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zhannawk [14.2K]
2 years ago
15

Find the length of each Arc. Do not round. Part 2. NO LINKS!! ​

Mathematics
1 answer:
Natali5045456 [20]2 years ago
5 0

Answer:

10)  95π/6 ft

12)  39π/4 ft

Step-by-step explanation:

10) 2π * 19 * 150/360 = 95π/6 ft

12)  2π* 13 * (3π/4)/(2π) = 39π/4 ft

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Please help i'm not doing good in math i will give you the crown
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Answer:

B

Step-by-step explanation:

The format for quadrant 2 would be (-,+) and answer choice be fits that because it is(-3,2)

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2 years ago
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Calculate δe, if q= 0.769 kj and w= -860 j .
Andreyy89
<span>Best Answer: 1) Q is positive therefore the process is endothermic. .769kj+-.810kj=-.041kj 2) Heat is released meaning it is exothermic. -66.9kj+45=-21.9kj 3) 7.29kj. Endothermic.</span>
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2 years ago
The average score on a standardized test is 750 points with a standard deviation of 50 points. What is the probability that a st
nalin [4]
The probability that student scored more than 850 we shall proceed as follows:
z=(x-μ)/σ
where:
x=850
μ=750
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thus
z=(850-750)/50
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Answer: P(x>850)=0.0228
4 0
3 years ago
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Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ&gt;0
eduard
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2787701

_______________


\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\&#10;\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\&#10;\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}


Square both sides:

\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\&#10;\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\&#10;\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\&#10;\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}

\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\&#10;\mathsf{17\,sin^2\,\theta=1}\\\\&#10;\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\&#10;\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\&#10;\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}


Since \mathsf{sin\,\theta} is positive, you can discard the negative sign. So,

\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}


Substitute this value back into \mathsf{(i)} to find \mathsf{cos\,\theta:}

\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\&#10;\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}


I hope this helps. =)


Tags:   <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

7 0
2 years ago
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