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2 years ago

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Kay and Allen sell cell phones. Last month Kay sold 17 more cell phones than Allen. Together they both sold 117 cell phones. How

many phones did each of them sell individually? Show the equation you used to determine the answers

1 answer:

mrs_skeptik [129]2 years ago

5 0

Answer:

Kay sold 67; Allen sold 50

Step-by-step explanation:

Let "a" represent the number of phones that Allen sold.

a + (a+17) = 117 . . . equation used to find the answer

2a = 100 . . . . . . . . subtract 17, collect terms

a = 50 . . . . . . . . . . divide by 2; the number Allen sold

a+17 = 67 . . . . . . . . Kay sold 17 more than Allen

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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

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3 years ago

on a recent test matthew got 18 out of 40 problems correct at the same rate how many problems would matthew get correct if there

Flauer [41]

Answer:

27

Step-by-step explanation:

18/40 = 27/60

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2 years ago

What is the product of -2x^3 + x -5 and x^3 – 3x?

lana66690 [7]

(-2x^3 +x -5) * (x^3 -3x)
= x^6*(-2*1) +x^4*(-2*-3 +1*1) +x^3*(-5) +x^2*(-3) +x(-5*-3)
= -2x^6 +7x^4 -5x^3 -3x^2 +15x

8 0

3 years ago

Sophia bought 18 packs of lemonade each pack has 24 cans answer key

Juliette [100K]

Answer is 432 cans in total
18 packs times 24 cans in each is 432

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2 years ago

Solve for c.<br> a(c+b) = d.

kipiarov [429]

Answer:

Step-by-step explanation:

Hello,

we assume that a is different from 0 otherwise there is not much we can say on c and then we can write

a(c+b)=d

<=> ac+ab=d

<=> ac = d - ab

<=> $c = \dfrac{d-ab}{a} =\dfrac{d}{a}-b$

Hope this helps

3 0

3 years ago