is the best representative.
8 0
3 years ago
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