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DiKsa [7]
3 years ago
13

Marty made a $220 bank deposit using $10 bills and $5 bills. She gave the teller a total of 38 bills, how many $5 bills were in

the deposit?
A) 6 five-dollar bills
B) 28 five-dollar bills
C) 32 five-dollar bills
D) 34 five-dollar bills
Mathematics
1 answer:
Luda [366]3 years ago
7 0

ANSWER: 32 five-dollar bills

======

EXPLANATION:

Let x be number of $5 bills

Let y be number of $10 bills

Since we have total of 38 bills, we must have the sum of x and y be 38

x + y = 38 (I)

Since the total amount deposited is $220, we must have the sum of 5x and 10y be 220 (x and y are just the "number of" their respective bills, so we multiply them by their value to get the total value):

5x + 10y = 220 (II)

System of equations:

\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ 5x + 10y &= 220 && \text{(II)} \end{aligned} \right.

Divide both sides of equation (II) by 5 so our numbers become smaller

\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ x + 2y &= 44 && \text{(II)} \end{aligned} \right.

Rearrange (I) to solve for y so that we can substitute into (II)

\begin{aligned} x + y &= 38 && \text{(I)} \\ y &= 38 - x \end{aligned}

Substituting this into equation (II) for the y:

\begin{aligned} x + 2y &= 44 && \text{(II)} \\ x + 2(38 - x) &= 44\\ x + 76 - 2x &= 44 \\ -x &= -32 \\ x &= 32 \end{aligned}

We have 32 five-dollar bills

======

If we want to finish off the question, use y = 38 - x to figure out number of $10 bills

y = 38 - 32 = 6

32 five-dollar bills and 6 ten-dollar bills

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3 years ago
Calculate the quotient below and give your answer in scientific notation. 0.00065/5*10^-2
swat32

Answer:

1.3x10^-6

Step-by-step explanation:

Use a calculator.

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5 0
3 years ago
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15% of amount = 112.79
Mazyrski [523]
To find the original amount, you divide the product by the percentage. 

112.79 / 0.15 ≈ 751.933 (the 3's repeat infinitely). 

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4 0
3 years ago
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Give EG = 5x+7 EF =5x and FG=2x-7
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Answer:

x = 7

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Step-by-step explanation:

  • EG = 5x+7
  • EF = 5x  
  • FG = 2x-7

Point F is on line segment of EG, therefore EG = EF + FG

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Then

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