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Aloiza [94]
3 years ago
14

What should I notice in every right triangle?

Mathematics
1 answer:
AURORKA [14]3 years ago
4 0

Answer:

Right triangles are triangles in which one of the interior angles is 90 degrees, a right angle. Since the three interior angles of a triangle add up to 180 degrees, in a right triangle, since one angle is always 90 degrees, the other two must always add up to 90 degrees (they are complementary).

The side opposite the right angle is called the hypotenuse. The sides adjacent to the right angle are the legs. When using the Pythagorean Theorem, the hypotenuse or its length is often labeled with a lower case c. The legs (or their lengths) are usually labeled a and b.

Step-by-step explanation:

pls mark as brainliest

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4 years ago
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gtnhenbr [62]

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<h3>How to graph the inequalities?</h3>

The system of inequalities is given as:

y > 2x + 4

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(0.667, 5.333)

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Vlada [557]
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Find the equation of the linear function represented by the table below in slope-
Harman [31]

Answer:

y = 4x+4

Step-by-step explanation:

Using the coordinate point (1, 8) and (2, 12) from the table

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c = 4

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y = 4x + 4

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8 0
3 years ago
Solve the system <br> 2x+2y+z=-2 <br> -x-2y+2z=-5<br> 2x+4y+z=0
drek231 [11]

The values of (x,y,z) are (3,-1,-2) , if the given are equations 2x+2y+z=- 2,-x-2y+2z=-5 and 2x+4y+z=0.

Step-by-step explanation:

The given is,

                          2x+2y+z=- 2.......................................(1)

                         -x-2y+2z=-5......................................(2)

                            2x+4y+z=0.........................................(3)

Step:1

           Equation (2) is multiplied by -1            ( Eqn(2) × -1 )

                                         x+2y-2z=5.............................(4)

          Subtract the equation (1) and (4)

                                        2x+2y+z=- 2

                                         x+2y-2z=5

                 ( - )

           (2x-x)+(2y-2y)+(z+2z)=(-2-5)

                                                  x+3z=-7......................(5)

Step:2

          Equation (2) is multiplied by -2                 ( Eqn(2) × -2)

                                        2x+4y-4z=10........................(6)

         Subtract equation (6) and (3),                  

                                        2x+4y-4z=10

                                         2x+4y+z=0

                   ( - )

       (2x-2x)+(4y-4y)+(-4z-z)= (10-0)

                                                     -5z=10

                                                         z = - \frac{10}{5}

                                                         z = -2

         From the equation (5),

                                                  x+3z=-7  

                                          x+(3)(-2)=-7

                                                           x = -7+6

                                                            x = -1

         From equation (1),

                                            2x+2y+z=- 2

          Substitute x and z values,

                               (2)(-1)+2y+(-2)=-2

                                                     2y - 4=2

                                                           2y=4-2

                                                           2y=2

                                                            y=\frac{2}{2}

                                                             y = 1

Step:3

                Check for solution,

                                  -x-2y+2z=-5

                Substitute x,y and z values,

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

                                         1-2-4=-5

                                                    -5 = -5

Result:

              The values of (x,y,z) are (3,-1,-2) , if the given are equations 2x+2y+z=- 2,-x-2y+2z=-5 and 2x+4y+z=0.

8 0
4 years ago
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