Write the related code using the look-at function to setup the model-view matrix.

I need help with programming in MATLAB. The following code transforms cartesian coordinates to the kepler elements. Can you give me the code for transforming kepler orbital elements to cartesian coordinates. The following code gives the 6 kepler elements. Transform those elements into cartesian coordinate that match the values under the Example Usage part of the code. Can you send a screenshot so I know the output of your code matches the input of the following code? I have asked this question twice and I have gotten the exact same code as the answer and it did not match. Please make sure the results match.   % Example usage:x = 1000;   y = 2000;   z = 3000;   vx = 4;     vy = -3;    vz = 2;      [a, ecc, inc, raan, argp, f] = cart2orb(x, y, z, vx, vy, vz); % Display the resultsdisp(['Semi-Major Axis (a): ', num2str(a), ' km']);disp(['Eccentricity (ecc): ', num2str(ecc)]);disp(['Inclination (inc): ', num2str(inc), ' degrees']);disp(['Right Ascension of Ascending Node (raan): ',…

12. Find the Laplace transform of the function t*e-2tcos(3t) using MATLAB and print it in the command window.

PYTHON JUPYTER NOTEBOOKS please show code clearly.  create a program for lower triangular system L x = b. The algorithm is described in the image for i and j with offset 1. Recall that NumPy and Python have offset 0 for their sequence data types.

Below is the questions

Need help in Matlab. How do I modify this program to produce eye shapes like the pink picture using scaling, rotating and translating? Thank you! figure(1) x(1:2,1)=[1;0]; plot(x(1), x(2),'*'); grid on; hold on; axis([-15 15 -15 15]); k=1; Inc=1; for j=1:Inc:360; x(1:2,k+1)=[cosd(Inc) -sind(Inc);sind(Inc) cosd(Inc)]*x(1:2,k); plot(x(1,k+1), x(2,k+1),'^'); k=k+1; %for j=1:Inc:360; %x(1:2,k+1)=[cosd(Inc) -sind(Inc);sind(Inc) cosd(Inc)]*x(1:2,k); %plot(x(1,k+1), x(2,k+1),'^'); %k=k+1; %pause(); end   for k=1:Inc:360; y1(1:2,k)=[1.5 0;0 3]*x(1:2,k); plot(y1(1,k),y1(2,k),'^') %for k=1:Inc:360; %y1(1:2,k)=[1.5 0;0 3]*x(1:2,k); %plot(y1(1,k),y1(2,k),'^') %pause(0.1); end   for k=1:Inc:360; y2(1:2,k)=[3 0;0 1.5]*x(1:2,k)+[7;5]; plot(y2(1,k),y2(2,k),'^') %for k=1:Inc:360; %y2(1:2,k)=[3 0;0 1.5]*x(1:2,k)+[7;5]; %plot(y2(1,k),y2(2,k),'^') %pause(0.1); end   for k=1:Inc:360; y3(1:2,k)=[3 0;0 1.5]*x(1:2,k)+[-7;5]; plot(y3(1,k),y3(2,k),'^') %for k=1:Inc:360; %y3(1:2,k)=[3 0;0 1.5]*x(1:2,k)+[-7;5];…

Given a vector of real numbers r = (r1, r2, ..., rm). We can standardize the vector using the formulation: vi = "im, where m ri-m is the mean of the vector r, and s is the standard deviation of r. The vector v = (v1, v2, ..., Un) will be the scaled vector. Write a Python function scale_vec (r) that takes the vector r as input and returns the scaled vector v. Sample inputs and outputs: ● Input: np.array([1, 3, 5]), output: [-1.22474487 0. 1.22474487] • Input: np. array([3.3, 1.2, -2.7, -0.6]), output: [1.35457092 0.40637128 -1.35457092 -0.40637128] Hint: Use numpy.mean and numpy.std with default parameters. # Write your function here. Let's test your function. [ ] import numpy as np print (scale_vec (np.array([1, 3, 5]))) print (scale_vec (np.array([3.3, 1.2, -2.7, -0.6])))

For input vector x, write a function to check whether there exists at most 5 elements in x whose values are less than 7.

Use Matlab to Plot the graph shown in image 2 using data in table on image 1, BMI= W*H Where W is the weight in kilograms and H is the height in meters. Or the four color curves, use the line width ‘3’. To display a BMI of a person use the following: circular MarkSize ‘8’, MarkerEdgeColor ‘cyan’. To display this you have to enter two values. these are the Height in meters and Weight in kilograms. And inform the outcome results to the person from the command window.

Attach code and plot screenshot.