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USPshnik [31]
3 years ago
12

The areas of the squares adjacent to two sides of a right triangle are 32 units^2 2 squared and 32 units^2 Find the length, xxx,

of the third side of the triangle.
Mathematics
1 answer:
NikAS [45]3 years ago
6 0

Answer:

8 units

Step-by-step explanation:

If the area of the squares are 32 units^2, we have that:

Area = Side * Side

Side^2 = 32

Side = sqrt(32) = 5.657 units

So as the squares are adjacent to two sides of the triangle, we have that the triangle has two sides of 5.657 units

As it is a right triangle, we can use the Pythagoras' theorem:

c^2 = a^2 + b^2

c^2 = 5.657^2 + 5.657^2

c^2 = 64

c = 8 units

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Step-by-step explanation:

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Which of the following expressions always represents how many more votes the winner received than the

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Unit 3 homework 6 Gina Wilson
Sonja [21]

Answer:

5) The equation of the straight line is   2 x - y + 1 =0

6) The equation of the straight line is   x + y -5 =0

7) The equation of the straight line is   5 x + 6 y - 24 =0

8) The equation of the straight line is  x - 4 y -4 =0

9) The equation of the parallel line is 3x + y -19 =0

Step-by-step explanation:

5)

The equation of the straight line is

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

          Slope of the line

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

        Given points are (1,3) , ( -3,-5)

         m = \frac{-5-3  }{-3-1 } = \frac{-8}{-4} = 2

The equation of the straight line is

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

                        y - 3 = 2 ( x - 1 )

                        y = 2x - 2 +3

                       2 x - y + 1 =0

The equation of the straight line is   2 x - y + 1 =0

  6)

The equation of the straight line is

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

          Slope of the line

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

        Given points are (1,4) , ( 6,-1)

         m = \frac{-1-(4)  }{6-1 } = \frac{-5}{5} = -1

The equation of the straight line is  

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

                        y - 1 = -1 ( x - 4 )

                        y - 1 = - x +4

The equation of the straight line is   x + y -5 =0

7)

The equation of the straight line is

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

          Slope of the line

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

 Given points are (-12 , 14) , ( 6,-1)

         m = \frac{-1-(14)  }{6+12 } = \frac{-15}{18} = \frac{-5}{6}

         y - 14  = \frac{-5}{6}  ( x - (-12) )

       6( y - 14 ) = - 5 ( x +12 )

      6 y - 84 = - 5x -60

       5 x + 6 y  -84 + 60 =0

      5 x + 6 y - 24 =0

8)

The equation of the straight line is

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

          Slope of the line

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

Given points are (-4 , -2) , ( 4 , 0)

              m = \frac{0+2}{4 +4}  = \frac{2}{8} = \frac{1}{4}

           y - (-2)  =\frac{1}{4} ( x - (-4) )

             4 ( y + 2) = x + 4

                x - 4 y -4 =0

  9)

The equation of the line y = 3x + 6  is parallel to the line

3x + y + k =0 is passes through the point ( 4,7 )

⇒   3x + y + k =0

⇒    12 + 7 + k =0

⇒    k = -19

The equation of the parallel line is 3x + y -19 =0

     

4 0
3 years ago
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