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Quiz 2
1. Solve

$x^2+5x+6=0$ .
Solution: The polynomial on the left-hand side factors:

$(x+3)(x+2)=0.$ Therefore by the zero-product property of the real numbers, we must conclude that

$x+3=0$ or

$x+2=0$ . Thus

$x=-3$ or

$x=-2$ .

2. Convert

$32^{\circ}$ to radians.
Solution: Calculate

$32^{\circ} = 32^{\circ} (1) = 32^{\circ} \left( \dfrac{\pi \mathrm{\hspace{2pt} rad}}{180^{\circ}} \right)=\dfrac{32\pi}{180}=\dfrac{8\pi}{45}.$ 3. Convert

$\dfrac{\pi}{8}$ radians to degrees.
Solution: Calculate

$\dfrac{\pi}{8} \mathrm{\hspace{2pt} rad} = \left( \dfrac{\pi}{8} \mathrm{\hspace{2pt} rad} \right) \left( \dfrac{180^{\circ}}{\pi \mathrm{\hspace{2pt} rad}} \right) = \left( \dfrac{45}{2} \right)^{\circ}.$ 4. What is the arc length subtende by an angle of

$\dfrac{\pi}{3}$ radians of a circle with radius

$7$ ?
Solution: In the formula

$s=r\theta$ , we were told

$\theta = \dfrac{\pi}{3}$ and

$r=7$ . Substituting these values in yields

$s=\left(7\right) \left( \dfrac{\pi}{3} \right) = \dfrac{7\pi}{3}.$ 5. What radius must a circle have if an angle of

$13^{\circ}$ subtends an arc length of

$27$ ?
Solution: In the formula

$s=r\theta$ , we are told that

$s=27$ and

$\theta=13^{\circ}$ . But the formula

$s=r\theta$ only works when

$\theta$ is in radians, so we must convert

$13^{\circ}$ to radians in order to use it. So, compute

$13^{\circ} = \left( 13^{\circ} \right) \left( \dfrac{\pi \mathrm{\hspace{2pt} rad}}{180^{\circ}} \right) = \dfrac{13\pi}{180} \mathrm{rad}.$ Now we plug these into

$s=r\theta$ to get

$27=r \left( \dfrac{13\pi}{180} \right)$ . Solve for

$r$ to get

$r = \dfrac{27(180)}{13\pi}=\dfrac{4860}{13\pi}$ .