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Chapter 1 Solutions
Developmental Mathematics (9th Edition)
- ) The set {1,2,..., 22} is to be split into two disjoint non-empty sets S and T in such a way that: (i) the product (mod 23) of any two elements of S lies in S; (ii) the product (mod 23) of any two elements of T lies in S; (iii) the product (mod 23) of any element of S and any element of T lies in T. Prove that the only solution is S = {1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18}, T= {5, 7, 10, 11, 14, 15, 17, 19, 20, 21, 22}.arrow_forwardPlease solve 6.31 by using the method of sections im stuck and need explanationarrow_forwardPlease don't use chatgpt.arrow_forward
- Solve Problem I, 4, from the Shushu jiuzhang, which is equivalent to N = 0 (mod 11), N = 0 (mod 5), N = 4 (mod 9), N = 6 (mod 8), N = 0 (mod 7).]arrow_forward19) Consider this initial value problem: y' + y = 2y = -21² + 2t+ 14, y(0) = 0, y (0) = 0 - What is the solution of the initial value problem?arrow_forward4) Consider the initial value problem " 8y +30y+25y = 0, y(0) = -2, y (0) = 8 What is the t-coordinate of the local extreme value of y = y(t) on the interval (0, ∞)? Enter your answer as a decimal accurate to three decimal places.arrow_forward
- Tips S ps L 50. lim x2 - 4 x-2x+2 51. lim 22 - X 52. 53. x 0 Answer lim x 0 lim 2-5 X 2x2 2 x² Answer -> 54. lim T - 3x - - 25 +5 b+1 b3b+3 55. lim X x-1 x 1 Answer 56. lim x+2 x 2 x 2 57. lim x²-x-6 x-2 x²+x-2 Answer-> 23-8 58. lim 2-22-2arrow_forward10) Which of the following is the general solution of the homogeneous second-order differential equation y + 8y + 52y=0? Here, C, C₁, and C2 are arbitrary real constants. A) y = C₁ecos(61) + C₂e*sin(61) + C B) y = et (sin(4t) + cos(6t)) + C C) y = C₁esin(6) + C₂e+ cos(6t) + C D) y = C₁esin(6) + C₂e+cos(6) E) y=e(C₁sin(61) + C₂cos(61))arrow_forward3) Consider the initial value problem ' y' + 8y = 0, y(0) = -4, y (0) = 16 What is the solution of this initial value problem? A) y = -4t - 2e8t D) y = -4 + 2e-8t B) y = -2 + 2e8t C) y = -2 -2e-8t E) y = -4+ 2e8t F) y = -2t-2e-8tarrow_forward
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