A carnival ride is in the shape of a wheel with a radius of 15 feet. The wheel has 20 cars equally spaced apart along the outside of the wheel. What is the area of a sector between any two cars? Round answer to the nearest hundredth if applicable. Numbers and Decimal only.

Each central angle will have measure \dfrac{2\pi}{24}=\dfrac\pi{12}\text{ rad}=15^\circ242π=12π rad=15∘ .

The arc length LL of each 1/24-th section of the wheel occurs in the following ratio with the whole wheel's circumference:

\dfrac{2\pi\text{ rad}}{2\pi(15)\text{ ft}}=\dfrac{\frac\pi{12}\text{ rad}}L\implies\dfrac1{15}=\dfrac\pi{12L}\implies L=\dfrac{15\pi}{12}\text{ ft}2π(15) ft2π rad=L12π rad⟹151=12Lπ⟹L=1215π ft

L\approx3.93\text{ ft}L≈3.93 ft

The area of the sector AA occurs in a similar ratio with the wheel's total area:

\dfrac{2\pi\text{ rad}}{\pi(15)^2\text{ ft}^2}=\dfrac{\frac\pi{12}\text{ rad}}A\implies\dfrac2{225}=\dfrac\pi{12A}\implies A=\dfrac{75\pi}8\text{ ft}^2π(15)2 ft22π rad=A12π rad⟹2252=12Aπ⟹A=875π ft2

A\approx29.45\text{ ft}^2A≈29.45 ft2

h who knows

step-by-step explanation:

4 and 5)16.97 or 12√2

6)15.59 or 9√3

7)y = 15.59 or 9√3     x =27

8)17.80