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coldgirl [10]
3 years ago
12

How do you find the equation of a line with Guassian elimination given two points?

Mathematics
1 answer:
Paraphin [41]3 years ago
7 0

Answer:

  The solution is similar to the 2-point form of the equation for a line:

  y = (y2 -y1)/(x2 -x1)·x + (y1) -(x1)(y2 -y1)/(x2 -x1)

Step-by-step explanation:

Using the two points, write two equations in the unknowns of the equation of the line.

For example, you can use the equation ...

  y = mx + b

Then for the points (x1, y1) and (x2, y2) you have two equations in m and b:

  b + (x1)m = (y1)

  b + (x2)m = (y2)

The corresponding augmented matrix for this system is ...

  \left[\begin{array}{cc|c}1&x1&y1\\1&x2&y2\end{array}\right]

____

The "b" variable can be eliminated by subtracting the first equation from the second. This puts a 0 in row 2 column 1 of the matrix, per <em>Gaussian Elimination</em>.

  0 + (x2 -x1)m = (y2 -y1)

Dividing by the value in row 2 column 2 gives you the value of m:

  m = (y2 -y1)/(x2 -x1)

This value can be substituted into either equation to find the value of b.

  b = (y1) -(x1)(y2 -y1)/(x2 -x1) . . . . . substituting for m in the first equation

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Give me a correct answer, My answer probably wrong.
Georgia [21]

Hello!

This is a problem about the general solution of a differential equation.

What we can first do here is separate the variables so that we have the same variable for each side (ex. dy with the y term and dx with the x term).

\frac{dy}{dx}=\frac{x-1}{3y^2}

3y^2dy=x-1dx

Then, we can integrate using the power rule to get rid of the differentiating terms, remember to add the constant of integration, C, to at least one side of the resulting equation.

y^3=\frac{1}{2}x^2-x+C

Then here, we just solve for y and we have our general solution.

y=\sqrt[3]{\frac{1}{2}x^2-x+C}

We can see that answer choice D has an equivalent equation, so answer choice D is the correct answer.

Hope this helps!

3 0
2 years ago
Solve the equation. 2x squared-3x+1=0
natali 33 [55]

2x^2-3x+1=0\\\\2x^2-2x-x+1=0\\\\2x(x-1)-1(x-1)=0\\\\(x-1)(2x-1)=0\iff x-1=0\ \vee\ 2x-1=0\ \ \ \ |+1\\\\x=1\ \vee\ 2x=1\ \ \ |:2\\\\\boxed{x=1\ \vee\ x=0.5}

4 0
3 years ago
Find the value of x.<br> If necessary, you may learn what the markings on a figure indicate.<br> 74°
marissa [1.9K]

Answer:

x = 61

Step-by-step explanation:

Left hand triangle containing 1 angle of 74

Label the other angle opposite the marked side also as 74

Find the third angle. Call it y.

y + 74 + 74 = 180           Combine like terms

y + 148 = 180                 Subtract 148 from both sides.

y = 180 - 148

y = 32

=============

Now work with the triangle on the right.

label the angle making up the right angle = z

32 + z = 90                             These two angles are complementary = 90

32 - 32 + z = 90 - 32              Subtract 32 from both sides

z = 58                                      Use 58 wherever you see z

x + x + z = 180                         Substitute

2x + 58 = 180                          Subtract 58 from both sides

2x = 122                                  Divide by 2

x = 61

5 0
3 years ago
Read 2 more answers
I NEED HELP ASAP!!!!!! 20 POINTS. Jake and Finn both worked hard over the summer. Together they earned a total of $400. Finn ear
Semenov [28]

Answer:

225 for finn jake gets 175

Step-by-step explanation:

400 divided by 2 = 200+25 witch gives Finn 225 dollars.  

J=175 f=225

4 0
2 years ago
Read 2 more answers
100 POINTS AND BRAINLIEST
zysi [14]

Answer:

8 square units and \frac{40}{3} square units

Step-by-step explanation:

The area of the triangle ABC is 24 square units.

1. Triangles ABC and FBG are similar with scale factor \frac{1}{3}, then

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2. Triangles ABC and DBE are similar with scale factor \frac{2}{3}, then

\dfrac{A_{\triangle DBE}}{A_{\triangle ABC}}=\dfrac{4}{9}\Rightarrow A_{\triangle DBE}=\dfrac{4}{9}\cdot 24=\dfrac{32}{3}\ un^2.

3. Thus, the area of the quadrilateral DFGE is

A_{DFGE}=A_{\triangle DBE}-A_{\triangle FBG}=\dfrac{32}{3}-\dfrac{8}{3}=8\ un^2.

and the area of the quadrilateral ADEC is

A_{ADEC}=A_{\triangle ABC}-A_{\triangle DBE}=24-\dfrac{32}{3}=\dfrac{40}{3}\ un^2.

4 0
3 years ago
Read 2 more answers
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