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valina [46]
2 years ago
6

Geometry worksheet 11.1-11.2 angles and arcs in a circle answer key

SAT
1 answer:
Andreyy892 years ago
6 0

An example of Geometry worksheet  question is "What is the difference between a minor arc and a major arc".

<h3>What is an arc?</h3>

In Mathematics, the term “arc”  connote a type of circular shape or a curve that merges two endpoints.

The difference between a minor arc and a major arc" is that:

  • The measure of a minor arc is known to be less than 180°
  • The measure of a major arc is known to be greater than 180°

Conclusively,  through the measurement of the arc, one can know if it is a major or a minor arc.

Learn more about Geometry from

brainly.com/question/12717588

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about 2 of them would not

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What does this excerpt tell us most about the cat? she sleeps a lot. She is content. She feels cold. She is unconcerned.
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1/2 (6x-4)-(3-x)=ax+x+b. What is the value of a-b
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1/2 (6x-4)-(3-x)=ax+x+b
3x - 2 - 3 + x = (a+1)x + b
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6 0
3 years ago
a helicopter flies parallel to the ground at an altitude of 1/2 kilometer and at a speed of 2 kilometers per minute. if the heli
Murrr4er [49]

The Pythagoras' theorem and the uniform motion allows to find the answer for the rate of change of distance is

               v = 32.3 m / s

The distance is the length of the segment that a white house wing with the helicopter, this can be found using the Pythagoras' theorem

               R = \sqrt{x^2 + y^2}

Where R is the distance, x is the horizontal distance where the helicopter flies and y is the vertical distance that corresponds to the flight height

y = 500 m

Since the helicopter flies at constant speed, we can use the uniform motion relation

              v_h = \frac{x}{t}

              x = v_ht

Where v_h is the speed of the helicopter and t is the time

We substitute  

                r = \sqrt {(V_h \ t)^2 + y^2}

Kinematics defines velocity as the change of position in the unit of time

            v = \frac{dR}{dt}

           v = \frac{1}{2} \  \frac{2 ( v_h t) v_h}{ \sqrt{(v_h t)^2 + y^2} }

           v = \frac{v_h^2 \ t}{ \sqrt{(v_h t)^2 + y^2} }

This expression is the change in distance as a function of time for a given speed. In the exercise, this speed is requested for a time of one minute, for the helicopter speed of

        v = 2km/min (\frac{1000m}{1km} ) ( \frac{1 min}{60s}) = 33.33 m / s

Let's calculate

            v = \frac{33.33^2 \ 60 }{ \sqrt{ )33.3 \ 60)^2 +500^2} }

         

            v = 32.3 m / s

In conclusion using the Pythagoras' theorem and the uniform motion we can find the answer for the rate of change in distance is

               v = 32.3 m / s

Learn more here: brainly.com/question/343682

4 0
3 years ago
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