Acash box of $1 and $5 bills is worth $45. the number of $1 bills is three more than the number of $5 bills. how many of each bill does it contain?

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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.

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.)

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

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step-by-step explanation: