Answer:
6=110
13=1101
18=10010
27=11011
Explanation:
A decimal number is converted to binary number by constantly dividing the decimal number by 2 till the number becomes zero and then write the remainders in reverse order of obtaining them.Then we will get our binary number.
I will provide you 1 example:-
18/2 = 9 the remainder =0
9/2 = 4 the remainder =1
4/2 = 2 the remainder =0
2/2 = 1 the remainder =0
1/2 = 0 the remainder =1
Writing the remainder in reverse order 10010 hence it is the binary equivalent of 18.
Answer:
The correct answer for the given question is " The Code fragment A runs fastly than the code fragment of B".
Explanation:
In this question there are some information is missing i. e options. The question does not give any options. The options for the given question is given below
(A.) The Code fragment A runs fastly than the code fragment of B.
(B.) The Code fragment B runs fastly than code fragment of A.
(C) The Code fragment A runs as fastly as code fragment of B.
So we conclude the answer i.e option(A) because As given in the question list1 is a MyArrayList and list2 is a MyLinkedList. , in list1 we fetching the data easily and fastly means that it remove the data easily as compare to list2 As MyArrayList is storing the list only and also we can fetch the data easily manner.
The list2 is an object of MyLinkedList means that it manipulating the data fastly as compared to MyArrayList but if we compared the fetching of data then MyArrayList is a better option so the code fragment runs fastly then code fragmented B.
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.
