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Llana [10]
2 years ago
5

What is the correct order of steps for bisecting angle ABC?

Mathematics
1 answer:
Marizza181 [45]2 years ago
8 0

The correct order for bisecting angle ABC is F D E B C A. Option D

<h3>Steps in bisecting angles</h3>

The steps involved in bisecting angles are;

  • Place compass point on the vertex of the angle (point B).
  • Stretch the compass to any length that will stay OF the angle
  • Swing an arc so the pencil crosses both sides (rays) of the given angle. You should now have two intersection points with the sides (rays) of the angle
  • Place the compass point on one of these new intersection points on the sides of the angle. Stretch the compass to a sufficient length to place your pencil well into the interior of the angle, this should be within the rays of the angle
  • Place an arc in this interior
  • Without changing the span on the compass, place the point of the compass on the other intersection point on the side of the angle and make a similar arc. The two small arcs in the interior of the angle should be intersecting
  • Connect the vertex of the angle (point B) to this intersection of the two small arcs

From the listed steps, the correct order for bisecting angle ABC is F D E B C A. Option D

Learn more about bisectors here:

brainly.com/question/11006922

#SPJ1

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154π cm²

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Radius (r) = 7 cm

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Step-by-step explanation:

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Solve 12^x^2+5x-4 = 12^2x+6
faust18 [17]

The solutions for ‘x’ are 2 and -5

<u>Step-by-step explanation:</u>

Given equation:

                    12^{x^{2}+5 x-4}=12^{2 x+6}

Since the base on both sides as ‘12’ are the same, we can write it as

                     x^{2}+5 x-4=2 x+6

                     x^{2}+5 x-2 x-4-6=0

                     x^{2}+3 x-10=0

Often, the value of x is easiest to solve by a x^{2}+b x+c=0 by factoring a square factor, setting each factor to zero, and then isolating each factor. Whereas sometimes the equation is too awkward or doesn't matter at all, or you just don't feel like factoring.

<u>The Quadratic Formula:</u> For a x^{2}+b x+c=0, the values of x which are the solutions of the equation are given by:

                       x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Where, a = 1, b = 3 and c = -10

                       x=\frac{-3 \pm \sqrt{(-3)^{2}-4(1)(-10)}}{2(1)}

                       x=\frac{-3 \pm \sqrt{9+40}}{2}

                       x=\frac{-3 \pm \sqrt{49}}{2}=\frac{-3 \pm 7}{2}

So, the solutions for ‘x’ are

                       x=\frac{-3+7}{2}=\frac{4}{2}=2

                       x=\frac{-3-7}{2}=\frac{-10}{2}=-5

The solutions for ‘x’ are 2 and -5

6 0
3 years ago
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