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Find all solutions of the given congruence: x5 + 3x2 +1 = 0(125). Show transcribed image...
Find all solutions to the congruence x2+ x+ 1≡0 mod 91. (Hint:factor the modulus, use trial and error to find the solutions modulo the factors, and the CRT to combine the results into solutions to the original equations.)
Please solve the above 4 questions. 1. Using the extended Euclidean Algorithm, find all solutions of the linear congruence 217x 133 (mod 329), where 0 x < 329 (Eg. if 5n, n 0,. ,6) 24 + 5n, п %3D 0, 1, . .., 6, type 24 + x< 11 2. Find all solutions of the congruence 7x = 5 (mod 11) where 0 (Eg. if 4,7 10, 13, type 4,7,10,13, none. or if there are no solutions, type I 3....
Q2 20 Find all solutions of the equati tanx=13 125 Find all solutions of the equat tanx=1 Find all solution of the equabon C)x= VE 123 Find all solutions of the equation 13 (22) use substitution to de berminé whether the given X value is a solution of the equation sinx_ ve x = 7 sin Xzi
Find the solutions of 2x1-3x2-7x3+5x4+2x5=-2 x1-2x2-4x5+3x4+x5=-2 2x1+0x2-4x3+2x4+x3=3 x1-5x2-7x3+6x4+2x5=-7
Solve the following system of equations and find all congruence class solutions if any exist; if no solution exists, explain why not: x24 (mod 35) 3.x 15 mod 21)
4. Suppose that p is a prime of the form 8k + 1 . Show that the congruence x4 has ether 0 solutions or 4 solutions. 2 (mod P) 4. Suppose that p is a prime of the form 8k + 1 . Show that the congruence x4 has ether 0 solutions or 4 solutions. 2 (mod P)
39. Suppose that the polynomial congruence f(x)0 (mod 7) has two distinct so- 0 lutions, what are the possible number of solutions of the congruence f(x) (mod 49)? 39. Suppose that the polynomial congruence f(x)0 (mod 7) has two distinct so- 0 lutions, what are the possible number of solutions of the congruence f(x) (mod 49)?
Find all solutions of the given equation 2- 6x3 + 3 = 0
differential equations 1 +.. 8 Find two power series solutions of the given differential equation about the ordinary point x = 0. (x2 + 1)" - 6y = 0 O Y1 = 1 + x2 + 3x4 xo and Y2 = x = x + 3x3 16 O x1 = 1 + 3x2 + x4 – xo + and y2 = x + x3 O Y1 = 1 + 3x2 + 5x* + 7x® + ... and y2 = x...
In Exercises use variation of parameters to find a particular solution, given the solutions of the complementary equation 11.xy" – 4xy' + 6y = x5/2, x > 0; yı = x, y2 = x3