Answer:
Step-by-step explanation:
If we take out the extra $3, we can group the bills into one each of $5 and $1, for a value of $6. There will be 7 such groups in the remaining $42.
That means there are 7 bills of the $5 denomination, and 3 more than that (10 bills) of the $1 denomination.
There are 7 $5 bills and 10 $1 bills.
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If you want to write an equation, it is usually best to let a variable stand for the most-valuable contributor. Here, we can let x represent then number of $5 bills. Then the value of the cash box is ...
   5x +(x+3) = 45
   6x = 42 . . . . . . . . subtract 3, collect terms
   x = 7 . . . . . . . . . . . there are 7 $5 bills
   x+3 = 10 . . . . . . . . there are 10 $1 bills
You may notice that this working parallels the verbal description above. (After we subtract $3, x is the number of $6 groups.)
 
        
             
        
        
        
Answer: Yes, it can.
Step-by-step explanation:
 
        
                    
             
        
        
        
Answer:
1. (+1)(+2) 
2. (−2)(+3) 
3. (−2)(+1)
Step-by-step explanation:
 
        
             
        
        
        
Log 2 over 3 = 0.10034333188
        
             
        
        
        
It doesnt say how many she buys at the coin show so we dont know but its going to be more than 6