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3 years ago

Solve the equation thanks!

Mathematics

2 answers:

AlladinOne [14]3 years ago

8 0

Answer:

c = $\frac{27}{40}$

Step-by-step explanation:

To eliminate the fractions, multiply all terms by the lowest common multiple of 8 and 5

The lowest common multiple of 8 and 5 is 40

5 + 40c = 32 ( subtract 5 from both sides )

40c = 27 ( divide both sides by 40 )

c = $\frac{27}{40}$

VMariaS [17]3 years ago

6 0

The answer to this problem is 27/40.

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A system of equations has infinitely many solutions. If 2y – 4x = 6 is one of the equations, which could be the other equation?

Molodets [167]

Hi there!

$\large\boxed{y = 2x + 6}$

For a system to have infinite solutions, the lines have to be the same. We can begin by rearranging the given equation into the format y = mx + b:

2y - 4x = 6

Move x to the opposite side:

2y = 4x + 6

Divide all terms by 2:

y = 2x + 6

The answer choice that matches this is the first one.

5 0

3 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:

Optimizing With Lagrange Multipliers

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

5 0

3 years ago

17-(5-p)=2(5p-16) Step by step explanation please

12345 [234]

Answer:

p= 44 /9

(44 over 9)

Step-by-step explanation:

3 0

2 years ago

What is the distance from the point (a, −b, −4) to the origin?

laiz [17]

Answer:

$\sqrt{a^2+b^2+16}$

Step-by-step explanation:

The distance (d) between two points in 3 dimensions is ...

d = √((x2 -x1)² +(y2 -y1)² +(z2 -z1)²)

Then the distance between (a, -b, -4) and (0, 0, 0) is ...

d = √((a -0)² +(-b -0)² +(-4 -0)²)

= √(a² +b² +16)

6 0

3 years ago

Particle 1 of charge q1 �� 5.00q and particle 2 of charge q2 �� 2.00q are fixed to an x axis. (a) as a multiple of distance

Naya [18.7K]

Assuming that the particle is the 3rd particle, we know that it’s location must be beyond q2; it cannot be between q1 and q2 since both fields point the similar way in the between region (due to attraction). Choosing an arbitrary value of 1 for L, we get

k q1 / d^2 = - k q2 / (d-1)^2

Rearranging to calculate for d:

(d-1)^2/d^2 = -q2/q1 = 0.4 
 d^2-2d+1 = 0.4d^2 
0.6d^2-2d+1 = 0
d = 2.72075922005613
d = 0.612574113277207

We pick the value that is > q2 hence,

d = 2.72075922005613*L

d = 2.72*L

3 0

3 years ago