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question:Find the Taylor series for f(x) = 10/x centered at a = -2.

answer:The Taylor series for f(x) = 10/x centered at a = -2 is given by: {eq}begin{align*} frac{10}{x} = sum_{n=0}^{infty} frac{(-1)^n10cdot n!}{(-1)^{n+1}2^{n+1} }(x+2)^n end{align*} {/eq} The radius of convergence of this series is R = 0, which means that the series converges only when x = -2.

question:Tim Dye, the CFO of Blackwell Automotive, Inc., is putting together this year's financial statements. He gathered the following balance sheet information: Cash: 23,015 Accounts payable: 163,257 Common stock: 313,900 Retained earnings: 512,159 Inventory: 214,100 Goodwill and other assets: 78,655 Net plant and equipment: 710,700 Short-term notes payable: 21,115 Accounts receivable: 141,258 Other current assets: 11,223 How much long-term debt does Blackwell Automotive have?

answer:To calculate the long-term debt, we need to first calculate the total assets and total liabilities of the company. Total assets = Current assets + Fixed assets Total assets = (23,015 + 214,100 + 141,258 + 11,223) + (710,700 + 78,655) Total assets = 389,596 + 789,355 Total assets = 1,178,951 Total liabilities = Current liabilities + Long-term debt Total liabilities = (163,257 + 21,115) + Long-term debt Total liabilities = 184,372 + Long-term debt Now, we can use the accounting equation to solve for long-term debt: Total assets = Total liabilities + Stockholders' equity 1,178,951 = 184,372 + Long-term debt + 826,059 Long-term debt = 1,178,951 - 184,372 - 826,059 Long-term debt = 168,520 Therefore, Blackwell Automotive has 168,520 in long-term debt. Blackwell Automotive has 168,520 in long-term debt.

question:Solve for y=f(x) given the parametric equations: y(t)=frac{4 left(1656 t^2+28980 t+127253right)^2}{117649}, quad x(t)=frac{144 t^2}{49}+frac{360 t}{7}+225

answer:To eliminate the parameter t and find the function y=f(x), we need to express t in terms of x and substitute into the equation for y(t). From x(t), we have: frac{144 t^2}{49}+frac{360 t}{7}+225 = x We can solve this quadratic equation for t: 144t^2 + 360 cdot 49t + 49 cdot 225 = 49x 144t^2 + 17640t + 11025 = 49x 16t^2 + 630t + 1225 = x (4t)^2 + 630(4t) + (35)^2 = x (4t + 35)^2 = x 4t + 35 = pmsqrt{x} 4t = pmsqrt{x} - 35 t = frac{pmsqrt{x} - 35}{4} Since we have two solutions for t, we'll consider both of them to express y(t) as a function of x: y = frac{4 left(1656 left(frac{pmsqrt{x} - 35}{4}right)^2+28980 left(frac{pmsqrt{x} - 35}{4}right)+127253right)^2}{117649} Simplifying, we obtain the Cartesian equation: y = frac{529 x^2}{49} + frac{874 x}{49} + frac{361}{49} Therefore, y=f(x) is given by: y = frac{529 x^2}{49} + frac{874 x}{49} + frac{361}{49}

question:In the context of Riemannian geometry, why is the open unit disk in the Euclidean plane considered complete as a metric space, yet not geodesically complete? And how does this relate to the Hopf–Rinow theorem?

answer:The open unit disk in the Euclidean plane is indeed complete as a metric space. This means that every Cauchy sequence converges within the disk. However, it is not geodesically complete because geodesics (lines in this case) can extend beyond the disk's boundary, and hence cannot be defined for all values of the parameter ( t ). The confusion may arise from the Hopf–Rinow theorem, which states that for a complete Riemannian manifold without boundary, geodesic completeness is equivalent to metric completeness. However, the closed unit disk is a manifold with boundary, and the theorem does not apply directly to such spaces. Therefore, the inapplicability of the theorem to the closed unit disk explains the difference between metric completeness and geodesic completeness in this case.