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Sedbober [7]
3 years ago
7

Attachment mathswatch!!!!! Please write answer only thank you!!!!!!!!

Mathematics
1 answer:
Tema [17]3 years ago
6 0

Answer:

A

Step-by-step explanation:

I think it’s a

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In 1994, the moose population in a park was measured to be 3000. By 1998, the population was measured
dem82 [27]

Answer:

  P(t) = 100t +2600

  4100 in 2005

Step-by-step explanation:

You are given two points for (year, population) = (t, p):

  (4, 3000), (8, 3400)

It is useful to use the two-point form of the equation for a line.

  p = (p2 -p1)/(t2 -t1)(t -t1) +p1

  p = (3400 -3000)/(8 -4)(t -4) +3000

  p = 400/4(t -4) +3000

  p = 100t +2600

  P(t) = 100t +2600 . . . . written in functional form

In 2005, the population is predicted to be ...

  P(15) = 100×15 +2600 = 4100

6 0
2 years ago
To finance her community college​ education, Sarah takes out a loan for ​$4300. After a year Sarah decides to pay off the​ inter
VashaNatasha [74]

Answer:

301

Step-by-step explanation:

4300*.07=301

8 0
3 years ago
Read 2 more answers
At the movie theater Leon can buy 4 large popcorns or 2 large popcorns and 3 candy bars for the same price. If the candy bars co
AlexFokin [52]
4p = 2p + 3c....p = popcorn, c = candy bars
c = 6

4p = 2p + 3(6)
4p = 2p + 18
4p - 2p = 18
2p = 18
p = 18/2
p = 9......whoa....9 bucks for popcorn....nothing is cheap anymore
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3 years ago
Find the circumference of each circle. Use 3.14 or 22/7 for π. Round to the nearest tenth, if necessary.
zhuklara [117]

c \:  = \pi \:  \times diameter \\  \:  \:  \:  \:  \:  \:  =  \frac{22}{7}  \times 20 \: inches \\  =  \: 62.9 \: inches

7 0
2 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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