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:
Step-by-step explanation:
500
We need an equation for this, theres no 'formula' for doing this (ignoring the quadratic formula, as that is a long-winded way of factorising a quadratic equation)
Inverse proportion is of the form:
y=k/x, in this case the variables are
z=k/x, and we are given the point (6,2) so we can solve for k
2=k/6
12=k, so our equation is:
z=12/x, so when x=13
z=12/13
