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choli [55]
3 years ago
13

Peter has been saving his loose change for several days. When he counted his quarters and dimes, he found they had a total value

of $13.10. The number of quarters was 15 more than 3 times the number of dimes. How many quarters and how many dimes did Peter have?

Mathematics
2 answers:
Serga [27]3 years ago
4 0

Peter had 48 quarters and 11 dimes

<h3>Further explanation</h3>

Simultaneous Linear Equations could be solved by using several methods such as :

  • <em>Elimination Method</em>
  • <em>Substitution Method</em>
  • <em>Graph Method</em>

If we have two linear equations with 2 variables x and y , then we need to find the value of x and y that satisfying the two equations simultaneously.

Let us tackle the problem!

Let :

<em>Number of quarters ( 25 cent coins ) = x</em>

<em>Number of dimes ( 10 cent coins ) = y</em>

<em>When he counted his quarters and dimes, he found they had a total value of $13.10.</em>

<h2>0.25x + 0.10y = 13.10</h2>

<em>The number of quarters was 15 more than 3 times the number of dimes.</em>

<h2>x = 15 + 3y</h2>

If we would like to use the Substitution Method , then second equations above could be substituted into first equations.

0.25x + 0.10y = 13.10

0.25 (15 + 3y) + 0.10y = 13.10

3.75 + 0.75y + 0.10y = 13.10

0.85y = 13.10 - 3.75

0.85y = 9.35

y = 9.35 / 0.85

<h3>y = 11</h3>

At last , we could find the value of x by substituting this y value into one of the two equations above :

x = 15 + 3y

x = 15 + 3(11)

x = 15 + 33

<h3>x = 48</h3>

<h3>Learn more</h3>
  • Perimeter of Rectangle : brainly.com/question/12826246
  • Elimination Method : brainly.com/question/11233927
  • Sum of The Ages : brainly.com/question/11240586

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Simultaneous Linear Equations

Keywords: Simultaneous , Elimination , Substitution , Method , Linear , Equations

jek_recluse [69]3 years ago
3 0

Answer: Peter had 48 quarters and 11 dimes.

Step-by-step explanation:

Let be "q" the number of quarters and "d" the number of dimes.

We know that $13.10 in cents is 1,310 cents. Then, we can set up the following system of equations:

\left \{ {{25q+10d=1,310} \atop {q=3d+15}} \right.

Applying the Substitution method, we can substitute the second equation into the first one and solve for "d":

25(3d+15)+10d=1,310\\\\75d+375+10d=1,310\\\\85d=1,310-375\\\\d=\frac{935}{85}\\\\d=11

Finally, we must substitute the value of "d" into the second equation to find the value of "q". Then:

q=3(11)+15\\\\q=33+15\\\\q=48

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Answer:

The frequency of the resulting harmonic motion is 0.000219 Hz

Step-by-step explanation:

We are going to calculate the time it takes for one single wave ocillation.

Frequency and the time taken to finish a single wave oscillation are inversely proportional. The formula for calculating frequency when given the time taken to complete a wave cycle is written as: f = 1 / T

In this formula, f represents frequency and T represents the time period or amount of time required to complete a single wave oscillation.

I consider the initial speed to be zero, because it is of no significance compared with the free fall into the earth, through the earth and back again.

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The diameter of the earth is 1.2742 * 10⁴ km which is 1.27 * 10⁷ m

2 times the radius = diameter, so the radius of the earth = (1.27 * 10⁷ m) /2 = 6.4 * 10⁶ m

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r = 6.4 * 10⁶ m

Now imagine the tunnel and the free fall.

1. Initially the rock has no speed.

2. Due to the gravitational accelleration, the rock will increase it's speed every second by a factor of 9.8.

3. The Rock gains speed untill it reached the centre of the earth. By then it will have reached it's maximum speed and it has travelled the distance r !

4. After this moment, the Rock will be slowed down because of the negative accelleration...

After it has travelled from the centre of the earth to the other end of the earth, it will have stopped completely, and again passing the distance r.

5. Now at the other end of the earth there is the same initial situation as described at point 1, only the Rock has travelled the distance equal to the diameter of the earth, (exactly 2 times r).

So basically, the samething happens once more, only this time it starts exactly from the other end of the earth...

6. Initially the rock has no speed.

7. Due to the gravitational accelleration, the rock will increase it's speed every second by a factor of 9.8.

8. The Rock gains speed untill it reached the centre of the earth. By then it will have reached it's maximum speed.

9. By now the Rock will be slowed down because of the negative accelleration... It is moving towards the initial starting point...

After it has travelled from the centre of the earth to the other end of the earth, it will have stopped completely.

10. Now finally the Rock is exactly at the starting position.

In reality there will have been some loss of speed due to friction, so the Rock will be slightly lower then the 100 m above the ground.

let's calculate the time it takes to free fall for the distance r.

initial speed =0 and after 6.4 * 10⁶ m it's speed will be maximum. We need to find out how much time passes before that distance is passed.

r = v*t + 0.5*a*t²

r = 0 + 0.5*a*t²

0.5*a*t² = r

t² = r / ( 0.5 * a )

t² = 6.4 *10⁶ / ( 0.5 * 9.8 )

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Now please confirm that in order for the Rock to move back to the initial starting point it has to travel 4 times as much time. It has to travel r to centre of the earth then another r to travel to to the other side of the earth, and back again. So indeed 4 times r.

The time it will take must be the same as 4 * 1142.86 s

now this is the time of one single wave ocillation.

Since T = 4571.43 s

f = 1 / 4571.43

f = 0.00021874993164 Hz

The frequency of the resulting harmonic motion is 2.19 *10-4

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