Beskou die stelsel van nie-lineêre vergelykings / Consider the system of nonlinear equations 4x} – 20x1 + 4 x3 = -1 2x1x

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Beskou Die Stelsel Van Nie Lineere Vergelykings Consider The System Of Nonlinear Equations 4x 20x1 4 X3 1 2x1x 1 (26.87 KiB) Viewed 59 times

Beskou die stelsel van nie-lineêre vergelykings / Consider the system of nonlinear equations 4x} – 20x1 + 4 x3 = -1 2x1x2 + 2x1 – 5x2 = 10 bestaande uit die implisiete krommes fi (x1, x)) = 0) en f2(x1, x)) = 0 onderskeidelik. / consisting of the implicit curves fi(x1, x)) = () and f2(x1, x)) = 0 respectively Question 1 of 5 1 Points Click to see additional instructions Hoeveel snypunte van die krommes f1 en fz is in die gebied waar X1, X2 € (-10,10] ? / How many intersection points of the curves f1 and f2 are in the region where X1, X2 € (-10,10] ? antwoord/answer: punte/points.
Question 2 of 5 1 Points Click to see additional instructions Bepaal die Jacobiaanse matriks J van die funksies f1 en fz in die punt (X1,x2)=(0,-3). / Determine the Jacobian matrix J of the functions fy and f2 at the point (X1,x2)=(0,-3). As/lf ) = (az az) bepaal die waardes van /determine the values of a, a, a3, 04. antwoord/answer: 2,= az .az a=
Click to see additional instructions Bepaal die benaderde oplossing van een iterasie naby (0,-3) van die bogenoemde stelsel met behulp van die Newtonmetode. Gee die waardes van die benaderde oplossing, afgerond tot vier desimale. / Detemine the approximate solution close to (0,-3) of the above system using one iteration of Newton's method. Give the values of the approximate solution rounded to four decimals. antwoord/answer: (x1,x2)=(
Question 4 of 5 3 Points Click to see additional instructions Bepaal die benaderde oplossings naby (0,-3), (5,-1.3) en (4.7,3) van die bogenoemde stelsel binne 'n toleransie van 10-6 deur die Newtonmetode te gebruik. Gee die waardes van die benaderde oplossings afgerond tot vier desimale. / Detemine the approximate solutions close to (0,-3), (5,-1.3) and (4.7,3) of the above system within a tolerance of 10-6 by using the Newton's method. Give the value of the approximate solutions rounded to four decimals. antwoord/answer Die oplossing naby / The solution close to (0,-3) is (X1,x2)=( Die oplossing naby / The solution close to (5,-1.3) is (X1,x2)=( Die oplossing naby / The solution close to (4.7,3) is (X1,x2)=(
Question 5 of 5 1 Points Click to see additional instructions Bepaal die oorblywende oplossing(s) van die bogenoemde stelsel binne 'n toleransie van 10- deur die Newtonmetode te gebruik. / Determine the remaining solution(s) of the above system within a tolerance of 10-6 by using the Newton's method. wenk: kies 'n beginpunt baie naby aan die vierde nul. / hint: choose an initial point very close to the fourth zero antwoord/answer: (X1,X2)=(