The answer is -6
So you need to drop the word sum on -6
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)
Answer: f(0) = 5.5 f(2) = 3.7
<u>Step-by-step explanation:</u>

When x = 0, f(x) = 5.5 <em>In other words, f(0) = 5.5</em>
When x = 2, f(x) = 3.7 <em>In other words, f(2) = 3.7</em>
Suppose y varies inversely with x and that y=2 when x=8 What is an equation for this inverse variation?
B. y= 2/8x
1 7/8 tons of sand will be required to fill the playground