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Crazy boy [7]
3 years ago
9

Parts of the proceeds from a garage sale was $435 worth of five dollars and $20 bills. If there were two more five dollar bills

and $20 bills, find the number of each the Denomination
Mathematics
1 answer:
Genrish500 [490]3 years ago
7 0

Answer: there were 19 five dollar bills and 17 20 dollar bills.

Step-by-step explanation:

Let x represent the number of five dollar bills that were received.

Let y represent the number of twenty dollar bills that were received.

Parts of the proceeds from a garage sale was $435 worth of five dollars and $20 bills. This means that

5x + 20y = 435 - - - - - - - - - - -1

If there were two more five dollar bills than $20 bills, it means that

x = y + 2

Substituting x = y + 2 into equation 1, it becomes

5(y + 2) + 20y = 435

5y + 10 + 20y = 435

25y = 435 - 10 = 425

y = 425/25 = 17

x = y + 2 = 17 + 2

x = 19

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The sum of the squares of three consecutive integers is 509. Determine the integers.
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Answer:

12, 13, 14

Step-by-step explanation:

Denote the integers as:

x

x+1

x+2

The sum of their squares, so that would be;

(x^(2)) + (( x + 1 )^(2)) + (( x + 2 )^(2)) = 509

write out the squares

x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = 509

combine like terms

3x^2 + 6x + 5 = 509

inverse operations

3x^2 + 6x + 5 = 509

                -5     -5

3x^2 + 6x = 504

factor

3x^2 + 6x = 504

3 ( x^2 + 2x ) =504

Inverse operations

3 ( x^2 + 2x ) = 504

/3                       /3

x^2 + 2x = 168

Factor again

x ( x + 2 ) = 168

At this point, it should be obvious that x is 12 (because 12 * 14 = 168)

So now substitute back into the consecutive numbers

x = 12

x + 1 = 13

x + 2 = 14

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