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Vlad1618 [11]
2 years ago
6

Daniel wrote the following two-column proof for the given information.

Mathematics
1 answer:
Pavel [41]2 years ago
8 0

The information given about the proof does that Daniel made an error on line 2.

<h3>How to illustrate the information?</h3>

Given:

1. AB = 3x +2;  BC = 4x + 8;  AC = 38

2. AB + BC = AC      incorrect (not an angle angle addition postulate)

3. 3x+2 + 4x + 8 = 38         correct

4. 7x + 10 = 38                    correct

5. 7x = 28                           correct

6. x = 4

Daniel made an error on line 2.

Here is the complete question:

Daniel wrote the following two-column proof for the given information. Given: AB = 3x + 2; BC = 4x + 8; AC = 38 Prove: x = 4 Statements Reason 1. AB = 3x + 2; BC = 4x + 8; AC = 38 1. Given 2. AB + BC = AC 2. Angle Addition Postulate 3. 3x + 2 + 4x + 8 = 38 3. Substitution Property of Equality 4. 7x + 10 = 38 4. Combining Like Terms 5. 7x = 28 5. Subtraction Property of Equality 6. x = 4 6. Division Property of Equality On which line, did Daniel make his error? line 2 line 3 line 4 line 5

Learn more about proof on:

brainly.com/question/4134755

#SPJ1

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Answer:

The coordinates of the midpoint of the segment are (-5.5,0,-8)

Step-by-step explanation:

In this question, we are tasked with calculating the midpoint of the segment PQ.

To calculate this, we employ the use of a mathematical formula as follows;

The coordinate of the midpoint are = {(x1+x2)/2, (y1+y2)/2 , (z1+ z2)/2}

Thus we have;

{(-7-4)/2, (3-3)/2 , (-7-9)/2} = (-11/2, 0/2, -16/2)

= (-5.5,0,-8)

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How do you do this problem?
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x = 125°

Step-by-step explanation:

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Find the orthocenter for the triangle described by each set of vertices.
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Answer:

The orthocentre of the given vertices ( 2 , -3.5)

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

The orthocentre is the intersecting point for all the altitudes of the triangle.

The point where the altitudes of a triangle meet is known as the orthocentre.

Given Points are K (3.-3), L (2,1), M (4,-3)

<em>The Altitudes are perpendicular line from one side of the triangle to the opposite vertex</em>

<em>The altitudes are  MN , KO , LP</em>

<u><em>step(ii):-</em></u>

<em>  </em>  Slope of the line  

                          KL = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  } = \frac{1-(-3)}{2-3} = -4

The slope of MN =

The perpendicular slope of KL

                           = \frac{-1}{m} = \frac{-1}{-4} = \frac{1}{4}

The equation of the altitude

                                 y - y_{1} = m( x-x_{1} )

                                y - (-3) = \frac{1}{4} ( x-4 )

                               4y +12 = x -4

                                x - 4 y -16 = 0 ...(i)

Step(iii):-

 Slope of the line  

                          LM = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  } = \frac{-3-1}{4-2} = -2

The slope of KO =

The perpendicular slope of LM

                           = \frac{-1}{m} = \frac{-1}{-2} = \frac{1}{2}

The equation of the altitude

                                 y - y_{1} = m( x-x_{1} )

The equation of the line passing through the point K ( 3,-3) and slope

m = 1/2

                                y - (-3) = \frac{1}{2} ( x-3 )

                                2y +6 = x -3

                             x - 2y -9 =0 ....(ii)    

Solving equation (i) and (ii) , we get

  subtracting equation (i) and (ii) , we get

                   x - 4y -16 -( x-2y-9) =0

                       - 2y -7 =0

                      -2y = 7

                        y = - 3.5

Substitute y = -3.5 in equation x -4y-16=0

               x - 4( -3.5) - 16 =0

               x +14-16 =0

                x -2 =0

                  x = 2

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