Consider the following function: f(x)=-12-21x+18x² - 2.75x3 Determine a root of the given function using the bisection method. Use initial guesses of x = -1 and x₁ = 0 and a stopping criterion of 10%. (Round the final answer to four decimal places. Include a minus sign if necessary.) Upload your iteration Table for which the first two iterations are provided. Explanation: Using bisection, the first iteration is = 140 =-0.5 f(-1)f(-0.5) 29. 75 (3.34375) = 99.47656 = Therefore, the root is in the second interval and the lower guess is redefined as xy=-0.5. The second iteration is Eg= -0.5+0 - -0.25 2 0.25 (0.5) 100% = 100% -0.25 f(-0.5)f(-0.25) = 3.34375 (-5.5820313) = -18.66492 Therefore, the root is in the first interval and the upper guess is redefined as x₁ = -0.25. The remainder of the iterations are displayed in the following table: x₁ f(x₁) Xu F(x) 1 -1 29.75 0 -12 2 -0.5 3.34375 0 -12 X, -0.5 -0.25 F(X) 3.34375 -5.5820313 100.00%

Consider the following function: f(x)=-12-21x+18x² - 2.75x3 Determine a root of the given function using the bisection method. Use initial guesses of x = -1 and x₁ = 0 and a stopping criterion of 10%. (Round the final answer to four decimal places. Include a minus sign if necessary.) Upload your iteration Table for which the first two iterations are provided. Explanation: Using bisection, the first iteration is = 140 =-0.5 f(-1)f(-0.5) 29. 75 (3.34375) = 99.47656 = Therefore, the root is in the second interval and the lower guess is redefined as xy=-0.5. The second iteration is Eg= -0.5+0 - -0.25 2 0.25 (0.5) 100% = 100% -0.25 f(-0.5)f(-0.25) = 3.34375 (-5.5820313) = -18.66492 Therefore, the root is in the first interval and the upper guess is redefined as x₁ = -0.25. The remainder of the iterations are displayed in the following table: x₁ f(x₁) Xu F(x) 1 -1 29.75 0 -12 2 -0.5 3.34375 0 -12 X, -0.5 -0.25 F(X) 3.34375 -5.5820313 100.00%

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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I have the answers but I want to know the steps with calculation to reach these answers. blank1 = 1.764 , blank2 = -2.18 , blank3= 43.307 , blank4 = 1.764
please write a matlab code
I have the answers but I want to know the steps. blank1 = 1.764 , blank2 = -2.18 , blank3= 43.307 , blank4 = 1.764
In Problem 3 use Euler’s method to obtain a four-decimalapproximation of the indicated value. First use h= 0.1 and then useh =0.05. Find an explicit solution for each initial-value problem andthen construct tables similar to Tables 2.6.3 and 2.6.4. 3) y'=y, y(0)=1; y(0)
MULTIPLE CHOICE -The answer is one of the options below please solve carefully and circle the correct option Please write clear .
I having a problem with my code in MATLAB. In the following code results for r is just a 1x3 matrix. Although inside the while loop r equals to multiple 1x3 matrices. I need r equal to one matrix that is of size 27032x3. So, I just need the multiple r matrices to merge together to get one big matrix. I need that matrix to retain its value outside the while loop as well. cc=0; % set line counter JD = 2460626.666667; fid = fopen('tle_catalog.txt'); % load the TLE tline2='gg'; while ischar(tline2) cc = cc+1; % counter name = fgets(fid);% for the ones with three lines tline1 = fgets(fid); % collect first line of two line elements tline2 = fgets(fid); % collect second line of two line elements if tline2>0 % stop at the end of the file % initialize the propagation [satrec, startmfe, stopmfe, deltamin] ... = twoline2rv(721, tline1, tline2, 'c', 'd'); time_JD = tline1(21:32); yeardayhour = str2double(regexp(tline1, '(\d{2})(\d{3})(\.\d+)', 'tokens', 'once')); dn =…
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I wanted to know how to create plots like these in MATLAB. I belive they were called herpolhode plots.
For all the following problems, 5. a) b) You need to show at least 3 iterations calculated manually with all steps. You do not need to include the M.files for the bisection method (bisect.m) and for false position (falspos.m). You must, however, show the command lines for the given functions with their variables and other parameters. Fanning friction factor For fluid flow in pipes, friction is described by a dimensionless number, the Fanning friction factor f. The Fanning friction factor is dependent on a number of parameters related to the size of the pipe and the fluid, which can all be represented by another dimensionless quantity, the Reynolds number Re. A formula that predicts ƒ given Re is the von Karman equation: 4log₁0 (Re√) - 0.4 = Typical values for the Reynolds number for turbulent flow are 10,000 to 500,000 and for the Fanning friction factor are 0.001 to 0.01. (a) Develop a function that uses bisection to solve for fgiven a user-supplied value of Re between 500 and…
In Problem 1 use Euler’s method to obtain a four-decimalapproximation of the indicated value. Carry out the recursion of(3) by hand, first using h= 0.1 and then using h= 0.05 (show all steps) 1) y'=2x-3y+1 y(1)=5; y(1.2)