Question

Can someone please answer. There is one problem. There's a picture. Thank you!

Answer 1

$\dfrac{95^o}{360^o}=\dfrac{19}{72}\\\\Perimeter\ of\ the\ circle:L=2\pi\cdot6=12\pi\approx12\cdot3.14=37.68\\\\\widehat{DG}=\dfrac{19}{72}\cdot37.68\approx9.95\\\\Answer:\boxed{9.95}$

Answer 2

Degrees: $s = \frac{\pi}{180}\theta r$
                $s = \frac{\pi}{180}(\frac{s}{r})(r)$
                $s = \frac{\pi}{180}(\frac{95}{6})(6)$
                $s = \frac{\pi}{180}(15.8\bar{3})(6)$
                $s = \frac{\pi}{180}(95)$
                $s = \frac{95\pi}{180}$
                $s = \frac{19\pi}{36}$
                $s = 95\°$

Radians: $s = \theta r$
               $s = \frac{95}{6}(6)$
               $s = 95$
               $s = \frac{19\pi}{36}$

Arc Length = $\frac{\theta}{360} * C$
Arc Length = $\frac{95}{360} * 2(3.14)(6)$
Arc Length = $\frac{19}{72} * 2(18.84)$
Arc Length = $\frac{19}{72} * 37.68$
Arc Length = $\frac{715.92}{72}$
Arc Length = $9.94\bar{3}$

The answer is A.