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Mashcka [7]
3 years ago
12

"A cash drawer contains 29 bills, all twenties and fifties. If the total value of the bills is $820, how many twenties and how m

any fifties are in the drawer?"
Mathematics
2 answers:
telo118 [61]3 years ago
7 0

Answer:there are 21 twenties and 8 fifties in the drawer.

Step-by-step explanation:

Let x represent the number of bills that are twenties in the drawer.

Let y represent the number of bills that are fifties in the drawer.

The cash drawer contains 29 bills. All twenties and fifties. This means that

x + y = 29

If the total value of the bills is $820, it means that

20x + 50y = 820 - - - - - - - - - - 1

Substituting x = 29 - y

20(29 - y) + 50y = 820

580 - 20y + 50y = 820

- 20y + 50y = 820 - 580

30y = 240

y = 240/30 = 8

x = 29 - y = 29 - 8

x = 21

maks197457 [2]3 years ago
3 0

Answer:

There are 21 twenties and 8 fifties in the drawer.

Step-by-step explanation:

This question can be solved by a system of equations.

I am going to say that:

x is the number of twenties

y is the number of fifties.

A cash drawer contains 29 bills.

This means that x + y = 29.

The total value of the bills is $820.

This means that 20x + 50y = 820.

So

x + y = 29

20x + 50y = 820

From the first equation, x = 29 - y. So:

20x + 50y = 820

20(29 - y) + 50y = 820

580 - 20y + 50y = 820

30y = 240

y = 8

And

x = 29 - y = 29 - 8 = 21

There are 21 twenties and 8 fifties.

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Answer:

Step-by-step explanation:

Law of sines is:

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Also, angles of a triangle add up to 180° or π.

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Since A+B+C = π, B+C = π−A:

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Using angle shift property:

sin(B−C) / sin A

Using angle sum/difference identity:

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Distribute:

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From law of sines, sin B / sin A = b / a, and sin C / sin A = c / a.

(b/a) cos C − (c/a) cos B

From law of cosines:

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2(c/a) cos B = 1 + (c/a)² − (b/a)²

(c/a) cos B = ½ + ½ (c/a)² − ½ (b/a)²

Substituting:

[ ½ + ½ (b/a)² − ½ (c/a)² ] − [ ½ + ½ (c/a)² − ½ (b/a)² ]

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(ii) a (cos B + cos C)

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b² = a² + c² − 2ac cos B

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Similarly:

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a cos C = 1/(2b) (a² + b² − c²)

Substituting:

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(b + c)/(2bc) (2bc + a² − b² − c²)

Distribute:

(b + c)/(2bc) (2bc) + (b + c)/(2bc) (a² − b² − c²)

(b + c) + (b + c)/(2bc) (a² − b² − c²)

From law of cosines, we know:

a² = b² + c² − 2bc cos A

2bc cos A = b² + c² − a²

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Common denominator:

(ab + ac) (b² + c² − a²) / (2abc) + (ab + bc) (a² + c² − b²) / (2abc) + (ac + bc) (a² + b² − c²) / (2abc)

[(ab + ac) (b² + c² − a²) + (ab + bc) (a² + c² − b²) + (ac + bc) (a² + b² − c²)] / (2abc)

We have to distribute, which is messy.  To keep things neat, let's do this one at a time.  First, let's look at the a² terms.

-a² (ab + ac) + a² (ab + bc) + a² (ac + bc)

a² (-ab − ac + ab + bc + ac + bc)

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(2a²bc + 2ab²c + 2abc²) / (2abc)

2abc (a + b + c) / (2abc)

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