Answer:
% here x and y is given which we can take as
x = 2:2:10;
y = 2:2:10;
% creating a matrix of the points
point_matrix = [x;y];
% center point of rotation which is 2,2 here
x_center_pt = x(2);
y_center_pt = y(2);
% creating a matrix of the center point
center_matrix = repmat([x_center_pt; y_center_pt], 1, length(x));
% rotation matrix with rotation degree which is 45 degree
rot_degree = pi/4;
Rotate_matrix = [cos(rot_degree) -sin(rot_degree); sin(rot_degree) cos(rot_degree)];
% shifting points for the center of rotation to be at the origin
new_matrix = point_matrix - center_matrix;
% appling rotation
new_matrix1 = Rotate_matrix*new_matrix;
Explanation:
We start the program by taking vector of the point given to us and create a matrix by adding a scaler to each units with repmat at te center point which is (2,2). Then we find the rotation matrix by taking the roatational degree which is 45 given to us. After that we shift the points to the origin and then apply rotation ans store it in a new matrix called new_matrix1.
Answer:
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Answer:
30 units of Food A and 45 units of Food B are to be purchased to keep costs at the minimum $105.
Explanation:
X = Amount of food A
Y = Amount of food B
Z= 2X+Y..... minimum cost equation
50X + 20Y > 2400 .................Vitamins .......(1)
30X + 20Y > 1800 ...................Minerals.......(2)
10X + 40Y > 1200 .................Calories ..........(3)
X > 0
y > 0
X=30 and Y = 45
Z= 2(30) + 45 = $105