Answer:
Let's use algebra to solve the problem.
Let x be the number of 50 cent pieces in the moneybox.
Then the number of 20 cent pieces in the moneybox is 600 - x (since there are 600 coins in total).
The value of x 50 cent pieces is 50x cents.
The value of (600 - x) 20 cent pieces is 20(600 - x) = 12000 - 20x cents.
The total value of the coins is the sum of these two values:
50x + (12000 - 20x) = 3000 + 30x
So there are 3000 + 30x cents in the moneybox.
To find the total number of cents, we need to evaluate this expression for x:
3000 + 30x
We don't know the value of x, but we do know that there are 600 coins in the moneybox, so we can set up another equation:
x + (600 - x) = 600
Simplifying this equation, we get:
x + 600 - x = 600
Simplifying further, we get:
600 = 600
This equation is always true, which means we can solve for x in terms of 600:
x = 600 - (600 - x)
x = 600 - (600 - x)
x = 600 - (600 - x)
x = 600 - 600 + x
x = x
So x can be any value between 0 and 600.
To find the total number of cents in the moneybox, we need to evaluate the expression 3000 + 30x for any value of x between 0 and 600.
The minimum value of this expression occurs when x = 0:
3000 + 30(0) = 3000
The maximum value of this expression occurs when x = 600:
3000 + 30(600) = 21000
So there are between 3000 and 21000 cents in the moneybox, depending on how many 50 cent pieces there are.
Hope This Helps!