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dedylja [7]
3 years ago
13

Show all work please!!!!!!

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
7 0
The midpoint formula is:
(point1 + point2)/2

1) (-2 + 12)/2 = 5

2) (-1 +5)/2 = 2
    (0+2)/2 = 1
     (2,1)

3) Sqrt of (x2-x1)^2 + (y2 - y1)^2
     sqrt of (7 -1)^2 + (-7 - 1)^2
     sqrt of (6)^2 + (-8)^2
     sqrt of 36 + 64
     sqrt of 100
     10

Hope this helps!
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Answer:

8090.54

Step-by-step explanation:

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4 0
2 years ago
the average rainfall in Clearview is 38 inches . this year 29.777 inches fell . How much less rain fell this year than falls in
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6 0
3 years ago
Which of the following is a polynomial function in factored form with zeros at –2, 5, and 8?
Usimov [2.4K]
(x + 2)(x - 5)(x - 8)
Substitute them into each other to get the whole equation)

7 0
3 years ago
A triangle has vertices at (1,10), (-5,2), and (7,2). What is its orthocenter. Show your work.
kondor19780726 [428]

Answer:

Question: What is the orthocenter of a triangle with the vertices (-1,2) (5,2) and (2,1)?

The coordinates of point A are (-1,2), point B are (5,2), and point C are (2,1).

The orthocent is the intersection of the three altitudes. An altitude goes from a vertex and is perpendicular to the line containing the opposite side.

In the coordinate plane the equations of the altitudes can be found and then a system of equations can be solved.

Altitude 1. From point C perpendicular to the line containing side AB.

Slope of line AB is 0 (horizontal line), a vertical line is perpendicular to a horizontal line. Thus, the equation of altitude 1 is  x=2 .

Altitude 2. From point B perpendicular to the line containing side AC.

Slope of line AC is  −13 , the slope of a line perpendicular to line AC is 3. The equation of altitude 2 is  y=3x−13  

Altitude 3. From point A perpendicular to the line containing side BC.

Slope of line BC is  13 , the slope of a line perpendicular to line BC is  −3 . The equation of altitude 3 is  y=−3x−1  

The orthocenter is the point where all three altitudes intersect.

x=2  

y=3x−13  

y=−3x−1  

Use substitution to solve the first two equations  y=3(2)−13=−7  

The orthocenter is the point  (2,−7)  

we did not need the third equation, but we can use it as a check, plug the coordinates into the third equation:

−7=−3(2)−1  

−7=−6−1  

−7=−7  it works.

3 0
3 years ago
Use Cramer's rule to find the solution to the following system of linear equations.
slavikrds [6]

Answer:

x = 53

y = 30

Step-by-step explanation:

Step(I):-

Given equations are

x -2y =-7 ...(I)

5x-9y =-5 ..(ii)

The matrix form  AX = B

                    \left[\begin{array}{ccc}1&-2\\ 5  & -9\\\end{array}\right] \left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{ccc}-7\\-5\\\end{array}\right]

The determinant

= \left|\begin{array}{ccc}1&-2\\5&-9\\\end{array}\right| = -9+10 =1

By using Cramer's Rule

Δ₁ =      \left[\begin{array}{ccc}-7&-2\\\\-5&-9\end{array}\right]



The determinant is     Δ₁ = -9 X -7 - (10 ) = 53

x = Δ₁ / Δ

x = 53

The determinant

Δ₂ =



Δ₂ = -5 +35

     

 

y = Δ₂/Δ =  30

     

5 0
2 years ago
Read 2 more answers
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