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2 years ago

How do u solve 5x+3=4 Y=-3x

2 answers:

6 0

The information from the first equation gives you the information needed for the second. To solve the first equation you must rearrange the equation to isolate X. In order to do that you can first move the 3 to the other side of the equation by subtracting it from both sides (5x + 3 - 3 = 4 - 3) and then simplify that to (5x = 4 - 3) and further to (5x = 1). Then to move the 5 you must divide both sides by 5 so you get (5x/5 = 1/5) which can be simplified to (x = 1/5)

From this you can use the X value and input it into the second equation
Y = -3(1/5) and then solve for Y.

Hope this helps!

balu736 [363]2 years ago

6 0

5x+3=4
5x=4-3
5x=1
x=1/5
so here we have the x
y=-3x replace the x
y=-3(1/5)
y=-3/5
good luck

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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:

$g: x_1+x_2+x_3=1000$

$g(x_1,x_2,x_3)= x_1+x_2+x_3-1000$

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

Consider this equation. cos(theta) = (-2√5)/5. If theta is an angle in quadrant II, what is the value of sin(theta)? NO LINKS!!!

vlada-n [284]

In second quadrant sin and cosec are positive

$\\ \tt\hookrightarrow sin^2\theta=1-cos^2\theta$

$\\ \tt\hookrightarrow sin^2\theta=1-(\dfrac{-2\sqrt{5}}{5})^2$

$\\ \tt\hookrightarrow sin^2\theta=1-\dfrac{20}{25}$

$\\ \tt\hookrightarrow sin^2\theta=\dfrac{5}{25}$

$\\ \tt\hookrightarrow sin\theta=\dfrac{\sqrt{5}}{5}$

$\\ \tt\hookrightarrow sin\theta=\dfrac{1}{\sqrt{5}}$

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2 years ago

Find the sum of nth term of the series 1+3/2+5/2^2+7/2^3+.......

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Answer:

an = (1 + 2·(n - 1))/2^(n - 1) = 2^(1 - n)·(2·n - 1)

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3 years ago

Factor. 4x2−49y2 Enter your answer in the box.

Makovka662 [10]

4x² - 49y² = (2x)² - (7y)² = (2x - 7)(2x + 7)

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3 years ago

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Ronch [10]

Answer:

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Step-by-step explanation:

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2 years ago