Acashier has a total of 30 bills consisting of ones, fives and twenties. the number of twenties is 5 less than the number of ones. the total value of money is $229. how many of each denomination of bill are there?

Let us say that:

a = ones

b = fives

c = twenties

So that the total money is:

1 * a + 5 * b + 20 * c = 229

=> a + 5b + 20c = 229                                       --> eqtn 1

We are also given that:

c = a – 5                                                                --> eqtn 2

a + b + c = 30                                                      --> eqtn 3

Rewriting eqtn 3 in terms of b:

b = 30 – a – c

Plugging in eqtn 2 into this:

b = 30 – a – (a – 5)

b = 35 – 2a                                                           --> eqtn 4

Plugging in eqtn 2 and 4 into eqtn 1:

a + 5(35 – 2a) + 20(a – 5) = 229

a + 175 – 10a + 20a – 100 = 229

11a = 154

a = 14

So,

b = 35 – 2a = 7

c = a – 5 = 9

Therefore there are 14 ones, 7 fives, and 9 twenties.

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

$5.41

step-by-step explanation:

$20 - $14.51 = $5.49.