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Norma-Jean [14]
3 years ago
13

A bag contains 77 coins consisting of quarters and nickels. The total value of the coins is $12.50. Write a system of equations

that can be used to determine the number of quarters, x, and the number of nickels, y, in the bag. a x + y = 77 0.25x + 0.5y = 12.5 b 0.25x +0.5 y = 77 x + y = 12.5 c x + y = 77 0.25x + 0.05y = 12.5 d 0.25x + 0.05y =77 x + y = 12.5
Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
8 0

Answer:

A

Step-by-step explanation:

because the answer needs both amounts of money and the beginning equation x + y = 77 can't have any other numbers.

777dan777 [17]3 years ago
7 0
I got this question too and I have no idea :(
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N 1)
a²<span> − 9a + 14 = 0 
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Rewrite as perfect squares

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the solution problem N 1 is the pair {7, 2}


N 2) 

a²<span> + 9a + 14 = 0
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Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² +9a+20.25)=-14+20.25

Rewrite as perfect squares

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a1=-4.5+√6.25-----> a1=-2

a2=-4.5-√6.25-----> a2=-7

the solution problem N 2 is the pair {-2,-7}

N 3) 

a² + 3a − 10 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 3a)=10

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 3a+2.25)=10+2.25

Rewrite as perfect squares

(a+1.5)²=12.25------> (a+1.5)=(+/-)√12.25

a1=-1.5+√12.25-----> a1=2

a2=-1.5-√12.25-----> a2=-5

the solution problem N 3 is the pair {2, -5}


N 4)

a²<span> + 5a − 14 = 0
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Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 5a) =14

Complete the square  Remember to balance the equation by adding the same constants to each side 

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Rewrite as perfect squares

(a+2.5)² =20.25-------> (a+2.5)=(+/-)√20.25

a1=-2.5+√20.25-----> a1=2

a2=-2.5-√20.25-----> a2=-7

the solution problem N 4 is the pair {2, -7}


N 5) 

a² − 5a − 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

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the solution problem N 5 is the pair {7, -2}

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