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tatuchka [14]
3 years ago
14

In circle "R" below, determine the length of arc AB if BR = 5 units.

Mathematics
2 answers:
ki77a [65]3 years ago
7 0
See the pic, I'd solved in it.

riadik2000 [5.3K]3 years ago
4 0
l=2 \pi r( \frac{degrees}{360} )

l=2 \pi × 5 × \frac{135}{360} = 11.78


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Help me with the right answer
Nadya [2.5K]

Answer:

x = 3

y = 1

Step-by-step explanation:

9x - y = 26  ---------------(I)

9x + y = 28 ---------------(II)

9x - y = 26

9x = 26 + y

Substitute 9x = 26 + y in equation (II)

26 + y + y = 28               {add like terms}

26 + 2y = 28             {subtract 26 from both sides}

         2y = 28 - 26

        2y = 2    {divide both sides by 2}

           y = 2/2

y = 1

Substitute y =1 in equation (I)

9x - 1 = 26

9x = 26 + 1

9x = 27

x = 27/9

x = 3

x = 3

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6 0
3 years ago
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Step2247 [10]

Answer: A : decreasing B : constant C: increasing D: Constant

Step-by-step explanation:

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Hoochie [10]

Answer:

\large\boxed{\sin2\theta=\dfrac{\sqrt3}{2},\ \cos2\theta=\dfrac{1}{2}}

Step-by-step explanation:

We know:

\sin2\theta=2\sin\theta\cos\theta\\\\\cos2\theta=\cos^2\theta-\sin^2\thet

We have

\sin\theta=\dfrac{1}{2}

Use \sin^2\theta+\cos^2\theta=1

\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}

\sin2\theta=2\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\sqrt3}{2}\\\\\cos2\theta=\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2}

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3 years ago
Brainliest plsss i need help
lesantik [10]

Answer:

I believe it is A

Step-by-step explanation:

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3 years ago
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Answer:

Step-by-step explanation:

formula for cylinder:V=πr2h

shape of the base: circle

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V=3.14x6.5^2x16

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3 years ago
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