Find the slope of the tangent line to the polar curve r=2−sin(θ)r=2−sin⁡(θ) at the point specified by θ=π/3

First recall that

x = rcosθ  and  y = rsinθ  

Thus

x = (2 - sinθ)(cosθ)  and  y = (2 - sinθ)(sinθ)

Now

dx/dθ = 2(sinθ)^2 - 2sinθ - 1  and  dy/dθ = 2cosθ - 2sinθcosθ

Now evaluating each derivative above @  θ = π/3,  we obtain

dx/dθ = (1 - 2√3)/2  and  dy/dθ = (2 - √3)/2

Now

dy/dx = (dy/dθ)(dθ/dx)

Thus @  θ = π/3,

dy/dx = [(2 - √3)/2][2/(1 - 2√3)]

dy/dx = (2 - √3)/(1 - 2√3)  is the slope of the required tangent

The answer is c because you turn 17.5% into a decimal which gives you .175 then you times that by $361.35 which gives you 63.23625 but then you round that which gives you 63.24. then you at 63.24 pulse 361.35 and you round it which gives you $425