Mathematics, 04.07.2019 20:30 111432

Astudent has some $1 and $5 bills in his wallet. he has a total of 31 bills that are worth $33. how many of each type of bill does he have?

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Answer from: kaleb6816

There can't be more than seven $5 bills, because they would be worth $33.

So, let's try all other cases and see which fits the requests:

If there are six $5 bills, they are worth $30. You need three more $1 bills to reach $33. So, you have a total of 9 bills, which is not what we want.If there are five $5 bills, they are worth $25. You need eight more $1 bills to reach $33. So, you have a total of 13 bills, which is not what we want.If there are four $5 bills, they are worth $20. You need thirteen more $1 bills to reach $33. So, you have a total of 17 bills, which is not what we want.If there are three $5 bills, they are worth $15. You need eighteen more $1 bills to reach $33. So, you have a total of 21 bills, which is not what we want.If there are two $5 bills, they are worth $10. You need twenty-three more $1 bills to reach $33. So, you have a total of 25 bills, which is not what we want.If there is one $5, it is worth $5. You need twenty-eight more $1 bills to reach $33. So, you have a total of 29 bills, which is not what we want.

So, there's no way you can have 31 bills worth $1 and $5 that are worth $33 in total.

Answer from: Quest

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

5×12+11

2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

60+11

3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

4 simplify \frac{5}{6}x

x to \frac{5x}{6}

3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

5 convert 1\frac{1}{3}1

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

1×3+1

6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

3+1

7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

to \frac{13.75}{3}+\frac{4x}{3}

13.75

\frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

13.75

9 simplify \frac{13.75}{3}

13.75

to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

−4.583333

11 simplify \frac{71}{12}-4.583333

−4.583333 to \frac{4}{3}

\frac{4x}{3}=\frac{4}{3}

12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

×3

13 simplify \frac{4}{3}\times 3

×3 to \frac{12}{3}

4x=\frac{12}{3}4x=

14 simplify \frac{12}{3}

to 44

4x=44x=4

15 divide both sides by 44

x=1

Answer from: Quest

answer: it is a.,b.,d.

step-by-step explanation:

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Astudent has some $1 and $5 bills in his wallet. he has a total of 31 bills that are worth $33. how...