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Newton's Method : If xn is an approximation a solution of f ( x ) = 0 and if given by, f ′ ( xn ) ≠ 0 the next approximation is given by
xn+1 = xn - f(xn)/f'(xn)
It has to lead to the question of while do we stop? How several times do we go through this procedure? One of the more common stopping points in the procedure is to continue till two successive approximations agree upon a given number of decimal places.
2/4 + 3/4 =
differentiate x to the power 3
Find out the determinant: Find out the determinant of the following 3 x 3 matrix, expanding about row 1. Solution:
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sq root of tan x dx
Solve the subsequent quadratic equation: Solve the subsequent quadratic equation through taking the square roots of both sides. 3x 2 = 100 - x 2 Solution: Step 1
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