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3 years ago

5

How do I write an equation with two points on the same line? (15,17) and (20,24) are on the same line. Write the equation of the

line.

1 answer:

Reil [10]3 years ago

5 0

Step-by-step explanation:

First, find the slope between the points.

m = (y₂ − y₁) / (x₂ − x₁)

m = (24 − 17) / (20 − 15)

m = 7/5

Next, write the equation in point-slope form, using either of the points.

y − 17 = 7/5 (x − 15)

If desired, simplify.

y − 17 = 7/5 x − 21

y = 7/5 x − 4

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Answer:

8 is how much you get paid ( sorry if I'm wrong)

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3 years ago

Two times a number minus six another number is six. Two times the first number added to three times the other number is twenty f

expeople1 [14]

Answer:

First Number: 2

Step-by-step explanation:

You need to work backwards, as an algebraic equation.

24 ÷ 3 = 8

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4 0

2 years ago

What is x?: 22+54x= -20 + 60x

DENIUS [597]

Answer:
x = 7

Explanation:
We are given
22 + 54x = -20 + 60x

We can add 20 on both sides using the Addition Property of Equality
42 + 54x = 60x

Subtract both sides by 54x using the Subtraction Property of Equality
42 = 6x

Divide both sides by 6 using the Division Property of Equality
x = 7

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3 years ago

I need help I’m in math class! Help please

AVprozaik [17]

Answer:

1350π $cm^3$ should be your answer

Step-by-step explanation:

15 x 15 x π = 225π

225π x 6 = 1350π

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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:

<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