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stealth61 [152]
2 years ago
11

How long is the arc intersected by a central angle of startfraction pi over 3 endfraction radians in a circle with a radius of 6

ft? round your answer to the nearest tenth. use 3.14 for pi.
Mathematics
1 answer:
vova2212 [387]2 years ago
4 0

The measure of the arc length is 6.28 units if the arc intersected by a central angle of  π/3 radians in a circle with a radius of 6 ft.

<h3>What is a circle?</h3>

It is described as a set of points, where each point is at the same distance from a fixed point (called the centre of a circle)

We have:

The radius of the circle r = 6 ft

The measure of the central angle θ = π/3

We know,

s = rθ

s is the arc length.

s = 6(π/3)

s = 2(3.14)

s = 6.28

Thus, the measure of the arc length is 6.28 units if the arc intersected by a central angle of  π/3 radians in a circle with a radius of 6 ft.

Learn more about circle here:

brainly.com/question/11833983

#SPJ1

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Step-by-step explanation:

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3 years ago
How do I complete these questions a bit confused . (extra credit work )​
otez555 [7]

Answer:

(1)

a = \frac{3\sqrt 3}{2}

b = \frac{3}{2}

(2)

a = \sqrt 6

b = \sqrt 2

Step-by-step explanation:

Solving (1):

Considering

\theta = 60^o

We have:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

This gives:

\sin(60^o) = \frac{a}{3}

Solve for a

a = 3 * \sin(60^o)

\sin(60^o) = \frac{\sqrt 3}{2}

So:

a = 3 * \frac{\sqrt 3}{2}

a = \frac{3\sqrt 3}{2}

To solve for b, we make use of Pythagoras theorem

3^2 = a^2 + b^2

This gives

3^2 = (\frac{3\sqrt 3}{2})^2 + b^2

9 = \frac{9*3}{4} + b^2

9 = \frac{27}{4} + b^2

Collect like terms

b^2 = 9 - \frac{27}{4}

Take LCM and solve

b^2 = \frac{36 - 27}{4}

b^2 = \frac{9}{4}

Take square roots

b = \frac{3}{2}

Solving (2):

Considering

\theta = 60^o

We have:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

This gives:

\sin(60^o) = \frac{a}{2\sqrt 2}

Solve for a

a = 2\sqrt 2 * \sin(60^o)

\sin(60^o) = \frac{\sqrt 3}{2}

So:

a = 2\sqrt 2 * \frac{\sqrt 3}{2}

a = \sqrt 2 * \sqrt 3

a = \sqrt 6

To solve for b, we make use of Pythagoras theorem

(2\sqrt 2)^2 = a^2 + b^2

This gives

(2\sqrt 2)^2 = (\sqrt 6)^2 + b^2

8 = 6 + b^2

Collect like terms

b^2 = 8 - 6

b^2 = 2

Take square roots

b = \sqrt 2

3 0
3 years ago
Use absolute value to express the distance between -10 and 16
Liono4ka [1.6K]

Answer:

The distance is 6

Step-by-step explanation:

the absolute value of -10 is simply 10.

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