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Butoxors [25]
3 years ago
5

What is the arc length of a circle that has a 6-inch radius and a central angle that is 65 degrees? Use 3. 14 for pi and round y

our answer to the nearest hundredth
Mathematics
1 answer:
noname [10]3 years ago
3 0
C=2πr

C=2π6

C=<span>37.6991118431
</span>
That's the arc length of the whole circle, i.e. the arc length of 360°.

65° is 18.055555555555555555555555555556% of 360°, (65/360*100)
so 18.055555555555555555555555555556% of 37.6991118431 is 
6.8067840827819444444444444444444.  Rounded to the nearest hundredth, the answer is 6.81 inches.

That's real pi, lets see if it makes a difference to use "stupid pi".

C=2πr

C=2π6

C=37.68

That's the arc length of the whole circle, i.e. the arc length of 360°.

65° is 18.055555555555555555555555555556% of 360°, (65/360*100)
so 18.055555555555555555555555555556% of 37.68 is 
6.803333333333333333333333.  Rounded to the nearest hundredth, the answer is 6.80 inches.
Yep makes a difference.  That's why you don't use stupid pi.  3.14159 is what we always used in engineering, or just the pi button and using a ton of digits.

Answer: 6.80 inches.
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&#10;f'(x) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \  0} \text{   (1)}

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f'(x) \ \textgreater \  0 \Leftrightarrow (x - 2)(x + 2)  \ \textgreater \  0

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f'(x) = 15x^2 - 60

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f'(x) \ \textless \  \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \  0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \  0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \  0} \text{ (2)}

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Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

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Therefore, f is concave up at (2, \infty).

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

f''(x) = 30x - 60

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f''(x) \ \textless \  0 &#10;\\ \\ \Leftrightarrow 30x - 60 \ \textless \  0 &#10;\\ \\ \Leftrightarrow 30(x - 2) \ \textless \  0 &#10;\\ \\ \Leftrightarrow x - 2 \ \textless \  0 &#10;\\ \\ \Leftrightarrow \boxed{x \ \textless \  2}

Therefore, f is concave down at (-\infty, 2).
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