$4 x number of visits (V)> price of membership (M)
V > M/4
Answer:
1, 4, 7
Explanation:
The instruction in the question can be represented as:
for i in range(1,10,3):
print i
What the above code does is that:
It starts printing the value of i from 1
Increment by 3
Then stop printing at 9 (i.e.. 10 - 1)
So: The sequence is as follows
Print 1
Add 3, to give 4
Print 4
Add 3, to give 7
Print 7
Add 3, to give 10 (10 > 10 - 1).
So, it stops execution.
The gear ratios would look like A:B=1:9 and C:D=1:32 if you are just needing the gear teeth ratio. Not a lot of information to go off of from the question though
Answer:
Following are the expression to this question:
if (young and famous==True):
Explanation:
For print, the given expression the code requires some modification that can be defined as follows:
young = True#defining a bool variable that holds a value True
famous = True#defining a bool variable that holds a value True
if (young and famous==True):#defining if block that check variable value
print('You must be rich!')#print message
else:#else block
print('There is always the lottery...')#print message
Output:
You must be rich!
Code explanation:
In the above-given code, two variable "young and famous" is declared, that hold a "True" which is a bool value, in this code, a conditional statement has used, that checks variable value, which can be defined as follows:
- In the if block, it uses the above declared variable with and gate to check its value is equal to true.
- If the condition is true, it will print the true block message, otherwise, go to the else block in this, it will print the else block message.
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.
