Keeping in mind that for a cost C(x) and profit P(x) and revenue R(x), the marginal cost, marginal profit and marginal revenue are respectively dC/dx, dP/dx and dR/dx, then
![\bf P(x)=0.03x^2-3x+3x^{0.8}-4400 \\\\\\ \stackrel{marginal~profit}{\cfrac{dP}{dx}}=0.06x-3+2.4x^{-0.2} \\\\\\ \cfrac{dP}{dx}=0.06x-3+2.4\cdot \cfrac{1}{x^{0.2}}\implies \cfrac{dP}{dx}=0.06x-3+2.4\cdot \cfrac{1}{x^{\frac{1}{5}}} \\\\\\ \cfrac{dP}{dx}=0.06x-3+\cfrac{2.4}{\sqrt[5]{x}}](https://tex.z-dn.net/?f=%5Cbf%20P%28x%29%3D0.03x%5E2-3x%2B3x%5E%7B0.8%7D-4400%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bmarginal~profit%7D%7B%5Ccfrac%7BdP%7D%7Bdx%7D%7D%3D0.06x-3%2B2.4x%5E%7B-0.2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06x-3%2B2.4%5Ccdot%20%5Ccfrac%7B1%7D%7Bx%5E%7B0.2%7D%7D%5Cimplies%20%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06x-3%2B2.4%5Ccdot%20%5Ccfrac%7B1%7D%7Bx%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06x-3%2B%5Ccfrac%7B2.4%7D%7B%5Csqrt%5B5%5D%7Bx%7D%7D)
Regrouping in subtraction means that if you can't subtract because the digit on the bottom has a greater value than the digit on the top (such as 3-6)m so in this case you would have to regroup twice.
bearing in mind that, on the III Quadrant, sine as well as cosine are both negative, and that hypotenuse is never negative, so, if the sine is -4/5, the negative number must be the numerator, so sin(x) = (-4)/5.
![\bf sin(x)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-(-4)^2}=a\implies \pm\sqrt{9}=a\implies \pm 3=a \\\\\\ \stackrel{III~Quadrant}{-3=a}~\hfill cos(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28x%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-4%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B5%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B5%5E2-%28-4%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B9%7D%3Da%5Cimplies%20%5Cpm%203%3Da%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7BIII~Quadrant%7D%7B-3%3Da%7D~%5Chfill%20cos%28x%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B-3%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B5%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)}\qquad \leftarrow \textit{let's use this one} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20tan%5Cleft%28%5Ccfrac%7B%5Ctheta%7D%7B2%7D%5Cright%29%3D%20%5Cbegin%7Bcases%7D%20%5Cpm%20%5Csqrt%7B%5Ccfrac%7B1-cos%28%5Ctheta%29%7D%7B1%2Bcos%28%5Ctheta%29%7D%7D%20%5C%5C%5C%5C%20%5Ccfrac%7Bsin%28%5Ctheta%29%7D%7B1%2Bcos%28%5Ctheta%29%7D%5Cqquad%20%5Cleftarrow%20%5Ctextit%7Blet%27s%20use%20this%20one%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B1-cos%28%5Ctheta%29%7D%7Bsin%28%5Ctheta%29%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

Answer:
this question doesn't make sense. Isn't there a speed?