In question 9, we want an equation for each situation, that is, how much is the monthly amount of each prize.
Let'use the variables in the table,, x for month, y₁ for monthly amount of A and y₂ monthly amount for B.
Assuming x = 0 is the first month, in A we will have 5000 plus 100 increase for each month thereafter.
Inthe first month, x = 0, there is no increase, and x = 1 has one increase, and so on.
So, the amount it had increases so far is x times 100. Plus the starting 5000, we will have the equation:

For prize B, we start with 2000 and have a 10% increase. 10% increase is the same as multiplying by 100% + 10% each month, that is, multiply by 1.10.
This multiplication accumulates over the months, so we have to exponentiate it by the number of months. The equation becomes:

We can see that in the first month, x = 0, so we will have the initial 2000 correctly.
Now, we can complete the table by substituting x into the equations:
x = 0:

x = 1:

x = 2:

And so one until x = 12.
We will get the table: