Display the seventh element in a 10 elements vector Solve: Code Will be, V(7) True False 2-Find the solution for x in: y + y' = sin(x)/2 Solve: dsolve('y+Dy =sin(x)/2') True False
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- Alphabet Random Walk• Write a program to generate a random walk that spans a 10*10 character array (The initial values of the elements are all.). The program must randomly walk from one element to another, moving one element position up, down, left or right each time. The elements that have been visited are labeled with the letters A through Z in the order in which they were visitedPlease design a data structure to allow the following three operations: 1. push (insert a new number) 2. find_Kmin (return the value of the K-th smallest number without removing it). 3. pop All operations should be done in 0(log k) time or better. PLEASE NOTE THAT YOU'RE GIVEN 'n' ELEMENTS BUT YOUR ALGORITHMS SHOULD TAKE NO MORE THAN log k time. PLEASE REMEMBER that it's 'k' and not 'n'.(Electrical eng.) Write a program that specifies three one-dimensional arrays named current, resistance, and volts. Each array should be capable of holding 10 elements. Using a for loop, input values for the current and resistance arrays. The entries in the volts array should be the product of the corresponding values in the current and resistance arrays (sovolts[i]=current[i]resistance[i]). After all the data has been entered, display the following output, with the appropriate value under each column heading: CurrentResistance Volts
- Plot. % MA TLAB code for range rng = linspace(-pi,pi,1000); % Plotting sin^2 plot(rng, sin(rng).^2) hold on % Plotting cos^2 plot(rng,cos(rng).^2) % Creating an array of character vectors lgd = {'sine', 'cosine'}; % Converting the array into cell array lgd=cell(lgd); % Adding legend legend(lgd).change this code to C++ #include <stdio.h> #include <stdlib.h> /** * structure to hold each point */ struct Point3D { float x; float y; float z; }; /** * Parses all the vertices from the path. * @param path The input file path. * @param points The points storage. * @param numPoints The number of points. * @return points Dynamically allocated array of points. * @return The center of mass. */ struct Point3D parseInput(const char *path, struct Point3D **points, int *numPoints) { // to hold each record char f_or_v; float x; float y; float z; // to store the file pointer FILE *inFile; // to iterate the points array int it; // structure to store the center of mass struct Point3D centerOfMass; // opening the file inFile = fopen(path, "r"); // counting the number of points *numPoints = 0; while(!feof(inFile)) { fscanf(inFile, "%c %f %f %f\n", &f_or_v, &x, &y, &z); *numPoints += f_or_v == 'v'; } // dynamically allocating space // for the points *points =…Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards, then backwards. End each loop with a newline. Ex: If courseGrades = {7, 9, 11, 10}, print:7 9 11 10 10 11 9 7 Hint: Use two for loops. Second loop starts with i = NUM_VALS - 1. Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element array (int courseGrades[4]). Also note: If the submitted code tries to access an invalid array element, such as courseGrades[9] for a 4-element array, the test may generate strange results. Or the test may crash and report "Program end never reached", in which case the system doesn't print the test case that caused the reported message. #include <iostream>using namespace std; int main() { const int NUM_VALS = 4; int courseGrades[NUM_VALS]; int i; for (i = 0; i < NUM_VALS; ++i) { cin >> courseGrades[i]; } /* Your solution goes…
- Array TypesObjective: Based on the given values and initializations, give what is being required of each statement.1. Given A[10], α=2000, esize=4 bytes:a) Find the number of elements.b) Find the address of the 6th element.c) Find the index no. of the 8th element.2. Given E[3][4], α=2020, esize=4 bytes:a) Find the total no. of elements.b) Find the address of the last element.c) Find the address of the 10th element.Write a for loop to print all NUM_VALS elements of vector courseGrades, following each with a space (including the last). Print forwards, then backwards. End with newline. Ex: If courseGrades = {7, 9, 11, 10}, print:7 9 11 10 10 11 9 7 Hint: Use two for loops. Second loop starts with i = courseGrades.size() - 1 (Notes)C++Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element vector (vector<int> courseGrades(4)). See "How to Use zyBooks".Also note: If the submitted code has errors, the test may generate strange results. Or the test may crash and report "Program end never reached", in which case the system doesn't print the test case that caused the reported message. #include <iostream>#include <vector>using namespace std; int main() { const int NUM_VALS = 4; vector<int> courseGrades(NUM_VALS); int i; for (i = 0; i < courseGrades.size(); ++i) { cin >> courseGrades.at(i);…Can you please explain how the source code works: //Gauss Elimination#include <iostream>#include <cmath>#include <vector>#include <fstream>#include <iterator>#include <string>#include <iomanip>using namespace std; int main() { double Mat[2][3]; vector<double> q(2); int Nx = 2, ii, jj, kk; Mat[1][1] = 2.0; Mat[1][2] = 3.0; Mat[1][3] = 2.5; Mat[2][1] = 1.3; Mat[2][2] = 3.0; Mat[2][3] = 2.5; cout << "Matrix: " << endl; for (ii = 0; ii < Nx; ii++) { for (jj = 0; jj < Nx + 1; jj++) { cout << Mat[ii][jj] << " "; cout << endl; } } // Triangularization for (ii = 0; ii < Nx - 1; ii++) for (kk = ii + 1; kk < Nx; kk++) { double t = Mat[kk][ii] / Mat[ii][ii]; for (jj = 0; jj <= Nx; jj++) Mat[kk][jj] = Mat[kk][jj] - t * Mat[ii][jj]; } // Resolution…
- Complete the following code: var myInput = $('#first_name'); console.log("The value of the input element with id #first_name is: ' + ___;If the elements in two arrays are related by their subscripts, the arrays are called _____________________ arrays. a. associated b. coupled c. matching d. parallelWrite a for loop to print all NUM_VALS elements of vector courseGrades, following each with a space (including the last). Print forwards, then backwards. End with newline. Ex: If courseGrades = {7, 9, 11, 10}, print:7 9 11 10 10 11 9 7 Hint: Use two for loops. Second loop starts with i = courseGrades.size() - 1 (Notes)Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element vector (vector<int> courseGrades(4)). See "How to Use zyBooks".Also note: If the submitted code has errors, the test may generate strange results. Or the test may crash and report "Program end never reached", in which case the system doesn't print the test case that caused the reported message. #include <iostream>#include <vector>using namespace std; int main() { const int NUM_VALS = 4; vector<int> courseGrades(NUM_VALS); int i; for (i = 0; i < courseGrades.size(); ++i) { cin >> courseGrades.at(i); }…
