A bank teller has 47 $20 and $5 bills in her cash drawer. The value of the bills is $490. How many $20 bills are
there?

Answers

Answer 1

Answer:

A bank teller has 47 $20 and $5 bills.

The value of the bills is $490.

Number of $20 bills --> x

Number of $5 bills --> y

x + y = 47 --> x = 47-y

20x + 5y = 490 --> 20x = 490-5y

so

20(47-y) = 490 -5y

940-20y = 490-5y

940-490 = -5y+20y

15y = 450

y= 30

so

x = 47-30

x = 17

IN TOTAL --> 17 $20 bills and 30 $5 bills.

Answer 2

Answer:

x = 17 $20 bills

y = 30 $5 bills

General Formulas and Concepts:

Order of Operations: BPEMDASMultivariable Systems

Step-by-step explanation:

Step 1: Define variables

x = # of $20 bills

y = # of $5 bills

Step 2: Set up systems of equations

[tex]\left \{ {{x+y=47} \atop {20x+5y=490}} \right.[/tex]

Step 3: Solve for x

Rewrite:                                        [tex]\left \{ {{y=47-x} \atop {20x+5y=490}} \right.[/tex]Substitute:                                     [tex]20x+5(47-x)=490[/tex]Distribute 5:                                   [tex]20x+235-5x=490[/tex]Combine like terms:                     [tex]15x+235=490[/tex]Subtract 235 on both sides:        [tex]15x=255[/tex]Divide both sides by 15:               [tex]x=17[/tex]

By solving for x, we now know that we have 17 $20 bills.

Step 4: Solve for y

Define:                                        x + y = 47Substitute:                                  17 + y = 47Subtract 17 on both sides:         y = 30

Now we know that we have 30 $5 bills.


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Answer:

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Answers

Answer:

29.05% of cookies sold were oatmeal cookies.

Step-by-step explanation:

Students sold chocolate chip cookies = 145

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Now, finding percent of the cookies sold that were oatmeal

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Answers

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

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Given the following data;

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Answers

A polar graph that is a limacon has a formula similar to [tex]r=a+bcos\theta[/tex]

Option B is correct.

A  limacon has a formula similar to [tex]r=a+bcos\theta[/tex]

Case 1 .  If  a < b  or  [tex]\frac{b}{a}>1[/tex]

Then the curve is limacon with an inner loop.

Case 2.  If  a>b   or   [tex]\frac{b}{a}<1[/tex]

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Answers

Answer:

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Step-by-step explanation:

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I hope this helps ^^

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Answers

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Step-by-step explanation:

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Answers

Answer:

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Answers

Answer:

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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Answers

Answer:

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Answers

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Step-by-step explanation:

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Answer:

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3 x y − 7 y 2 − 6 x + 20 y + 5

Step-by-step explanation:

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Liam had a container of oatmeal that contained 5 and one-half cups of oatmeal. Liam ate StartFraction 4 over 9 EndFraction cup of oatmeal every morning before school for a week. Which is the best estimate of the number of cups of oatmeal that Liam had left in the container at the end of the five-day week?
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Answers

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Answers

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Answer:

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A donor gives $100,000 to a university, and specifies that it is to be used to give annual scholarships for the next 20 years. If the university can earn 4% interest, how much can they give in scholarships each year?

Answers

9514 1404 393

Answer:

  $7,358

Step-by-step explanation:

Assuming the interest is compounded annually, the amortization formula is useful here.

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A jogger is running home. His distance from home, as a function of time, is modeled by y=-7x +8.
Which statement best describes the function?
A. The function is nonlinear.
B. Not enough information is given.
C. The function is linear at some points and nonlinear at other points.
D. The function is linear.

Answers

Answer: the function is a liner

Step-by-step explanation: trust

Consider the feasible region in the xy-plane defined by the following linear inequalities.
x≥0
y ≥0
x ≤ 10
x +y≥ 5
x + 2y ≤ 18
Part 2 Exercises:
1. Find the coordinates of the vertices of the feasible region. Clearly show how each vertex is determined and which lines form the vertex.
2. What is the maximum and the minimum value of the function Q = 60x+78y on the feasible region?

Answers

Answer:

1. (0,5), (0,9), (10,4), (10,0), (5,0)

2. [tex]Q_{max}=912[/tex]

[tex]Q_{min}=300[/tex]

Step-by-step explanation:

1.

In order to determine the coordinates of the vertices of the feasible region, we must first graph each of the inequalities. The feasible region is the region where all the inequalities cross each other. In this case it's the region shaded on the attached picture.

The first point is the intercept between the equations x=0 and x+y=5 so in order to find this first coordinate we need to substitute x=0 and solve for y.

0+y=5

y=5

(0,5)

The next point is the intercept between the equations x=0 and x+2y=18, so again, we substitute x for zero and solve for y:

0+2y=18

[tex]y=\frac{18}{2}[/tex]

y=9

(0,9)

The next coordinate is the intercept between the lines x=10 and x+2y=18, so we substitute x for 10 and solve for y:

10+2y=18

2y=18-10

2y=8

[tex]y=\frac{8}{2}[/tex]

y=4, so the oint is

(10,4)

The next point is the intercept between the lines x=10 and y=0, so the point is:

(10,0)

The final point is the intercept between the equations: y=0 and x+y=5. We substitute y for zero and solve for x:

x+0=5

x=5

so the point is:

(5,0).

2. In order to determine the maximum and minimum value of the function Q=60x+78y on the feasible region, we must evaluate it for each of the points found on part 1.

(0,5)

Q=60(0)+78(5)

Q=390

(0,9)

Q=60(0)+78(9)

Q=702

(10,4)

Q=60(10)+78(4)

Q=912

(10,0)

Q=60(10)+78(0)

Q=600

(5,0)

Q=60(5)+78(0)

Q=300

So now we compare the answers and pick the minimum and maximum results.

We get that:

[tex]Q_{max}=912[/tex]

when x=10 and y=4

and

[tex]Q_{min}=300[/tex]

When x=5 and y=0

The feasible region is the possible set of a constraint

The vertices are: [tex]\mathbf{(x,y) \to (0,5), (0,9), (5,0)}[/tex]The maximum and the minimum values are 702 and 300, respectively.

(a) The coordinates of the vertices at the feasible region

The constraints are given as:

[tex]\mathbf{x \ge 0}[/tex]

[tex]\mathbf{y \ge 0}[/tex]

[tex]\mathbf{x \le 0}[/tex]

[tex]\mathbf{x + y\ge 5}[/tex]

[tex]\mathbf{x + 2y\ge 18}[/tex]

See attachment for the graph of the constraints

From the graph, the vertices are:

[tex]\mathbf{(x,y) \to (0,5), (0,9), (5,0)}[/tex]

(b) The minimum and the maximum values of objective function Q

The objective function is:

[tex]\mathbf{Q=60x +78y}[/tex]

Substitute [tex]\mathbf{(x,y) \to (0,5), (0,9), (5,0)}[/tex] in Q

[tex]\mathbf{Q=60(0) +78(5) = 390}[/tex]

[tex]\mathbf{Q=60(0) +78(9) = 702}[/tex]

[tex]\mathbf{Q=60(5) +78(0) = 300}[/tex]

Hence, the maximum and the minimum values are 702 and 300, respectively.

Read more about feasible regions at:

https://brainly.com/question/7243840

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