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)
![\bf a)\qquad \cfrac{dP}{dx}=0.06(200)-3+ \cfrac{2.4}{\sqrt[5]{200}} \\\\\\ b)\qquad \cfrac{dP}{dx}=0.06(2000)-3+\cfrac{2.4}{\sqrt[5]{2000}} \\\\\\ c)\qquad \cfrac{dP}{dx}=0.06(5000)-3+\cfrac{2.4}{\sqrt[5]{5000}} \\\\\\ d)\qquad \cfrac{dP}{dx}=0.06(10000)-3+\cfrac{2.4}{\sqrt[5]{10000}}](https://tex.z-dn.net/?f=%5Cbf%20a%29%5Cqquad%20%0A%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06%28200%29-3%2B%20%5Ccfrac%7B2.4%7D%7B%5Csqrt%5B5%5D%7B200%7D%7D%0A%5C%5C%5C%5C%5C%5C%0Ab%29%5Cqquad%20%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06%282000%29-3%2B%5Ccfrac%7B2.4%7D%7B%5Csqrt%5B5%5D%7B2000%7D%7D%0A%5C%5C%5C%5C%5C%5C%0Ac%29%5Cqquad%20%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06%285000%29-3%2B%5Ccfrac%7B2.4%7D%7B%5Csqrt%5B5%5D%7B5000%7D%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%29%5Cqquad%20%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06%2810000%29-3%2B%5Ccfrac%7B2.4%7D%7B%5Csqrt%5B5%5D%7B10000%7D%7D)
Evaluate the algebraic expression for the given values 6+5(x-6)³ for X=8
1 answer:
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Answer:
Therefore, option C is true.
Step-by-step explanation:
From the graph, we get the sequence
7, 5, 3, 1, ...
Here,
a₁ = 7 is the first element.
An arithmetic sequence has a constant difference 'd' and is defined by
computing the differences of all the adjacent terms
The difference between all the adjacent terms is the same and equal to
substituting a₁ = 7 and d = -2 in the nth term of the sequence
Therefore, option C is true.