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.)
1 is not. 2 is. 3 is not. 4 is. 5 is. 6 is. 7 is not. 8 is.
Y = <span>b^x
when x = 1
y = b^1
y = b
Therefore, the value of b is the same as the value of y when x =1
From the graph,
When x = 1, y = 0.5
Therefore, b = 0.5
To confirm this
From the graph,
When x = -1, y = 2
Since </span>y = b^x<span>
2 = </span>b^-1
2 = 1/b
2b = 1
b = 0.5
When x = -2, y = 4
Since y = b^x
4 = b^-2
4 = 1/(b^2)
b^2 = 1/4
b = √(1/4)
b = 1/2
b = 0.5
Therefore, it is conformed that b = 0.5
Answer:
704
Step-by-step explanation:
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