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By Xu Y., Bao W.
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Additional resources for A 1/3 Pure Subharmonic Solution and Transient Process for the Duffings Equation
D) The inequality 0 x Ϫ 3 0 Ն 1 says that x is at least one unit away from 3 on the number line. This means that either x Ն 4 or x Յ 2, as shown in Figure 5. [Here’s an alternative way of thinking about this: The numbers satisfying the given inequality are precisely those numbers that do not satisfy the inequality in part (c). ] _2 _1 0 1 2 Figure 2 |x| Ͻ 2 _2 _1 0 1 2 Figure 3 |x| Ͼ 2 2 3 4 Figure 4 |x Ϫ 3| Ͻ 1 . 2 A In Exercises 1–16, evaluate each expression. 1. 0 3 0 2. 3 ϩ 0 Ϫ3 0 3. 0 Ϫ6 0 4.
B) With a graphing utility, just as in graphing by hand, we first need to solve the given equation for one variable in terms of the other. From part (a) we have F ϭ 95 C ϩ 32. Because most graphing calculators (and some types of computer software) require that we use x and y for the variables, we’ll rewrite the equation as y ϭ 95 x ϩ 32. Then we can enter it in the graphing utility as F ؍95 C ؉ 32 C F 0 5 10 15 Ϫ5 Ϫ10 32 41 50 59 23 14 9 y ϭ a b * x ϩ 32 5 The symbol * denotes multiplication.
A) 23. (a) ϭ ϭ 3x x xϪ2 9x 3 2 3 5 (b) ϭ (b) ϭ 3x xϩ1 xϪ2 9x Ϫ 2 3 3 5 2 ϭ ϩ1 ϭ5 (c) (c) 3x x xϪ2 x Ϫ2 3 19. In Exercises 24 – 33, solve each equation by factoring. 25. x 2 Ϫ 5x ϭ Ϫ6 24. x 2 Ϫ 5x Ϫ 6 ϭ 0 26. 10z2 Ϫ 13z Ϫ 3 ϭ 0 27. 3t2 Ϫ t Ϫ 4 ϭ 0 28. (x ϩ 1)2 Ϫ 4 ϭ 0 29. x 2 ϩ 3x Ϫ 40 ϭ 0 30. x(2x Ϫ 13) ϭ Ϫ6 31. x(3x Ϫ 23) ϭ 8 32. x(x ϩ 1) ϭ 156 33. x 2 ϩ (2 15)x ϩ 5 ϭ 0 In Exercises 34 – 41, use the quadratic formula to solve each equation. In Exercises 34 – 39, give two forms for each solution: an expression containing a radical and a calculator approximation rounded off to two decimal places.
A 1/3 Pure Subharmonic Solution and Transient Process for the Duffings Equation by Xu Y., Bao W.