MATH 126 Final: MATH 126 UW Final Exam Winter 2013 Solutions
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One possible justi cation is: the plane through (2, 2, 2), (1, 0, 0), and (2, 3, 7) is. 2 ) + 2(z 4 and lim t . 2. t + 2 (a) t(t) = (cid:28) 2 (b) 2(x 2) + (y 1. The curve gets straighter as t . 3. (a) i. point; ii. ellipse; iii. pair of lines; iv. hyperbola; v. pair of lines; vi. hyperbola. (b) cone (c) t, t: x + y + z = 3. 5. (a) z = 3(x 2) + 8(y 2) 6 (b) 6. 43: 18 inches by 18 inches by 36 inches. 3 sin 1 (b) 2 (cid:20) 1000 2(50)3/2. Xk=0 x2k+3 k! for all real x (b) t5(x) = x3 + x5 (c) g(x) = Xk=0 x2k+4 (2k + 4)k! for all real x. 9. (a) t2(x) = 2 + 1 (b) |f (x) t2(x)| 10.
